Real Analysis - Vanderbilt University · negative integers (including zero), the integers, the rational numbers, the real numbers, and the complex numbers. If Ais a collection of

Real Analysis

Jesse Peterson

February 1, 2017

2

Contents

1 Preliminaries 71.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Countability . . . . . . . . . . . . . . . . . . . . . . . . . 81.1.2 Transfinite induction . . . . . . . . . . . . . . . . . . . . . 91.1.3 The axiom of choice . . . . . . . . . . . . . . . . . . . . . 121.1.4 Ordinals and Cardinals . . . . . . . . . . . . . . . . . . . 131.1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.2 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.3 Normed spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.3.1 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.3.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Measure and integration 252.1 Measurable sets and functions . . . . . . . . . . . . . . . . . . . . 26

2.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.2 Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.2.1 Outer measures . . . . . . . . . . . . . . . . . . . . . . . . 332.2.2 Caratheodory’s extension theorem . . . . . . . . . . . . . 342.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.3 Borel measures on R . . . . . . . . . . . . . . . . . . . . . . . . . 362.3.1 Lebesgue measure on R . . . . . . . . . . . . . . . . . . . 382.3.2 Regularity of Borel measures . . . . . . . . . . . . . . . . 402.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.4 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.4.1 Integrable functions . . . . . . . . . . . . . . . . . . . . . 432.4.2 Properties of integration . . . . . . . . . . . . . . . . . . . 452.4.3 Functions which agree almost everywhere . . . . . . . . . 472.4.4 Convergence properties . . . . . . . . . . . . . . . . . . . 472.4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.5 Product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

2.6 Signed and complex measures . . . . . . . . . . . . . . . . . . . . 552.6.1 Signed measures . . . . . . . . . . . . . . . . . . . . . . . 56

3

4 CONTENTS

2.6.2 Complex measures . . . . . . . . . . . . . . . . . . . . . . 582.6.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2.7 The Radon-Nikodym Theorem . . . . . . . . . . . . . . . . . . . 592.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3 Point set topology 653.1 Topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.2 Continuous maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.3 Compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 763.4 The Stone-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . 77

3.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.5 The Stone-Cech compactification . . . . . . . . . . . . . . . . . . 79

3.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 823.6 The property of Baire . . . . . . . . . . . . . . . . . . . . . . . . 82

3.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 843.7 Cantor spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

3.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 883.8 Standard Borel spaces . . . . . . . . . . . . . . . . . . . . . . . . 89

3.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

4 Differentiation and integration 974.1 The Lebesgue differentiation theorem . . . . . . . . . . . . . . . . 97

4.1.1 Vitali’s covering lemma . . . . . . . . . . . . . . . . . . . 974.1.2 The Lebesgue differentiation theorem . . . . . . . . . . . 984.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

4.2 Functions of bounded variation . . . . . . . . . . . . . . . . . . . 1024.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

4.3 Absolutely continuous and singular functions . . . . . . . . . . . 1064.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

5 Lp spaces 1095.1 Holder’s and Minkowski’s inequalities . . . . . . . . . . . . . . . 109

5.1.1 Minkowski’s integral inequality . . . . . . . . . . . . . . . 1125.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

5.2 The dual of Lp-spaces . . . . . . . . . . . . . . . . . . . . . . . . 1135.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

6 Functional analysis 1176.1 Topological vector spaces . . . . . . . . . . . . . . . . . . . . . . 117

6.1.1 Locally convex spaces . . . . . . . . . . . . . . . . . . . . 1176.1.2 The open mapping and closed graph theorems . . . . . . 1196.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6.2 The Hahn-Banach theorem . . . . . . . . . . . . . . . . . . . . . 122

CONTENTS 5

6.2.1 Separating convex sets . . . . . . . . . . . . . . . . . . . . 1256.2.2 The Krein-Milman theorem . . . . . . . . . . . . . . . . . 1276.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

6.3 Hilbert space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1296.3.1 Inner product spaces . . . . . . . . . . . . . . . . . . . . . 1296.3.2 Orthogonal subspaces and the Riesz representation theorem1316.3.3 Orthonormal bases and dimension . . . . . . . . . . . . . 1336.3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

6 CONTENTS

Chapter 1

Preliminaries

1.1 Sets

We assume that the reader is familiar with the basic language and concepts ofset theory. We use the notation N,Z,Q,R,C to denote respectively the non-negative integers (including zero), the integers, the rational numbers, the realnumbers, and the complex numbers. If A is a collection of sets then we denotetheir union by ∪A∈AA = a | a ∈ A for some A ∈ A, and their intersectionby ∩A∈AA = a | a ∈ A for all A ∈ A. If the family of sets is indexedA = Aii∈I then we also denote the union and intersection respectively by∪i∈IAi and ∩i∈IAi. The difference of two sets A and B is A \ B = a | a ∈A and a 6∈ B, and their symmetric difference is A∆B = (A \B) ∪ (B \A).If A is a subset of a set X, and we write Ac for the complement of A in X,i.e., Ac = X \A.

The power set of a set X is denoted by 2X and is the collection of allsubsets of X, i.e., 2X = A | A ⊂ X. The Cartesian product X × Y of twosets X and Y consists of all ordered pairs (x, y) such that x ∈ X and y ∈ Y .A function (or mapping) f : X → Y from X to Y is a a subset of X × Ywhich has the property that for each x ∈ X there exists a unique y ∈ Y suchthat the pair (x, y) is contained in this subset. In this case we write y = f(x)(or sometimes y = fx) for each x ∈ X. If A ⊂ X and B ⊂ Y , then the imageof A is denoted by f(A) = f(a) | a ∈ A, and the inverse image of B isdenoted by f−1(B) = x ∈ X | f(x) ∈ B. If f : X → Y and g : Y → Z,the composition of f and g is denoted by g f , and is defined by the formula(g f)(x) = g(f(x)).

A function f is injective (or 1-1) if f(x) = f(y) only when x = y, and fis surjective (or onto) if f(X) = Y . f is bijective if it is both injective andsurjective, and in this case f has a unique inverse map f−1 : Y → X suchthat f−1 f and f f−1 are the identity maps on X and Y respectively.

A sequence in a set X is a function from N to X. If f : N → X is asequence and g : N → N is such that g(n) < g(m) whenever n < m, then we

7

8 CHAPTER 1. PRELIMINARIES

say that f g is a subsequence of f . Through abuse of notation, we will oftenidentify a sequence with its range, for instance, we may say “let ann∈N ⊂ Xbe a sequence”.

If A is a family of sets, their Cartesian product∏A∈AA consists of all

functions f : A → ∪a∈AA such that f(A) ∈ A for each A ∈ A. Similar to unionsand intersections, if A is an indexed family A = Aii∈I then the Cartesianproduct is written

∏i∈I Ai. If X =

∏i∈I Ai, and i ∈ I, then the coordinate

map πi : X → Ai is given by πi(x) = xi, and we call xi the ith coordinate of x.If each Ai is a fixed set A, then we denote πi∈IAi by AI . If I = 1, 2, . . . , n,then we denote AI by An and identify this with the set of ordered n-tuples ofelements of A.

1.1.1 Countability

If X and Y are sets we write |X| ≤ |Y | (resp. |X| = |Y |) if there existsan injective (resp. bijective) map f : X → Y . We also write |X| < |Y | if|X| ≤ |Y |, and there is no bijection from X to Y .

Theorem 1.1.1 (The Cantor-Schroder-Bernstein Theorem). If |X| ≤ |Y | and|Y | ≤ |X| then |X| = |Y |.

Proof. Suppose f : X → Y , and g : Y → X are both injective. Set B =∪n∈N(f g)n(Y \ f(X)), and set A = X \ g(B). Then we have g(B) = X \ A,and

f(A) = f(X) \ (f g)(B) = Y \ ((Y \ f(X)) ∪ (f g)(B)) = Y \B.

Hence if we define θ : X → Y by θ(x) =

f(x) if x ∈ A,g−1(x) if x ∈ Y \A = g(B),

then θ gives a bijection.

A set X is countable if |X| ≤ |N|. We say that X is uncountable if it isnot countable.

Proposition 1.1.2. 1. If X and Y are countable, then so is X × Y .

2. If I is countable and Xi is countable for each i ∈ I then ∪i∈IXi is alsocountable.

Proof. We let p : N → N be the map which takes n, to the nth prime number.Suppose f : X → N, and g : Y → N are injective, and consider h : X×Y → N byh(x, y) = p(f(x))g(y). If p(f(x1))g(y1) = h(x1, y1) = h(x2, y2) = p(f(x2))g(y2)

then by uniqueness or prime factorization we have p(f(x1)) = p(f(x2)) andg(y1) = g(y2). As p, f , and g are injective we then have x1 = x2 and y1 = y2.Thus, h is injective.

Similarly, if I is countable, and Xi is countable for each i ∈ I, then considerf : I → N injective, and for each i ∈ I consider fi : Xi → N injective. Wedefine g : ∪i∈IXi → N, by setting g(x) = p(f(i))fi(x) where f(i) is the smallestnumber so that x ∈ Xi. Then similar to above it is easy to check that g isinjective and hence ∪i∈IXi is countable.

1.1. SETS 9

Corollary 1.1.3. Z and Q are countable.

Proof. We have Z = N ∪ 0 ∪ −N showing that Z is countable. Also, writingany rational number in reduced fraction form a/b with a ∈ Z and b ∈ N \ 0,defines an injective function f(a/b) = (a, b) ∈ Z ×N. Since Z×N is countable,so is Q.

Proposition 1.1.4 (Cantor’s diagonalization method). Let X be a set, then|X| < |2X |.

Proof. The injective map f : X → 2X given by f(x) = x shows that we have|X| ≤ |2X |. Now, suppose we have an injective function g : X → 2X . We letA = x ∈ X | x 6∈ g(x). Then, if x ∈ X and x ∈ g(x) we have x 6∈ A and henceg(x) 6= A. Similarly, if x ∈ X and x 6∈ g(x) then x ∈ A and hence g(x) 6= A.We therefore have produced a set which is not in the range of g showing that gis not surjective. As g was arbitrary we then have |X| 6= |2X |.

Proposition 1.1.5. |R| = |2N|, and hence R is uncountable.

Proof. Note that we have |2N| = |2Z|. Writing each real number in its binaryexpansion (If there is ambiguity we choose the representation which ends inzeros) gives an injective map from R to 2Z. On the other hand, each sequencein 2N we may view as a decimal expansion, and this gives an injective map from2N into R.

1.1.2 Transfinite induction

A relation on X is a subset R ⊂ X ×X. We write xRy to mean (x, y) ∈ R. Arelation R is an equivalence relation if the following properties hold:

• xRx for each x ∈ X.

• If xRy then yRx.

• If xRy and yRz then xRz.

A relation ≺ is a partial ordering if the following properties hold:

• x ≺ x for each x ∈ X.

• If x ≺ y and y ≺ z then x ≺ z.

• If x ≺ y and y ≺ x then x = y.

We write x y if x ≺ y and x 6= y. An order isomorphism between twopartially ordered sets is a bijection which preserves the partial orderes. A partialordering ≺ is linear (or total) if for each x, y ∈ X we have either x ≺ y ory ≺ x.

If X is partially ordered by ≺, a maximal element of X is an elementx ∈ X such that if x ≺ y then we have x = y. If E ⊂ X, then an upperbounded for E is an element x ∈ X such that y ≺ x for each y ∈ E. We may


similarly define minimal elements and lower bounds. A linear ordering is saidto be well ordered if every nonempty subset of X has a minimal element. Forexample, N is well ordered by its usual ordering. If (X,≤) is a well ordered setand x ∈ X we define the initial segment of x to be Ix = y ∈ X | y < x. Theelements of Ix are called predecessors of x. Note that either Ix ∪ x = X, orelse Ix ∪ x = Iy where y is the minimal element in X \ (Ix ∪ x).

Proposition 1.1.6 (The principle of transfinite induction). Let X be a wellordered set. If A ⊂ X is such that x ∈ A whenever Ix ⊂ A, then A = X.

Proof. By contraposition, if A 6= X we let x ∈ X \ A be the minimal element.Then we have x 6∈ A, and Ix ⊂ A from the definition of x.

Lemma 1.1.7. Let X be a well ordered set and A ⊂ X, then ∪x∈AIx is eitheran initial segment or A = X.

Proof. If Ac is nonepmty then let x be the minimal element in Ac. It’s theneasy to see that A = Ix.

Proposition 1.1.8 (The principle of transfinite recursion). Let X be a wellordered set, Y a set, and let F = ∪x∈XY Ix denote the space of all functionsfrom initial segments of X to Y . If G : F → Y , then there exists a uniquefunction g : X → Y so that g(x) = G(g|Ix) for each x ∈ X.

Proof. We let E denote the family of functions f : I → Y such that I = Xor is an initial segment in X, and f satisfies the formula f(x) = G(f|Ix) forall x ∈ I. If f ′ : I ′ → Y is another such function in F and I ⊂ I ′, then setA = x ∈ I | f ′(x) = f(x). If x ∈ I and Ix ⊂ A then we have

f ′(x) = G(f ′|Ix) = G(f|Ix) = f(x),

hence x ∈ A. It then follows by transfinite induction that A = I and hencef ′|I = f .

We may then consider J the union of all I such that there is f : I → Y withf ∈ E . By Lemma 1.1.7 either J = X, or J is an initial segment. We define afunction g : J → Y by letting g(x) = f(x), where f : I → Y is in E and x ∈ I.By our remarks above it follows easily that g is well defined and g ∈ E .

If J 6= X then J = Ix for some x ∈ X. We could then extend the functiong to g : J ∪ x → Y such that g|J = g and g(x) = G(g). We would then haveg ∈ E which would contradict the maximality of g. Thus, we conclude thatJ = X and g : X → Y is our desired function. Uniqueness follows easily fromthe remarks above.

Intuitively, the previous result states that if we have an initial value (G(∅)),and a procedure for choosing a new value based on the ones previously chosen(this is the function G). Then we can defines resursively a unique function onall of X.

1.1. SETS 11

Lemma 1.1.9. Let (X,≤) and (Y,≺) be well ordered sets, and suppose f : X →Y is an order isomorphism, then f(Ix) = If(x) for each x ∈ X. Conversely, iff : X → Y is such that f(Ix) = If(x) for each x ∈ X, then f(X) is either Y oran initial segment in Y , and f is an order isomorphism onto its image.

Proof. Suppose first that f : X → Y is an order isomorphism. If x ∈ X anda < x, then f(a) < f(x) and hence f(Ix) ⊂ If(x). Considering the inverse of fgives the reverse inclusion If(x) = f(f−1(If(x))) ⊂ f(Ix).

Now suppose f : X → Y such that f(Ix) = If(x) for each x ∈ X. Then ify ∈ f(X) we have Iy ⊂ f(X) and hence f(X) is either Y , or an inital segment inY . If x1 < x2 then we have f(x1) ∈ f(Ix2

) = If(x2) and so f(x1) < f(x2), andthis also shows that f is injective. If f(x1) ≤ f(x2) then f(x2) 6∈ If(x1) = f(Ix1

)hence x1 ≤ x2. Thus, f is an order isomorphism onto its image.

Proposition 1.1.10. Suppose (X,≤) is a well ordered set and I ⊂ X is eitheran initial segment, or is equal to X. If f : X → I is an order isomorphism,then I = X and f is the identity map.

Proof. Let A = x ∈ X | f(x) = x. If x ∈ X is such that Ix ⊂ A, thenLemma 1.1.9 shows that Ix = f(Ix) = If(x), hence x = f(x) showing thatx ∈ A. The result then follows by transfininte induction.

Theorem 1.1.11. If X and Y are well ordered, then exactly one of the followingholds:

1. X is order isomorphic to Y ;

2. X is order isomorphic to an initial segment in Y ;

3. Y is order isomorphic to an initial segment in X.

Moreover, in each of the cases the order isomorphism is unique.

Proof. We may assume Y is non-empty. Suppose that Y is not order isomorphicto an initial segment inX. If x ∈ X and g : Ix → Y with g(Ix) 6= Y then letG(g)denote the smalles element in Y \G(g), otherwise let G(g) be the initial elementin Y . By the principal of transfinite recursion there is a function f : X → Y sothat g(x) = G(gIx) for each x ∈ X.

We set A = x ∈ X | g(Ix) = Ig(x). If Ia ⊂ A then from Lemma 1.1.9we have that g(Ia) is either Y , or an inital segment in Y , and we have that gdefines an order isomorphism from Ia onto g(Ia). As Y is not order isomorphicto an initial segment in X it follows that g(Ia) is an initial segment in Y ,say g(Ia) = Iy. Then from the definition of G we have g(a) = y and henceg(Ia) = Ig(a), showing that a ∈ A. By the principle of transfinite inductionwe then have A = X and from Lemma 1.1.9 it follows that X is either orderisomorphic to Y or to an initial segment in Y .

Thus, one of the three cases above must hold, and Proposition 1.1.10 showsthat no more than one can hold, and that the order isomorphism is unique.


1.1.3 The axiom of choice

Consider the following four principles:

AC (The Axiom of Choice): If Aii∈I is a nonempty collection of nonemptysets then

∏i∈I Ai is nonempty.

WO (The Well Ordering Principle) Every set X can be well ordered.

ZL (Zorn’s Lemma) If X is a nonempty partially ordered set and every linearlyordered subset of X has an upper bound, then X has a maximal element.

HM (The Hausdorff Maximal Principle) Every partially ordered set has a max-imal linearly ordered subset.

These principles are logically equivalent, and after this section we will usethem in these notes without explicit reference. In this section we show that wehave the implications AC =⇒ WO =⇒ ZL =⇒ HM. The reverse implicationsare easier and we leave them as exercises.

Proposition 1.1.12. The axiom of choice implies the well ordering principle.

Proof. Let X be a nonempty set. By the axiom of choice there exists f ∈∏Y ∈2X\X(X \ Y ). That is, f : 2X \X → X is a function such that f(Y ) 6∈ Y

for each Y ( X.

We define an f -string to be a well ordered set (A,≤) such that A ⊂ Xand a = f(Ia) for all a ∈ A. We let F denote the set of f -strings. If (A,≤)and (B,≤′) are f -strings, and h : A → B is an order isomorphism, then letE = a ∈ A | h(a) = a. If x ∈ A is such that Ix ⊂ E, then by Lemma 1.1.9 wehave

x = f(Ix) = f(h(Ix)) = f(Ih(x)) = h(x).

It then follows from transfinite induction that B = A, and h is the identity map.

It then follows from Theorem 1.1.11 that given any two distinct f -stringswe must have that one is the initial segment of the other, so that F is linearlyordered by inclusion. We let A denote the union over all sets in F and we let≺ denote the induced relation on A. Since F is linearly ordered by inclusion itfollows easily that each (A,≤) is well ordered and and each initial segment ofA is an f -string. From this it then follows that (A,≤) itself is an f -string, andis then the unique maximal f -string.

If A 6= X then we could create the larger f -string (A∪f(A),≤′) where ≤′agrees with ≤ when restricted to A, and a ≤′ f(A) for all a ∈ A. This wouldthen contradict the maximality of (A,≤) and so we conclude that A = X, andhence X is well ordered by ≤.

Proposition 1.1.13. The well ordering principle implies Zorn’s lemma.

1.1. SETS 13

Proof. Let (X,≺) be a nonempty partially ordered set such that every linearlyordered subset has an upper bound. We let C denote the set of all linearlyordered subsets of X. By the well ordering principle there exist well orders ≤1

and ≤2 on X and 2X respectively. To simplify notation we set Y = 2X .We define f : C → X as follows: If C ∈ C and C does not contain a maximal

element in X, then we let f(C) be the ≤1-least element which is not in C, butis a ≺-upper bound for C, otherwise if C contains a maximal element x0 ∈ Xthen we set f(C) = x0.

We now recursively define a function g : Y → X such that g(y) ≺ g(z)whenever y ≤2 z. Indeed, suppose y ∈ Y and that g has been defined on Iy,then g(Iy) is a well ordered (and hence linearly ordered) subset of X and wemay set g(y) = f(g(Iy)).

By Cantor’s diagonalization argument g cannot be injective. Thus, thereexists some y ∈ Y such that g(y) = g(z) for some z ∈ Iy. It then follows thatf(g(Iy)) = g(y) ∈ g(Iy). By construction of f we must have that f(g(Iy)) is amaximal element.

Proposition 1.1.14. Zorn’s lemma implies the Hausdorff maximal principle.

Proof. Let X be a nonempty partially ordered set and let C denote the space ofall linearly ordered subsets. Then C is ordered by inclusion. If C0 ⊂ C is a subsetwhich is linearly ordered by inclusion then we may consider C = ∪C0∈C0C0.Then C ⊂ X is also linearly ordered and is an upper bound for C0 with respedto the inclusion order. We may then apply Zorn’s Lemma to C to produce amaximal linearly ordered subset.

1.1.4 Ordinals and Cardinals

Proposition 1.1.15. Suppose X 6= ∅, then |X| ≤ |Y | if and only if there existsa surjection from Y to X.

Proof. Suppose that f : X → Y is injective, and take x ∈ X. We may then de-

fine a surjective function g : Y → X, by letting g(y) =

f−1(x) if x ∈ F (X)x otherwise.

Conversely, if g : Y → X is surjective, then for each x ∈ X we may choosef(x) ∈ Y so that g(f(x)) = x. If f(x) = f(y) then x = g(f(x)) = g(f(y)) = y.Thus, f : X → Y is injective.

Proposition 1.1.16. For any sets X,Y , either |X| ≤ |Y |, or |Y | ≤ |X|.

Proof. If we well order X and Y , then this follows immediately from Theo-rem 1.1.11.

We would like to define an ordinal as an order isomorphic equivalence classof well ordered sets, the previous proposition would then show that we have alinear ordering on the collection of ordinals. We should be somewhat careful herethough as there is no reason that the collection of all well ordered sets shouldbe a set itself, and we have only discussed equivalence relations and orderings


on sets. One option to make this precise is to work outside of the universe ofsets, that is, we can call the collection of all well ordered sets a “class”, and thenwe can introduce equivalence relations and well orderings for classes. Anotheroption to make this notion precise is the following definition proposed by vonNeumann:

A set α is an ordinal if every element of α is also a proper subset of α, andif α is well ordered with respect to set inclusion.

The first few ordinals are then 0 = ∅, 1 = 0, 2 = 0, 1, 3 = 0, 1, 2. Thefirst infinite ordinal is ω = 0, 1, 2, 3, 4, . . .. For each ordinal α we may considerthe set α∪α, which is again an ordinal, this is the successor ordinal which wedenote by α + 1. Any ordinal which is not a successor ordinal is called a limitordinal. If two ordinals are order isomorphic, then it follows easily by inductionthat they must be the same. More generally, if α and β are ordinals, then byTheorem 1.1.11 either α and β are order isomorphic, in which case α = β, orelse one (say α) is isomorphic to (and hence equal to) an initial segment of theother, in which case we have α ⊂ β.

The following proposition shows that von Neumann’s definition captures allorder isomorphism classes.

Proposition 1.1.17. Let X be a well ordered set, then there exists a uniqueordinal α such that X and α are order isomorphic.

Proof. We let A denote the subset of X consisting of all points x such thatthere exists an ordinal αx, and an order isomorphism fx : Ix → αx. Note thatif x, y ∈ A with x < y, then we obtain an order isomorphism from αx to aninitial segment in αy, hence we have αx ⊂ αy. We then have αy = ∪x≤yαx, andfy |Ix = fx.

If Iz ⊂ A, and z is not a successor, then we set α = ∪x∈Izαx. Note that α isagain an ordinal. We define the function f : Iz → α by setting f(x) = fy(x) forsome x < y < z. (note that such a y exists since z is not a successor). Then f iswell defined and implements an order isomorphism between Iz and α, showingthat z ∈ A.

Similarly, if Iz ⊂ A, and z is a successor to y ∈ A, then we consider theordinal α = αy ∪ αy, and we define the function f : Iz → α by f(x) = fy(x)for x < y, and f(y) = αy. We then again see that z ∈ A.

By transfinite induction we then have A = X, and the result then followseasily.

Just as there is a first infinite ordinal, there is also a first uncountable ordinal:

Proposition 1.1.18. There is a unique uncountable ordinal ω1 such that Ix iscountable for each x ∈ ω1.

Proof. We let ω1 be the set of all countable ordinals α. Then ω1 is orderedby inclusion. If α ∈ ω1 then α is a countable ordinal and hence so are all theordinals contained in α, thus α ⊂ ω1. We therefore have that ω1 is an ordinal,and Ix is countable for each x ∈ ω1. Note that ω1 cannot be countable sinceω1 6∈ ω1.

1.1. SETS 15

If Ω′ were another such ordinal, then we could not have Ω′ ⊂ ω1 since ω1

is not countable. We similarly could not have ω1 ⊂ Ω′. Hence we must haveω1 = Ω′.

If we fix a set X, then by the well ordering principle there exists a wellordering on X and hence a bijection between X and some ordinal α. Thecardinality of X is the smallest ordinal such that there exists such a bijection.We denote the cardinality of X by |X|, and note that this is consistent with ournotation above.

1.1.5 Exercises

Exercise 1.1.19. If X is countably infinite then there is a bijection from Xonto N.

Exercise 1.1.20. We have |RN| = |R|.

Exercise 1.1.21. Let 2<N denote the set of finite sequences in 0, 1, then 2<N

is countable.

Exercise 1.1.22. Let X be a countably infinite set. There exists an uncount-able family F ⊂ 2X so that for any distinct pair A,B ∈ F we have that A ∩Bis finite. (Hint: It may help to consider the case X = 2<N.)

A complex number α is algebraic if it is the solution of a polynomial havingrational coefficients.

Exercise 1.1.23. The set of algebraic numbers is countable.

Exercise 1.1.24. The Hausdorff maximal principle implies Zorn’s lemma.

Exercise 1.1.25. Zorn’s lemma implies the well ordering principle.

Exercise 1.1.26. The well ordering principle implies the axiom of choice.

Let X be a set, and ≤ be a linear ordering on X. We say that the linearorder is dense if for all x < y there exists z ∈ X such that x < z < y.

Exercise 1.1.27 (Cantor’s back-and-forth method). Let (X,≤) and (Y,≤) becountable dense linear orderings which do not have upper or lower bounds.Enumerate X = x1, x2, . . ., and Y = y1, y2, . . ..

1. There exist increasing sequences of finite sets An ⊂ X, Bn ⊂ Y , andorder preserving bijections fn : An → Bn such that xn ∈ An, yn ∈ Bn,and fn+1|An = fn, for all n ≥ 1.

2. There exists an order preserving bijection f : X → Y .


A (undirected) graph consists of a pair (V,E) where V is a set (the vertexset) and E ⊂ V×V (the edge set) such that (v, w) ∈ E if and only if (w, v) ∈ E.A subgraph is a graph (V0, E0) with V0 ⊂ V , and E0 ⊂ E.

If (V,E) is a graph, two vertices v, w ∈ V are adjacent if (v, w) ∈ E. Welet N(v) denote the set of vertices which are adjacent to v. A graph (V,E) islocally finite if |N(v)| < ∞ for each v ∈ V . A finite simple path is aninjective function p : 1, 2, . . . , n → V , such that (p(k), p(k + 1)) ∈ E for all1 ≤ k < n; we say that n is the length of the path. A ray is an injectivefunction p : N → V such that (p(k), p(k + 1)) ∈ E for all 1 ≤ k. A graph isconnected if for any distinct vertices v, w ∈ V , there exists a finite simple pathp : 1, 2, . . . , n → V such that p(1) = v and p(n) = w.

Exercise 1.1.28 (Konig’s lemma). If a locally finite connected graph (V,E)has infinitely many vertices, then (V,E) admits a ray.

1.2 Metric spaces

Let X be a set. A semimetric on X is a function d : X × X → [0,∞) suchthat for all x, y, z ∈ X the following properties hold:

1. d(x, y) = d(y, x).

2. d(x, z) ≤ d(x, y) + d(y, z).

If, in addition, we have that x = y if and only if d(x, y) = 0, then d is a metric.A metric space is a pair (X, d) consisting of a set X and a metric d on X.When d is understood we will sometimes refer to the metric space X.

Examples of metric spaces include:

1. Euclidean space Rn with metric d(x, y) = ‖x− y‖.

2. If X is any set and for all x, y ∈ X we have d(x, y) =

0 if x = y;1 if x 6= y,

then (X, d) is a metric space.

3. If (X, d) is a metric space and A ⊂ X, then (A, d|A×A) is a metric space.

4. If (X1, d1) and (X2, d2) are metric spaces, then X1×X2 is a metric spacewith metric d((x1, x2), (y2, y2)) = max(d1(x1, y1), d2(x2, y2)).

5. If (X, d) is a metric space and f : [0,∞) → [0,∞) is a strictly increasingfunction satisfying f(0) = 0, and f(s + t) ≤ f(s) + f(t) for all s, t ∈ R(e.g., f(t) =

√t, or f(t) = t

t+1 ), then (X, f d) is again a metric space.

If (X, d) is a metric space, x ∈ X and r > 0, then the ball of radius r aboutx is B(r, x) = y ∈ X | d(x, y) < r; we call any such set B(r, x) an r-ball. Aset A ⊂ X is open if for each x ∈ A there exists r > 0 such that B(r, x) ⊂ A.A set A ⊂ X is closed if Ac is open. Both ∅ and X are open. If a set is both

1.2. METRIC SPACES 17

closed and open then we say it is clopen. Note that the collection of open setsis closed under finite intersections and arbitrary unions. Taking complementsshows that the collection of closed sets is closed under finite unions and arbitraryintersections. If E ⊂ X then the closure E of E is the intersection of all closedsets which contian E. We say the E is dense if E = X.

If (X1, d1) and (X2, d2) are metric spaces, and f : X1 → X2, then f is

1. isometric if d2(f(x), f(y)) = d1(x, y), for all x, y ∈ X1;

2. Lipschitz continuous if there exists K ≥ 0 (a Lipschitz constant),such that for all x, y ∈ X1 we have d2(f(x), f(y)) ≤ Kd1(x, y);

3. contractive if f is Lipschitz continuous with Lipschitz constant 1;

4. uniformly continuous if for all ε > 0, there exists δ > 0 so thatf(B(δ, x)) ⊂ B(ε, f(x)) for all x ∈ X1;

5. continuous at x ∈ X1 if for each ε > 0, there exists δ > 0 so thatf(B(δ, x)) ⊂ B(ε, f(x)).

6. continuous if f is continuous at each point x ∈ X1.

7. a homeomorphism if f is bijective and both f and f−1 are continuous.

If X is a metric space, a sequence xnn∈N ⊂ X has a limit point x ∈ X iffor all ε > 0, there exists N ∈ N so that xn ∈ B(ε, x) for n ≥ N . We say thatxnn converges to x and write limn→∞ xn = x if this is the case. xnn∈N isconvergent if it converges to some point x ∈ X.

Proposition 1.2.1. If (X1, d1) and (X2, d2) are metric spaces, the followingare equivalent:

1. f : X1 → X2 is continuous;

2. for any convergent sequence xnn∈N we have that f(xn)n∈N is alsoconvergent and limn→∞ f(xn) = f(limn→∞ xn);

3. f−1(O) is open for each open set O ⊂ X2.

Proof. First, suppose f : X1 → X2 is continuous, and xnn∈N is a sequencesuch that x = limn→∞ xn. Fix ε > 0. Since f is continuous there exists δ > 0so that f(B(δ, x)) ⊂ B(ε, f(x)). Since x = limn→∞ xn, there exists N ∈ N sothat xn ∈ B(δ, x) for all n ≥ N . Hence, f(xn) ∈ f(B(δ, x)) ⊂ B(ε, f(x)) for alln ≥ N . Since ε > 0 was arbitrary we have f(x) = limn→∞ f(xn).

Next, suppose that O ⊂ X2 is open, but f−1(O) is not open. Then thereexists x ∈ f−1(O) such that for all n ∈ N we have B(1/n, x) 6⊂ f−1(O). Foreach n ∈ N choose xn ∈ B(1/n, x). Since O is open there exists ε > 0 so thatB(ε, f(x)) ⊂ O. We then have x = limn→∞ xn, and f(xn) 6∈ O ⊃ B(ε, f(x)) forall n ∈ N, hence f(xn)n∈N does not converge to f(x).


Finally, suppose that f−1(O) is open whenever O ⊂ X2 is open. Fix x ∈ Xand ε > 0. Then B(ε, f(x)) is open and hence so is f−1(B(ε, f(x))). There-fore, there exists δ > 0 so that B(δ, x) ⊂ f−1(B(ε, f(x))). Hence, we havef(B(δ, x)) ⊂ B(ε, f(x)) showing that f is continuous.

A sequence of functions fn : X → Y is said to converge pointwise to afunction f : X → Y , if f(x) = limn→∞ fn(x) for each x ∈ X. The sequencefnn∈N converges uniformly to f if limn→∞ supx∈X |f(x)− fn(x)| = 0.

Proposition 1.2.2. Suppose fn : X → Y are continuous, and fnn∈N con-verges to f : X → Y uniformly, then f is continuous.

Proof. Fix ε > 0, and x ∈ X. Since fn → f uniformly, there exists n ∈ N sothat supy∈X |f(y) − fn(y)| < ε/3. Since fn is continuous there exists an openset O containing x so that for y ∈ O we have |fn(y)− fn(x)| < ε/3. For y ∈ Owe then have

|f(y)− f(x)| ≤ |f(y)− fn(y)|+ |fn(y)− fn(x)|+ |fn(x)− f(x)| < ε.

Thus, f is continuous.

A sequence xnn∈N is Cauchy if for all ε > 0, there exists N ∈ N so thatd(xn, xm) < ε for all n,m ≥ N . A metric space is complete if every Cauchysequence is convergent.

Proposition 1.2.3. Rn with its Euclidean metric is complete.

Proof. A sequence in Rn is Cauchy if and only if its coordinates are Cauchy, and,similarly, a sequence converges if and only if its coordinates converge. Thus, thegeneral result follows from R. Suppose xnn∈N ⊂ R is Cauchy. Then thereexists N ∈ N so that |xN − xm| < 1 for all m ≥ N . Hence xnn∈N is bounded.We let x = lim supn→∞ xn.

Fix ε > 0, then there exists N ∈ N so that |xn−xm| < ε/2 for all n,m ≥ N .Also, there exists n ≥ N so that |x − xn| < ε/2. Then for all m ≥ N we have|x− xm| ≤ |x− xn|+ |xn − xm| < ε. It then follows that x = limn→∞ xn.

Proposition 1.2.4. A closed subset of a complete metric space is complete,and a complete subspace of an arbitrary metric space is closed.

Proof. Suppose (X, d) is complete and F ⊂ X is closed. Let xnn∈N ⊂ F beCauchy. Then it is also Cauchy in X and by completeness there exists x ∈ Xso that xn → x. If ε > 0 then there exists N ∈ N so that xn ∈ B(ε, x) for alln ≥ N . In particular we have B(ε, x) ∩ F 6= ∅ and hence x 6∈ F c as this is openand ε > 0 was arbitrary.

Conversely, Suppose F ⊂ X is a subspace which is not closed. Then F c isnot open and hence there exists x ∈ F c, so that B(1/n, x)∩F 6= ∅ for all n ∈ N.Take xn ∈ B(1/n, x) ∩ F for each n ∈ N. Then x = limn→∞ xn and hencexnn∈N is Cauchy. However, x 6∈ F and hence xnn∈N does not converge to apoint in F . Thus, F is not complete.


If (X, d) is a metric space, then we let Cauchy(X) denote the set of all Cauchysequences inX. On Cauchy(X)2 we define the function d by d(xnn∈N, ynn∈N) =limn→∞ d(xn, yn). We leave it as an exercise to verify that this limit actuallyexists. Using the properties of the metric d it is then easy to see that wehave d(s, t) = d(t, s) and d(s, r) ≤ d(s, t) + d(t, r), for all Cauchy sequencess, t, r. We define an equivalence relation on Cauchy(X) by s ∼ t if and only ifd(s, t) = 0. It’s easy to see that this is indeed an equivalence relation, and thatd(s, t) only depends on the equivalence classes that s and t lie in. Thus, settingX = Cauchy(X)/ ∼, we may view d as a function on X×X where it then givesa metric. We also have a natural isometric embedding π : X → X which takesa point x ∈ X to the constant sequence π(x) = xn∈N. We call the metricspace (X, d) the completion of (X, d), and we usually view X as a subspaceby identifying X with π(X).

Proposition 1.2.5. (X, d) is complete, and X is a dense subspace.

Proof. We first show that π(X) is a dense subspace. Suppose that xnn∈N isCauchy and ε > 0 is given. Then there exists N ∈ N so that |xn − xm| < ε forall n,m ≥ N . In particular, we have that |xN − xm| < ε for all m ≥ N . Hence,we have d(π(xN ), xnn∈N) ≤ ε.

Next we show that (X, d) is complete. Note that if snn∈N, tnn∈N ⊂ Xsuch that 0 = limn→∞ d(sn, tn), then snn∈N is Cauchy (resp. convergent) ifand only if tnn∈N is Cauchy (resp. convergent). Thus, it is enough to con-sider Cauchy sequences which are valued in the dense subspace π(X). Supposetherefore that π(xn)n∈N is Cauchy, with xn ∈ X. If we set s = xnn∈Nthen it follows easily that we have 0 = limn→∞ d(s, π(xn)). Hence (X, d) iscomplete.

If E ⊂ X, then E is bounded if there exists K > 0, such that d(x, y) ≤ K,for all x, y ∈ E. If Vii∈I is a family of subsets of X such that E ⊂ ∪i∈IVi,then Vii∈I is a cover of E. E is totally bounded if for any ε > 0, there isa finite collection of ε-balls which cover E. Note that totally bounded sets arealso bounded.

Lemma 1.2.6. A metric space (X, d) is totally bounded if and only if everysequence has a Cauchy subsequence.

Proof. Suppose (X, d) is totally bounded and xnn∈N is a sequence. Fix ε > 0.We will inductively define a decreasing sequence of infinite subsets Aj ⊂ N,such that d(xn, xm) < 2/j for all n,m ∈ Aj , and j ≥ 2. We first set A1 = N.Suppose now that Aj−1 has been chosen for j ≥ 2. Since E is totally bounded,there exist a finite collection of 1/j-balls O1, . . . , Ok which cover E. Since Aj−1

is infinite for some Oi we must have that Aj = k ∈ Aj−1 | xk ∈ Oi is infinite.We new choose a subsequence by taking nj ∈ Aj so that nj is strictly

increasing. If ε > 0, and j ∈ N so that 2/j < ε then we have d(xnk , xnl) <2/j < ε for all k, l ≥ j. Therefore we have that the subsequence xnjj∈N isCauchy.


Conversely, if E is not totally bounded then there exists ε0 > 0 so thatthere is no cover of E by finitely many ε0-balls. We may therefore inductivelyconstruct a sequence xn ∈ E so that d(xn, xm) ≥ ε0 for all n,m ∈ N. We thenhave that no subsequence of xnn∈N is Cauchy.

Lemma 1.2.7. Let (X, d) be a totally bounded metric space, then every opencover has a countable subcover.

Proof. Suppose that Vii∈I is an open cover. For each n ∈ N take xn1 , . . . , xnkn ⊂X, so that ∪knj=1B(1/n, xj) = X. We then let On,m be the collection of allopen balls B(1/m, xj) which are contained in some Vi, for i ∈ I. We setO = ∪n,m∈NOn,m.

If x ∈ E then we have x ∈ Vi for some i ∈ I. We then have B(1/n, x) ⊂ Vifor some n ∈ N. For some 1 ≤ j ≤ k2n we then have x ∈ B(1/2n, x2n

j ) ⊂ Vi.Thus, we see that O covers X and is countable. Moreover, each set in O iscontained in Vi for some i ∈ I, thus a countable subcollection of Vii∈I mustcover X.

Theorem 1.2.8. If E ⊂ X, the following are equivalent:

1. E is complete and totally bounded.

2. (The Bolzano-Weierstrass Property) Every sequence in E has a subse-quence which converges to a point in E.

3. (The Heine-Borel Property) If Vii∈I is a cover of E by open sets, thenthere exists a finite set F ⊂ I such that Vii∈F is also a cover of E.

Proof. (1 =⇒ 2) Suppose that E ⊂ X is complete and totally bounded. Letxnn∈N ⊂ E be a sequence. By Lemma 1.2.6 there exists a Cauchy subsequencexnjj∈N, and since E is complete we must have that this subsequence converges.Therefore E satisfies the Bolzano-Weierstrass property.

(2 =⇒ 1) If E is not totally bounded then Lemma 1.2.6 shows that Xhas a sequence which has no Cauchy (and hence no convergent) subsequence.Similarly, if E is not complete then there exists a Cauchy sequence xnn∈Nwhich does not converge, and it then follows easily that no subsequence canconverge either. We have therefore shown the equivalence between the Bolzano-Weierstrass property and being complete and totally bounded.

(3 =⇒ 1) If E is not totally bounded then there exists ε0 > 0 so that thereis no cover of E by finitely many ε0-balls. However, all ε0-balls cover E andhence E does not have the Heine-Borel Property. Also, if E is not complete,then we may consider the completion E and take a point x ∈ E \ E. Thenconsider On = y ∈ E | d(y, x) > 1/n. We then have that Onn∈N is anincreasing sequence of open sets, such that ∪n∈NOn = E. However, On 6= E forany n ∈ N since E is dense in E. Hence, we again have shown that E does nothave the Heine-Borel property.

(1 =⇒ 3) Suppose that E is totally bounded and Vii∈I is an opencover which does not have a finite subcover. By Lemma 1.2.7 we may pass


to a countable cover so that we may assume Vii∈I is countable and thensequence this as Vnn∈N. We inductively define a sequence xnn∈N by takingxn 6∈ ∪nk=1Vn. Note that by construction, each open set Vn can contain at mostfinitely many elements in the sequence xnn∈N. By Lemma 1.2.6 there existsa Cauchy subsequence xnjj∈N. If this subsequence converged to some pointx ∈ E, then x would be contained in some open set Vn and it would follow thatinfinitely many xnj ’s would belong to Vn contradicting our remark above. Thus,we must have that xnjj∈N is a Cauchy sequence which does not converge andhence X is not complete.

Any set E which satisfies the conditions of the previous theorem is called acompact set. Note that homeomorphisms preserve open sets and hence fromthe Heine-Borel Property we see that homeomorphisms preserve compact sets.This is not the case however for complete sets.

Proposition 1.2.9. Let (X, d) and (Y, ρ) be metric spaces with X compact.Suppose that f : X → Y is continuous. Then f(X) is compact and f is uni-formly continuous.

Proof. If Oii∈I is an open cover of f(X), then since f is continuous we havethat f−1(Oi)i∈I is an open cover of X. By the Heine-Borel property thereexists a finite subcover f−1(O1), . . . , f−1(On). We then have that O1, . . . , Oncovers f(X) and so by the Heine-Borel property we have that f(X) is compact.

To see that f is uniformly continuous we fix ε > 0. Since f is continuous,for each x ∈ X there exists δx > 0 so that f(B(δx, x)) ⊂ B(ε/2, f(x)). ThenB(δx/2, x)x∈X covers X and by the Heine-Borel property there is a finitesubcover B(δx1/2, x1), . . . , B(δxn/2, xn).

Set δ = min1≤i≤nδxi/2. Then if 1 ≤ i ≤ n and x ∈ B(δxi/2, xi) we haveB(δ, x) ⊂ B(δxi , xi) and hence

f(B(δ, x)) ⊂ f(B(δxi , xi)) ⊂ B(ε/2, f(xi)).

Therefore, f(B(δ, x)) ⊂ B(ε, f(x)). Since B(δx1/2, x1), . . . , B(δxn/2, xn) covers

X it follows that f is uniformly continuous.

1.2.1 Exercises

Exercise 1.2.10. Suppose that X is a set and d : X × X → [0,∞) is asemimetric on X. We define a relation ∼ on X by x ∼ y if d(x, y) = 0. Then∼ is an equivalence relation on X and we have a well defined metric on X/ ∼given by d([x], [y]) = d(x, y).

Exercise 1.2.11. There are two homeomorphic metric spaces (X1, d1) and(X2, d2) such that (X1, d1) is complete, while (X2, d2) is not.

A metric space (X, d) is separable if it contains a countable dense set.

Exercise 1.2.12. A compact metric space is separable.


We let `∞(N) denote the set of uniformly bounded sequences from N to C.We consider this as a complete metric space whose metric is given by d(f, g) =‖f − g‖∞ = supn∈N |f(n)− g(n)|.

Exercise 1.2.13 (Kuratowski). Every bounded seprable metric space is iso-metric to a subspace of `∞(N).

1.3 Normed spaces

We assume the reader is familiar with the basic properties of vector spaces. LetK = R, or K = C, and suppose that V is a K-vector space. A seminorm on Vis a map V 3 v 7→ ‖v‖ ∈ [0,∞) which satisfies

1. ‖v + w‖ ≤ ‖v‖+ ‖w‖;

2. ‖kv‖ = |k|‖v‖, for k ∈ K, and v, w ∈ V .

If, in addition, we have that ‖v‖ = 0 if and only if v = 0, then we say that ‖ · ‖is a norm. Associated with a (pre)norm is a (pre)metric d which is given byd(v, w) = ‖v + w‖. A normed space is a pair (V, ‖ · ‖) where V is a vectorspace and ‖ · ‖ is a norm on V . If the associated metric is complete then thenormed space is a Banach space. Examples of Banach spaces include:

1. `1n = Kn, with norm ‖(α1, . . . , αn)‖1 =∑nk=1 |αk|.

2. More generally, if 1 ≤ p < ∞, `pn = Kn, with norm ‖(α1, . . . , αn)‖p =

(∑nk=1 |αk|p)

1/p.

3. `∞n = Kn, with norm ‖(α1, . . . , αn)‖∞ = max|αk| | 1 ≤ k ≤ n.

If (V, ‖ · ‖V ), and (W, ‖ · ‖W ), are normed spaces, and T : V → W is alinear operator, then we say that T is bounded if there exists K > 0 so that‖Tv‖W ≤ K‖v‖V for all v ∈ V . We let B(V,W ) (or B(V ) if V = W ) denote theset of bounded linear operators. Then B(V,W ) is a K-vector space, where thevector space structure is taken pointwise, i.e., (T + S)(v) = T (v) + S(v), and(kT )(v) = k(T (v)), for k ∈ K, v ∈ V , and T, S ∈ B(V,W ). If T ∈ B(V,W ) thenthe operator norm of T is given by ‖T‖B(V,W ) = supv∈V,‖v‖V ≤1 ‖Tv‖V . Thespace B(V,W ), together with its operator norm, is a normed space.

1.3.1 Algebras

We again let K = R, or K = C. A K-algebra is a K-vector space A, togetherwith a binary operation A × A 3 (a, b) 7→ ab ∈ A (called multiplication, orcomposition), such that

1. (ab)c = a(bc);

2. α(ab) = (αa)b = a(αb);

1.3. NORMED SPACES 23

3. a(b+ c) = (ab) + (ac);

4. (a+ b)c = (ac) + (bc), for α ∈ K, a, b, c ∈ A.

Examples of algebras include:

1. The vector space of n × n matrices Mn(K), together with matrix multi-plication.

2. The space of K-polynomials with its usual vector space structure andmultiplication.

3. `∞n where multiplication is taken coordinate-wise (α1, . . . , αn)·(β1, . . . , βn) =(α1β1, . . . , αnβn).

A normed algebra is an algebra A, which also has a norm ‖ · ‖ whichsatisfies ‖ab‖ ≤ ‖a‖‖b‖, for a, b ∈ A. A Banach algebra is a normed algebrawhere the norm is complete.

The space `∞n (K) is a normed algebra. Also, if V and W are normed spacesthen B(V,W ), with its operator norm, is a normed algebra.

If (X, d) is a metric space, we let Cb(X) denote the space of all complex-valued continuous functions which are uniformly bounded (If X is compactthen the boundedness is automatic and we use the notation C(X) instead). Forf ∈ Cb(X), the uniform norm of f is given by ‖f‖∞ = supx∈X |f(x)|.

Proposition 1.3.1. Let (X, d) be a metric space, then Cb(X), endowed withthe uniform norm, is a Banach algebra.

Proof. First, note that Cb(X) is clearly an algebra, and ‖·‖∞ is clearly a norm onCb(X). If f, g ∈ Cb(X), then ‖fg‖∞ = supx∈X |f(x)g(x)| ≤ supx,y∈X |f(x)g(y)| =‖f‖∞‖g‖∞ so that Cb(X) is a normed algebra.

Suppose fnn∈N ⊂ Cb(X) is Cauchy. Therefore, for each x ∈ X the se-quence fn(x)n∈N is Cauchy and hence converges to some f(x) ∈ C. We thenhave that 0 = limn→∞ ‖fn − f‖∞, and f ∈ Cb(X) by Proposition 1.2.2. There-fore Cb(X) is complete and hence is a Banach algebra.

Note that if X is a set and d is the metric d(x, y) =

0 if x = y;1 if x 6= y,

then

every function is continuous and hence Cb(X) is the space of all uniformlybounded functions.

1.3.2 Exercises

In the following we consider vector spaces over a field K, where K = R, orK = C.

Recall that if V is a vector space and V0 ⊂ V is a subspace, then the quotientspace V/V0 is defined to be the set of cosets v+ V0 | v ∈ V . This is naturallya vector space whose vector space operations satisfy α(v1 + V0) + (v2 + V0) =(αv1 + v2) + V0, for all v1, v2 ∈ V , and scalar α.


Exercise 1.3.2. Suppose that V is a K-vector space and ‖ · ‖0 is a seminormon V . Set V0 = v ∈ V | ‖v‖0 = 0. Then V0 is a linear subspace and we havea well defined norm on V/V0 given by ‖v + V0‖ = ‖v‖0, for each v ∈ V .

Exercise 1.3.3. Let (V, ‖·‖V ) and (W, ‖·‖W ) be normed spaces, and T : V →Wa linear operator. Then T is bounded if and only if T is continuous.

Exercise 1.3.4. Let (V, ‖ · ‖V ) and (W, ‖ · ‖W ) be normed spaces, then theoperator norm on B(V,W ) is indeed a norm, and that with this norm B(V ) is anormed algebra. Moreover, B(V,W ) is a Banach space if W is a Banach space.

Exercise 1.3.5. Let V be a finite dimensional K-vector space, and suppose‖ · ‖1, and ‖ · ‖2 are norms on V . Then the identity map from (V, ‖ · ‖1) to(V, ‖ · ‖2) is a homeomorphism.

Exercise 1.3.6. Let V be a finite dimensional normed space. Then the closedunit ball B(1, 0) is compact, and V is a Banach space.

Exercise 1.3.7 (Riesz’ lemma). Let (V, ‖ · ‖) be a normed space, W ⊂ V aproper closed subspace, and fix 0 < α < 1. Then there exits x 6∈W with ‖x‖ = 1so that infy∈W ‖x−y‖ ≥ α. (Hint: Start with x0 6∈W , set d = infy∈W ‖x0−y‖,take x1 ∈ W so that ‖x0 − x1‖ ≥ d − ε for some suitably chosen ε > 0, andshow that x = ‖x0 − x1‖−1(x0 − x1) works.)

Exercise 1.3.8. Let V be a normed space such that the closed unit ball B(1, 0)is compact. Then V is finite dimensional.

If (V, ‖ · ‖) is a normed space, a series∑∞n=1 xn is said to converge if the

partial sums∑kn=1 xn converge as k tends to infinite. A series

∑∞n=1 xn is said

to converge absolutely if∑∞n=1 ‖xn‖ <∞.

Exercise 1.3.9. Let (V, ‖ · ‖) be a normed space over K. Then V is a Banachspace if and only if every absolutely convergent series converges.

Exercise 1.3.10. Let (V, ‖ · ‖) be a normed space over K. Then the metricspace completion V of of V has a vector space structure which extends the vectorspace structure of V . Thus, every normed space is a dense linear subspace of aBanach space.

Chapter 2

Measure and integration

Suppose we wanted to assign the notion of size (or measure) to a collection Mof certain subsets of a Rn. That is, for a subset E ∈ M we want to assign anumber 0 ≤ µ(E) ≤ ∞ which tells us in some sense how large E is. Then wemight want the following properties to hold:

(a) M = 2Rn

.

(b) µ is countably additive: If Ej∞j=1 is a sequence of disjoint sets in M,

then µ(∪∞j=1Ej) =∑∞j=1 µ(Ej).

(c) If E can be transformed to F using translations, rotations, and reflections,then µ(E) = µ(F ).

(d) µ assigns a finite, nonzero value to the unit cube.

Unfortunately, these conditions are mutually inconsistent. This was firstnoticed by Vitali in 1905. Suppose we had such a function µ : 2R → [0,∞].Consider the equivalence relation on [0, 1) which is given by s ∼ t if t− s ∈ Q.Take E ⊂ [0, 1) so that E contains exactly one element from each equivalenceclass. For each t ∈ Q consider the set Et = E + t mod 1, i.e.,

Et = ((E + t) ∩ [t, 1)) ∪ ((E + t− 1) ∩ [0, t)).

Then Ett∈Q is a countable family of pairwise disjoint sets which cover [0, 1).Note that for each t ∈ Q we have

µ(Et) = µ((E + t) ∩ [t, 1)) + µ((E + t− 1) ∩ [0, t))

= µ(E ∩ [0, 1− t)) + µ(E ∩ [1− t, 1)) = µ(E).

Hence µ([0, 1)) = µ(∪t∈QEt) =∑t∈Q µ(Et) =

∑t∈Q µ(E), so that µ([0, 1)) ∈

0,∞. A contradiction then follows easily.We must therefore compromise of some of the conditions above. Conditions

(c) and (d) seem essential to having a good notion of size, thus we look to weakenconditions (a) or (b). One thing we might try is to weaken countable additivityto finite additivity:

25

26 CHAPTER 2. MEASURE AND INTEGRATION

(b’) If Ejkj=1 is a finite sequence of disjoint sets in M, then µ(∪∞j=1Ej) =∑kj=1 µ(Ej).

The question of whether there exists a function µ satisfying (a), (b′), (c), and (d)is quite interesting, and we’ll come back to this later. (It turns out that such aµ exists when n ≤ 2, and does not exist otherwise!)

Another possibility is to not try to measure every subset of Rn, but ratheronly a certain nice class M which excludes Vitali’s set above. We would wantM to contains all intervals, and to be closed under taking countable unions andcomplements. In this case, one can indeed obtain such a µ, as was first shownby Lebesgue in 1901 (his dissertation!). Before we present Lebesgue’s proof wefirst take a detour to the abstract setting.

2.1 Measurable sets and functions

Let X be a nonempty set. An algebra of subsets of X is a nonempty collectionA of subsets of X which is closed under finite unions and complements. Aσ-algebra is a nonempty collection E of subsets of X which is closed undercountable unions and complements. Observe that σ-algebras are also closedunder countable intersection. Also, observe that we have ∅, X ∈ E .

Note that the intersection of any family of σ-algebras is again a σ-algebra.It follows that if A is any collection of subsets of X, then there is a uniquesmallest σ-algebraM(A) which contains A. M(A) is the σ-algebra generatedby A. If X is a metric space, then the Borel σ-algebra is the σ-algebra B(X)generated by the open subsets of X.

A measurable space is a pair, consisting of a set X, together with a σ-algebra of subsets of X. Let (X,M) and (Y,N ) be two measurable spaces.A function f : X → Y is measurable if f−1(E) ∈ M for all E ∈ N . Wedenote by M(X;Y ) the set of all measurable functions from X to Y (with theunderlying σ-algebras implicit). We denote by M(X) = M(X;C) where C isendowed with the Borel σ-algebra. Thus, f ∈M(X) if and only if f−1(E) ∈Mfor any Borel set E ⊂ C.

Lemma 2.1.1. Suppose (X,M), (Y,N ), and (Z,P) are measurable spaces andf : X → Y , g : Y → Z are measurable, then g f : X → Z is measurable.

Proof. If E ∈ P then (g f)−1(E) = f−1(g−1(E)) and the result is immediate.

Proposition 2.1.2. Suppose N is generated as a σ-algebra by E ⊂ N . Afunction f : X → Y is measurable if and only if f−1(E) ∈M for all E ∈ E.

Proof. We let A = E ⊂ Y | f−1(E) ∈M. Then E ⊂ A and hence it is enoughto show that A is a σ-algebra. Note that ∅ ∈ A. If E ∈ A, then f−1(Ec) =f−1(E)c and hence Ec ∈ A. Also, if Enn∈N ⊂ A, then f−1(∪n∈NEn) =∪n∈Nf−1(En) and hence ∪n∈NEn ∈ A. Therefore, A is a σ-algebra.

2.1. MEASURABLE SETS AND FUNCTIONS 27

Corollary 2.1.3. Suppose X and Y are metric spaces and f : X → Y iscontinuous, then f is measurable with respect to the Borel σ-algebras.

Proof. Since a function is continuous if and only if the inverse images of opensets are open, and since the Borel σ-algebra is generated by open sets this followsfrom the previous proposition.

Proposition 2.1.4. Let (X, d) be a separable metric space, then B(X) is gen-erated by the open balls B(r, x), for x ∈ X and r > 0.

Proof. Let O ⊂ X be open and let xnn∈N ⊂ O be a countable dense subset.For each n ∈ N we let rn denote the supremum over all r > 0 so that B(r, xn) ⊂O. Then B(rn, xn) ⊂ O, and by density it follows that ∪n∈NB(rn, xn) = O.

Thus, any open set is contained in the σ-algebra generated by open ballsand hence this is also true for any Borel set.

Corollary 2.1.5. Suppose f : X → R. Then the following conditions areequivalent:

1. f ∈M(X;R).

2. f−1(O) ∈M for any open set O ⊂ R.

3. f−1((a, b)) ∈M for any a, b ∈ R.

4. f−1((−∞, b)) ∈M for any b ∈ R.

Proposition 2.1.6. Let (X,M) be a measurable space.

1. If f ∈M(X), and φ : C→ C is continuous, then φ f ∈M(X).

2. If f, g ∈M(X), and α ∈ C, then αf, f+g, fg, |f |,Re (f), Im (f) ∈M(X).

3. If f, g ∈M(X;R) then maxf, g,minf, g ∈ M(X;R).

Proof. The first assertation follows from Lemma 2.1.1 and Corollary 2.1.3. Itthen follows that if f is measurable then so is αf , |f |, Re (f), and Im (f), sincemultiplication by α, absolute value, and taking real and imaginary parts arecontinuous functions.

More generally, consder C2 with the metric d((a, b), (x, y)) = max|a−x|, |b−y|. Then we have B(r, (a, b)) = B(r, a)×B(r, b), and if f, g ∈M(X) then thefunction given by (f, g)(x) = (f(x), g(x)) satisfies

(f, g)−1(B(r, (a, b))) = f−1(B(r, a)) ∩ g−1(B(r, b))

and hence is measurable. Then if φ : C2 → C is continuous we must have thatφ (f, g) is again measurable.

Since addition and multiplication are continuous on C, and since maximumand minimum are continuous on R, it then follows that if f, g ∈ M(X) thenf+g, fg ∈M(X), and if f, g ∈M(X;R) then maxf, g,minf, g ∈ M(X;R).


Proposition 2.1.7. Suppose fnn∈N ⊂ M(X), and f : X → C so thatfn(x)→ f(x), for each x ∈ X. Then f ∈M(X).

Proof. Since a function f is measurable if and only if its real and imaginaryparts are measurable we may assume that fn ∈M(X;R) for each n ∈ N.

If r ∈ R, then f(t) < r if and only if there exists k,N ∈ N so that fn(t) <r − 1/k for all n ≥ N . Hence,

f−1((−∞, r)) = ∪k∈N ∪N∈N ∩n≥Nf−1n ((−∞, r − 1/k)),

and thus f−1((−∞, r)) is measurable. It then follows that f is measurable fromCorollary 2.1.5.

If E ∈ M, then the characteristic (or indicator) function on E is the

function 1E : X → C given by 1E(x) =

1 if x ∈ E,0 if x 6∈ E. Clearly, character-

istic functions are measurable. A simple function is a finite complex linearcombination of characteristic functions. Simple functions are also measurable byProposition 2.1.6, and from the previous proposition we have that any pointwiselimit of simple functions is then measurable.

Proposition 2.1.8. If f ∈ M(X), then f is a pointwise limit of simple func-tions. If f is bounded then f is a uniform limit of simple functions.

Proof. By considering the real and imaginary parts it is enough to considerthe case f ∈ M(X;R). For N ∈ N, and −N2 ≤ k ≤ N2 we let EN,k =f−1

([kN ,

k+1N

)), and set

fN =∑

−N2≤k≤N2

k

N1EN,k .

Then if f(x) ∈ [−N,N) we have |f(x)− fN (x)| ≤ 1/N , and the result follows.

2.1.1 Exercises

Exercise 2.1.9. Suppose we have an algebra C ⊂ 2X with the proeprty that ifEn ∈ C and En ⊂ En+1, for n ≥ 1, then ∪∞n=1En ∈ C. Then C is a σ-algebra.

Exercise 2.1.10. Suppose M ⊂ 2X is an algebra, such that if En∞n=1 ⊂ Mare pairwise disjoint then ∪∞n=1En ∈M. Then M is a σ-algebra.

Exercise 2.1.11. Suppose (X,M) is a measurable space and f, g ∈M(X;R).The sets

x ∈ X | f(x) < g(x) and x ∈ X | f(x) = g(x)

are measurable.

2.2. MEASURES 29

Exercise 2.1.12. Suppose (X,M) is a measurable space and fnn∈N ⊂M(X;R).Set

C = x ∈ X | fn(x)n∈N converges .Then C is measurable.

Exercise 2.1.13. Suppose (X,M, µ) is a measure space, (Y,N ) is a measurablespace and θ : X → Y is measurable. For each set E ⊂ Y we set θ∗µ(E) =µ(θ−1(E)). Then θ∗µ is a measure on (Y,N ) called the push forward measureof µ with respect to θ.

Exercise 2.1.14. The set Borel σ-algebra B ⊂ 2R has cardinality |R|. Hint:To show |B| ≤ |R| let B0 denote the set of open intervals, and inductivelydefine for each ordinal α < ω1 the set Bα+1 to consist of all sets of the form(∪∞i=1Ei) ∪ (∪∞j=1F

cj ), where Ei, fi ∈ Bα, and for each limit ordinal α < ω1 set

Bα = ∪β<αBβ . Then show that |Bα| ≤ |R| for each α < ω1 and B = ∪α<ω1Bα.

2.2 Measures

If (X,M) is a measurable space, then a measure on (X,M) is a set functionµ :M→ [0,∞] that satisfies

1. µ(∅) = 0.

2. µ is countably additive: if En∞n=1 is a sequence of disjoint sets inM,then µ(∪∞n=1En) =

∑∞n=1 µ(En).

A measure space is a triple (X,M, µ) where (X,M) is a measurable spaceand µ is a measure on (X,M).

Here are some basic properties of measures:

Proposition 2.2.1. Let (X,M, µ) be a measure space.

1. (Monotonicity) If E,F ∈M and E ⊂ F , then µ(E) ≤ µ(F ).

2. (Subadditivity) If En∞n=1 ⊂M, then µ(∪∞n=1En) ≤∑∞n=1 µ(En).

3. (Continuity from below) If En∞n=1 ⊂ M and E1 ⊂ E2 ⊂ · · · thenµ(∪∞n=1En) = limn→∞ µ(En).

4. (Continuity from above) If En∞n=1 ⊂ M and E1 ⊃ E2 ⊃ · · · , withµ(E1) <∞, then µ(∩∞n=1En) = limn→∞ µ(En).

Proof. If E,F ∈M with E ⊂ F , then we have µ(F ) = µ(E) +µ(F \E) ≥ µ(E)showing monotonicity.

If En∞n=1 ⊂ M, then setting Fn = En \ (∪k<nEk) we have Fn ⊂ En,Fn∞n=1 ⊂M are pairwise disjoint, and ∪∞n=1Fn = ∪∞n=1En. From monotonic-ity we then obtain subadditivity:

µ(∪∞n=1En) = µ(∪∞n=1Fn) =

∞∑n=1

µ(Fn) ≤∞∑n=1

µ(En).


If En∞n=1 ⊂ M and E1 ⊂ E2 ⊂ · · · , then setting F1 = E1, and Fn =En \ En−1 for n > 1 we have that Fn∞n=1 are pairwise disjoint and hence

µ(∪∞n=1En) = µ(∪∞n=1Fn) =

∞∑n=1

µ(Fn)

= limN→∞

N∑n=1

µ(Fn) = limN→∞

µ(∪Nn=1Fn) = limN→∞

µ(EN ).

If En∞n=1 ⊂ M and E1 ⊃ E2 ⊃ · · · , then taking Fn = E1 \ En, we haveF1 ⊂ F2 ⊂ · · · . By continuity from below we have µ(∪∞n=1Fn) = limn→∞ µ(Fn).Since µ(E1) = µ(En)+µ(Fn) = µ(∪∞n=1Fn)+µ(∩∞n=1En), and since µ(E1) <∞we have

µ(∪∞n=1En) = µ(E1)− µ(∪∞n=1Fn) = limn→∞

µ(E1)− µ(Fn) = limn→∞

µ(En).

A measure µ is finite if µ(X) < ∞. µ or (X,M, µ) is σ-finite, if X =∪∞n=1En where En ∈ M with µ(En) < ∞. µ or (X,M, µ) is semifinite iffor all E ∈ M with µ(E) > 0 there exists A ∈ M with A ⊂ E such that0 < µ(A) <∞.

Note that if (X,M, µ) is σ-finite then it must also be semifinite. Indeed, ifX = ∪∞n=1En with En ∈ M, µ(En) < ∞, and if E ∈ M with µ(E) > 0, thenµ(E ∩ En) ≤ µ(En) <∞, for all n and we have 0 < µ(E ∩ En) for at least onen since 0 < µ(E) ≤

∑∞n=1 µ(E ∩ En).

A measure µ or measure space (X,M, µ) has the essential suprema prop-erty if for any E ⊂M there exists E ∈M such that µ(A\E) = 0 for all A ∈ E ,and if E0 ∈M is any other measurable set which satisfies µ(A \E0) = 0 for allA ∈ E then we also have µ(E\E0) = 0. A measure µ or measure space (X,M, µ)is localizable if it is semifinite and has the essential suprema property.

Here are some examples of measure spaces:

1. If X is a set and M = 2X , then the counting measure on X is givenby µ(E) = |E| if E is finite, and µ(E) = ∞ if E is infinite. It’s not hardto check that this space is always localizable and it is σ-finite if and onlyif X is countable.

2. If X is a nonempty set andM = 2X , then the Dirac measure (or pointmass) at x0 ∈ X is given by µ(E) = 1 if x0 ∈ E, and µ(E) = 0 if x0 6∈ E.

3. If (X,M, µ) is a measure space and E ∈ M then we may consider anew measure µE on (X,M) given by µE(F ) = µ(F ∩ E). This is therestriction measure on E.

4. Suppose X is a set and M consists of all sets E ⊂ X such that eitherE or Ec is countable. Then counting measure restricted to M gives ameasure. This measure is always semifinite and satisfies the essentialsuprema property if and only if X is countable.

2.2. MEASURES 31

5. Suppose (X,M) is a measurable space and N ⊂M is a non-empty collec-tion of subsets such that N is closed under countable union and wheneverwe have E ∈ N and F ∈ M with F ⊂ E then F ∈ N . We define the

measure µ∞ on M by setting µ∞(E) =

0 if E ∈ N ,∞ if E 6∈ N , Then µ gives

a measure on M. This will be semifinite if and only if N =M.

Given a measure space (X,M, µ), we say a set E ⊂ X is σ-finite if E =∪∞n=1En where En ∈ M with µ(En) < ∞. We say that E is a null set ifµ(E) = 0. We say that E is conull if Ec is null. A property is said to holdalmost everywhere (or µ-almost everywhere) if it holds on a conull set.The collection of null sets is non-empty, and closed under countable unionsand taking measurable subsets, therefore given any measure space (X,M, µ)we may consider the corresponding measure µ∞ as described above. Then µ∞

will satisfy the essential suprema property if and only if µ does.In practice most interesting measure spaces one encounters are localizable.

In part because these are the spaces in which a nice integration theory can bedeveloped. The latter two examples above show that there do exist more generalmeasure spaces, however we shall view these spaces as pathalogical.

Lemma 2.2.2. Suppose (X,M, µ) is a measure space and Fn∞n=1 ⊂ M is acountable partition of X so that µFn has the essential suprema property for eachn ≥ 1, then µ has the essential suprema property.

Proof. Suppose E ⊂M, and for each n take En ∈M so that µ(Fn∩(A\En)) = 0for all A ∈ E , and if E0 ∈ M is such that µ(Fn ∩ (A \ E0)) = 0 for all A ∈ Ethen we have µ(Fn ∩ (En \ E0)) = 0.

Set E = ∪∞n=1(En ∩ Fn). If A ∈ E then we have

µ(A \ E) =

∞∑n=1

µ(Fn ∩ (A \ E)) =

∞∑n=1

µ(Fn ∩ (A \ En)) = 0.

Also, if E0 ∈M such that µ(A \ E0) = 0 for all A ∈ E then we have

µ(E \ E0) =

∞∑n=1

µ(Fn ∩ (E \ E0)) =

∞∑n=1

µ(Fn ∩ (En \ E0)) = 0.

Proposition 2.2.3. Suppose (X,M, µ) is a σ-finite measure space, then (X,M, µ)is localizable.

Proof. We already noted above that µ is semifinite, thus we only need to showthat it satisfies the essential suprema property. By the previous lemma it isenough to consider the case when µ is finite. Suppose E ⊂M, and let

E+ = E ∈M | µ(A \ E) = 0 for all A ∈ E.


Note that X ∈ E+ and E+ is closed under countable intersection. Let a =infµ(E) | E ∈ E+, then there exists a sequence Ek∞k=1 ⊂ E+ so thatµ(Ek)→ a. We set E = ∩∞k=1Ek so that E ∈ E+ and µ(E) ≤ infk→∞ µ(Ek) =a ≤ µ(E).

If E0 ∈ E+, then E0 ∩ E ∈ E+ and hence ∞ > µ(E0 ∩ E) ≥ a = µ(E). Wethen have 0 = µ(E)− µ(E0 ∩ E) ≥ µ(E \ E0).

Proposition 2.2.4. If (X,M, µ) is a semifinite measure space, then for allE ∈M we have

µ(E) = supµ(A) | A ⊂ E,A ∈M, and µ(A) <∞.

Proof. If µ(E) <∞ then this is obvious, therefore we may assume that µ(E) =∞. We let a = supµ(A) | A ⊂ E,A ∈ M, and µ(A) < ∞, and take An ⊂E,An ∈ M such that µ(An) → a. We set Bk = ∪kn=1An and B = ∪∞n=1An.Then µ(Bk) → µ(B), and Bk ⊂ E, hence a ≥ µ(Bk) ≥ µ(Ak) → a, so thatµ(B) = a. If we had a <∞ then µ(E \B) =∞ and so by semifiniteness thereexists A0 ∈M, A0 ⊂ E \B so that 0 < µ(A0) <∞. We then have An∪A0 ⊂ Eand µ(An ∪A0) = µ(An) + µ(A0), therefore

a ≥ µ(An ∪A0) = µ(An) + µ(A0)→ a+ µ(A0) > a,

which cannot happen. Thus, we must have a =∞ = µ(E).

Proposition 2.2.5. Suppose (X,M, µ) has the essential suprema property, andF ⊂M(X, [0,∞]). Then there exists h ∈M(X, [0,∞]) so that

µ(x ∈ X | h(x) < f(x)) = 0

for each f ∈ F , and if h ∈M(X, [0,∞]) is any other function with this propertythen we have

µ(x ∈ X | h(x) < h(x)) = 0.

Proof. For each k ≥ 0, n ≥ 1 consider the collection Ek,n = f−1([k/n,∞] | f ∈F, and let Ek,n be such that µ(A \ Ek,n) = 0 for all A ∈ Ek,n and if E0 isanother measurable set with this property then we have µ(E \E0) = 0. It thenfollows that µ(Ek,n \ Ek′,n′) = 0 whenever k/n ≥ k′/n′.

Let h(x) = supk≥0,n≥1k/n | x ∈ Ek,n. Then for each f ∈ F we have

µ(x ∈ X | h(x) < f(x)) = µ(∪k≥0,n≥1x ∈ X | h(x) ≤ k/n < f(x)) = 0.

Moreover, if h is another measurable function with this property and if for eachk ≥ 0, n ≥ 1 we set En,k = h−1([k/n,∞]), then we have µ(f−1(k/n,∞]) \En,k) = 0 for every f ∈ F and hence µ(Ek,n \ Ek,n) = 0. It then follows that

µ(x ∈ X | h(x) < h(x)) = 0.

We call a function h in the previous proposition an essential supremumof F .

2.2. MEASURES 33

2.2.1 Outer measures

An outer measure is a set function µ∗ : 2X → [0,∞] that satisfies

1. µ∗(∅) = 0.

2. (Monotonicity) µ∗(A) ≤ µ∗(B) if A ⊂ B.

3. (Subadditivity) µ∗(∪n∈NAn) ≤∑∞n=1 µ

∗(An).

Proposition 2.2.6. Suppose S ⊂ 2X and µ0 : S → [0,∞] is such that ∅ ∈ S,and µ0(∅) = 0. For E ⊂ X define

µ∗(A) = inf

∞∑n=1

µ0(En) | En ∈ S and A ⊂ ∪∞n=1En

.

Then µ∗ is an outer measure.

Proof. Since ∅ covers itself we clearly have µ∗(∅) = 0.If A ⊂ B ⊂ X, then as any cover of B also covers A it follows that the set

for which we are taking the infimum for B is contained in the corresponding setfor A. Therefore µ∗(A) ≤ µ∗(B).

Fix ε > 0. If Ann∈N ⊂ 2X , then for each n ∈ N there exists Enj j∈N ⊂ Sso that

∑j∈N µ0(Enj ) < µ∗(An) + ε2−n. We then have ∪n∈NAn ⊂ ∪n,j∈NEnj ,

and soµ∗(∪n∈NAn) ≤

∑n,j∈N

µ0(Enj ) < ε+∑n∈N

µ∗(An).

As ε > 0 was arbitrary it then follows that

µ∗(∪n∈NAn) ≤∑n∈N

µ∗(An).

The outer measure µ∗ in the previous proposition is called the outer mea-sure associated to µ0.

If µ∗ is an outer measure on X, then a set A ⊂ X is µ∗-measurable if

µ∗(S) = µ∗(S ∩A) + µ∗(S ∩Ac) for all S ⊂ X.

Theorem 2.2.7 (Caratheodory). Suppose µ∗ is an outer measure on X, thenthe collection M of all µ∗-measurable sets is a σ-algebra, and the restriction ofµ∗ to M is a measure.

Proof. Since µ∗(∅) = 0 we have µ∗(S) = µ∗(∅) + µ∗(S) for each S ⊂ X, hence∅ ∈ M. Also, note that M is clearly closed under taking complements.

If A,B ∈M, then for each S ⊂ X we have

µ∗(S) = µ∗(S ∩A) + µ∗(S ∩Ac)= µ∗(S ∩A ∩B) + µ∗(S ∩A ∩Bc) + µ∗(S ∩Ac ∩B) + µ∗(S ∩Ac ∩Bc)≥ µ∗(S ∩ (A ∪B)) + µ∗(S ∩ (A ∪B)c) ≥ µ∗(S).


We therefore have that A ∪ B ∈ M and if A and B are disjoint then takingS = A ∪B we have

µ∗(A ∪B) = µ∗((A ∪B) ∩A) + µ∗((A ∪B) ∩Ac) = µ∗(A) + µ∗(B).

It then follows easily thatM is closed under unions of finite families, and henceM is an algebra. To show thatM is a σ-algebra it is then enough to show thatM is closed under taking countable unions of pairwise disjoint families.

If Ann∈N ⊂ M is a sequence of pairwise disjoint sets, then set Bn =∪nk=1Ak and B = ∪∞k=1Ak. Since An ∈M, if S ⊂ X, and n > 1 we have

µ∗(S ∩Bn) = µ∗(S ∩Bn ∩An) + µ∗(S ∩Bn ∩Acn)

= µ∗(S ∩An) + µ∗(S ∩Bn−1).

By induction it then follows easily that

µ∗(S ∩Bn) =

n∑k=1

µ∗(S ∩An).

Hence,

µ∗(S) = µ∗(S ∩Bn) + µ∗(S ∩Bcn)

≥n∑k=1

µ∗(S ∩Ak) + µ∗(S ∩Bc).

Taking n→∞ then gives

µ∗(S) ≥∞∑k=1

µ∗(S ∩Ak) + µ∗(S ∩Bc)

≥ µ∗(S ∩B) + µ∗(S ∩Bc) ≥ µ∗(S). (2.1)

Thus, B ∈M, showing that M is a σ-algebra. Taking S = B in (2.1) shows

µ∗(B) =

∞∑k=1

µ∗(Ak).

Hence µ∗ defines a measure on M.

2.2.2 Caratheodory’s extension theorem

If A ⊂ 2X is an algebra, a function µ0 : A → [0,∞] is a premeasure if

1. µ0(∅) = 0.

2. Whenever Enn∈N ⊂ A are disjoint such that ∪∞n=1En ∈ A, then we haveµ0(∪∞n=1En) =

∑∞n=1 µ0(En).

2.2. MEASURES 35

Theorem 2.2.8 (Caratheodory’s extension theorem). Suppose A ⊂ 2X is analgebra, µ0 : A → [0,∞] is a premeasure, and µ∗ is the associated outer measure,then every set E ∈ A is µ∗-measurable and we have µ∗(E) = µ0(E).

Moreover, if M denotes the σ-algebra generated by A, and if µ∗ defines asemifinite measure on M, then µ∗ is the unique measure on M which extendsµ0.

Proof. If E,An ∈ A, for n ≥ 1 with E ⊂ ∪∞n=1An, then setting Bn = E ∩ (An \(∪n−1k=1Ak)) we have that Bn ⊂ An, and Bn∞n=1 is a family of pairwise disjoint

sets in A such that E = ∪∞n=1Bn. We therefore have µ0(E) =∑∞n=1 µ0(Bn) ≤∑∞

n=1 µ0(An). Thus, it follows that µ0(E) ≤ µ∗(E) ≤ µ0(E). Hence, µ∗ is anextension of µ0.

If A ∈ A, S ⊂ X, and ε > 0, then we may take An∞n=1 ⊂ A so thatS ⊂ ∪∞n=1An and

∑∞n=1 µ0(An) ≤ µ∗(S) + ε. Since µ0 is finitely additive on A

it then follows that

µ∗(S) + ε ≥∞∑n=1

(µ0(An ∩A) + µ0(An ∩Ac))

≥ µ∗(S ∩A) + µ∗(S ∩A∗) ≥ µ∗(S).

As this was for ε > 0 arbitrary we then have that A is µ∗-measurable.

Suppose now thatM is the σ-algebra generated by A, and let ν be anothermeasure on (X,M) so that ν(A) = µ0(A) for all A ∈ A. Then for E ∈ M, ifE ⊂ ∪∞n=1An with An ∈ A we have

ν(E) ≤∞∑n=1

ν(An) =

∞∑n=1

µ0(An),

and it follows that ν(E) ≤ µ∗(E).

If we have E ∈ M such that µ∗(E) < ∞, and if ε > 0, then there existAn∞n=1 ⊂ A so that µ∗(E) + ε > µ∗(∪∞n=1An), and hence setting A = ∪∞n=1Anwe have µ∗(A \ E) < ε. Therefore,

µ∗(E) ≤ µ∗(A) = ν(A) = ν(E) + ν(A \ E)

≤ ν(E) + µ∗(A \ E) < ν(E) + ε.

Since ε > 0 was arbitrary we then have µ∗(E) ≤ ν(E) whenever µ∗(E) < ∞.If µ∗ gives a semifinite measure on M then by Proposition 2.2.4 it follows thatfor all E ∈M we have

ν(E) ≥ supν(A) | A ⊂ E,A ∈M, and ν(A) <∞≥ supµ∗(A) | A ⊂ E,A ∈M, and µ∗(A) <∞ = µ∗(E),

and hence in this case we have ν(E) = µ∗(E) for all E ∈M.


2.2.3 Exercises

A measure space (X,M, µ) is complete if every subset of a null set is measur-able (and hence also null).

Exercise 2.2.9. Suppose (X,M, µ) is a measure space and let N = E ∈M | µ(E) be the space of null sets. We let M = E ∪ F | E ∈ M and F ⊂N for some N ∈ N. Then M is a σ-algebra and there is a unique extension µof µ to a complete measure on M.

The measure space (X,M, µ) from the previous theorem is called the com-pletion of (X,M, µ).

Exercise 2.2.10. If µ∗ is an outer measure on X, M is the collection of allµ∗-measurable sets, and µ is the restriction of µ∗ to M, then (X,M, µ) is acomplete measure space.

Let (X,M, µ) be a measure space. A function f ∈ M(X) is essentiallybounded if there exists M ∈ [0,∞) such that µ(x ∈ X | |f(x)| > M) = 0.We let L∞(X,µ) denote the space of all (complex valued) essentially boundedfunctions, and for f ∈ L∞(X,µ) we set

‖f‖ = infM ∈ [0,∞) | µ(x ∈ X | |f(x)| > M) = 0.

For clarity, we sometimes may write ‖f‖∞ instead of ‖f‖.

Exercise 2.2.11. L∞(X,µ) is an algebra and ‖ · ‖∞ gives a seminorm onL∞(X,µ).

We let L∞(X,µ) be the normed algebra obtained from L∞(X,µ) by iden-tifying two functions f and g when ‖f − g‖∞ = 0, i.e., when f = g almosteverywhere.

Exercise 2.2.12. If fnn∈N ⊂ L∞(X,µ) is Cauchy with respect to ‖ · ‖∞,then there exists f ∈ L∞(X,µ) such that ‖f − fn‖∞ → 0, hence L∞(X,µ) is aBanach algebra.

Exercise 2.2.13. Let (X,M, µ) be a finite measure space and for E,F ∈ Mset ρ(E,F ) = µ(E∆F ). Then ρ gives a semimetric on M.

Exercise 2.2.14. Let (X,M, µ) be a finite measure space and ρ defined asabove. Then ρ is a complete semimetric. Hint: If Enn∈N is Cauchy, bypassing to a subsequence we may suppose µ(En∆Em) ≤ max2−n, 2−m, andin this case setting Fm = ∪k≥mEk, we have that Fmm∈N is again Cauchy, andµ(Fm∆En) < 2−n+4 for m > n.

2.3 Borel measures on RBy a Borel measure on a metric space, we mean a measure on the Borelσ-algebra.

2.3. BOREL MEASURES ON R 37

Lemma 2.3.1. Let F : R → R be increasing and right continuous. Let I bethe collection of intervals of the form (a, b], for −∞ ≤ a < b < ∞, or of theform (a,∞) for −∞ ≤ a < ∞, and set µ0((a, b]) = F (b)− F (a) if b < ∞, andµ0((a,∞)) = F (∞) − F (a), where F (±∞) = limt→±∞ F (t). We let A denotethe algebra consisting of finite unions of intervals in I. Then µ0 extends to apremeasure on A.

Proof. Note that if (a, b] = ∪nk=1(ak, bk], then after rearranging we may assumethat a = a1 < b1 = a2 < b2 = · · · < bk−1 = ak < bk = b, and we have that

µ0((a, b]) = F (b)− F (a) =

n∑k=1

F (bk)− F (ak) =

n∑k=1

µ0((ak, bk]).

We similarly have that if I1, . . . , In ∈ I are disjoint and ∪nk=1Ik = (a,∞), thenµ0((a,∞)) =

∑nk=1 µ0(Ik). From this it then follows easily that we obtain a well

defined finitely additive set function µ0 : A → [0,∞] by setting µ0(∪nk=1Ik) =∑nk=1 µ0(Ik) for pairwise disjoint sets I1, . . . , In ∈ I.We will now show that µ0 is a premeasure on A. Suppose that Ij∞j=1 is

a pairwise disjoint sequence of intervals in A, such that ∪∞j=1Ij = I ∈ I. Thenwe have

µ0(I) = µ0(∪nj=1Ij) + µ0(I \ ∪nj=1Ij) ≥ µ0(∪nj=1Ij) =

n∑j=1

µ0(Ij).

Taking a limit as n→∞ we see that µ0(I) ≥∑∞n=1 µ0(Ij).

For the reverse inequality we first assume that I = (a, b], where a and bare finite. Fix ε > 0. As F is right continuous there exists δ > 0 so thatF (a + δ) − F (a) < ε. Similarly, if Ij = (aj , bj ] then there exist δj > 0 sothat F (bj + δj) − F (bj) < ε2−j . Since the open intervals (aj , bj + δj) coverthe compact set [a + δ, b] there exists n ∈ N and j1, . . . , jn, so that [a + δ, b] ⊂∪ni=1(aj1 , bji + δji). We may further assume that no subcollection also covers[a + δ, b] and by reordering and reindexing j1, . . . , jn as 1, . . . , n we may thenassume that

a1 < a+ δ ≤ a2 < b1 + δ1 ≤ a3 < · · · ≤ an < bn−1 + δn−1 ≤ b < bn + δn.

We then have

µ0(I) < F (b)− F (a+ δ) + ε

≤ F (bn + δn)− F (a1) + ε

= F (bn + δn)− F (an) +

n−1∑j=1

(F (aj+1)− F (aj)) + ε

≤ F (bn + δn)− F (an) +

n−1∑j=1

(F (bj + δj)− F (aj)) + ε

≤n∑j=1

F (bn)− F (an) + 2ε ≤∞∑j=1

µ0(Ij) + 2ε.


As ε > 0 was arbitrary it then follows that µ0(I) ≤∑∞j=1 µ0(Ij).

For the case of a general interval I ∈ I, we can easily check that have µ0(I) =lima→−∞,b→∞ µ0(I∩[a, b)) and

∑∞j=1 µ0(Ij) = lim a→ −∞, b→∞

∑∞j=1 µ0(Ij∩

[a, b)). Using the case −∞ < a < b < ∞ above and taking limits then showsthat µ0(I) =

∑∞j=1 µ0(Ij).

If we now consider general sets E,Ej ∈ A, such that Ej∞j=1 is pairwisedisjoint and E = ∪∞j=1Ej , then writing each set as a finite union of disjoint inter-

vals, and using finite additivity of µ0 it then follows that µ0(E) =∑∞j=1 µ0(Ej).

Hence, µ0 is a premeasure on A.

Theorem 2.3.2. Let F : R → R be increasing and right continuous. Thenthere is a unique Borel measure µF on R such that µF ((a, b]) = F (b)−F (a) forall a, b ∈ R.

Conversely, if µ is a Borel measure on R which is finite on all bounded Borelsets and we define

F (x) =

µ((0, x]) if x > 0,0 if x = 0,

−µ((−x, 0]) if x < 0,

then F is increasing, right continuous, and µ = µF .

Proof. We let A denote the algebra in Lemma 2.3.1, and note that the σ-algebra generated by A is the Borel σ-algebra. By Lemma 2.3.1 there existsa premeasure µ0 on A so that µ0((a, b]) = F (b) − F (a) for a, b ∈ R. ByCaratheodory’s extension theorem there then exists a Borel measure µf on Rsuch that µF ((a, b]) = F (b) − F (a). Moreover, since µF is σ-finite it then alsofollows from Caratheodory’s extension theorem that µF is the unique Borelmeasure with this property.

If µ is a Borel measure on R which is finite on all bounded Borel sets andif we define F as above, then it follows from monotonicity that F is increasingand from continuity from above/below that F is right continuous. Moreover,we see easily that for a, b ∈ R we have µ((a, b]) = F (b) − F (a). By uniquenessof the measure µF it then follows that µ = µF .

Given F : R → R increasing and right continuous, the completion of thecorresponding measure µF is called the Lebesgue-Stieltjes measure associ-ated to F , and F is called a distribution function associated to µF . It’s easyto check that two distribution functions associated to the same measure mustdiffer by a constant.

2.3.1 Lebesgue measure on RThe Lebesgue-Stieltjes measure corresponding to the function F (x) = x is calledLebesgue measure on R and usually denoted by λ. A set E ⊂ R is Lebesguemeasurable if it is λ∗-measurable where λ∗ is the outer measure corresopndingto λ. Note that by Exercise 2.2.10 if E ⊂ R satisfies λ∗(E) = 0, then E isLebesgue measurable.


Theorem 2.3.3. If E ⊂ R is Borel, then so is E + s and rE for all s, r ∈ R.Moreover, we have λ(E + s) = λ(E) and λ(rE) = |r|λ(E).

Proof. Since addition and multiplication are continuous it follows from Corol-lary 2.1.3 that E+ s and rE are Borel. By uniqueness of the Lebesgue-Stieltjesmeasure to show that λ(E+ s) = λ(E) and λ(rE) = |r|λ(E), it suffices to showthese equalities when E is a half open interval, in which case this is obvious.

Note that every point x ∈ R has Lebesgue measure zero. It follows thatevery countable set has Lebesgue measure zero. There are also uncountablesets with Lebesgue measure zero. The Cantor set C is the set of all x ∈ [0, 1]that have a base-3 expansion x =

∑∞n=1 an3−n with an 6= 1 for all n (note that

such an expansion, if it exists, must be unique). We may may obtain C bystarting with the unit interval [0, 1] and removing the open middle third ( 1

3 ,23 ),

then removing the open middle thirds ( 19 ,

29 ) and ( 7

9 ,89 ) of the two remaining

interval, etc.

Proposition 2.3.4. The Cantor set C is compact, contains no non-trivial openinterval, and has no isolated points. Moreover, C has cardinality |R| and satis-fies λ(C) = 0.

Proof. C is obtained by removing open intervals, thus C is a decreasing unionof closed subsets of [0, 1], and so C itself is a closed subset of [0, 1] which mustthen be compact. If x =

∑∞n=1 an3−n ∈ C with an ∈ 0, 2 for all n, then for

each n ∈ N consider xn ∈ C which has the same expansion as x except for thenth coefficient, which is either 0 if an = 2, or 2 if an = 0. Then xnn∈N ⊂ Cis an infinite sequence such that xn → x. Hence, C has no isolated points.

By considering the lengths of the intervals removed from C we have

λ(C) = 1−∞∑n=1

2n−1

3n= 0.

If x =∑∞n=1 an3−n ∈ C with an 6= 1 for all n, then set f(x) =

∑∞n=1 bn2−n

where bn = an/2. Then the series describing f(x) is a base-2 expansion andevery number in [0, 1] can be expressed in this way, thus f : C → [0, 1] is asurjection which shows that |C| = |R|.

Corollary 2.3.5. Let L ⊂ 2R denote the σ-algebra of Lebesgue measurablesubsets, then |L| = |2R|. Hence, B(R) ( L ( 2R.

Proof. We clearly have |L| ≤ |2R|, and if C is the Cantor set then λ(C) = 0,hence any subset of C is Lebesgue measurable and so we have |2R| = |2C | ≤ |L|.

By Exercise 2.1.14 we have |B(R)| = |R|, hence B(R) ( L. Also, Vitali’s setE constucted at the beginning of the chapter cannot be Lebesgue measurablehence L ( 2R.

Note that the function f : C → [0, 1] in the proof of Proposition 2.3.4 ismonotone increasing. Moreover, if x, y ∈ C, with x < y, then f(x) = f(y)


only if x and y are the endpoints of one of the intervals removed from [0, 1]. Inthis case we have f(x) = f(y) = m2−n where m,n are integers. Thus we mayextend f on the interval (x, y) by letting it be constant m2−n. In this way weextend f to a monotone increasing function f : [0, 1]→ [0, 1]. The function f iscalled the Cantor function.

Theorem 2.3.6. Let f : [0, 1] → [0, 1] be the Cantor function. Then the fol-lowing hold:

1. f is continuous.

2. The derivative of f exists, and equals zero, almost everywhere.

3. There exists a Lebesgue measurable set E ⊂ [0, 1] such that f(E) is notLebesgue measurable.

Proof. Note that f is surjective and hence cannot have any jump discontinuities.Since f is monotone increasing it must then be continuous.

f is constant on each middle third, and hence the derivative of f exists andequals zero, on each middle third, and as we say above, the union of these openintervals is conull.

If we consider Vitali’s example of a non-measurable set E ⊂ [0, 1], thensetting E0 = f−1(E) ∩ C we have that E0 is contained in a measure zero setand hence must be measurable. Since f(C) = [0, 1] we have f(E0) = E.

2.3.2 Regularity of Borel measures

Theorem 2.3.7. Suppose µ is a finite Borel measure on a metric space (X, d).Then µ is regular: For E ⊂ X Borel we have

µ(E) = infµ(G) | E ⊂ G and G is open= supµ(F ) | F ⊂ E and F is closed.

Proof. We let Σ denote the family of Borel sets E which satisfy the conclusionof the theorem. If E ⊂ X is closed then Gn = x ∈ X | d(x,E) < 1/n isopen for each n ∈ N and we have ∩∞n=1Gn = E. By continuity from above ofmeasures we have that µ(E) = limn→∞ µ(Gn). Hence it follows that E ∈ Σ. Itis also clear that Σ is closed under taking complements.

If En∞n=1 ⊂ Σ and ε > 0, then there exist Fn, Gn ⊂ X with Fn closed andGn open such that Fn ⊂ En ⊂ Gn and µ(Gn \ Fn) < ε2−n. If we set G = ∪∞n=1

and F = ∪∞n=1Fn then we have F ⊂ ∪∞n=1En ⊂ G, and G \ F ⊂ ∪∞n=1(Gn \ Fn)hence µ(G \ F ) ≤

∑∞n=1 µ(Gn \ Fn) < ε. By continuity from above we have

limn→∞ µ(G \ (∪nk=1Fk)) = µ(G \ F ) < ε. Hence, for n large enough we haveµ(G \ (∪nk=1Fk)) < ε. Since G is open and ∪nk=1Fk ⊂ F ⊂ E is closed it thenfollows that ∪∞n=1En ∈ Σ. Thus, Σ is a σ-algebra which contains the closed setsand hence must contain all Borel sets.

A set E ⊂ X is a Gδ-set if it a countable intersection of open sets. A setE ⊂ X is an Fσ-set if it is a countable union of closed sets.


Corollary 2.3.8. Suppose µ is a σ-finite Borel measure on a metric space(X, d). Then for every Borel set E ⊂ X there exists an Fσ-set F , and a Gδ-setG such that F ⊂ E ⊂ G and µ(G \ F ) = 0.

Proof. This follows easily from Theorem 2.3.7 when µ is finite. If µ is σ-finitewe may write X as a disjoint union X = ∪nn=1En where En ⊂ X is Boreland µ(En) < ∞. Suppose E ⊂ X is Borel and ε > 0. For each n ≥ 1consider the Borel measure µn on X given by µn(A) = µ(A∩En), then µn is afinite measure and so there exist Fσ-sets F 1

n ⊂ E ∩ En, F 2n ⊂ Ec ∩ En so that

µ((E ∩ En) \ F 1n) = µ((Ec ∩ En) \ F 2

n) = 0.

If we set F 1 = ∪∞n=1F1n ⊂ E and F 2 = ∪∞n=1F

2n ⊂ Ec, then we have µ(E \

F 1) = µ(Ec \ F 2) = 0. Then G = (F 2)c is Gδ and satisfies E ⊂ G, andµ(G \E) = µ(Ec \ F 2) = 0. Hence, µ(G \ F 1) = µ(G \E) + µ(E \ F 1) = 0.

Corollary 2.3.9. Suppose µF is a Lebesgue-Stieltjes measure on R, and E ⊂ Ris a Borel set such that µF (E) <∞. Then for every ε > 0, there exists G ⊂ Rsuch that G is a finite union of intervals and µF (E∆G) < ε.

Proof. Suppose E ⊂ R is Borel and ε > 0. Take t > 0 so that µ(E \ (−t, t)) <ε/3. By the previous Corollary there exists a sequence of open set Gn such thatµ(E∆∩∞n=1Gn) = 0. Thus, settingG′n = Gn∩(−t, t) we have limk→∞ µ((∩kn=1G

′n)\

E) = 0, and hence for some k we have µ((∩kn=1G′n) \ E) < ε/3. Since ∩kn=1G

′n

is open it is a countable union of intervals, hence there exists a set G which isa finite union of intervals such that µ((∩kn=1G

′n) \G) < ε/3. We then have

µ(E∆G) ≤ µ(E \ (−t, t)) + µ((∩kn=1G′n) \ E) + µ((∩kn=1G

′n) \G) < ε.

Theorem 2.3.10. Suppose µ is a finite Borel measure on a complete separablemetric space (X, d). Then µ is tight: For E ⊂ X Borel we have

µ(E) = supµ(K) | K ⊂ A and K is compact.

Proof. Since µ is regular it is enough to consider the case when E is closed, andthen restricting to E we might as well assume that E = X. So it suffices toshow µ(X) = supµ(K) | K is compact.

Fix ε > 0. Let xi∞i=1 be a countable dense set in X. If n ≥ 1 then∪∞i=1B(1/n, xi) = X and hence limk→∞ µ(∪ki=1B(1/n, xi)) = µ(X). We takekn so that µ(X \ ∪kni=1B(1/n, xi)) < ε2−n. Let K = ∩∞n=1 ∪

kni=1 B(1/n, xi).

Then K is closed and totally bounded, and hence compact. We also haveµ(X \K) ≤

∑∞n=1 µ(X \ ∪kni=1B(1/n, xi)) < ε.

Theorem 2.3.11 (Lusin’s Theorem). Suppose µ is a finite Borel measure ona metric space (X, d), and f ∈ M(X). For each ε > 0 there exists a closed setF ⊂ X such that µ(F c) < ε and f|F is continuous.


Proof. Fix ε > 0, and take an enumeration of the rationals Q = qn∞n=1.Then En,k = f−1((qn, qk)) is measurable and hence there exist Fn,k closed andVn,k opne so that Fn,k ⊂ En,k ⊂ Vn,k with µ(Vn,k \ Fn,k) < ε2−n−k. SetU = ∪n,k(Vn,k \ Fn,k) and F = U c. Then µ(U) < ε, and F is closed. Moreover,f−1((qn, qk))∩F = Vn,k ∩F . Since every open set is a union of sets of the form(qn, qk) it then follows easily that f|F is continuous.

Corollary 2.3.12. Suppose µF is a Lebesgue-Stieltjes measure on R, and f ∈M(R) is such that f vanishes outside a finite measure set. Then for all ε > 0there exists a continuous function g ∈ C0(R) so that

µF (x ∈ R | f(x) 6= g(x)) < ε.

Proof. Fix ε > 0 and take t0 > 0 so that µF (((−∞,−t0) ∪ (t0,∞)) ∩ x ∈R | f(x) 6= 0) < ε/4. By considering the restriction of µF to [−t0, t0] theprevious theorem gives a closed set E ⊂ [−t0, t0] so that f|E is continuous andµF ([−t0, t0] \ E) < ε/4.

We let a = inf E and b = supE, and take a′ < a, and b′ > b so thatµF ([a′, a)) + µF ((b, b′]) < ε/2. If t ∈ Ec, a < t < b we let (t1, t2) denote thelargest interval in Ec which contains t. We define g so that

g(t) =

f(t) if t ∈ E,0 if t < a′, or t > b′,t−a′a−a′ f(a) if ∈ [a− 1, a),b′−tb′−bf(b) if t ∈ (b, b+ 1],t−t1t2−t1 f(t2) + t2−t

t2−t1 f(t1) if t ∈ (t1, t2).

We then have that g ∈ C0(R) and g agrees with f on E, hence it follows easilythat µF (x ∈ R | f(x) = g(x)) < ε.

2.3.3 Exercises

Exercise 2.3.13. For every ε > 0, there exists a compact set K ⊂ [0, 1] whichcontains no isolated points and no non-trival open interval such that λ(K) >1− ε.

Exercise 2.3.14. Let E ⊂ R be a Borel set such that λ(E) < ∞. Then themaps R 3 t 7→ λ(E∆tE), and t 7→ λ(E∆(E + t)) are continuous.

Exercise 2.3.15. If (X,M, µ) is a measure space and Aj∞j=1 ⊂ M, we setlim infj→∞Aj = ∪∞N=1 ∩∞k=N Ak. Then µ(lim inf Aj) ≤ lim inf µ(Aj).

Exercise 2.3.16. There exists a measurable function f : [0, 1]→ R so that forall (a, b) ⊂ [0, 1] and (c, d) ⊂ R we have

λ(x ∈ (a, b) | f(x) ∈ (c, d)) > 0.

Exercise 2.3.17. Show that there exists a Borel set A ⊂ [0, 1] such that 0 <m(A ∩ I) < m(I) for every subinterval I of [0, 1].

2.4. INTEGRATION 43

2.4 Integration

2.4.1 Integrable functions

Let (X,M, µ) be a measure space. If E ∈ M has finite measure, and f =∑kn=1 αn1En is a simple function with respect to some measurable partition

Enkn=1 of E, then we define the integral of f to be∫f =

k∑n=1

αnµ(En).

Note that if we have another representation f =∑lm=1 βm1Fm then by writing

Fm = ∪kn=1Fm ∩ En we see that∑kn=1 αnµ(Fm ∩ En) = βmµ(Fm). Summing

over m then shows that the integral is well defined. Depending on the situationwe also use the following notation for the integral∫

f dµ;

∫X

f dµ;

∫X

f(x) dµ(x);

∫X

f(x) dx

If A ⊂ X is measurable we write∫Af for the integral

∫X

1Af .We let L1

0(X) denote the set of all simple functions having a decomposition

f =∑kn=1 αn1En where the Enkn=1 gives a partition of a finite measure set E.

If I ⊂ C we denote by L10(X; I) those functions in L1

0(X) which take values in I,and we also set L1

0(X)+ = L10(X; [0∞)). We note that L1

0(X) is a vector spaceover C. We also note that the integral defines a linear functional on L1

0(X).If f ∈ L1

0(X)+ then we clearly have∫f ≥ 0. Linearity then shows that for

f, g ∈ L10(X;R) we have ∫

f ≤∫g, if f ≤ g. (2.2)

Also, note that for f ∈ L10(X) the triangle inequality in C shows that∣∣∣∣∫ f

∣∣∣∣ ≤ ∫ |f |. (2.3)

Similarly, if we have f, g ∈ L10(X), then taking a partition of a set of finite mea-

sure such that both f and g are simple functions with respect to this partition,the triangle inequality in C shows that∫

|f + g| ≤∫|f |+

∫|g|. (2.4)

Therefore, we may define the L1-seminorm on L10(X) as

‖f‖1 =

∫|f |. (2.5)

Finally, note that ‖f‖1 = 0 if and only if f = 0 almost everywhere.


We let L1(X) be the Banach space completion of L10(X) after identifying

functions which agree almost everywhere. Depending on the situation we mayuse the terminology

L1(X); L1(µ); L1(X,µ); L1(X,M, µ).

Note that since |∫f | ≤

∫|f | it follows that the integral is continuous with

respect to the seminorm ‖ · ‖1. Hence, the integral extends continuously toL1(X), and we use the same terminology here.

Every vector in L1(X) is the limit of a Cauchy sequence of functions inL1

0(X). We now wish to find a more tractable realization of vectors in L1(X).

Lemma 2.4.1. Let fnn∈N ⊂ L10(X) be a Cauchy sequence, then there exists a

subsequence fnkk∈N which converges almost everywhere to a measurable func-tion f , and such that for all ε > 0 there exists a measurable set A ⊂ X withµ(A) < ε, such that fnkk∈N converges to f uniformly on Ac.

Proof. Suppose fnn∈N ⊂ L10(X) is Cauchy. Passing to a subsequence we may

assume that ‖fn − fn+1‖1 ≤ 4−n, for all n ∈ N. Set

En = x ∈ X | |fn(x)− fn+1(x)| ≥ 2−n.

Then we have |fn − fn+1| ≥ 2−n1En , and hence it follows that

4−n ≥∫|fn − fn+1| ≥

∫2−n1En = 2−nµ(En).

For N ∈ N set AN = ∪n≥NEn, and set A = ∩N∈NAN . Then

µ(AN ) ≤∑n≥N

µ(En) ≤∑n≥N

2−n < 2−N+1.

Hence µ(A) = 0.If x ∈ AcN and n ≥ N then |fn(x) − fn(x + 1)| < 2−n. It therefore follows

that fn(x)n∈N is Cauchy and hence converges to some f(x) ∈ C. There-fore there exists a function f : Ac → C so that fn(x) → f(x) for all x ∈ Ac.Note that we also have that fn converges uniformly to f on each set AN , andlimN→∞ µ(AN ) = 0. From Proposition 2.1.7 we see that f : Ac → C is measur-able, and we may extend f to a measurable function on X by setting f(x) = 0for all x ∈ A.

Lemma 2.4.2. Suppose fnn∈N, gnn∈N ⊂ L10(X) are Cauchy sequences, and

f ∈ M(X) such that both sequences converge almost everywhere to f . Thenlimn→∞ ‖fn − gn‖1 = 0.

Proof. If we consider hn = fn − gn, then hn∞n=1 is Cauchy in L10(X) and

satisfies hn → 0 almost everywhere. We must show that ‖hn‖1 → 0. Fix ε > 0,and take N ∈ N so that ‖hn − hm‖1 < ε for all n,m ≥ N . Using Lemma 2.4.1and passing to a subsequence we may assume that there exists a measurable set

2.4. INTEGRATION 45

A ⊂ X with µ(A) < ε1+‖hN‖∞ , such that hn → 0 uniformly on Ac. Let E ⊂ X

be a set of finite measure such that fN vanishes outside of E. Then for n largewe have ∫

A∪Ec|hn| ≤

∫A∪Ec

|hN |+∫A∪Ec

|hn − hN |

≤∫A

|hN |+ ‖hn − hN‖1

≤ µ(A)‖hN‖∞ + ‖hn − hN‖1 < 2ε.

Hence,

lim supn→∞

‖hn‖1 = lim supn→∞

(∫A∪Ec

|hn|+∫Ac∩E

|hn|)

≤ 2ε+ µ(E) lim supn→∞

‖hn|Ac‖∞ = 2ε.

Since ε > 0 was arbitrary we have that ‖hn‖1 → 0.

We let L1(X) denote the set of all measurable functions f ∈M(X) such thatf is the almost everywhere limit of a Cauchy sequence of functions in L1

0(X).Functions in L1(X) are said to be integrable. Clearly, L1(X) is a vector space.Lemma 2.4.2 shows that we have a well defined map Ξ : L1(X)→ L1(X) whichassigns to each function f ∈ L1(X) a Cauchy sequence which converges almosteverywhere to f . The map Ξ is clearly linear and the kernel consists of allmeasurable functions which are zero almost everywhere. We extend the integralto a linear map on L1(X) by setting

∫f =

∫Ξ(f). In other words, given a

function f ∈ L1(X) we have∫f = limn→∞

∫fn where fnn∈N is a Cauchy

sequence in L10(X) which converges almost everywhere to f .

Lemma 2.4.1 shows that the map Ξ is surjective. Thus, we may think of thespace L1(X) as the space of integrable functions where we identify functionswhich agree almost everywhere.

2.4.2 Properties of integration

Lemma 2.4.3. If fnn∈N ⊂ L10(X) is Cauchy, and f ∈ L1(X) such that

fn → f almost everywhere, then |fn|n∈N is also Cauchy, and |fn| → |f |almost everywhere.

Proof. This follows easily from the inequality ||a| − |b|| ≤ |a − b| for all a, b ∈C.

Theorem 2.4.4. If f ∈ L1(X) then |f | ∈ L1(X). Moreover, the inequalities(2.2), (2.3), (2.4), and (2.5) hold for all functions in L1(X).

Proof. The fact that |f | ∈ L1(X) if f ∈ L1(X) follows from the previous lemma,which also shows that the function f 7→ |f | is uniformly continuous on boundedsubsets of L1

0(X). Since f 7→∫f and f 7→ ‖f‖1 are also uniformly continuous


on bounded subsets of L10(X), it follows that these maps are also continuous on

the completion L1(X). The inequalities (2.2), (2.3), (2.4), and (2.5) then followby continuity, since they hold on the dense subspace L1

0(X).

Recall that if we let L∞(X) denote the space of essentially bounded func-tions, then we have a complete semi-norm on L∞(X) given by ‖f‖∞ = infM ∈[0,∞) | µ(x ∈ X | |f(x)| > M) = 0. An essentially bounded function f sat-isfies ‖f‖∞ = 0 if and only if f is zero almost everywhere. Thus, if we identifyfunctions which agree almost everywhere then we obtain a Banach space L∞(X).This is also a Banach algebra since ‖fg‖∞ ≤ ‖f‖∞‖g‖∞. We let L∞0 (X) denotethe space of simple functions. Since any function f ∈ L∞(X) agrees almost ev-erywhere with a bounded function it follows from Proposition 2.1.8 that L∞0 (X)is dense in L∞(X).

Theorem 2.4.5. Suppose f ∈ L1(X), and g ∈ L∞(X), then gf ∈ L1(X) and‖gf‖1 ≤ ‖g‖∞‖f‖1.

Proof. Suppose first that f ∈ L10(X) and g ∈ L∞0 (X), say f =

∑Nn=1 αn1En

with µ(En) < ∞, and g =∑Mm=1 βm1Fm . We may assume that EnNn=1 and

FmMm=1 are each pariwise disjoint. We then have

gf =

N∑n=1

M∑m=1

βmαn1Fm∩En ∈ L10(X),

and

‖gf‖1 =

N∑n=1

M∑m=1

|βmαn|µ(Fm ∩ En)

≤N∑n=1

M∑m=1

‖g‖∞|αn|µ(Fm ∩ En)

≤N∑n=1

‖g‖∞|αn|µ(En) = ‖g‖∞‖f‖1.

For the general case, suppose that f ∈ L1(X) and g ∈ L∞(X). Take fn ∈L1

0(X) so that fn∞n=1 is Cauchy in L10(X) and fn → f almost everywhere.

Also, take gn ∈ L∞0 (X) so that ‖gn − g‖∞ → 0. Then gnfn → gf almosteverywhere, and from the triangle inequality and the argument above we have

‖gnfn − gmfm‖1 ≤ ‖gn‖∞‖fn − fm‖1 + ‖gn − gm‖∞‖fm‖1.

Therefore gnfn∞n=1 is Cauchy in L10(X). We then have that gf ∈ L1(X) and

‖gf‖1 = limn→∞

‖gnfn‖1 ≤ limn→∞

‖gn‖∞‖fn‖1 = ‖g‖∞‖f‖1.

2.4. INTEGRATION 47

Corollary 2.4.6. If f ∈ L1(X), and h ∈ M(X) such that |h| ≤ |f |, thenh ∈ L1(X). In particular, f ∈ L1(X) if and only if |f | ∈ L1(X).

Proof. We let g(x) = 0 if f(x) = 0, and g(x) = h(x)/f(x) otherwise. Then g ismeasurable and ‖g‖∞ ≤ 1. Therefore h = gf ∈ L1(X).

Corollary 2.4.7. If µ(X) <∞ then L∞(X) ⊂ L1(X), and ‖f‖1 ≤ ‖f‖∞µ(X).

Proof. If µ(X) < ∞ then 1X ∈ L1(X) with ‖1X‖1 = µ(X). Therefore forf ∈ L∞(X) we have ‖f‖1 ≤ ‖f‖∞‖1X‖1 = ‖f‖∞µ(X).

We set L1(X)+ = L1(X) ∩M(X; [0,∞)), and we set L∞(X)+ = L∞(X) ∩M(X)+.

2.4.3 Functions which agree almost everywhere

Let (X,M, µ) be a measure space. So far we have introduced M(X), L∞(X),and L1(X), as the spaces of all measurable, essentially bounded, and integrablefunctions respectively. It is often the case that we are interested in functionsonly up to measure zero, and so we consider the spaces M(X), L∞(X), andL1(X) which are respectivley the quotient of the above spaces where we haveidentified functions which agree almost everywhere. Note that M(X) does notdepend on the measure µ, however M(X) does.

The elements in M(X) are equivalence classes of functions, however it iscumbersome to state this explicitly each time. Thus, in the sequel when wewrite f ∈ M(X) (or f ∈ L∞(X), f ∈ L1(X)) we mean that we can take f tobe any function in M(X) which represents this equivalence class. Similarly, ifwe write an expression, e.g., f ≤ g with f, g ∈M(X), then this is expression ismeant to be understood as occuring almost everywhere.

As an example, we might say fn∞n=1 ⊂ M(X), and f ∈ M(X) such thatfn → f almost everywhere. This is unambiguous as the countable union ofmeasure zero sets has measure zero, and thus replacing fn and f by functionswhich agree almost everywhere does not change the fact that fn → f almosteverywhere. As long as we restrict to countably many functions/operations ata time this will not cause any difficulty.

2.4.4 Convergence properties

We begin this subsection by improving Lemma 2.4.1 to the case when fn ∈L1(X).

Theorem 2.4.8. Let fnn∈N ⊂ L1(X) be a Cauchy sequence, then there ex-ists a subsequence fnkk∈N which converges almost everywhere to a measurablefunction f , and such that for all ε > 0 there exists a measurable set A ⊂ X withµ(A) < ε, such that fnkk∈N converges to f uniformly on Ac.


Proof. In the proof of Lemma 2.4.1 the only reason we needed fn ∈ L10(X) was

so that ‖fn‖1 was defined and satisfied ‖fn‖1 =∫fn, and that if g ≤ h then∫

g ≤∫h. By Theorem 2.4.4 we also have these facts now for general functions

in L1(X). Thus, the proof follows verbatim as in Lemma 2.4.1.

Theorem 2.4.9 (The Monotone Convergence Theorem). Let (X,M, µ) be ameasure space. Suppose fn∞n=1 ⊂ L1(X)+ is a sequence such that fn ≤ fn+1

for all n, and such that∫fn is bounded. Then fn∞n=1 converges in L1, and

almost everywhere to a function f ∈ L1(X)+.

Proof. Suppose a = sup∫fn <∞. Since fn∞n=1 is increasing so is

∫fn∞n=1,

and for n ≤ m we have ‖fm − fn‖1 =∫

(fm − fn). Since∫fn → a it then

follows that fn∞n=1 is Cauchy in L1. By the previous theorem there thenexists a subsequence which converges in L1 and almost everywhere to a functionf ∈ L1. Since we have an increasing sequence it then follows that fn∞n=1

converges almost everywhere and in L1 to f .

Corollary 2.4.10. Let (X,M, µ) be a measure space. Suppose fn∞n=1 ⊂L1(X)+ is a sequence such that fn+1 ≤ fn for all j. Then fn∞n=1 convergesin L1, and almost everywhere to a function f ∈ L1(X)+.

Proof. We apply the monotone convergence theorem to the sequence f1 −fn∞n=1.

Lemma 2.4.11 (Fatou’s Lemma). If fn∞n=1 ⊂ L1(X)+, is such that lim infn→∞∫fn <

∞, then lim infn→∞ fn(x) exists for almost every x ∈ X. Moreover, lim infn→∞ fnis a measurable function which is in L1(X), and we have∫

lim infn→∞

fn ≤ lim infn→∞

∫fn.

Proof. Fix k and consider the decreasing sequence gm∞m=1 where

gm = inffk, fk+1, . . . , fm.

Then gm∞m=1 decreases to infm≥k fm, and applying the previous corollary wehave that infm≥k fm is in L1(X), and∫

infm≥k

fm ≤ infm≥k

∫fm.

By hypothesis we have that infm≥k∫fm∞m=1 is bounded, and since infm≥k fm∞k=1

is increasing we then have from the monotone convergence theorem that lim infn→∞ fnexists almost everywhere, is in L1(X), and satisfies∫

lim infn→∞

fn = limn→∞

∫infm≥n

fm ≤ lim infn→∞

∫fn.

2.4. INTEGRATION 49

Theorem 2.4.12 (The Fatou-Lebesgue Theorem). Let fn∞n=1 ⊂ L1(X;R).If there exists a function g ∈ L1(X)+ such that |fn| ≤ g for all n ≥ 1 thenlim infn→∞ fn and lim supn→∞ fn exist almost everywhere, are integrable, andwe have ∫

lim infn→∞


∫fn ≤ lim sup

n→∞

∫fn ≤

∫lim supn→∞

fn.

Proof. The first inequality follows from linearity of the integral and by applyingFatou’s lemma to the non-negative functions fn + g. The second inequality isobvious. The third inequality follows by applying Fatou’s lemma to the non-negative functions g − fn.

Theorem 2.4.13 (Lebesgue’s Dominated Convergence Theorem). Suppose fnis a sequence in L1(X), such that fn → f almost everywhere. If there existsg ∈ L1(X)+ such that |fn| ≤ g for all n ∈ N, then f ∈ L1(X) and

∫fn →

∫f .

Proof. Note that |f | ≤ g almost everywhere and so by Corollary 2.4.6 we havethat f ∈ L1(X).

By considering separately the real and imaginary parts of fn we see that itis enough to consider the case when fn is real valued. In this case it followsfrom the Fatou-Lebesgue theorem that∫

f =

∫lim infn→∞


∫fn

≤ lim supn→∞

∫fn

≤∫

lim supn→∞

fn =

∫f,

and the result then follows.

Theorem 2.4.14 (Egorov’s Theorem). Let (X,M, µ) be a finite measure space,and suppose fn∞n=1 ⊂M(X), and f ∈M(X) such that fn → f almost every-where. Then for each ε > 0 there exists A ⊂ X measurable such that µ(A) < εand fn → f uniformly on Ac.

Proof. We let En,k = x ∈ X | |fn(x) − f(x)| ≥ 1/k. Then En,k ∈ M andsince fn(x)→ f(x) for almost every x ∈ X we have that µ(∩∞N=1 ∪∞n=N En,k) =0, for every k ∈ N. Thus, there exists Nk so that for each k ∈ N we haveµ(∪∞n=Nk

En,k) < ε2−k. We set A = ∪∞k=1 ∪∞n=NkEn,k, so that µ(A) < ε.

If k ∈ N, and n ≥ Nk we have |fn(x)−f(x)| < 1/k for all x ∈ Ac. Thereforefn → f unifomrly on Ac.

We extend the integral to certain real valued functions as follows: If g ∈L1(X;R) and f ∈M(X; [0,∞)) is not integrable, then we write

∫(f + g) =∞,

and∫−(f+g) = −∞. Many of the results above extend to this setting, although

the inequalities become trivial in the case when f is not integrable.


2.4.5 Exercises

Exercise 2.4.15 (Chebyshev’s inequality). Let (X,M, µ) be a measure spaceand suppose f ∈ L1(X,µ), then for each α > 0 we have

µ(x ∈ X | |f(x)| > α) ≤ 1

α‖f‖1.

Exercise 2.4.16. There does not exist a metric d on L∞([0, 1], λ) so that asequence of functions fn∞n=1 ⊂ L∞([0, 1], λ) converge almost eveywhere to afunction f ∈ L∞([0, 1], λ) if and only if d(fn, f) → 0. Hint: Find a sequencefn∞n=1 ⊂ L∞([0, 1], λ) which does not converge almost everywhere to anyfunction but such that every subsequence has a further subsequence which doesconverge almost everywhere to some function.

Exercise 2.4.17. Let (X,M, µ) be a σ-finite measure space and suppose f ∈M(X; [0,∞)). Define

I1(f) = sup

∫g | g ∈ L1

0(X; [0,∞)), g ≤ f,

I2(f) = inf

∫g | g ∈ L1

0(X; [0,∞)), f ≤ g.

Show that f is integrable if and only if I1(f) < ∞, and this this case we haveI1(f) = I2(f) =

∫f .

Exercise 2.4.18. Suppose (X,M, µ) is a measure space and f ∈M(X, [0,∞)).Set F (λ) = µ(f−1([λ,∞))). Then F is measurable and F ∈ L1([0,∞)) if andonly if f ∈ L1(X). Moreover, in this case we have

∫fdµ =

∫Fdλ.

Exercise 2.4.19. Suppose (X,M, µ) is a measure space, (Y,N ) is a measurablespace, and θ : X → Y is measurable. Then for all f ∈ M(Y ) we have f θ ∈L1(X,µ) if and only if f ∈ L1(Y, θ∗µ), and in this case we have∫

f θ dµ =

∫f d(θ∗µ).

2.5 Product spaces

If (X,M) and (Y,N ) are measurable spaces, we let M⊗ N ⊂ 2X×Y denotethe σ-algebra generated by sets of the form E × F where E ∈ M and F ∈ N .In other words, M⊗N is the smallest σ-algebra so that the projection mapsπX : X × Y → X and πY : X × Y → Y are measurable. More generally, if(Xi,Mi)i∈I is a family of measurable spaces then we denote by ⊗i∈IMi thesmallest σ-algebra so that the projection maps are measurable.

Proposition 2.5.1. Let (X,M, µ) and (Y,N , ν) be measure spaces, then thereis measure ζ on M⊗N so that ζ(E×F ) = µ(E)ν(F ), for E ∈M and F ∈ N .Here we use the convention 0 ·∞ = 0. If µ and ν are σ-finite then this measureis unique.

2.5. PRODUCT SPACES 51

Proof. We let A denote the algebra generated by sets of the from E × F whereE ∈ M and F ∈ N . If E × F = ∪∞n=1En × Fn, where E,En ∈ M, andF, Fn ∈ N , with µ(E), ν(F ) <∞. Then for x ∈ X and y ∈ Y we have

1E(x)1F (y) = 1E×F (x, y) =

∞∑n=1

1En×Fn(x, y) =

∞∑n=1

1En(x)1Fn(y).

Integrating with respect to x, and using the monotone convergence theoremgives

µ(E)1F (y) =

∞∑n=1

µ(En)1Fn(y).

If we then integrate with respect to y we obtain

µ(E)ν(F ) =

∞∑n=1

µ(En)ν(Fn).

If A ∈ A, then we may write A as a finite disjoint union A = ∪ni=1Ei × Fiwhere Ei ∈ M and Fi ∈ N . From above we then see that setting ζ0(A) =∑ni=1 µ(Ei)ν(Fi) gives a well defined premeasure on A. Caratheodory’s exten-

sion theorem then shows that this extends to a measure ζ, and if µ and ν areσ-finite then so is ζ and hence this measure is unique.

The measure constructed by Caratheodory’s extension theorem in the pre-vious proof is called the product measure and denoted by µ×ν (or µ2 if µ = ν).We can of course generalize the above proposition easily to any finite numberof measure spaces.

If E ⊂ X × Y , and x ∈ X, y ∈ Y then we define the x-section Ex andy-section Ey by Ex = y ∈ Y | (x, y) ∈ E, and Ey = x ∈ X | (x, y) ∈ E.Also, if f : X × Y → C we define the x-section fx and y-section fy byfx(y) = fy(x) = f(x, y).

Proposition 2.5.2. Let (X,M) and (Y,N ) be measurable spaces. If E ∈M ⊗ N then Ex ∈ N and Ey ∈ M for all x ∈ X and y ∈ Y . Also, iff : X × Y → C is M ⊗ N -measurable then fx is N -measurable and fy isM-measurable for all x ∈ X and y ∈ Y .

Proof. We let Σ denote the collection of subsets E ⊂ X × Y such that Ex ∈ Nand Ey ∈M for all x ∈ X and y ∈ Y . Then Σ contains all sets of the form A×Bwith A ∈ M and B ∈ N . Since (∪∞n=1En)x = ∪∞n=1(En)x and (Ec)x = (Ex)c,(and similarly for y) it follows that Σ is a σ-algebra and hence must containM⊗N .

If f : X × Y → C is M⊗N -measurable, then as (fx)−1(C) = (f−1(C))x(and similarly for y) it then follows that fx and fy are measurable.

If X is a set, then a family E ⊂ 2X is called a monotone class if X ∈ E andE is closed under countable monotone unions and intersections, i.e., whenever


En∞n=1 ⊂ E with E1 ⊂ E2 ⊂ · · · then we have ∪∞n=1En ∈ E , and wheneverEn∞n=1 ⊂ E with E1 ⊃ E2 ⊃ · · · then we have ∩∞n=1En ∈ E . Given a familyof monotone classes it is clear that the intersection is again a monotone class,thus for any collection of sets E0 there exists a smallest monotone class whichcontains E0, we call this the monotone class which is generated by E .

Lemma 2.5.3 (The monotone class lemma). Suppose A ⊂ 2X is an algebra,then the monotone class generated by A coincides with the σ-algebra generatedby A.

Proof. We letM denote the monotone class generated by A. Since a σ-algebrais a monotone class then it suffices to show that M is a σ-algebra. For this itsuffices to show that M is closed under taking complements and finite unionssince if En∞n=1 ⊂ M then ∪Nn=1En is a monotone increasing sequence andhence if these finite unions are in M then so is ∪∞n=1En.

For E ∈ M we set K(E) = F ∈ M | E \ F, F \ E,E ∪ F ∈ M. We willshow that K(E) = M for each E ∈ M. This is shown by the following sixsteps, each of which is easily verified:

1. If F ∈ K(E) then E ∈ K(F ). (This follows from symmetry in the defini-tion of K(E) and K(F ).)

2. If E ∈ A then A ⊂ K(E). (This follows since A ⊂M).

3. K(E) is a monotone class for all E ∈ M. (This follows since M is amonotone class).

4. If E ∈ A then K(E) =M. (This follows from (2) and (3) since M is thesmallest monotone class which contains A).

5. A ⊂ K(E) for all E ∈M. (This follows from (4) and (1)).

6. K(E) =M for all E ∈M. (This follows from (5) and (3)).

Lemma 2.5.4. Suppose (X,M, µ) and (Y,N , ν) are σ-finite measure speaces,and A ∈M⊗N . Then the functions x 7→ ν(Ex) and y 7→ µ(Ey) are measurableand

µ× ν(E) =

∫ν(Ex) dµ(x) =

∫µ(Ey) dν(y).

Proof. We first consider the case when µ and ν are finite. We let Σ denote thefamily of sets inM⊗N such that the conclusion of the proposition holds. Thenfrom the arguemnt in the proof of Proposition 2.5.1 we see that Σ contains thealgebra A generated by sets of the form E × F with E ∈M and F ∈M.

If En∞n=1 ⊂ Σ such that En ⊂ En+1 for all n ∈ N, then we have that∪∞n=1(En)x = (∪∞n=1En)x and so by the monotone convergence theorem we

2.5. PRODUCT SPACES 53

have ∫ν((∪∞n=1En)x) dµ(x) = lim

N→∞

∫ν((EN )x) dµ(x)

= limN→∞

µ× ν(EN )

= µ× ν(∪∞n=1En).

And we similarly have∫µ((∪∞n=1En)y) dν(y) = µ×ν(∪∞n=1En). Thus, ∪∞n=1En ∈

Σ. Since µ and ν are finite a similar argument shows that if En ⊃ En+1 for alln ∈ N, then ∩∞n=1En ∈ Σ. Therefore, Σ is a monotone class which contains Aand by the monotone class lemma we have that M = Σ.

If µ and ν are σ-finite, sayX = ∪∞n=1Xn and Y = ∪∞n=1Yn with µ(Xn), ν(Yn) <∞, then the result follows by first restricting to Xn×Yn then using the monotoneconvergence theorem as we did above.

Theorem 2.5.5 (The Fubini-Tonelli Theorem). Suppose (X,M, µ) and (Y,N , ν)are measure spaces, and f : X × Y → C is M⊗N -measurable. Consider thefollowing conditions:

1. f ∈ L1(X × Y, µ× ν).

2. For almost every x ∈ X, fx ∈ L1(Y, ν), and the function fL1(Y )(x) =‖fx‖L1(Y ) is in L1(X).

3. For almost every y ∈ Y , fy ∈ L1(X,µ), and the function fL1(X)(y) =‖fy‖L1(X) is in L1(Y ).

Then (1) implies both (2) and (3), and if µ and ν are σ-finite then all threeconditions are equivalent.

Moreover, if (1) (and hence also (2) and (3)) is satisfied then we have∫f d(µ× ν) =

∫ (∫f(x, y) dν(y)

)dµ(x) (2.6)

=

∫ (∫f(x, y) dµ(x)

)dν(y).

Proof. We first consider the case when µ and ν are σ-finite. If f is a character-istic function then the result follows from Lemma 2.5.4. By linearity we thenhave the result for simple functions. Suppose now that f ∈M(X × Y )+. Thenthere exists an increasing sequence of simple functions ϕn which are valued inthe non-negative reals so that ϕn(x, y)→ f(x, y) for all (x, y) ∈ X × Y . By themonotone convergence theorem we then have∫

f d(µ× ν) = limn→∞

∫ϕn d(µ× ν)

= limn→∞

∫ (∫ϕn(x, y) dν(y)

)dµ(x)

=

∫ (∫f(x, y) dν(y)

)dµ(x).


We similarly have∫f d(µ × ν) =

∫ (∫f(x, y) dµ(x)

)dν(y). Thus, for non-

negative valued functions we see that the three conditions above are equivalentand that (2.6) holds. From linearity we then get the result for general measur-able functions when µ and ν are σ-finite.

If µ or ν is not σ-finite but f ∈ L1(X × Y, µ × ν), then we see that Gn =(x, y) | |f(x, y)| ≥ 1/nmust have finite measure for all n ≥ 1. Therefore, thereexist Ek ∈M and Fk ∈ N so that Gn ⊂ ∪∞k=1Ek ×Fk ⊂ (∪∞k=1Ek)× (∪∞k=1Fk),and

∑∞k=1 µ(Ek)ν(Fk) <∞. In otherwords, we have Gn ⊂ E ×F where E and

F are σ-finite. It then follows that there exist σ-finite sets E, and F so that(x, y) ∈ X × Y | f(x, y) 6= 0 = ∪∞n=1Gn ⊂ E × F . Restricting to the σ-finitemeasure spaces (E, µE) and (F , νF ) we see that the result then follows from theσ-finite case.

We give the following 2 examples which show how the hypotheses of theFubini-Tonelli theorem are necessary:

Example 2.5.6. Consider [0, 1] with Lebesgue measure, and let f(x, y) =x2−y2

(x2+y2)2 . Then fixing x 6= 0 we have∫ 1

0

f(x, y) dy =

∫ 1

0

x2 + y2 − 2y2

(x2 + y2)2dy

=

∫ 1

0

1

x2 + y2dy +

∫ 1

0

y d

(1

x2 + y2

)=

∫ 1

0

1

x2 + y2dy +

y

x2 + y2

∣∣∣∣1y=0

−∫ 1

0

1

x2 + y2dy

=1

x2 + 1.

We therefore have∫ 1

0

(∫ 1

0

x2 − y2

(x2 + y2)2dy

)dx = tan−1(x)|1x=0 =

π

4,

and as f(y, x) = −f(x, y) we have∫ 1

0

(∫ 1

0x2−y2

(x2+y2)2 dx)dy = −π

4 . We must

therefore have that x2−y2(x2+y2)2 6∈ L

1([0, 1]2, λ2).

Example 2.5.7. Consider [0, 1] with its Borel σ-algebra B, and consider Lebesguemeasure λ on [0, 1], and also counting measure µ on [0, 1]. In the product space([0, 1]2,B ⊗ B, λ× µ) we may consider the measurable subset ∆ = (x, x) | x ∈[0, 1], and set f = 1∆. Then for every x ∈ [0, 1] we have fx ∈ L1([0, 1], µ)and

∫ (∫f(x, y) dλ(y)

)dµ(x) = 0. Similarly, for every y ∈ [0, 1] we have

fy ∈ L1([0, 1], λ) and∫ (∫

f(x, y) dµ(x))dλ(y) = 1. So that also in this case the

iterated integrals do not agree. We must therefore have that λ×µ(∆) =∞, andwe see that for non-σ-finite spaces the iterated integrals need not agree even forfunctions valued in the non-negative reals.

2.6. SIGNED AND COMPLEX MEASURES 55

If λ is Lebesgue measure on R and n ≥ 1, then Lebesgue measure on Rnis defined to be λn. When there is no danger of confusion we will just write λfor λn.

Theorem 2.5.8. If E ⊂ Rn is Borel and t ∈ Rn, then λ(E + t) = λ(E).

Proof. If E is a disjoint union of products of intervals than the formula λ(E +t) = λ(E) clearly holds. As such sets form an algebra which generates the Borelσ-algebra, and since E 7→ λ(E + t) gives a Borel measure, the proposition thenfollows from uniqueness in Caratheodory’s extension theorem.

Theorem 2.5.9. If T ∈ GLn(R) then T∗λ = |detT |−1λ, i.e., for all Borel setsE ⊂ Rn we have

λ(T−1(E)) = |detT |−1λ(E). (2.7)

Proof. Since every invertible matrix can be row reduced to the identity matrixit follows that every linear transformation T ∈ GLn(R) is a composition ofelementrary matrices of the following types:

1. T1(x1, . . . , xj , . . . , xn) = (x1, . . . , cxj , . . . , xn) with c 6= 0.

2. T2(x1, . . . , xj , . . . , xn) = (x1, . . . , xj + cxi, . . . , xn) with i 6= j.

3. T3(x1, . . . , xi, . . . , xj , . . . , xn) = (x1, . . . , xj , . . . , xi, . . . , xn).

Also, if we can show that (2.7) holds for matrices of the above type then asthe determinant is multiplicative it also holds for their composition. Thus, itis enough to verity (2.7) for matrices of the above type. These all follow easilyfrom the Fubini-Tonelli theorem. Writing

λ(E) =

∫· · ·∫

1E(x1, . . . , xn) dλ(x1) . . . dλ(xn)

we see that (2.7) holds for matrices of the first and second type by their cor-responding formulas in one dimensions, while matrices of the third type justcorrespond to changing the orders of integration.

2.5.1 Exercises

Exercise 2.5.10. Consider N with the counting measure, and consider the func-tion f : N2 → C given by f(n,m) = 1 if n = m, f(n,m) = −1 if n = m+ 1, andf(n,m) = 0 otherwise. Then

∑∞n=1 (

∑∞m=1 f(n,m)) and

∑∞m=1 (

∑∞n=1 f(n,m))

both exist but are not equal.

2.6 Signed and complex measures

In this section we extend the notion of a measure to allow set functions whichmay give negative, or even complex, values. The Hahn and Jordan decom-position theorems below, together with the polar decomposition theorem forcomplex measures, give the main tools to relate this more general setting to thenon-negative valued case we have already considered.


2.6.1 Signed measures

A signed measure on a measurable space (X,M) is a function ν : M →[−∞,∞] such that

1. at most one of the values in −∞,∞ are obtained;

2. ν(∅) = 0;

3. if En∞n=1 are pairwise disjoint measurable sets then

ν(∪∞n=1En) =

∞∑n=1

ν(En),

where this series converges absolutely if ν(∪∞n=1En) is finite.

If µ1 and µ2 are measures on (X,M), at least one of which is finite, thenwe obtain a signed measure µ1 − µ2 on (X,M) by setting (µ1 − µ2)(E) =µ1(E)− µ2(E) for any E ∈M.

If ν is a signed measure on (X,M), and E ∈ M, then we say that E ispositive (resp. negative, null) with respect to ν if for all F ⊂ E measurablewe have ν(F ) ≥ 0 (resp. ≤ 0, = 0). Note that positive (resp. negative, null) setsare preserved under taking measurable subsets, and also under taking countableunions.

If µ is a measure on (X,M) and E ∈M with µ(E) <∞, the we may obtaina signed measure on (X,M) by setting ν(F ) = µ(F ∩ Ec)− µ(F ∩ E). In thiscase E is a negative set, Ec is a positive set, and a set is null for ν if and onlyif it is null for µ. The Hahn decomposition theorem shows that every signedmeasure (or its negative) arrises in this way.

Lemma 2.6.1. Let ν be a signed measure on (X,M), and suppose E ∈M suchthat −∞ < ν(E) < 0. Then there exists a negative set N ⊂ E with ν(N) < 0.

Proof. Note first that there does not exist a measurable subset F ⊂ E withν(F ) = ∞, since otherwise we would have ν(E) = ν(F ) + ν(E \ F ) = ∞. Weinductively define a non-decreasing sequence of number nk∞k=0 ⊂ N ∪ ∞,and pairwise disjoint subsets Ek∞k=0 of E as follows: Set E0 = ∅ and n0 =1. Having defined n0, . . . , nk and E0, . . . , Ek we let nk+1 denote the smallestinteger such that there exists a measurable subset Ek+1 ⊂ E \ (∪kj=0Ej) withν(Ek+1) ≥ 1/nk+1. If no such number exists we set nk+1 =∞ and Ek+1 = ∅.

We then have that∞ > ν(∪∞k=0Ek) =∑∞k=0 ν(Ek) ≥

∑∞k=0

1nk≥ 0 (here we

use the convention 1∞ = 0). Since this series converges we must have nk →∞.

We set N = E \ (∪∞k=0Ek) then we have ν(N) ≤ ν(E) < 0. If F ⊂ N ismeasurable then by our choice of nk we must have ν(F ) ≤ 1/(nk − 1) for eachk. Since nk →∞ this then shows that ν(F ) ≤ 0, and hence N is negative.

Theorem 2.6.2 (The Hahn decomposition theorem). Let (X,M) be a measur-able space and let ν be a signed measure on (X,M). Then there exists a positiveset P ∈M so that N = P c is a negative set. Moreover if P is another positiveset such that P c is a negative set, then we have that P∆P is null.

2.6. SIGNED AND COMPLEX MEASURES 57

Proof. We may assume that the value −∞ is never obtained (otherwise consider−ν). We let a = infν(E) | E ∈ M, E negative ≤ 0. We take negative setsEn ∈ M so that ν(En) → a, and we set N = ∪∞n=1En. Then N is a negativeset, and we have ν(N) = ν(En) + ν(N \ En) ≤ ν(En) for each n ≥ 1, henceν(N) = a.

We claim that N c is positive. Otherwise, there would exist a measurable setE ⊂ N c with −∞ < ν(E) < 0, and by the previous lemma there would thenexist a negative set N0 ⊂ E with ν(N0) < 0. However, we would then havethat N ∪N0 is negative and ν(N ∪N0) = ν(N) + ν(N0) < a, contradicting ourdefinition of a. This then finishes the existence part of the theorem.

Suppose now that N is another negative set such that N c is positive, andlet F ⊂ N \N be measurable. Then F ⊂ N hence ν(F ) ≤ 0, and F ⊂ N c henceν(F ) ≥ 0. Therefore ν(F ) = 0 and N \N is a null set. We similarly have thatN \ N is null and hence so is N∆N .

If (X,M) is a measurable space, then two (signed) measures µ, η on (X,M)are singular, which we write as µ ⊥ η, if there exists E ∈ M so that E isa conull set for µ and Ec is a conull set for η. Note that this is a symmetricrelation.

Theorem 2.6.3 (The Jordan decomposition theorem). Let (X,M) be a mea-surable space and let ν be a signed measure on (X,M). Then there exist uniquesingular measures ν−, ν+ on (X,M), at least one of which is finite, so thatν = ν+ − ν−.

Proof. From the Hahn decomposition theorem there exists P ∈ M a positiveset so that N = P c is a negative set. We define ν+ by ν+(E) = ν(E ∩ P ) forall E ∈ M and we define ν− by ν−(E) = −ν(E ∩ N) for all E ∈ M. Thatthese define measures is easily seen from the definition of a signed measure.Moreover, we have ν(E) = ν(E ∩ P ) + ν(E ∩ N) = ν+(E) − ν−(E) for allE ∈M, hence at least one of ν− or ν+ is finite and we have ν = ν+−ν−. Since,ν+(N) = ν(N ∩ P ) = −ν−(P ) = 0 we have ν+ ⊥ ν−.

Suppose now that η1, η2 are singular measure on (X,M), at least one ofwhich is finite, such that ν = η1− η2. Take E ∈M so that E is a conull set forη1 and Ec is a conull set for η2. Then we clearly have that E is a positive setfor ν, and Ec is a negative set for ν. Therefore, P∆E is a null set for ν by theuniqueness part of the Hahn decomposition theorem. If we have F ∈ M thenwe have

η1(F ) = η1(F ∩ E) = ν(F ∩ E) = ν(F ∩ P ) = ν+(F ).

Hence, η1 = ν+. We similarly have that η2 = ν− which then shows uniqueness.

If ν is a signed measure on (X,M) then the measures ν+ and ν− are calledrespectively the positive and negative variations of ν. The measure |ν| =ν+ + ν− is called the absolute variation of ν. We also set ‖ν‖ = |ν|(X) and


call this the total variation of ν. It’s easy to see that the absolutely variationsatisfies

|ν|(A) = sup

∞∑k=1

|ν(Ek)|, (2.8)

where the supremum is taken over all measurable partitions of A.

2.6.2 Complex measures

A complex measure on a measurable space (X,M) is a set function ν :M→C such that

1. ν(∅) = 0;

2. if En∞n=1 are pairwise disjoint measurable sets then

ν(∪∞n=1En) =

∞∑n=1

ν(En),

where this series converges absolutely.

Given a complex measure ν, we may consider the real and imaginary partsRe (ν), Im (ν), which give signed measures such that ν(E) = Re (ν)(E) +iIm (ν)(E) for each E ∈ M. The absolute variation of ν is the set func-tion |ν| : M → [0,∞] given by equation (2.8). The total variation of ν isgiven by ‖ν‖ = |ν|(X).

Proposition 2.6.4. Let ν be a complex measure on (X,M), then |ν| is a mea-sure on (X,M), ‖ν‖ <∞, and for all A ∈M we have |ν(A)| ≤ |ν|(A).

Proof. Suppose An∞n=1 ⊂M is a sequence of pairwise disjoint sets. If Ek∞k=1

gives a measurable partition of ∪∞n=1An, then for each n ≥ 1 we have a measur-able partition of An given by Ek ∩An∞k=1. We therefore have

∞∑n=1

|ν|(An) ≥∞∑n=1

∞∑k=1

|ν(Ek ∩An)| ≥∞∑k=1

∣∣∣∣∣∞∑n=1

ν(Ek ∩An)

∣∣∣∣∣ =

∞∑k=1

|ν(Ek)|.

Taking supremums over all such partitions then gives∑∞n=1 |ν|(An) ≥ |ν|(∪∞n=1An).

Also, if ε > 0, and if Enk ∞k=1 is a measurable partition of An such that∑∞k=1 |ν(Enk )|+ ε2−n ≥ |ν|(An), then we have

|ν|(∪∞n=1An) + ε ≥∞∑

n,k=1

|ν(Enk )|+ ε ≥∞∑n=1

|ν|(An).

Therefore, |ν|(∪∞n=1An) ≥∑∞n=1 |ν|(An), and hence |ν| is a measure.

We have ‖ν‖ ≤ ‖Re (ν)‖ + ‖Im (ν)‖, and from the Hahn decompositiontheorem we see that ‖Re (ν)‖ + ‖Im (ν)‖ < ∞. Hence, ‖ν‖ < ∞. Also, ifE ∈M then the inequality |ν(E)| ≤ |ν|(E) follows easily from the definition of|ν|.

2.7. THE RADON-NIKODYM THEOREM 59

2.6.3 Exercises

Exercise 2.6.5. Consider [0, 1] with the Borel σ-algebra. Let ν be countingmeasure and µ be Lebesgue measure on [0, 1], then there do not exist Borelmeasures ν0, ν1 on [0, 1] so that ν = ν0 + ν1, ν0 ⊥ µ, and ν1 µ.

If µ is a measure on (X,M) and f ∈ L1(X,µ) then we obtain a complexvalued measure fµ by (fµ)(E) =

∫Ef dµ.

Exercise 2.6.6. If f ∈ L1(X,µ) then |fµ| = |f |µ, and ‖fµ‖ = ‖f‖1.

We let Mb(X) denote the space of all complex valued measures on (X,M).

Exercise 2.6.7. The map ν 7→ ‖ν‖ gives a norm on Mb(X), and with this normMb(X) is a Banach space.

2.7 The Radon-Nikodym Theorem

If µ and ν are measures on (X,M), then ν is absolutely continuous withrespect to µ (and we write ν µ) if every µ-null set is also a ν-null set. Theterminology is justified by the following proposition:

Proposition 2.7.1. Suppose µ and ν are measures on (X,M) with ν finite,then ν µ if and only if for every ε > 0 there exists δ > 0 so that if E ∈ Mwith µ(E) < δ then we have ν(E) < ε.

Proof. Clearly the above condition implies that ν is absolutely continuous withrespect to µ, thus we only need to show the converse. Suppose therefore thatthe condition above does not hold. Then there exists ε > 0 and En ∈ M sothat µ(En) < 2−n, and ν(En) ≥ ε for all n ≥ 1. We let Fk = ∪∞n=kEn, andset F = ∩∞k=1Fk. Then Fk is decreasing and µ(Fk) → 0, so that µ(F ) = 0.However, ν(Fk) ≥ ε for all k ≥ 1, hence ν(F ) ≥ ε showing that ν is notabsolutely continuous with respect to µ.

Theorem 2.7.2 (The Lebesgue decomposition theorem). Suppose µ and ν aremeasures on (X,M) such that ν is σ-finite, then there exist unique measures ν0

and ν1 on (X,M) so that ν = ν0 + ν1, ν0 ⊥ µ, and ν1 µ.

Proof. We first consider the case when ν(X) <∞. We let N denote the spaceof µ-null sets, and we set a = supν(E) | E ∈ N. We take En ∈ N so thatν(En) → a and we set E = ∪∞n=1En. Then E ∈ N and ν(E) ≥ ν(En) for alln ≥ 1, hence ν(E) = a.

We let ν0 be the measure given by ν0(F ) = ν(F ∩ E), and we let ν1(F ) =ν(F ∩Ec). Then we clearly have ν = ν0 + ν1, and we also have ν0 ⊥ µ since Eis a µ-null set and Ec is a ν0-null set. If F ∈ N , then F ∪ E ∈ N and hencea ≥ ν(F ∪ E) = ν(F ∩ Ec) + ν(E) ≥ a, therefore we must have ν(F ∩ Ec) = 0.It therefore follows that ν1(F ) = ν(F ∩ Ec) = 0 and hence ν1 µ.

If ν = ν0 + ν1 is another decomposition with ν0 ⊥ µ and ν1 µ, then forall F ∈ N we have ν0(F ) = ν(F ) = ν0(F ), and since ν0 ⊥ µ we then have that


ν0 = ν0, and it follows that ν1 = ν1. This then finishes the theorem in the casewhen ν(X) <∞.

For the general case, we write X = sup∞n=1En where En∞n=1 ⊂ M is apairwise disjoint sequence with ν(En) < ∞. We consider the restriction of νto En and from above there are unique measures νn0 , ν

n1 which have the conull

set En and such that νn0 ⊥ µ and νn1 µ. If we set ν0 =∑∞n=1 ν

n0 and

ν1 =∑∞n=1 ν

n1 then it is then easy to see that ν0 and ν1 are then the unique

measures which satisfy the conclusion of the theorem.

Lemma 2.7.3. Suppose η and µ are measures on (X,M), with µ finite, η 6= 0and such that η µ. Then there exists δ > 0, and E ∈ M so that µ(E) > 0and η ≥ δµ on E, i.e., η(F ) ≥ δµ(F ) for all F ∈M, F ⊂ E.

Proof. For each n ∈ N we consider a Hahn decomposition X = Pn ∪ Nn ofη − 1

nµ. We set P = ∪∞n=1Pn and N = ∩∞n=1Nn, so that P = N c. Since N is anegative set for η − 1

nµ for each n, we have 0 ≤ η(N) ≤ 1nµ(N) for each n, and

hence η(N) = 0, so that η(P ) > 0. Since η µ we than also have µ(P ) > 0.Therefore, for some n we have µ(Pn) > 0, and as Pn is a positive set for

η − 1nµ we have η(F ) ≥ 1

nµ(F ) for all F ∈M, F ⊂ Pn.

Theorem 2.7.4 (The Radon-Nikodym theorem). Let µ and ν be σ-finite mea-sures on (X,M) such that ν µ. Then there exists a unique f ∈ M(X,µ) sothat for all E ∈M we have

ν(E) =

∫E

f dµ. (2.9)

Proof. We first consider the case when µ and ν are finite. We set

F =

f ∈M(X, [0,∞)) |

∫E

f dµ ≤ ν(E), for all E ∈M,

and

a = supf∈F

∫f dµ.

Note that if f, g ∈ F then h = maxf, g ∈ F , since if we set F0 = x ∈ X |f(x) ≥ g(x), then for E ∈ M we have

∫Eh dµ ≤

∫F0∩E f dµ +

∫F c0∩E

g dµ ≤ν(F0 ∩ E) + ν(F c0 ∩ E) = ν(E).

We choose fn ∈ F so that∫fn dµ → a. Setting hn = maxf1, . . . , fn,

we then have hn ∈ F , and hn∞n=1 is an increasing sequence. If we set h =limn→∞ hn then for each E ∈ M it follows from the monotone convergencetheorem that

∫Eh dµ = limn→∞

∫Ehn dµ ≤ ν(E). Therefore h ∈ F , and we

have a ≥∫h dµ = limn→∞

∫fn dµ = a. So that

∫h dµ = a.

We claim that ν = µh. If not, then setting η = ν − µh, we have η ν µ,and η 6= 0, so that by Lemma 2.7.3 there exists δ > 0 and E ∈ M withµ(E) > 0 so that η ≥ δµ on E, or equivalently, µh + δµ ≤ ν on E. This wouldthen show that h + δ1E ∈ F , and hence a ≥

∫(h + δ1E) dµ = a + δµ(E) > a,


giving a contradiction. If h were another such function then we would have∫Eh dµ =

∫Eh dµ for all E ∈ M and from this it follows that h = h, µ-almost

everywhere.

In general, since µ and ν are σ-finite, we may decompose X as a countabledisjoint union of measurable sets X = ∪∞n=1Xn, such that µ(Xn), ν(Xn) < ∞.By the finite measure case above, there then exists a measurable function fn ∈M(X,µ) so that for all E ∈ M we have ν(E ∩ Xn) =

∫E∩Xn fn dµ. If we set

f =∑∞n=1 fn then it is easy to see that for all E ∈M we have ν(E) =

∫Ef dµ.

Uniqueness follows similar to the finite case above.

The function f ∈ M(X,µ) in the previous theorem is called the Radon-Nikodym derivative of ν with respect to µ and denoted by dν

dµ .

There is also a Radon-Nikodym theorem for complex measures:

Theorem 2.7.5 (The Radon-Nikodym theorem for complex measures). Let µbe a σ-finite measure on (X,M) and let ν be a complex measure on (X,M)which is absolutely continuous with respect to µ, then there exists a unique f ∈L1(X,µ) so that ν(E) =

∫Ef dµ for all E ∈M.

Proof. By considering the real and imaginary parts separately it is enough toconsider the case when ν is a finite signed measure. We let ν = ν+ − ν− bethe Jordan decomposition. Then ν+ and ν− are both absolutely continuouswith respect to µ and so by the Radon-Nikodym theorem there exists f+, f− ∈M(X,µ) so that ν± = f±µ. Note that since ν+, and ν− are finite measureswe have that f+, f− ∈ L1(X,µ). We then have ν = fµ and f ∈ L1(X,µ).Uniqueness follows just as in the previous theorem.

Corollary 2.7.6 (Polar decomposition for complex measures). Let µ be a com-plex measure on (X,M), then there exists f : X → T measurable such thatµ = f |µ|. Moreover, if g : X → T is a measurable function such that µ = g|µ|then g = f |µ|-almost everywhere.

Proof. Since µ |µ| it follows from the Radon-Nikodym theorem that thereexists a unique f ∈ M(X, |µ|) so that µ = f |µ|. We just need to show thatf(x) ∈ T for |µ|-almost every x ∈ X.

Suppose this were not the case. Then there exists α 6∈ T so that α isin the essential range of f . We take δ > 0 so that 2δ < d(α,T). We setE = f−1(B(δ, α)) so that |µ|(E) > 0.

Assume first that |α| > 1. Since µ = f |µ| it then follows that for any measur-able set F ⊂ E we have |µ(F )| = |f |µ|(F )| ≥ (α− δ)|µ|(F ). If ∪∞n=1En gives ameasurable partition of E then we have |µ|(E) ≥ (α− δ)

∑∞n=1 |µ|(En). Taking

a supremum over all partitions gives |µ|(E) ≥ (α− δ)|µ|(E), a contradiction.

If we had |α| < 1, then a similar computation would show that |µ|(E) ≤(α + δ)|µ|(E) which is again a contradiction. We must therefore have thatf(x) ∈ T for |µ|-almost every x ∈ X.


Lemma 2.7.7. A measure space (X,M, µ) is semifinite if and only if for allf ∈M(X,µ) we have

‖f‖∞ = sup

∣∣∣∣∫ fg dµ

∣∣∣∣ | g ∈ L1(X,µ), ‖g‖1 ≤ 1

.

(Where we set ‖f‖∞ =∞ if f 6∈ L∞(X,µ).)

Proof. Suppose first that µ is semifinite. If f ∈M(X,µ), take w ∈ L∞(X,µ;T)so that wf = |f |. Set α = ‖f‖∞ = ‖|f |‖∞ and fix ε > 0. If α < ∞ thenα is in the essential range of |f | and hence F = |f |−1((α − ε, α]) has positivemeasure. We let E ⊂ F be a measurable set with finite positive measure (whichexists by semifiniteness) and set g = 1

µ(E)1E then ‖wg‖1 = ‖g‖1 = 1 and by

Theorem 2.4.5 we have

α ≥∣∣∣∣∫ fwg dµ

∣∣∣∣ =1

µ(E)

∫E

|f | dµ ≥ α− ε.

So that ‖f‖∞ = sup|∫fg dµ| | g ∈ L1(X,µ), ‖g‖1 ≤ 1.

Similarly, if α =∞ then F = |f |−1((N,∞)) has positive measure for all N >0, and the same argument above then shows that ‖f‖∞ =∞ = sup|

∫fg dµ| |

g ∈ L1(X,µ), ‖g‖1 ≤ 1.Conversely, suppose µ is not semifinite. Then there exists E ∈ M so that

µ(E) =∞, and for any measurable subset F ⊂ E we have µ(F ) ∈ 0,∞, henceif g ∈ L1(X,µ) we must have g(x) = 0 for almost every x ∈ E. Setting f = 1Ewe then have ‖f‖∞ = 1, while sup

∣∣∫ fg dµ∣∣ | g ∈ L1(X,µ), ‖g‖1 ≤ 1

= 0.

Theorem 2.7.8. Let (X,M, µ) be a measure space and consider the map Ψ :L∞(X,µ) → L1(X,µ)∗ given by Ψ(f)(g) =

∫fg dµ. Then Ψ is isometric if µ

is semifinite, and Ψ is surjective if µ has the essential suprema property.

Proof. That Ψ maps into L1(X,µ)∗ follows from Theorem 2.4.5. From Lemma 2.7.7we see that this map is injective if and only if µ is semifinite.

Suppose µ has the essential suprema property, and ϕ ∈ L1(X,µ)∗. Fix E ⊂X so that µ(E) <∞. Then F 7→ ϕ(1F∩E) defines a complex measure on (X,M)which is absolutely continuous with respect to µ and so by the Radon-Nikodymtheorem there exists a function fE ∈ M(X,µ) so that ϕ(1F∩E) =

∫F∩E fE dµ

for all F ∈M. Since µ(E) <∞ we have from Lemma 2.7.7 that ‖fE‖∞ ≤ ‖ϕ‖.Note that by uniqueness in the Radon-Nikodym we have that if E1, E2 ∈M

have finite measure then fE1and fE2

agree almost everywhere on E1∩E2. If welet f denote an essential supremum of fE | µ(E) <∞ as in Proposition 2.2.5,then ‖f‖∞ ≤ ‖ϕ‖ and for each E ∈ M with µ(E) < ∞ we have f(x) = fE(x)for almost every x ∈ E.

It then follows that for every function g ∈ L1(X,µ) such that µ(x ∈ X |g(x) 6= 0) <∞ we have ϕ(g) =

∫fg dµ. Since functions of this type are dense

in L1(X,µ) it follows that ϕ(g) =∫fg dµ for each g ∈ L1(X,µ).


2.7.1 Exercises

Exercise 2.7.9. Suppose λ ν µ, then

dµ

dλ=dµ

dν

dν

dλµ− almost everywhere.

Exercise 2.7.10. If ν µ and µ ν then

dµ

dν=

(dν

dµ

)−1

µ− almost everywhere.

Exercise 2.7.11. If ν µ and g ∈ M(X), then g is µ-integrable if and onlyif g dνdµ is ν-integrable, and in this case we have∫

g dµ =

∫gdµ

dνdν.


Chapter 3

Point set topology

3.1 Topological spaces

Let X be a set. A topology on X is a family T of subsets of X, whichcontains ∅ and X, and is closed under finite intersections and arbitrary unions.A topological space is a pair (X, T ) consisting of a set X, together with atopology T on X. When T is understood we sometimes refer to the topologicalspace X. The following are examples of topological spaces:

1. If X is any set then 2X and ∅, X are both topologies on X, called thediscrete and trivial (or indiscrete) topologies respectively.

2. If (X, d) is a metric space and T consists of all open subsets of X then(X, T ) is a topological space. In this case we call (X, T ) metrizable.

3. If X is a set then T = ∅∪U ⊂ X | U c is finite gives a topology on X.

Generalizing the case of metric spaces, we call the sets in T , open sets, andwe call a set closed if its complement is open. If a set is both open and closedthen we say it is clopen. A set A ⊂ X is a Gδ-set if A is the intersection ofcountably many open sets, and a set B ⊂ X is an Fσ-set if B is the countableunion of closed sets.

If A ⊂ X, then the closure A of A is the intersection of all closed setscontaining A, and hence is the smallest closed set containing A. The interiorAo of A is the union of all open sets contained in A. The difference A\Ao is theboundary of A and denoted by ∂A. If A = X then A is dense and if Ao = ∅then A is nowhere dense. A topological space (X, T ) is separable if it has acountable dense subset.

Given two topologies T1 and T2 on X such that T1 ⊂ T2, we say that T1 isweaker (or coarser) than T2, and T2 is stronger ( or finer) than T1. Thus,the trivial topology is the coarsest topology, while the discrete topology is thefinest topology. If E ⊂ 2X , then the intersection of all topologies on X which

65

66 CHAPTER 3. POINT SET TOPOLOGY

contain E is clearly a topology and is denoted by T (E). It is called the topologygenerated by E . For example:

1. if K = Rn, or Cn, then the Zariski topology on K is the weakesttopology such that the zero set k ∈ K | p(k) = 0 of any polynomial p isclosed.

2. the Sorgenfrey line Rl is the space R, together with the topology gen-erated by all half-open intervals [a, b).

3. the Moore plane is the (closed) upper half plane Γ = (x, y) ∈ R2 | y ≥0, together with the topology generated by Euclidean open sets, and allsets of the form (x0, 0) ∪ (O \ (x, 0) | x ∈ R) where O is an openneighborhood of x0 in the Euclidean sense.

4. If (X,≤) is a linearly ordered set, the order topology on X is the topol-ogy generated by the open sets x ∈ X | a < x < b for all pair (a, b) ∈ X2

such that a < b.

If x ∈ X then a neighborhood of x is a set A ⊂ X so that x ∈ O ⊂ Afor some open set O ∈ T . A point x ∈ X is an accumulation point (orcondensation point, or limit point) of a set E if every neighborhood of xhas nonempty intersection with E \x. A neighborhood base for x ∈ X is afamily Oii∈I of open neighborhoods of x such that for any open neighborhoodO of x there is some Oi so that Oi ⊂ O. A base for the topology T is a familyOii∈I of open sets which contains a neighborhood base for any point x ∈ X.

Proposition 3.1.1. If E ⊂ 2X then T (E) consists of all unions of finite inter-sections of E.

Proof. If we let T denote the unions of all finite intersections of E then we mustshow that T is a topology. Clearly T is closed under arbitrary unions. SupposeU1, . . . , Un ∈ T . Then we may write Ui = ∪j∈JiOj,i where Oj,i is a finiteintersection of sets in E . Therefore ∩ni=1Ui = ∪j1∈J1,...,jn∈Jn (∩ni=1Oji,i) ∈ T , sothat T is also closed under finite intersections.

A topological space (X, T ) is first countable if each point has a countableneighborhood base which is countable. (X, T ) is second countable if it has acountable base.

A topological space is:

1. T1 if x is closed for each point x ∈ X;

2. Hausdorff (or T2) if for each x 6= y, there exist disjoint open sets U, V ∈T , such that x ∈ U and y ∈ V ;

3. regular (or T3) if it is T1 and for each closed set A ⊂ X and x ∈ Ac thereexist disjoint open sets U, V with x ∈ U and A ⊂ V ;

4. normal (or T4) if it is T1, and for any disjoint closed sets A,B ⊂ X thereare disjoint open sets U, V ⊂ X so that A ⊂ U , and B ⊂ V .

We leave it to the reader to check the implications T4 =⇒ T3 =⇒ T2 =⇒ T1.

3.2. CONTINUOUS MAPS 67

3.1.1 Exercises

Exercise 3.1.2. Show that a metric space X is separable, if and only if X issecond countable.

Exercise 3.1.3. Show that in a first countable space, singletons x are Gδ.

Exercise 3.1.4. Prove that every metric space is normal and first countable.

Exercise 3.1.5. Prove that a metric space is separable if and only if it is secondcountable.

Exercise 3.1.6. Let X = R and let T be the family of all sets of the formU ∪ (V ∩Q) where U and V are open sets in the usual sense. Show that T givesa topology on R which is Hausdorff but not regular.

Exercise 3.1.7. Suppose (X, d) is a metric space. Show that closed subsets ofX are Gδ.

Exercise 3.1.8. Let (X, d) be a metric space and consider the bounded metric

d′(x, y) = d(x,y)1+d(x,y) . Show that (X, d′) describes the same topology on X.

Exercise 3.1.9 (Frechet). Let A,B ⊂ R be two countable dense sets, showthat there is a homeomorphism θ : R → R so that θ(A) = B. Hint: UseExercise 1.1.27.

3.2 Continuous maps

Let (X, T ) and (Y,S) be two topological spaces. A map f : X → Y is contin-uous if f−1(U) is open for any open set U ⊂ Y . Note that this agrees with ourterminology for metric spaces. We say that f is open if F (U) is open for all Uopen. We say that f is a homeomorphism if it is bijective, continuous, andopen.

Proposition 3.2.1. Suppose E generates the topology on Y , then f : X → Yis continuous if and only if f−1(U) is open for each U ∈ E.

Proof. If f is continuous then we trivially have that f−1(U) is open for eachU ∈ E . conversely, if f−1(U) is open for each U ∈ E , then as the inverseimage of a function distributes over unions and intersections it follows thatf−1(O) is open whenever O is a union of finite intersections of sets in E . ByProposition 3.1.1 every open set is of this form and hence f is continuous.

A directed set is a set A, together with a binary relation ≤ such that

1. x ≤ x for all x ∈ A.

2. if x ≤ y and y ≤ z then x ≤ z.

3. for each x, y ∈ A there exists some z ∈ A so that x ≤ z and y ≤ z.


A net in X is a function f : A→ X from a nonempty directed set A into X. Weusually prefer to think of a net as being indexed by A and so we write this asfαα∈A, and we sometimes abuse notation by identifying a net with its image,so that we might say “let xαα∈A ⊂ X be a net”. Note that sequences are justnets when the index set is N, nets were first introduced by Moore and Smith in1922 as a generalization of sequences. A net xαα∈A ⊂ X has a limit x ∈ X iffor every open neighborhood U of x there exists α ∈ A so that xβ ∈ U for allβ ≥ α. If x is a limit of a net xαα∈A then we say that this net is convergentand we write limα→∞ xα = x. In general, limits need not be unique, however,it’s easy to see that if X is Hausdorff then limits must be unique.

Proposition 3.2.2. Suppose X and Y are topological spaces and f : X → Y ,then f is continuous if and only if for any convergent net xαα∈A such that x =limα→∞ xα, we have that f(xα)α∈A is also convergent and limα→∞ f(xα) =f(x). Moreover, if X is first countable, then one may consider only sequencesrather than nets.

Proof. Suppose first that f is continuous and xαα∈A is a net such that x =limα→∞ xα. If we fix an open neighborhood U of f(x) then f−1(U) is an openneighborhood of x and since x = limα→∞ xα there then exists a ∈ A so thatxβ ∈ f−1(U) for all β ≥ a. Therefore, f(xβ) ∈ U for all β ≥ a and hencelimα→∞ f(xα) = f(x).

Conversely, suppose that for any net xαα∈A such that x = limα→∞ xα, wehave that f(xα)α∈A is also convergent and limα→∞ f(xα) = f(x).

Conversely, suppose that f is not continuous and let U be an open set in Ysuch that f−1(U) is not open in X. Therefore there exists a point x ∈ f−1(U)so that f−1(U) contains no open neighborhood of x. We let A denote the setof open neighborhoods of x, and note that this is a directed set when orderedby reverse inclusion. For each O ∈ A we take xO ∈ O \ f−1(U). Then we havelimO→∞ xO = x. However, f(xO) 6∈ U for each O ∈ A and hence f(xO)O∈Adoes not converge to f(x).

If X is a set and fi : X → Yii∈I is a family of maps from X into topologicalspaces Yi then there is a unique weakest topology on X making each of themaps fi continuous. We call this topology the weak topology on X generatedby fii∈I . For example if Xii∈I is a family of topological spaces then wemay endow

∏i∈I Xi with the weak topology generated by the coordinate maps

πi :∏j∈I Xj → Xi, we always consider

∏i∈I Xi with this topology unless

otherwise stated.

Proposition 3.2.3. If Xi is Hausdorff for each i ∈ I then∏i∈I Xi is Hausdorff.

Proof. Suppose x, y ∈∏i∈I Xi such that x 6= y. Then for some coordinate

i ∈ I we have πi(x) 6= πi(y). Since Xi is Hausdorff there exists disjoint openneighborhoodsO and U of πi(x) and πi(y) respectively, then π−1

i (O) and π−1i (U)

give disjoint open neighborhoods of x and y respectively.

Proposition 3.2.4. If Y is a topological space and f : Y →∏i∈I Xi, then f is

continuous if and only if πi f is continuous for each i ∈ I.


Proof. Since composition of continuous maps are continuous we see that if f iscontinuous then πi f is continuous for each i ∈ I. Conversely, suppose πi fis continuous for each i ∈ I. Then for any i ∈ I and open set O ⊂ Xi we havethat f−1(π−1

i (O)) is open. Since these sets generate the topology on∏i∈I Xi

we then have that f is continuous.

Proposition 3.2.5. A net xαα∈A converges to x ∈∏i∈I Xi if and only if

for each i ∈ I the net πi(xα)α∈A converges to πi(x).

Proof. If xαα∈A is a net which converges to x. Then for each i ∈ I we havethat πi(xα)α∈A converges to πi(x) by Proposition 3.2.2. Conversely, if foreach i ∈ I we have πi(xα)α∈A converges to πi(x), then for each i1, . . . , in ∈ Iand O1, . . . , On open neighborhoods of πi1(x), . . . , πin(x) respectively, we havethat there exists a ∈ A so that πik(xα) ∈ Ok for each α ≥ a. Since sets of theform ∩nk=1π

−1ik

(Ok) form a base for the topology it then follows that xαα∈Aconverges to x.

In the case when each Xi is equal to some fixed space X, then we areconsidering the function space XI , and the topology we are considering is thetopology of pointwise convergence; a net fαα∈A converges to f : I → Xif and only if for each i ∈ I we have limα→∞ fα(i) = f(i).

If X is a topological space then we denote by Cb(X) the space of all con-tinuous functions with bounded image. We consider the uniform norm of fas

‖f‖∞ = supx∈X|f(x)| | x ∈ X.

The function d(f, g) = ‖f − g‖∞ gives a metric on Cb(X), which we call theuniform metric.

Proposition 3.2.6. The space Cb(X) is a Banach algebra when endowed withthe uniform metric, and pointwise operations.

Proof. The arguments in Propositions 1.3.1 and 1.3.1 when X is a metric spacework equally well here.

If K ⊂ C is closed, then we denote by Cb(X;K) the subspace of all functionswhich take values in K. It is easy to see that Cb(X;K) is a closed subspace ofCb(X) in the uniform norm.

Lemma 3.2.7. Let X be a normal space. Suppose that A and B are disjointclosed sets in X, and let D = k2−n | n ≥ 1 and 0 < k < 2n be the set ofdyadic rationals in (0, 1). There is a family Ur | r ∈ D of open sets in X suchthat A ⊂ Ur ⊂ Bc for all r ∈ D and Ur ⊂ Us for r < s.

Proof. Set U0 = A and U1 = W c. As X is normal there exist disjoint opensets V and W such that A ⊂ V and B ⊂ W . We set U1/2 = V , so that

A ⊂ U1/2 ⊂ U1/2 ⊂ W c ⊂ Bc. We now select Ur for r = k2−n by induction onn. Suppose N ≥ 2 and we have chosen Ur for r = k2−n, for 0 < k < 2n, and


1 ≤ n < N . Then for each r = (2j+1)2−N , 0 ≤ j < 2N−1, we have that Uj21−N

and (U(j+1)21−N )c are disjoint closed sets and so as above we may choose Ur sothat

A ⊂ Uj21−N ⊂ Ur ⊂ Ur ⊂ U(j+1)21−N ⊂ Bc.

Lemma 3.2.8 (Urysohn’s Lemma). Let X be a normal space. If A and B aredisjoint closed sets in X, then there exists f ∈ C(X; [0, 1]) such that f|A = 0,and f|B = 1.

Proof. Let Urr∈D be as in the previous lemma. Set U1 = X and for x ∈ Xdefine f(x) = infr | x ∈ Ur. Since A ⊂ Ur for 0 < r < 1 we have f|A = 0. Wealso have f|B = 1 and clearly f(X) ⊂ [0, 1], so all that remains is to show thatf is continuous. Towards this end note that f(x) < a if and only if x ∈ Ur forsome r < α, thus f−1((−∞, a)) = ∪r<αUr is open. Also, f(x) > a if and onlyif x 6∈ Ur for some r > a, and hence if and only if x 6∈ Us for some s > a (sinceUs ⊂ Ur for s < r). Thus f−1((a,∞)) = ∪s>α(Ux)c is open. Since half-linesgenerate the topology on R it follows that f is continuous.

A space X is Tychonoff (or T31/2) if it is T1 and for every closed set A andpoint x ∈ Ac, there exists a continuous function f ∈ C(X; [0, 1]) so that f|A = 0,and f(x) = 1. Note that by Urysohn’s lemma we have that T4 =⇒ T31/2. It’salso easy to check that T31/2 =⇒ T3.

Theorem 3.2.9 (The Tietze Extension Theorem). Let X be a normal space.If A is a closed subset of X and f : A → R is continuous, then there existsF : X → R continuous, such that F|A = f . Moreover, if f is bounded then wemay choose F so that ‖F‖∞ = ‖f‖∞.

Proof. We first consider the case when f is bounded. We may assume f isnonconstant. If we set a = inf f(A) and b = inf f(A), then by replacing f with(f−a)/(b−a) we may assume that f : X → [0, 1]. We will inductively constructa sequence of continuous functions gn : X → [0, 1] so that

f(x)−n∑i=0

2i

3i+1gi(x) ∈ [0, (2/3)n+1]

for all x ∈ A. Then Proposition 3.2.6 shows that F =∑∞i=1 gi defines a contin-

uous function with ‖F‖ ≤ 1, such that F agrees with f on A.To construct g0 we set E = f−1([0, 1/3]) and F = f−1([2/3, 1]). Then E and

F are disjoint closed sets and so by Urysohn’s lemma there exists g0 : X → [0, 1]continuous so that g0(x) = 0 for x ∈ E and g0(x) = 1 for x ∈ F . We then havef(x) − 1

3g0(x) ∈ [0, 2/3] for all x ∈ A. Now suppose g0, . . . , gn−1 have beenconstructed so that

f(x) = f(x)−n−1∑i=0

2i

3i+1gi(x) ∈ [0, (2/3)n]


for all x ∈ A. Then as above we set E = f−1([0, 2n/3n+1]) and set F =f−1([2n+1/3n+1, (2/3)n], and we take gn : X → [0, 1] continuous so that gn(x) =0 if x ∈ E and gn(x) = 1 if x ∈ F . Then, we have

f(x)−n∑i=0

2i

3i+1gi(x) = f(x)− 2n/3n+1gn(x) ∈ [0, (2/3)n+1],

finishing the induction step.For the case when f is not bounded we take a homeomorphism θ : R →

(−1/2, 1/2) and consider θ f : A→ (−1/2, 1/2). Then from above there existsa continuous function F : X → [−1/2, 1/2] so that F agrees with θf on A. Welet E = F−1(−1/2, 1/2). Then E is a closed set which is disjoint from A andso by Urysohn’s lemma there exists g : X → [0, 1] so that g|E = 0 and g|A = 1.

Then F = gF also agrees with θ f on A and sastisfies F : X → (−1/2, 1/2).Therefore θ−1 F gives the desired continuous function.

Corollary 3.2.10. If X is normal, A ⊂ X is closed, and f ∈ Cb(A), then thereexists F ∈ Cb(X) such that F|A = f , and ‖F‖∞ = ‖f‖∞.

Proof. We assume f 6= 0. By considering the real and imaginary parts sepa-rately, it then follows from the previous theorem that there exists a boundedcontinuous function F0 such that F0 agrees with f on A. We let E = x ∈ X ||F0(x)| ≥ ‖f‖∞. Then

h(x) =

‖f‖∞/|F0(x)| if x ∈ E;1 if x 6∈ E;

gives a continuous function and hF0 agrees with f on A and satisfies ‖hF0‖∞ =‖f‖∞.

3.2.1 Exercises

A topological space is disconnected if there exists nonempty disjoint open setsU, V which cover X; otherwise X is connected. A subset E ⊂ X is connectedor disconnected if this is the case in the relative topology.

Exercise 3.2.11. (a) Show that if Aii∈I is a family of connected subsetssuch that ∩i∈IAi 6= ∅ then ∪i∈IAi is connected.

(b) Show that if A ⊂ X is connected then A is also connected.

(c) Show that every point x ∈ X is contained in a unique maximal connectedsubset of X, and this subset is closed. (This is the connected componentof x).

A topological space is totally disconnected if x is the connected com-ponent of x, for each x ∈ X.


Exercise 3.2.12. Show that the continuous image of a connected set is con-nected.

A topological space (X, T ) is arc-connected if for each x, y ∈ X thereexists a continuous function f : [0, 1]→ X so that f(0) = x and f(1) = y.

Exercise 3.2.13. Show that arc-connected spaces are connected. Also, find anexample of a connected space which is not arc-connected.

3.3 Compact spaces

Generalizing the case for metric spaces, a topological space (X, T ) is compactif every open cover has a finite subcover. A subset E ⊂ X is compact if itis compact with respect to the relative topology. We say a subset E ⊂ X isprecompact if E is compact.

A family of subsets F of X has the finite intersection property if for anyF1, . . . , Fn ∈ F , with n ≥ 1 we have ∩ni=1Fi 6= ∅.

Proposition 3.3.1. A topological space X is compact if and only if, whenever Fis a non-empty family of closed subsets which has the finite intersection propertythen we have ∩F∈FF 6= ∅.

Proof. By contraposition a space X is compact if and only if whenever G is afamily of open sets which does not have a finite subfamily covering X, then Gitself does not cover X. Taking complements of the sets in G, then gives thecriterion for compactness above.

Proposition 3.3.2. Suppose X and Y are topological spaces with X compact.If f : X → Y is continuous then f(X) is compact.

Proof. Suppose G is an open cover for f(X), then f−1(O) | O ∈ G is an opencover for X and hence has a finite subcover f−1(O1), . . . , f−1(On). It thenfollows that O1, . . . , On is a finite subcover of f(X). Hence, f(X) is compact.

By the previous proposition, any continuous map from a compact space toC is bounded, thus for compact spaces we write C(X) for Cb(X).

Proposition 3.3.3. A closed subset of a compact space is compact. Also, acompact subset of a Hausdorff space is closed.

Proof. First, suppose X is compact, and F ⊂ X is closed, if G is an open coverof F , then G ∪ F c is an open cover for X and hence by compactness there isa finite subcover G0, then G0 \ F c is a finite subcover of G which covers F ,showing that F is compact.

Next, suppose X is Hausdorff and F ⊂ X is compact. Fix x0 6∈ F . SinceX is Hausdorff, for each x ∈ X there exist disjoint open neighborhoods Ox,and Gx of x and x0 respectively. We have that Oxx∈F covers F and so by

3.3. COMPACT SPACES 73

compactness there is a finite subcover Ox1 , . . . , Oxn. If we set U = ∩ni=1Gxithen we have that U is open and is disjoint from F ⊂ ∪ni=1Oxi . Thus, x 6∈ F ,and as x was arbitrary it then follows that F = F is closed.

Corollary 3.3.4. Suppose X and Y are topological spaces with X compact andY Hausdorff, and suppose f : X → Y is a continuous bijection. Then f is ahomeomorphism.

Proof. Suppose O ⊂ X is open. By Proposition 3.3.3 we have that Oc is com-pact. By Proposition 3.3.2 we then have that f(Oc) is compact and henceclosed. Since f is a bijection we then have that f(O) = f(Oc)c is open. Thus,f−1 is continuous and hence f is a homeomorphism.

Proposition 3.3.5. Suppose X is a Hausdorff topological space, and E,F ⊂ Xare disjoint compact subsets, then there exist disjoint open sets U, V ⊂ X sothat E ⊂ U and F ⊂ V .

Proof. We first consider the case when E is a singleton E = x. Since X isHausdorff, for each y ∈ F there exists disjoint open sets Oy and Vy so thaty ∈ Oy and x ∈ Vy. We then have that Oyy∈F forms an open cover of F andby compactness there exists a finite subcover Oyini=1. If we set U = ∪ni=1Oyiand V = ∩ni=1Vyi , then U and V are disjoint open sets such that F ⊂ U andx ∈ V .

We now consider the general case. From above, for each y ∈ E there existdisjoint open sets Oy and Vy so that y ∈ Oy and F ⊂ Vy. Again by compactnessthere exists a finite collection Oyini=1 which covers E. Then U = ∪ni=1Oyi andV = ∩ni=1Vyi are disjoint open sets and we have E ⊂ U , while F ⊂ V .

Corollary 3.3.6. A compact Hausdorff space X is normal.

Proof. This follows directly from Propositions 3.3.3 and 3.3.5.

Theorem 3.3.7 (Tychonoff’s Theorem). If Xii∈I is a family of compacttopological spaces, then

∏i∈I Xi is also compact.

Proof. Suppose F is a family of closed subsets of∏i∈I Xi with the finite in-

tersection property. By Zorn’s lemma there exists a maximal family E of (notnecessarily closed) subsets with the finite intersection property such that F ⊂ E .Note that E itself must then be closed under finite intersections, and if E ∈ Eand E ⊂ F , then F ∈ E .

For each i ∈ I the family πi(E) | E ∈ E has the finite intersection propertyand hence by compactness we have ∩E∈Eπi(E) 6= ∅. Take xi a point in thisintersection. We let x be the point in

∏i∈I Xi whose ith coordinate is xi.

We claim that all neighborhoods of x are contained in E . To prove this it isenough to show that the neighborhoods of the form ∩nk=1π

−1ik

(Eik) are containedin E , and since E is closed under finite intersections it is then enough to showthat neighborhoods of the form π−1

i (Ei) are contained in E . To see this note

that since xi ∈ πi(E) for any E ∈ E it follows that π−1i (Ei) ∩ E 6= ∅ for all


E ∈ E . This then shows that E ∪ π−1i (Ei) has the finite intersection property

and hence π−1i (Ei) ∈ E by maximality of E . Thus, arbitrary neighborhoods of x

are contained in E and hence have non-trivial intersection with an arbitrary setE ∈ E . Thus, x ∈ E for all E ∈ E , and hence x ∈ ∩E∈EE ⊂ ∩F∈FF , showingthat

∏i∈I Xi is compact.

If X is a Banach space, then the weak∗-topology on X∗ is defined to be thecoarsest topology so that the maps X∗ 3 ϕ 7→ ϕ(x) are continuous for eachx ∈ X.

Theorem 3.3.8 (The Banach-Alaoglu theorem). Let X be a Banach space.Then the closed unit ball in X∗ is compact in the weak∗-topology.

Proof. Let D =∏x∈X B(‖x‖, 0). Since closed balls in Euclidean space are

compact, it then follows from Tychonoff’s theorem that D is compact. We letK denote the closed unit ball in X∗ and consider the map π : K → D whereπ(ϕ) has coordinates π(ϕ)x = ϕ(x). Note that since ϕ is in the unit ball wehave |π(ϕx)| = |ϕ(x)| ≤ ‖x‖, so that π is well defined. Also note that π isinjective since if π(ϕ) = π(ψ) then for each point x ∈ X we have ϕ(x) = ψ(x).

If ϕαα∈A is a net in K then ϕαα∈A converges to ϕ if and only if for eachx ∈ X we have ϕα(x)→ ϕ(x), and this is also if and only if π(ϕα)→ π(ϕ) in D.Therefore π defines a homeomorphism from D onto its image. K is thereforecompact if and only if the image of π is closed.

Note that D consists of functions from X to K (where K = R or K = C), andas these functions take x to an element in the ball B(‖x‖ < x) it follows thatthey are bounded functions. Thus, the image of π consists of those functionswhich are linear, i.e.

π(K) = ∩x1,x2∈X,α∈Kf ∈ D | fx1+αx2 = fx1 + αfx2.

As an intersection of closed sets is closed it then follows that π(K) is closed.

A property P of topological spaces is said to hold locally for a space Xif each x ∈ X has a neighborhood which satisfies P. For example a locallycompact space X is one in which each point x ∈ X has a compact neighborhood.Euclidean spaces Rn are examples of locally compact spaces.

A subset F ⊂ Cb(X) is equicontinuous at x ∈ X if for each ε > 0 there isa neighborhood U of x such that |f(y)− f(x)| < ε for all u ∈ U and f ∈ F . Fis equicontinuous if it is equicontinuous at each point. Also, F is pointwisebounded if f(x) | f ∈ F is bounded for each x ∈ X.

Theorem 3.3.9 (The Arzela-Ascoli Theorem). Let X be a compact Hausdorffspace. If F ⊂ C(X) is equicontinuous and pointwise bounded, then F is totallybounded in the uniform metric, and F is precompact.

Proof. Fix ε > 0. Since F is equicontinuous, for each x ∈ X there exists anopen neighborhood Ox of x such that |f(y) − f(x)| < ε/4 for all f ∈ F , andy ∈ Ox. The family Oxx∈X is an open cover, and since X is compact it iscovered by a finite collection Ox1

, . . . , Oxn .

3.3. COMPACT SPACES 75

Set K = supf∈F,1≤k≤n |f(xk)|. Since F is pointwise bounded we have

K < ∞. We cover the closed ball B(K, 0) ⊂ C with finitely many ε/4 ballsB(ε/4, z1), . . . , B(ε/4, zm).

We consider the finite set

F = φ : 1, . . . , n → 1, . . . ,m | there exists f ∈ F such that

f(xi) ∈ B(ε/4, zφ(i)) for all 1 ≤ i ≤ n,

and for each φ ∈ F , we choose fφ ∈ F which realizes the fact that φ ∈ F .

If f ∈ F , then as B(ε/4, z1), . . . , B(ε/4, zm) cover B(K, 0), there exists afunction φ ∈ F so that f(xi) ∈ B(ε/4, zφ(i)), for 1 ≤ i ≤ n. If y ∈ Oxi , we thenhave

|f(y)− fφ(y)| ≤ |f(y)− f(xi)|+ |f(xi)− fφ(xi)|+ |fφ(xi)− fφ(y)|< ε/4 + ε/2 + ε/4 = ε.

Thus ‖f − fφ‖∞ < ε, showing that F is covered by finitely many ε-balls, and istherefore totally bounded. It then follows that F is precompact in the uniformnorm by the Heine-Borel property.

A topological space is σ-compact if it is a countable union of compactsubsets.

Lemma 3.3.10. Suppose X is a σ-compact, locally compact Hausdorff space.Then there is a sequence Un∞n=1 of precompact open sets such that Un ⊂ Un+1

for each 1 ≤ n <∞, and ∪∞n=1Un = X.

Proof. We have X = ∪∞n=1Fn where Fn are compact sets. Since X is locallycompact each point x ∈ Fn has a precompact open neighborhood Ox, thenOxx∈Fn covers Fn and hence has a finite subcover Ox1

, . . . , Oxk . SettingVn = ∪ki=1Oxi , we then have that Vn is open, precompact, and Fn ⊂ Vn for each1 ≤ n <∞. Setting Un = ∪nm=1Vn then produces the desired sequence.

Note that if Un∞n=1 are as in the previous lemma and F ⊂ X is compact,then Un∞n=1 covers F and hence by compactness there exists a finite subcover.However, since Un∞n=1 is an increasing union it then follows that F ⊂ Un forsome 1 ≤ n <∞.

Theorem 3.3.11. Let X be a σ-compact, locally compact Hausdorff space. Iffn∞n=1 is a sequence which is equicontinuous and pointwise bounded, then thereexists a continuous function f : X → C and a subsequence of fn∞n=1 whichconverges to f uniformly on compact sets.

Proof. We write X = ∪∞n=1Fn where each Fn = Un as in the previous lemma.Set F0 = ∅, g0 = ∅, and f0

n∞n=1 = fn∞n=1. For n ≥ 1 we inductively choosegk : Fk → C, and subsequences fkn∞n=1 of fk−1

n ∞n=1 so that gk|Fk−1= gk−1 as

follows: We suppose that gk and fkn∞n=1 have already been chosen for 0 ≤ k <∞. Restricting fkn∞n=1 to Fk+1 we have an equicontinuous, pointwise bounded


family and hence this is precompact in the uniform norm by the Arzela-AscoliTheorem. Therefore there exists a subsequence fk+1

n ∞n=1 such that fk+1n ∞n=1

converges uniformly on Fk+1 to a continuous function gk+1 : Fk+1 → C. Asfk+1n ∞n=1 is a subsequence of fkn∞n=1 we have that gk+1|Fk = gk.

We may then define g : X → C by g(x) = gk(x) for x ∈ Fk. Then g is a welldefined continuous function. If we consider the diagonal subsequence fnn ∞n=1

then we have that fnn ∞n=1 converges to g uniformly on Fk for any 1 ≤ k <∞.Since any compact set is covered by some Fk it follows that fnn ∞n=1 convergesto g uniformly on compact sets.

3.3.1 Exercises

Exercise 3.3.12. Show that a metric space X is compact if and only if everycontinuous real valued function on X is bounded.

Exercise 3.3.13. Let (X, T ) be a compact Hausdorff space. Show that if T ′is any weaker topology then (X, T ′) is not Hausdorff. Show that if T ′ is anystronger topology then (X, T ′) is not compact.

Let X be a locally compact topological space, and fix a point ω 6∈ X. Onthe space X = X ∪ω we define a new topology whose open sets consist of theopen sets in X, together with the compliments (in X) of compact subsets of X.The space X is called the one-point compactification of X.

Exercise 3.3.14. Show that X is a compact Hausdorff space and that therelative topology from X ⊂ X agrees with the topology on X.

Let X be a locally compact topological space, a function f : X → C is saidto vanish at infinity if for every ε > 0, there exists a compact set K ⊂ X sothat |f(x)| < ε for all x ∈ Kc. We denote by C0(X) the space of all continuousfunctions which vanish at infinity.

Exercise 3.3.15. Show that C0(X) is a closed subspace of Cb(X).

Exercise 3.3.16 (Compare this with Exercise 3.1.3). Suppose X is a compactHausdorff space such that singletons x are Gδ.

1. For each x ∈ X find a countable open cover O of X \ x so that x 6∈ Ofor all O ∈ O.

2. Show that X is first countable.

If (V,E) is a graph, and k ∈ N, a k-coloring of the graph (V,E) is anassignment f ∈ 1, 2, . . . , kV such that for all (v, w) ∈ E we have f(v) 6= f(w).

Exercise 3.3.17. Prove the De Bruijn-Erdos theorem: If (V,E) is a graph suchthat a k-coloring exists for every finite subgraph, then a k-coloring exists for(V,E). Hint: For each finite subgraph (V0, E0) consider

F(V0,E0) = f ∈ 1, . . . , kV | f|V0gives a k − coloring of (V0, E0),

then show that the family of all such F(V0,E0) has the finite intersection property.(This approach is due to Gottschalk.)

3.4. THE STONE-WEIERSTRASS THEOREM 77

3.4 The Stone-Weierstrass Theorem

A subset A ⊂ Cb(X) (resp. Cb(X;R)) is an algebra if it is a complex (resp. real)vector subspace such that fg ∈ A for all f, g ∈ A. A is said to separate pointsif for all x 6= y there exists f ∈ A such that f(x) 6= f(y). A subset A ⊂ Cb(X;R)is a lattice if it is a real vector subspace such that f ∨ g = maxf, g ∈ A, andf ∧ g = minf, g ∈ A for all f, g ∈ A.

Lemma 3.4.1. For any ε > 0 there is a polynomial p on R such that p(0) = 0and ||x| − p(x)| < ε for x ∈ [−1, 1].

Proof. Consider the Maclaurin series 1 −∑∞n=1 cnt

n for (1 − t)1/2. This seriesconverges absolutely and uniformly on [−1, 1] and its sum is (1−t)1/2. Therefore,given any ε > 0 we may take a suitable partial sum to obtain a polynomial qso that |(1 − t)1/2 − q(t)| < ε/2 for t ∈ [−1, 1]. Setting r(x) = q(1 − x2), wethen obtain a polynomial r such that ||x|− r(x)| < ε/2 for x ∈ [−1, 1]. If we setp(x) = r(x)−r(0), then p is a polynomial such that p(0) = 0 and ||x|−p(x)| < εfor all x ∈ [−1, 1].

Proposition 3.4.2. Let X be a topological space. If A ⊂ Cb(X;R) is a closedsubalgebra, then A is a lattice.

Proof. As f ∨ g = 12 (f + g+ |f − g|) and f ∧ g = 1

2 (f + g− |f − g|) it is enoughto show that |f | ∈ A whenever f ∈ A. Suppose f ∈ A and ε > 0 is given.We may assume f 6= 0. Since f/‖f‖∞ maps into [−1, 1] it follows from theprevious lemma that there exists a polynomial p on R so that p(0) = 0 and|(p f)/‖f‖∞ − |f |/‖f‖∞| < ε. Since p(0) = 0 it follows that p has 0 for itsconstant coefficient, thus since A is an algebra we have pf ∈ A, and since ε > 0was arbitrary it then follows that |f |/‖f‖∞ ∈ A and hence also |f | ∈ A.

Lemma 3.4.3. Let X be a compact space. Suppose A ⊂ Cb(X;R) is a closedlattice and f ∈ C(X;R). If for every x, y ∈ X there exists g ∈ A so thatg(x) = f(x) and g(y) = f(y), then f ∈ A.

Proof. Fix ε > 0. For each x, y take gx,y ∈ A so that gx,y(x) = f(x) andgx,y(y) = f(y). Let Ux,y = z ∈ X | f(z) < gx,y(z) + ε. Fix y; then Ux,yx∈Xis an open cover of X and so there is a finite subcover Uxi,yni=1. Set gy =maxgx1,y, . . . , gxn,y ∈ A. Then f < gy + ε on X and f(y) = gy(y) so that insome neighborhood Vy of y we have that f > gy−ε. We then have that Vyy∈Xcovers X and so there is a finite subcover Vyjmj=1. Set g = mingy1 , . . . , gym ∈A. Then ‖f − g‖∞ < ε, g ∈ A, and since ε > 0 was arbitrary we then havef ∈ A.

Theorem 3.4.4 (The Stone-Weierstrass Theorem). Let X be a compact Haus-dorff space. If A ⊂ C(X;R) is a closed algebra which separates points theneither A = C(X;R) or else there exists x0 ∈ X such that A = f ∈ C(X;R) |f(x0) = 0.


Proof. We first consider the case when X is a two point set x, y, so thatC(x, y) is two dimensional. If A is two dimensional then we are done. Also,since A separates points we have a function f ∈ A with f(x) 6= f(y), so that wemay assume A is one dimensional. Then f2 = cf for some c ∈ R, so that f(x)and f(y) distinct roots of the polynomial t2 − ct. It then follows that eitherf(x) = 0 in which case A = g ∈ C(x, y) | g(x) = 0, or else f(y) = 0 inwhich case A = g ∈ C(x, y) | g(y) = 0.

We now consider the general case. Suppose x, y ∈ X, with x 6= y. Con-sidering the restriction map from A we obtain an algebra Ax,y ⊂ C(x, y;R).Note that since A separates points so does Ax,y. If there exists x0 ∈ X sothat f(x0) = 0 for every f ∈ A, then as A separates points there can be atmost one such x0 and from above we then have Ax,y = C(x, y) wheneverx0 6∈ x, y. It then follows from Proposition 3.4.2 and Lemma 3.4.3 thatA = f ∈ C(X) | f(x0) = 0. Otherwise Ax,y = C(x, y) for all x, y ∈ Xin which case it again follows from Proposition 3.4.2 and Lemma 3.4.3 thatA = C(X).

Corollary 3.4.5. Let K ⊂ Rn be compact, and f ∈ C(K;R), then for everyε > 0 there exists a polynomial p : Rn → R so that |f(k) − p(k)| < ε for allk ∈ K.

Proof. Since the space of polynomials forms an algebra which contains the con-stant functions and separate points it follows that this space is dense in C(K;R)by the Stone-Weierstrass theorem.

Theorem 3.4.6 (The Complex Stone-Weierstrass Theorem). Let X be a com-pact Hausdorff space. If A ⊂ C(X) is a closed algebra which is closed undercomplex conjugation and separates points then either A = C(X) or else thereexists x0 ∈ X such that A = f ∈ C(X;R) | f(x0) = 0.

Proof. Since Re f = (f+f)/2 and Im f = (f−f)/2i it follows that the set of realand imaginary parts of functions in A is an algebra AR in C(X;R). Moreover itis easy to see that this separates points and hence the Stone-Weierstrass theoremapplies. Since A = f + ig | f, g ∈ AR the complex version then follows.

3.4.1 Exercises

Exercise 3.4.7. Suppose X and Y are compact Hausdorff spaces and f ∈C(X×Y ). Show that for all ε > 0 there exist g1, . . . , gn ∈ C(X) and h1, . . . , hn ∈C(Y ) so that |f(x, y)−

∑ni=1 gi(x)hi(y)| < ε for all (x, y) ∈ X × Y .

Exercise 3.4.8. Let X be a compact Hausdorff space. An ideal in C(X) is asubalgebra I ⊂ C(X), such that fg ∈ I whenever f ∈ C(X) and g ∈ I.

1. If I ⊂ C(X) is an ideal, let h(I) = x ∈ X | f(x) = 0 for all f ∈ I, thehull of I. Show that h(I) is closed.

3.5. THE STONE-CECH COMPACTIFICATION 79

2. If A ⊂ X, let k(A) = f ∈ C(X) | f(x) = 0 for all x ∈ A, the kernelof A. Show that k(A) is a closed ideal in C(X) which is closed underconjugation.

3. Show that k(h(I)) = I for any ideal I ⊂ C(X) which is closed underconjugation, and h(k(A)) = A for any subset A ⊂ X.

Given a topological space X and an equivalence relation R on X, we letX/R denote the set of equivalence classes and we consider q : X → X/R thequotient map q(x) = [x]. We endow X/R with the weakest topology so that qis continuous.

Exercise 3.4.9. Let X be a compact Hausdorff space.

1. Show that X/R is Hausdorff if and only if R is a closed subset of X ×X.

2. If R ⊂ X ×X is closed, consider AR = f q | f ∈ C(X/R). Show thatAR is a closed subalgebra of C(X) which contains the constant functionsand is closed under complex conjugation.

3. Show that R 7→ AR gives a bijection between equivalence relations on Xwhich are closed in X×X, and closed subalgebras of C(X) which containthe constant functions and are closed under complex conjugation.

3.5 The Stone-Cech compactification

Let X be a topological space. A map χ : Cb(X) → C is a homomorphism ifit is linear, and satisfies χ(fg) = χ(f)χ(g) for f, g ∈ Cb(X), and χ(1) = 1. Wedenote by σ(Cb(X)) the space of all such homomorphisms and we endow thiswith the topology of pointwise convergence inherited from CCb(X). Note that ifϕ, χ ∈ σ(Cb(X)) then ϕ = χ if and only if ker(ϕ) = ker(χ).

Lemma 3.5.1. σ(Cb(X)) is compact.

Proof. Suppose ϕ ∈ σ(Cb(X)), and f ∈ Cb(X) with ‖f‖∞ ≤ 1. We claim that|ϕ(f)| ≤ 1. If this were not the case then the function g(x) = ϕ(f)−f(x) wouldsatisfy |g(x)| ≥ |ϕ(f)| − 1 for each x ∈ X and hence the function h(x) = 1

g(x)

would be in Cb(X). However, we would then have 1 = ϕ(1) = ϕ(g)ϕ(h) = 0 acontradiction.

Also, restricting to Cb(X)1 = f ∈ Cb(X) | ‖f‖ ≤ 1, gives the same topol-ogy of pointwise convergence and hence we may view σ(Cb(X)) as a subspaceof DB where D is the closed unit disc in C. Since DB is compact it is thenenough to show that σ(Cb(X)) is a closed subspace.

Suppose therefore that ϕαα∈A is a net of homomorphisms which convergepointwise to a function ϕ : Cb(X) → C. As addition, scalar multiplication arecontinuous in C it then follows that ϕ is linear, and as multiplication is jointlycontinuous in C it follows that ϕ(fg) = ϕ(f)ϕ(g) for all f, g ∈ Cb(X). Thereforeϕ is a homomorphism.


Theorem 3.5.2 (Stone). Let X be a topological space. For each x ∈ X denoteby βx : Cb(X) → C the homomorphism given by βx(f) = f(x), then X 3x 7→ βx ∈ σ(C(X)) is a continuous map with dense image which satisfies theuniversal property that if π : X → K is any continuous map into a compactHausdorff space K, then there exists a unique continuous map βπ : σ(C(X))→K, such that for x ∈ X we have π(x) = βπ(βx). In particular, if X is a compactHausdorff space then β is a homeomorphism.

Proof. If xi ⊂ X is a net such that xi → x, then for any f ∈ Cb(X) we haveβxi(f) = f(xi) → f(x) = βx(f), hence βxi → βx. Thus, x 7→ βx is continuous.To show that this map has dense image we suppose by way of contradiction thatϕ ∈ σ(Cb(X)) is not in the closure of β(X), and set I = ker(ϕ).

If ψ ∈ β(X), then there exists fψ ∈ I such that fψ 6∈ ker(ψ). Hence, forsome cψ > 0, and an open neighborhood Oψ of ψ, we have that |ψ′(f)| >cψ for all ψ′ ∈ Oψ. As β(X) is compact we may take a finite subcover ofthe cover Oψψ∈β(X)

. Thus, we obtain f1, . . . , fn ∈ I, and c > 0 such that∑ni=1 ψ(|f |2) > c for all ψ ∈ β(X). In particular we have

∑ni=1 |f |2(x) =

βx(∑ni=1 |f |2) > c, for all x ∈ X. Thus, if we consider the function g(x) =

1/(∑n

i=1 |f |2), then g ∈ Cb(X) and we have fg = 1. We would then have

1 = ϕ(1) = ϕ(fg) = ϕ(f)ϕ(g) = 0, a contradiction. Thus, we must have thatβ(X) = σ(Cb(X)).

If X is a compact Hausdorff space then β is surjective since the image is denseand compact. Moreover, β is injective since Cb(X) separates points. Hence, βis a homoeomorphism, being a continuous bijection between compact Hausdorffspaces.

In general, to see that β : X → σ(Cb(X)) satisfies the above universalproperty, suppose that K is a compact Hausdorff space and π : X → K iscontinuous. We then obtain a continuous map π∗ : C(K) → Cb(X) given byπ∗(f)(x) = f(π(x)). Thus, we obtain the continuous map π : σ(Cb(X)) →σ(C(K)) by π(ϕ)(g) = ϕ(π∗(g)). Since K is compact and Hausdorff we haveestablished above that βK : K → σ(Cb(K)) is a homeomorphism. Thus, we

obtain a continuous map βπ : σ(Cb(X)) → K by setting βπ = βK−1 π. If

x ∈ X, and g ∈ C(K) then we compute directly

π(βx)(ϕ)(g) = βx(π∗(g)) = π∗(g)(x) = g(π(x)) = βKπ(x)(g).

Hence, βπ(βx) = π(x).

If X is a topological space, then the Stone-Cech compactification of Xconsists of a compact Hausdorff space βX, together with a continuous map β :X → βX, which satisfies the universal property given in the previous theorem.If follows easily that, up to homeomorphism, this is uniquely defined by itsuniversal property. The previous theorem shows that βX exists and may beidentified with σ(Cb(X)). The following easy consequence (implicit already inTychonoff’s work) was obtained independently by Cech using different methods:

Corollary 3.5.3 (Stone, Cech). Let X be a topological space, then β : X → βXis a homeomorphism onto its image if and only if X is a Tychonoff space.

3.5. THE STONE-CECH COMPACTIFICATION 81

Proof. From the previous theorem we have that β : X → βX is continuous.Since X is Tychonoff we have, in particular, that Cb(X) separates points, andit then follows that β is injective. Thus, we just need to show that β is anopen map into β(X). Suppose that F ⊂ X is closed and x ∈ X \ F . As X isTychonoff there exists f : X → [0, 1] continuous so that f|F = 0, and f(x) = 1.

Thus, we have βx(f) = 1 while βy(f) = 0 for all y ∈ F , and hence βx 6∈ β(F ).

Since x 6∈ F was arbitrary it follows that β(F )∩ β(X) = β(F ), and hence β(F )is closed in β(X). As β is injective, taking complements shows that β is an openmap into β(X).

As a subspace of a Tychonoff space is again Tychonoff, and compact Haus-dorff spaces are normal and hence Tychonoff by Corollary 3.3.6, the previouscorollary gives the following characterization of Tychonoff spaces.

Corollary 3.5.4. A topological space X is Tychonoff if and only if X is home-omorphic to a subspace of a compact Hausdorff space.

Considering the one-point compactification gives the following:

Corollary 3.5.5. Locally compact spaces are Tychonoff.

Theorem 3.5.6 (The Tietze Extension Theorem for Tychonoff spaces). Let Xbe a Tychonoff space, K ⊂ U ⊂ X, with K compact, and U open. If f ∈ C(K)then there exists F ∈ Cb(X), with ‖F‖∞ = ‖f‖∞, such that F|K = f , andF|Uc = 0.

Proof. Since X is Tychonoff, the map β : X → βX is a homeomorphism onto itsimage. Thus β(K) ⊂ βX is compact, and there exists V ⊂ βX open such thatβ(K) ⊂ β(U) = V ∩ βX ⊂ V . We consider the function g : β(K) ∪ V c → C bysetting g(β(k)) = f(k) for k ∈ K, and g(x) = 0 for x ∈ V c. Since β(U) and V c

are disjoint closed sets, and βX is normal, they can be separated so that they areboth clopen in the relative topology. Thus, g is continuous and by the TietzeExtension Theorem for compact Hausdorff spaces there is then a continuousfunction G ∈ C(βX), with ‖G‖∞ = ‖g‖∞ = ‖f‖∞ so that G|β(K) = g, andG|V c = 0. Taking F = G β then gives the desired function.

Lemma 3.5.7. If X is normal and second countable then there exists a count-able family F ⊂ Cb(X; [0, 1]) which separates points.

Proof. Let E be a countable base for X. For each U, V ∈ E such that U ⊂ V wemay use Urysohn’s lemma to construct a continuous function fU,V : X → [0, 1]so that f|U = 0 and f|V c = 1. If we let F be the collection of all such fU,V andclaim that F separates points. Indeed, if x, y ∈ X with x 6= y, then as X isnormal there exist disjoint closed neighborhoods E and F of x and y respectively.Then there must exist U, V ∈ E neighborhoods of x and y respectively such thatU ⊂ E and V ⊂ F . We then have that fU,V (x) = 0 while fU,V (y) = 1.

Proposition 3.5.8. Every second countable normal space is homeomorphic toa subspace of the Hilbert cube [0, 1]N


Proof. Suppose X is normal and second countable. Then by Lemma 3.5.7 thereis a countable family F ⊂ Cb(X) which separates points in F . Consider theevaluation map e : X → [0, 1]F given by e(x)(f) = f(x). Then this is continuousand since F separates points it is injective. Since [0, 1]F is compact e extendsto a continuous map βe : βX → [0, 1]F . If F ⊂ X is closed then F ⊂ βX iscompact and satisfies F ∩ X = F . As βe is continuous we have that βe(F ) iscompact, and hence e(F ) = βe(F ) ∩ e(X) is closed in e(X). Thus, the map eonto its image preserves closed sets and hence e is a homeomorphism of X ontoits image in [0, 1]F .

Theorem 3.5.9 (The Urysohn Metrization Theorem). Every second countablenormal space is metrizable.

Proof. The Hilbert cube is metrizable. Indeed, the explicit metric d(f, g) =∑∞n=1 2−n|f(n) − g(n)| is easily seen to give the topology on [0, 1]N. Since

subspaces of metrizable spaces are again metrizable, the result then followsfrom Proposition 3.5.8

3.5.1 Exercises

Exercise 3.5.10. Let X be a compact Hausdorff space. Show that X is asecond countable if and only if C(X) is separable.

Exercise 3.5.11. Suppose that a topological space X has a countable basis ofclopen sets, show that X embedds into 0, 1N.

Exercise 3.5.12. Let X and Y be compact Hausdorff spaces and suppose φ :C(X)→ C(Y ) is a (unital) homomorphism, i.e., φ is complex linear, φ(1) = 1,and φ(fg) = φ(f)φ(g) for all f, g ∈ C(X). Show that there exists a uniquecontinuous map π : Y → X so that φ(f) = f π for all f ∈ C(X). Moreover,show that π is bijective if and only if φ is bijective.

3.6 The property of Baire

A topological space X is completely metrizable if there is a complete metricon X which gives the topology.

Proposition 3.6.1. A Gδ subset A of a completely metrizable space X is com-pletely metrizable in the relative toplogy.

Proof. Suppose that d is a complete metric giving the topology on X.We consider first the case when A is open. In this case we may consider the

metric d1 on A given by d1(x, y) = d(x, y) +∣∣∣ 1d(x,Ac) −

1d(y,Ac)

∣∣∣. Then it is easy

to check that d1 is a complete metric on A which gives the relative topology.Next suppose that A = ∩n∈NOn where each On ⊂ X is open. for each n ∈ N

we let dn be a complete metric on On which gives the relative topology on On.

Replacing dn(x, y) with dn(x,y)1+dn(x,y) we assume that dn(x, y) < 1 for all x, y ∈ On.

3.6. THE PROPERTY OF BAIRE 83

We define the metric d by d(x, y) =∑∞n=1 2−ndn(x, y). It is then easy to see

that d gives a complete metric on A which gives the relative topology.

Theorem 3.6.2 (Kuratowski’s Extension Theorem). Let X be a Hausdorffspace and A ⊂ X a dense subset of X. Suppose that Y is a completely metrizablespace and f : A → Y is continuous, then there exists a continuous extensionf : B → Y where B ⊂ X is Gδ with A ⊂ B.

Proof. We fix a complete metric d on Y . For x ∈ X we set

oscf (x) = infdiamf(U ∩A) | U an open neighborhood of x.

Then Bn = x ∈ X | ocsf (x) < 1/n is open for each n ∈ N, and henceB = ∩nBn = x ∈ X | oscf (x) = 0 is Gδ. Since f is continuous on A we haveA ⊂ B.

We define f : B → Y by f(x) = limα→∞ f(xα) where xαα ⊂ A is any netsuch that xα → x. Since B = x ∈ X | oscf (x) = 0 and Y is a complete space

it follows easily that f is a well defined continuous extension of f .

Corollary 3.6.3. Let X be a Hausdorff space and A ⊂ X a dense subset suchthat A is completely metrizable, then A is a Gδ-set in X.

Proof. If we consider the identity map on A then from Kuratowski’s theoremthere exists a Gδ-set G ⊂ X with A ⊂ G and a continuous extension f : G →A ⊂ G. Since A is dense in G and f agrees with the identity on A it then followsthat f is the identity map, hence A = G.

Corollary 3.6.4. A subspace F of a completely metrizable space X is completelymetrizable if and only if F is Gδ.

Proof. Replacing X with F , we may assume that F is dense. The result thenfollows from Proposition 3.6.1 and Corollary 3.6.3.

A Polish space is a topological space X which is separable and completelymetrizable.

Corollary 3.6.5. A topological space X is Polish if and only if X is homeo-morphic to a Gδ subset of a second countable compact Hausdorff space.

Proof. If X is Polish, then Proposition 3.5.8 shows that X is homeomorphic toa subset of a second countable compact Hausdorff space, and Corollary 3.6.3shows that this subset must be Gδ.

Conversely, second countable compact Hausdorff spaces are completely metriz-able by Urysohn’s Metrization Theorem, hence if X is a Gδ subset then X iscompletely metrizable by Corollary 3.6.4.

A topological space X is Cech-complete if it is homeomorphic to a Gδ-subset of a compact Hausdorff space.


Corollary 3.6.6. Let X be a completely metrizable space, then X is Cech-complete.

Proof. Let β : X → βX be the Stone-Cech compactification of X. As X isTychonoff, β is a homeomorphism onto its image. Corollary 3.6.3 shows thatthe image must be a Gδ-set.

Note that, by considering the one point compactification, any locally com-pact Hausdorff space is also Cech-complete.

Let X be a topological space. A subset A ⊂ X is meager if it is a countableunion of nowhere dense sets. A subset B ⊂ X is comeager (or residual) if itscomplement is meager. We say that X is a Baire space if every comeager setis dense. Equivalently, X is a Baire space if whenever Onn∈N is a sequenceof open dense sets, we have that ∩nOn is dense. The following lemma is left tothe reader.

Lemma 3.6.7. Let X be a Baire space, and Y ⊂ X a dense Gδ-subset, then Yis a Baire space.

Theorem 3.6.8 (Baire). Cech-complete spaces are Baire.

Proof. By the previous lemma it is enough to show that compact Hausdorffspaces are Baire. Thus, suppose X is a compact Hausdorff space and Onn∈Nis a sequence of open dense sets. Let U be any non-empty open set in X. Wenow inductively define a decreasing sequence of closed sets Fnn, and non-empty open sets Gnn such that Gn ⊂ Fn ⊂ U ∩ O1 ∩ · · ·On: Since O1 isdense, we have O1 ∩ U 6= ∅. Let F1 be a non-empty closed subset of O1 ∩ Uwith non-empty interior Gn. Now suppose F1, . . . , Fn and G1, . . . , Gn have beenconstructed. Since On−1 is dense we have On−1 ∩ Gn is non-empty and hencewe may take Fn+1 to be any closed subset of On−1∩Gn with non-empty interiorGn+1.

By construction we then have ∩nFn ⊂ U ∩ ∩nOn, and by compactness wehave that ∩nFn is not empty. Since U was an arbitrary non-empty open subsetit follows that ∩nOn is dense in X.

3.6.1 Exercises

Note that from Corollary 3.6.5 the space of irrationals R \Q with its subspacetopology is Polish, even though the usual metric is far from complete. The nexttwo exercises give an explicit way to see this.

Exercise 3.6.9. Suppose Xn∞n=1 is a sequence of completely metrizable spaces.Show that

∏∞n=1Xn is completely metrizable. Moreover, show that

∏∞n=1Xn is

separable if each Xn is separable.

Exercise 3.6.10. Show that R\Q and the Baire space NN are homeomorphic.Hint: Consider continued fraction expansions.

3.6. THE PROPERTY OF BAIRE 85

Exercise 3.6.11. Let X be a compact Hausdorff space and suppose |X| =∞.Show that, as a complex vector space, C(X) has no countable basis.

Exercise 3.6.12. Suppose fn : R → R are continuous, and f : R → R suchthat fn(x)→ f(x) for each x ∈ R.

1. Show that if I is an open interval of positive length then f−1(I)∩f−1(I)c

is Fσ and nowhere dense.

2. Show that if f is not continuous at a point x then there exists an openinterval I with rational endpoints such that x ∈ f−1(I) ∩ f−1(I)c.

3. Show that f is continuous on a dense set of points in R.

Exercise 3.6.13. Show that there exists a function f ∈ C([0, 1]) so that f isnot monotone on any interval of positive length.

Exercise 3.6.14 ([hg]). Let f : R → R be an infinitely differentiable functionsuch that at each point x ∈ R there is a derivative f (n) so that f (n)(x) = 0. Let

Y = x ∈ R |f|O = p|O for some polynomial p

and some neighborhood O of x,

and let X = Y c. Suppose that X 6= ∅. For each n ≥ 0 let Sn = x ∈ R |f (n)(x) = 0.

(a) Show that X is a closed set without isolated points.

(b) Show that there exists an interval (a, b) such that ∅ 6= (a, b) ∩X ⊂ Sn.

(c) Reach a contradiction by showing that f (n)(x) = 0 for all x ∈ (a, b).

(d) Conclude that, in fact, X = ∅, and deduce from this that f agrees with apolynomial on R.

Exercise 3.6.15. For each n,m ∈ N let

An,m =

f ∈ C([0, 1]) | there exists x ∈ [0, 1] such that

∣∣∣∣f(t)− f(x)

t− x

∣∣∣∣ ≤ n if 0 < |x− t| < 1

m

.

1. Show that if f ∈ C([0, 1]) is differentiable at some point in [0, 1] thenf ∈ An,m for some n,m ∈ N.

2. Show that An,m is closed in C([0, 1]).

3. Show that An,m is nowhere dense in C([0, 1]).

4. Show that the set of functions f ∈ C([0, 1]) which are nowhere differen-tiable is dense in C([0, 1]).


3.7 Cantor spaces

A Cantor space is a non-empty compact Hausdorff space without isolatedpoints and having a countable basis consisting of clopen sets. For example,suppose Fn∞n=1 is a sequence of finite sets with |Fn| ≥ 2, which we consideras discrete topological spaces, then X =

∏∞n=1 Fn is a Cantor space. Indeed, by

Tychonoff’s theorem X is non-empty compact Hausdorff. A countable basis ofclopen sets are given by finite intersections of sets of the form π−1

n (En) whereEn ⊂ Fn is a non-empty set. It is also easy to see that since |Fn| ≥ 2, X hasno isolated points.

Given a set A, we let A<N denote the set of finite sequences in A, i.e., A<N

consists of the empty set, together with the disjoint union of An, for n ∈ N. Ifs = (s1, . . . , sn) ∈ A<N, and k ∈ A, we denote by s k the sequence (s1, . . . , sn, k).If s ∈ AN, and n ∈ 0 ∪ N then we denote by s|n the sequence which consistsof the first n entires of s.

A Cantor scheme on a set X is a family Ass∈0,1<N of subsets of Xsuch that

1. Asˆ0 ∩Asˆ1 = ∅, for s ∈ 0, 1<N,

2. Asî ⊂ As, for s ∈ 0, 1<N.

If (X, d) is a metric space and we additionally have limn→∞ diam(As|n) = 0 for

any s ∈ 0, 1N then we say that Ass∈0,1<N has vanishing diameter.

Lemma 3.7.1. Suppose (X, d) is a complete metric space and we have a Can-tor scheme Ass∈0,1<N which has vanishing diameter and such that As is

nonempty open for each s ∈ 0, 1<N. Then for each s ∈ 0, 1<N there is aunique element f(s) in ∩n∈NAs|n, and the map f : 0, 1N → X is a continuousembedding.

If, moreover, we have that A∅ = X and As = Asˆ0 ∪Asˆ1, for s ∈ 0, 1<N,then f is a homeomorphism.

Proof. Since Ass∈0,1<N has vanishing diameter and (X, d) is complete it

follows that f is well defined. Moreover, since Asˆ0 ∩Asˆ1 = ∅, for s ∈ 0, 1<N,it follows that f is injective.

A sequence sn∞n=1 ⊂ 0, 1N converges to a point s if and only if for eachk ∈ N we have sn|k = s|k for large enough n, thus it follows that f(sn) ∈ As|kfor large enough n, since k is arbitrary we then have limn→∞ f(sn) = f(s).Therefore f is continuous.

If A∅ = X and As = Asˆ0 ∪ Asˆ1, for s ∈ 0, 1<N, then it follows that f issurjective and hence a homeomorphism since 0, 1N is compact.

Theorem 3.7.2 (Brouwer). Any two Cantor spaces are homeomorphic.

Proof. Let C be a Cantor space. By Lemma 3.7.1, to prove the theorem it isenough to produce a Cantor scheme Ass∈0,1<N which has vanishing diameterand satisfies

3.7. CANTOR SPACES 87

1. A∅ = X;

2. As is open nonempty;

3. As = Asˆ0 ∪Asˆ1, for s ∈ 0, 1<N.

We construct Ass∈0,1<N as follows: Fix a compatible metric d on Xand decompose X as a disjoint union X = X1 ∪ X2 ∪ · · · ∪ Xn so that eachXi is nonempty clopen and has diameter at most 1/2. Using the notationak = aa . . . a (k times), we define the sets A0kˆ1 = Xk+1, for 0 ≤ k < n − 1,and A0k = Xi+1 ∪ · · · ∪Xn, for 0 ≤ k < n. We now repeat this process withineach Xi, using sets of diameter at most 1/3. We then continue this process byinduction.

Proposition 3.7.3. Let X be a nonempty Polish space without isolated points,then there exists an embedding of 0, 1N into X.

Proof. Let d be a compatible metric on X. By Lemma 3.7.1 it is enough toproduce a Cantor scheme Uss∈0,1<N in X such that

1. Us is open nonempty for each s;

2. diam(Us) ≤ 2−length(s);

3. Usî ⊂ Us, for s ∈ 0, 1<N.

We construct such a scheme by induction on the length. Let U∅ be any opennonempty set with diameter at most 1. Given Us, as X has no isolated pointsthere exist distinct points x, y ∈ Us. We then let Usˆ0 and Usˆ1 be disjoint openneighborhoods in Us of x and y respectively so that each has diameter at most2−length(s)−1.

Theorem 3.7.4 (The Cantor-Bendixson theorem). Let X be a Polish space.Then X can be written uniquely as P ∪C, where P has no isolated points, andC is countable open.

Proof. We let P denote the set of points x ∈ X such that any neighborhood ofx has uncountably many points, and we let C = P c. If Onn∈N is a countableopen basis, then C is the union of all countable On, hence, C is countable andopen. Each neighborhood in X of each point in P is uncountable, and since Cis countable, this also holds for each neighborhood in P , thus P has no isolatedpoints.

For uniqueness, suppose that X = Q ∪D where D is countable open and Qhas no isolated points. Since D is countable open we clearly have that D ⊂ C.If x ∈ C \ D were isolated as C is open we would have that x is also isolatedin Q, however, Q has no isolated points and hence we conlude that C \D alsohas no isolated points. By Proposition 3.6.1 we then have that C \ D is acountable Polish space without isolated points. Proposition 3.7.3 then showsthat C \D = ∅ and hence C = D, and P = Q.


Lemma 3.7.5. Suppose A ⊂ 0, 1N is nonempty closed, then there exists acontinuous map f : 0, 1N → A so that f is the identity on A.

Proof. For each s ∈ 0, 1N we define f(s) to be the point in A which has thelongest common initial segment with s. It’s easy to see that f is well defined,and if s, t ∈ 0, 1N and k ∈ N such that s|k = t|k then f(s)|k = f(t)|k. Since asequence sn∞n=1 ⊂ 0, 1N converges to a point s if and only if for each k ∈ Nwe have sn|k = s|k for large enough n, it then follows that f is continuous.

In Proposition 3.5.8 we saw that every compact metric has a continuous in-jective map into the Hilbert cube. The following result gives a nice complementto this result.

Theorem 3.7.6 (The Hausdorff-Alexandroff theorem). Every nonempty com-pact metric space X is a continuous image of the Cantor space.

Proof. Set C = 0, 1N. We first prove the theorem in the case when X is theHilbert cube. Note that the map f(x) =

∑∞n=0 xn2−n−1 maps C continuously

onto [0, 1], hence CN maps continuously onto [0, 1]N. Since CN is homeomorphicto C we are done.

We now consider the general case. Since X is a compact metric space,Proposition 3.5.8 shows that we may assume X ⊂ [0, 1]N. From above we knowthat there is a continuous surjection g : C → [0, 1]N. Then g−1(X) ⊂ C is closedand from Lemma 3.7.5 there is a continuous surjection h : C → g−1(X). Themap g h then gives a continuous surjection of C onto X.

3.7.1 Exercises

Exercise 3.7.7. Give an explicit homoemorphism between the Cantor space0, 1N and the usual Cantor set C ⊂ [0, 1].

A Souslin scheme on a set X is a family Ass∈N<N of subsets of X. ALusin scheme on X is a Souslin scheme such that

1. Asî ∩Asˆj = ∅, for s ∈ N<N, i 6= j.

2. Asî ⊂ As, for s ∈ N<N.

If (X, d) is a metric space and we additionally have limn→∞ diam(As|n) = 0 for

any s ∈ 0, 1N then we say that Ass∈N<N has vanishing diameter. In thiscase we let D = s | ∩n∈NAs|n 6= ∅, and for s ∈ D we define f(s) ∈ X so thatf(s) = ∩n∈NAs|n. The map f : D → X is the associated map.

Exercise 3.7.8. Suppose (X, d) is a metric space and we have a Souslin schemeAss∈N<N which has vanishing diameter, and associated map f : D → X.

1. Show that f is continuous.

2. Show that f is open if each As is open and As ⊂ ∪n∈NAsˆn, for all s ∈ N<N.

3.8. STANDARD BOREL SPACES 89

3. Show that f is injective if Ass∈N<N is a Lusin scheme.

4. Show that f is surjective if A∅ = X, and As = ∪n∈NAsˆn, for all s ∈ N<N.

Exercise 3.7.9. Show that every nonempty Polish space without isolated pointsX is the continuous image of the Baire space NN.

Exercise 3.7.10. Let (X, d) be a complete metric space such that every com-pact subset of X has empty interior. Show that for each nonempty open setA ⊂ X, there exists ε > 0, so that if A ⊂ ∪n∈NBn and diam(Bn) < ε, thenBn 6= ∅ for infinitely many n ∈ N.

Exercise 3.7.11. Prove the Alexandrov-Urysohn Theorem: If X is a Polishspace which has a countable basis of clopen sets and such that any compactsubset of X has empty interior, then X is homeomorphic to NN.

Exercise 3.7.12. Show that if (X, d) is a nonempty countable metric spacewithout isolated points, then there is an embedding F : X → NN which hasdense image. Hint: Find a Lusin scheme on X so that the associated mapf : D → X is open, and bijective, with D dense, and then let F = f−1.

Exercise 3.7.13. Prove Sierpinski’s Theorem: Let (X, d) be a nonempty count-able metric space without isolated points, then X is homeomorphic to Q withits usual topology. Hint: Combine Exercises 3.1.9, 3.6.10, and 3.7.12.

Exercise 3.7.14. Consider Q2 with the lexicographical ordering (q, r) ≤(s, t) if and only if either q < s, or q = s and r ≤ t. Show that Q2 with thecorresponding order topology is homeomorphic to Q2 with its usual producttopology.

Exercise 3.7.15 (Benyamini). Show that there exists f ∈ Cb(R) such thatgiven any doubly infinite sequence ynn∈Z there is a point t ∈ R so thatyn = f(t + n) for n ∈ Z. Hint: If C ⊂ [0, 1/2] is homeomorphic to the Cantorset, first construct a continuous surjection f : C → [0, 1]Z. Then define g :∪n∈ZC + n → R by g(t + n) = πn(f(t)) for t ∈ C. Then extend g to acontinuous function in Cb(R).

3.8 Standard Borel spaces

A standard Borel space is a measurable space (X,M) so that M is theBorel σ-algebra for some Polish topology on X. A standard measure spaceis a σ-finite measure space (X,M, µ) whose underlying measurable structure(X,M) is a standard Borel space. If, in addition, we have µ(X) = 1, then wesay that (X,M, µ) is a standard probability space. A Lebesgue space isa standard probability space which has no atoms, e.g., X = [0, 1] with its Borelstructure and Lebesgue measure.

Lemma 3.8.1. Let X be a Polish space with topology T , and suppose A ⊂ Xclosed, then T ∪ A is again a Polish topology.


Proof. Since X is Polish there exists a complete metric d on X such that (X, d)

gives the topology T on X. Replacing d with d(x,y)1+d(x,y) we may assume that

diamd(X) ≤ 1.Since A ⊂ X is closed, d restricts to a complete metric on A, and from Propo-

sition 3.6.1 we have a complete metric d1 on Ac, which satisfies diamd1(Ac) ≤ 1,and gives the topological structure to Ac. We may then define a metric on Xby

d(x, y) =

d(x, y) if x, y ∈ A,d1(x, y) if x, y ∈ Ac,

1 otherwise.

Then d is a complete metric on X, and the corresponding topology is T ∪A.

Lemma 3.8.2. Let X be a Polish space with topology T , and suppose thatTn∞n=1 is a sequence of Polish topologies on X such that T ⊂ Tn for eachn ∈ N, then the topology T∞ generated by ∪n∈NTn is again a Polish topology onX. Moreover, if B(Tn) = B(T ) for all n ∈ N, then B(T∞) = B(T ).

Proof. Let Xn = X for n ∈ N. Consider the map ϕ : X →∏∞n=1Xn given

by ϕ(x) = (x, x, . . .). Then ϕ gives a homeomorphism between (X, T∞) andϕ(X) ⊂

∏∞n=1Xn. Thus, to show that (X, T∞) is Polish, it is enough to show

that ϕ(X) ⊂∏∞n=1Xn is closed. Suppose (xn) 6∈ ϕ(X), then for some i < j we

have xi 6= xj . We let U and V be disjoint open sets in T (hence also in Ti andTj) so that xi ∈ U and xj ∈ V , then

(xn) ∈ π−1i (U) ∩ π−1

j (V ) ⊂ ϕ(X)c.

Since Polish spaces are separable, given any set G which generates the topol-ogy, we have that any open set is a countable union of finite intersections in G.Thus, if G is in a given σ-algebraM, then this σ-algebra contains all Borel sets.

If Tn ⊂ B(T ), then ∪n∈NT ⊂ B(T ) and this generates the Polish topologyT∞. Thus, from the remark above we have that B(T∞) ⊂ B(T ).

Theorem 3.8.3. Let X be a Polish space, and Enn∈N a countable collectionof Borel subsets, then there exists a finer Polish topology on X with the sameBorel structure, such that for each n ∈ N, En is clopen in this new topology.

Proof. We first consider the case of a single Borel subset E ⊂ X. We let Adenote the set of subsets which satisfy the conclusion of the theorem and we letB be the σ-algebra of Borel subsets of X.

Lemma 3.8.1 shows that A contains all closed subsets of (X, d). It is alsoclear that A is closed under taking complements. Thus, to conclude that B ⊂ Ait is then enough to show that A is closed under countable intersections. IfAn ∈ A, and Tn are finer Polish topologies on X, with Borel structure B, suchthat An is clopen in Tn for each n ∈ N, then by Lemma 3.8.2 there is a finerPolish topology T∞ which generates B and such that An is clopen in T∞, for each


n ∈ N. We then have that ∩n∈NAn is closed in T∞, and hence by Lemma 3.8.1we have that ∩n∈NAn ∈ A.

Having established the result for a single Borel set E, we may then applyLemma 3.8.1 to obtain the result for a sequence of Borel sets Enn∈N.

Corollary 3.8.4. Let (X,B) be a standard Borel space, and E ∈ B a Borelsubset, then (E,B|E) is a standard Borel space.

Proof. By the previous theorem we may assume X is Polish and E ⊂ X isclopen, and hence Polish. We then have that B|E is the associated Borel struc-ture on E and hence (E,B|E) is standard.

Corollary 3.8.5. Let X be a standard Borel space, Y a Polish space, andf : X → Y a Borel map, then there exists a Polish topology on X which generatesthe same Borel structure and such that f is continuous with respect to thistopology.

Proof. Let En be a countable basis for the topology on Y . By Theorem 3.8.3there exists a Polish topology on X which generates the same Borel structureand such that f−1(En) is clopen for each n ∈ N. Hence, in this topology f iscontinuous.

Lemma 3.8.6. Let X be a Polish space, then there exists a Souslin schemeEss∈N<N consisting of Borel subsets such that the the following conditions aresatisfied:

(i) E∅ = X.

(ii) For each s ∈ N<N, Es = tk∈NEsˆk.

(iii) For each s ∈ NN the set ∩n∈NEs|n consists of at most one element.

(iv) For each s ∈ NN, ∩n∈NEs|n = x 6= ∅ if and only if Es|n 6= ∅ for alln ∈ N, and in this case for any sequence xn ∈ Es|n we have xn → x.

Proof. Let d be a complete metric on X which generates the Polish topologyon X, and such that X has diameter at most 1. We will inductively constructEss∈N<N so that for s ∈ Nn the diameter of Es is at most 2−n. First, we setE∅ = X. Now suppose Es has been constructed for each s ∈ ∅ ∪kn=1 Nn. Ifs ∈ Nk, let xnn∈N be a countable dense subset of Es (note that any subspaceof a separable metric space is again separable).

We define Esî = Es ∩ (B2−k−1(xi) \ ∪j<iB2−k−1(xj)), where Br(x) denotesthe open ball of radius r centered at x. It is then easy to see that for eachs ∈ N<N, we have Es = tk∈NEsˆk. Moreover, for each s ∈ NN, we have thatEs|n has diameter at most 2−n, hence ∩n∈NEs|n contains at most one element.

Finally, if Es|n 6= ∅ for all n ∈ N, then as the diameter of Es|n converges to

0, it follows from completeness, that there exists x ∈ ∩n∈NEs|n, and for eachsequence xn ∈ Es|n we have xn → x.


If X is a standard Borel space and A,B ⊂ X are disjoint, then we say thatA and B are Borel separated if there exists a Borel subset E ⊂ X such thatA ⊂ E, and B ⊂ X \ E.

Lemma 3.8.7. Let X be a standard Borel space and suppose that A = ∪n∈NAn,and B = ∪m∈NBm, are such that An and Bm are Borel separated for eachn,m ∈ N, then A and B are Borel separated.

Proof. Suppose En,m is a Borel subset which separates An and Bm for eachn,m ∈ N. Then E = ∪n∈N ∩m∈N En,m separates A and B.

If X is a Polish space, a subset E ⊂ X is analytic if there exits a Polishspace Y and a continuous function f : Y → X such that E = f(Y ). Note thatit follows from Corollary 3.8.5 that if f : Y → X is Borel then f(Y ) is analytic.In particular, it follows that all Borel sets are analytic. If X is a standard Borelspace then a subset E ⊂ X is analytic if it is analytic for some (and hence all)Polish topologies on X which give the Borel structure.

Theorem 3.8.8 (The Lusin Separation Theorem). Let X be a standard Borelspace, and A,B ⊂ X two disjoint analytic sets, then A and B are Borel sepa-rated.

Proof. We may assume that X is a Polish space, and that there are Polish spacesY1, and Y2, and continuous functions fi : Yi → X such that A = f1(Y1) andB = f2(Y2).

Let Ess∈N<N (resp. Fss∈N<N) be a Souslin scheme for Y1 (resp. Y2) whichsatisfies the conditions in Lemma 3.8.6. If A and B are not Borel separated thenby Lemma 3.8.7 we may recursively define sequences s, r ∈ NN such that f1(Es|n)and f2(Fr|n) are not Borel separated for each n ∈ N. In particular, we have thatEs|n and Fr|n are non-empty for each n ∈ N, hence there exists a ∈ Y1, b ∈ Y2

such that ∩n∈NEs|n = a, ∩n∈NFr|n = b.If V,W ⊂ X are disjoint open subsets with f1(a) ∈ V , and f2(b) ∈W , then

by continuity of fi, for large enough n we have f1(Ex|n) ⊂ V , and f2(Fy|n) ⊂W .Hence V separates Ex|n from Fy|n for large enough n, a contradiction.

Corollary 3.8.9. If X is a standard Borel space then a subset E ⊂ X is Borelif and only if both E and X \ E are analytic.

Corollary 3.8.10. let X be a standard Borel space, and let Ann∈N be asequence of disjoint analytic subsets, then there exists a sequence Enn∈N ofdisjoint Borel subsets such that An ⊂ En for each n ∈ N.

Proof. It is easy to see that the countable union of analytic sets is analytic.Hence, by Lusin’s separation theorem we may inductively define a sequenceof Borel subsets Enn∈N such that An ⊂ En, while (∪k>nAk) ∪ (∪k<nEk) ⊂X \An.

Theorem 3.8.11 (Lusin-Souslin). Let X and Y be standard Borel spaces, andf : X → Y an injective Borel map, then f(X) is Borel, and f implements anisomorphism of standard Borel spaces between X and f(X).


Proof. We first show that f(X) is Borel. By Corollary 3.8.5 we may assumethat X and Y are Polish spaces and f is continuous. Let Ess∈N<N be a Souslinscheme for X which satisfies the conditions of Lemma 3.8.6. Then f(Es)s∈N<N

gives a Souslin scheme of analytic sets for Y , and since f is injective it followsthat for each s ∈ Nn we have that f(Esˆk)k∈N are pairwise disjoint. Thus, byCorollary 3.8.10 there exist pairwise disjoint Borel subsets Ysˆkk∈N such thatf(Esˆk) ⊂ Ysˆk for each k ∈ N.

We inductively define a new Souslin scheme Css∈N<N for Y by setting

C∅ = Y , and Csˆk = Cs ∩ f(Esˆk) ∩ Ysˆk for all s ∈ N<N, and k ∈ N. Then foreach s ∈ N<N we have that Cs is Borel, and also

f(Es) ⊂ Cs ⊂ f(Es).

We claim that f(X) = ∩k∈N ∪s∈Nk Cs, from which it then follows that f(X) isBorel.

If y ∈ f(X), then let x ∈ X be such that f(x) = y. There exists s ∈ NN

such that x ∈ ∩k∈NEs|k, and hence y ∈ ∩k∈Nf(Es|k). Thus, y ∈ ∩k∈NCs|k ⊂∩k∈N ∪s∈Nk Cs. Conversely, if y ∈ ∩k∈N ∪s∈Nk Cs, then there exists s ∈ NN such

that y ∈ Cs|k ⊂ f(Es|k) for each k ∈ N. Hence Es|k 6= ∅ for each k ∈ N and thus

∩k∈NEs|k = x for some x ∈ X. We must then have that f(x) = y, since ifthis were not the case there would exists an open neighborhood U of f(x) suchthat y 6∈ U . By continuity of f we would then have that f(Es|k) ⊂ U for large

enough k, and hence y ∈ ∩k∈Nf(Es|k) ⊂ U , a contradiction.Having established that f(X) is Borel, the rest of the theorem follows easily.

We have that f gives a bijection from X to f(X) which is Borel, and if E ⊂ Xis Borel, then from Corollary 3.8.4 and the argument above we have that f(E)is again Borel. Thus, f−1 is a Borel map.

Corollary 3.8.12. Suppose X and Y are standard Borel spaces such that thereexists injective Borel maps f : X → Y , and g : Y → X, then X and Y areisomorphic as standard Borel spaces.

Proof. Suppose f : X → Y , and g : Y → X are injective Borel maps. FromTheorem 3.8.11 we have that f and g are Borel isomorphisms onto their imageand hence we may apply an argument used for the Cantor-Schroder-Bernsteintheorem. Specifically, if we set B = ∪n∈N(f g)n(Y \ f(X)), and we set A =X \ g(B), then we have g(B) = X \A, and

f(A) = f(X) \ (f g)(B) = Y \ ((Y \ f(X)) ∪ (f g)(B)) = Y \B.

Hence if we define θ : X → Y by θ(x) =

f(x) if x ∈ A,g−1(x) if x ∈ Y \A = g(B),

then we have that θ is a bijective Borel map whose inverse is also Borel.

Theorem 3.8.13 (Kuratowski). Any two uncountable standard Borel spaces areisomorphic. In particular, two standard Borel spaces X and Y are isomorphicif and only if they have the same cardinality.


Proof. Let X be an uncountable standard Borel space, we’ll show that X isisomorphic as Borel spaces to the Polish space C = 2N. Note that by Corol-lary 3.8.12 it is enough to show that there exist injective Borel maps f : X → C,and g : C → X. Note that such a map g exists by Proposition 3.7.3 and theCantor-Bendixson theorem, so we only need to construct f .

To construct f , fix a metric d on X such that d gives the Borel structureto X and such that the diameter of X is at most 1. Let xn be a countabledense subset of (X, d), and define f0 : X → [0, 1]N by (f0(x))(n) = d(x, xn).The function f0, is clearly injective and continuous, thus to construct f it isenough to construct an injective Borel map from [0, 1]N to C, and since CN ishomeomorphic to C, it is then enough to construct an injective Borel map from[0, 1] to C, and this is easily done. For example, if y ∈ [0, 1) then we mayconsider its dyadic expansion y =

∑∞k=1 bk2−k, where in the case when y is a

dyadic rational we take the expansion such that bk is eventually 0. Then it iseasy to see that [0, 1) 3 y 7→ bkk ∈ C gives an injective function which iscontinuous except at the countable family of dyadic rational, hence is Borel.We may then extend this map to [0, 1] by sending 1 to (1, 1, 1, · · · ) ∈ C.

Theorem 3.8.14 (von Neumann). Let (X,M, µ) be a standard probability spacewhich has no atoms, then there exists an isomorphism of standard Borel spacesθ : X → [0, 1] so that θ∗µ is Lebesgue measure on [0, 1].

Proof. By Kuratowski’s theorem we may assume that (X,M) is [0, 1] with itsBorel σ-algebra. We then consider f : [0, 1]→ [0, 1] the cumulative distributionfunction f(x) = µ([0, x]). Then f is a monotone nondecreasing function whichis continuous on the right. Since f(x)− limt→x− f(t) = µ(x) = 0 we see thatf is continuous. Moreover, f(0) = 0, and f(1) = 1 so f is surjective. For eachy ∈ [0, 1] we have that f−1(y) is a closed set and since f is monotone thismust be a closed interval. We let M denote the set of y’s so that f−1(y) isan interval of positive length. Then M is countable and the set N = f−1(M)has µ-measure zero. g = f|X\N then gives a Borel isomorphism from X \N to[0, 1] \M such that g∗µ is Lebesgue measure on [0, 1] \M .

We fix C ⊂ [0, 1] an uncountable Borel set with Lebesgue measure zero,e.g., C the usual Cantor set. Then B = g−1(C) has µ-measure zero and isagain uncountable. Thus, N = N ∪ B, and M = M ∪ C are both uncountablestandard Borel spaces by Corollary 3.8.4 and so by Kuratowski’s theorem thereis a Borel isomorphism h : N → M .

We then define θ : X → [0, 1] by setting θ(x) =

g(x) if x ∈ X \ M ;

h(x) if x ∈ M.Then θ is a Borel isomorphism and θ∗µ is Lebesgue measure.

If (X,M, µ) is a measure space, then we may consider on M the equiv-alence relation ∼ given by E ∼ F if µ(E∆F ) = 0. We denote the set of

equivalence classes by M = M/ ∼. We transfer the operations of comple-

ments, countable unions and countable intersections to M respectively as as[E]′ = [Ec], ∨∞n=1[En] = [∪∞n=1En], and ∧∞n=1[En] = [∪∞n=1En]. Note that these


are well defined operations. We also denote by µ the function on M given byµ([E]) = µ(E), where [E] denotes the equivalence class of E. If (Y,N , ν) isanother measure space, then a measure algebra homomorphism is a mapα : N → M such that for E,E1, E2, . . . ∈ N we have

1. α([∅]) = [∅];

2. α([E]′) = α([E])′;

3. α(∨∞n=1[En]) = ∨∞n=1α([En]);

4. µ(α([E])) = ν([E]).

As an example, if we have a measure preserving map θ : X → Y , then we obtaina measure algebra homomorphism θ by setting θ([E]) = [θ−1(E)].

Theorem 3.8.15 (von Neumann). Let (X,M, µ), and (Y,N , µ) be probability

spaces without atoms such that Y is standard. Suppose α : N → M is a measurealgebra homomorphism, then there exists a measurable map θ : X → Y so thatα = θ.

Proof. By Theorem 3.8.14 we may assume that Y = [0, 1], endowed with theBorel σ-algebra and Lebesgue measure.

For each rational r ∈ Q ∩ [0, 1] we choose a measurable set Xr ⊂ X suchthat [Xr] = α([[0, r)]) = α([[0, r]]). We may assume that X0 = ∅, and X1 =X. By replacing Xr with ∪s∈Q∩[0,1],s<rXs we may further assume that Xr =∪s∈Q∩[0,1],s<rXs, and hence Xr ⊂ Xs for all r, s ∈ Q ∩ [0, 1] with r ≤ s.

We define π : X → Y , by π(x) = infr ∈ Q ∩ [0, 1] | x ∈ Xr. Notethat for each t ∈ [0, 1] we have π−1([0, t)) = ∪r∈Q∩[0,1],r<tXr and hence π is ameasurable map, which satisfies

π([[0, t)]) = π([∪r∈Q∩[0,1],r<tXr]) = π([Xt]) = α([[0, t)])

for all t ∈ Q ∩ [0, 1]. Using the properties of a measure algebra homomorphismit then follows that π([E]) = α([E]) for any set E in the σ-algebra generated by[0, t) | t ∈ Q ∩ [0, 1]. As this generates the Borel σ-algebra N it follows thatα = π.

3.8.1 Exercises

Exercise 3.8.16. Let (X,M, µ), and (Y,N , µ) be probability spaces such thatY is standard, and suppose θ, φ : X → Y are measure preserving maps. Showthat θ = φ if and only if θ and φ agree almost everywhere.

Exercise 3.8.17. Suppose (X,M, µ), and (Y,N , µ) are standard probability

spaces without atoms, and α : N → M is a measure algebra homomorphism.Show that if α is surjective, then there exists an isomorphism of standard Borelspaces θ : X → Y such that θ = α.


Chapter 4

Differentiation andintegration

4.1 The Lebesgue differentiation theorem

4.1.1 Vitali’s covering lemma

If (X, d) is a metric space, B ⊂ X is a (open or closed) ball in X, and c > 0,then we denote by cB the (open or closed respectively) ball in X having thesame center as B, and having radius satisfy rad(cB) = crad(B).

Lemma 4.1.1 (Vitali’s covering lemma). Let (X, d) be a metric space and letF be a collection of balls in X, having positive radii, such that

suprad(B) | B ∈ F <∞,

then there exists a pairwise disjoint subcollection G ⊂ F such that for all B ∈ Fthere exists C ∈ G with B ∩ C 6= ∅, and with B ⊂ 5C.

Proof. Set R = suprad(B) | B ∈ F and partition F as F = ∪∞n=0Fn whereB ∈ Fn if and only if rad(B) ∈ (2−n−1R, 2−nR]. We inductively define sub-collections Gn ⊂ Fn as follows: Set H0 = F0. By Zorn’s lemma there existsG0 ⊂ F0, a maximal (with respect to inclusion) pairwise disjoint family. Havingdefined G0, . . . ,Gn−1 we let

Hn = C ∈ Fn | C ∩B = ∅ for all B ∈ ∪n−1k=0Gk,

and we again use Zorn’s lemma to find Gn ⊂ Hn which is a maximal pairwisedisjoint family. We set G = ∪∞n=0Gn, and note that G is pairwise disjoint.

If B ∈ F , then B ∈ Fn for some n ≥ 0. Either B 6∈ Hn, in which casethere exists C ∈ ∪n−1

k=0Gk such that B ∩ C 6= ∅, or else B ∈ Hn, in which caseby maximality of Gn there exists C ∈ Gn so that B ∩ C 6= ∅. Since B ∈ Fm wehave rad(B) ≤ 2−nR, and so in either case there exists C ∈ G with B ∩ C 6= ∅and with rad(C) > 2−n−1R ≥ 1

2 rad(B), hence B ⊂ 5C.

97

98 CHAPTER 4. DIFFERENTIATION AND INTEGRATION

If E ⊂ Rd, and V is a collection of closed balls in Rd, then we say that V isa Vitali covering of E if for each x ∈ E, and ε > 0 there exists B ∈ V withx ∈ B such that rad(B) < ε.

Theorem 4.1.2 (Vitali’s covering theorem). Let E ⊂ Rd be a Borel set withfinite Lebesgue measure, and suppose V is a Vitali covering of E, then thereexists a pairwise disjoint (hence countable) subcollection G ⊂ V so that

λ(E \ ∪C | C ∈ G) = 0.

Proof. By considering the subcollection of V consisting of balls with radius atmost 1 we may assume that suprad(B) | B ∈ V ≤ 1. By Vitali’s coveringlemma there then exists a pairwise disjoint subcollection G ⊂ V such that forevery B ∈ V there exists C ∈ G with B ∩ C 6= ∅ and with B ⊂ 5C.

Fix r > 0 and set Z = (E \ ∪C | C ∈ G) ∩ B(r, 0). We let G denote thesubcollection consisting of balls which intersect B(r, 0), and hence are containedin B(r+ 2, 0). Partition G as G = ∪∞n=0Gn where C ∈ Gn if and only if rad(C) ∈(2−n−1, 2−n]. Since G is pairwise disjoint we then have

∞∑n=0

∑C∈Gn

λ(C) = λ(∪C∈GC) ≤ λ(B(r + 2, 0)) <∞.

Fix ε > 0. Then there exists N ≥ 0 so that∑∞n=N

∑B∈Gn λ(B) < ε. Set

K = ∪N−1n=0 ∪C∈Gn C which is compact as it is a finite union of closed balls. If

z ∈ Z, then z 6∈ K and as V is a Vitali covering of E there then exists B ∈ Vsuch that z ∈ B, B ⊂ B(r, 0), and B ∩K = ∅. As B ∈ V and B ⊂ B(r, 0) thereexists C ∈ G so that B ∩ C 6= ∅ and z ∈ B ⊂ 5C. Since B ∩ C 6= ∅ we musthave C 6⊂ K and hence C ∈ ∪∞n=NGn. Since z ∈ Z was arbitrary we have thenshown that

Z ⊂ ∪∞n=N ∪C∈Gn 5C,

hence

λ(Z) ≤∞∑n=N

∑C∈Gn

λ(5C) < 5dε.

Since ε > 0 was arbitrary we conclude that λ(Z) = 0, and since r > 0 wasarbitrary we then conclude that λ(E \ ∪C | C ∈ G) = 0.

4.1.2 The Lebesgue differentiation theorem

A function f : Rd → C is locally integrable if∫K|f | dλ <∞ for any compact

set K ⊂ Rd. We let L1loc(Rd) denote the space of all locally integrable functions.

Theorem 4.1.3 (The Lebesgue differentiation theorem). Let f ∈ L1loc(Rd),

then for almost every x ∈ Rd we have

f(x) = limr→0

1

λ(B(r, x))

∫B(r,x)

f dλ.

4.1. THE LEBESGUE DIFFERENTIATION THEOREM 99

Note that Lebesgue’s differentiation theorem is obvious when f is continuous.Our strategy to prove the theorem in general will be to approximate f in L1(Rd)by a continuous function. It then becomes necessary to control the the size ofthe set on which the averages 1

λ(B(r,x))

∫B(r,x)

|f | dλ can be large in terms of the

L1-norm of the function f . This is achieved by the following lemma:

Lemma 4.1.4. Suppose f ∈ L1(Rd) with compact support, and for each x ∈ Rdset

f(x) = lim supr→0

1

λ(B(r, x))

∫B(r,x)

|f | dλ, (4.1)

then for α > 0 we have

λ(x ∈ Rd | f(x) > α) ≤ 1

α‖f‖1.

Proof. Since f has compact support, so does f and hence the set

E = x ∈ Rd | f(x) > α

has finite measure. Let V be the collection of closed balls B such that

1

λ(B)

∫B

|f | dλ > α.

From the definition of f we see that V is a Vitali covering of E. By Vitali’s cover-ing theorem there exists a pairwise disjoint family G, so that λ(E \∪C∈GC) = 0.Hence,

λ(E) ≤∑C∈G

λ(C) ≤∑C∈G

1

α

∫C

|f | dλ ≤ 1

α‖f‖1.

Proof of Theorem 4.1.3. First note that if the theorem holds for 1B(n,0)f foreach n ∈ N then the theorem also holds for f , therefore we may assume that f ∈L1(Rd) with compact support. The set of continuous functions with compactsupport is dense in L1(Rd). Indeed, it is enough to approximate characteristicfunctions, which can be easily done by combining the regularity of Lebesguemeasure (Corollary 2.3.8) with Urysohn’s lemma. Let gn be a sequence ofcontinuous functions with compact support such that ‖f − gn‖1 → 0.

For x ∈ Rd and n ∈ N we then have

lim supr→0

∣∣∣∣∣ 1

B(r, x)

∫B(r,x)

f dλ− f(x)

∣∣∣∣∣ ≤ lim supr→0

∣∣∣∣∣ 1

B(r, x)

∫B(r,x)

(f − gn) dλ

∣∣∣∣∣+ lim sup

r→0

∣∣∣∣∣ 1

B(r, x)

∫B(r,x)

gn dλ− gn(x)

∣∣∣∣∣+ |gn(x)− f(x)|.


Since gn is continuous the second term on the right vanishes. Thus,

lim supr→0

∣∣∣∣∣ 1

B(r, x)

∫B(r,x)

f dλ− f(x)

∣∣∣∣∣ ≤ f − gn(x) + |gn(x)− f(x)|,

where f − gn is defined as in (4.1).If ε > 0 and we let E be the set of points where the left hand side is greater

than ε then we have

E ⊂ x ∈ Rd | f − gn(x) > ε/2 ∪ x ∈ Rd | |gn(x)− f(x)| > ε/2.

Using Lemma 4.1.4 together with Chebyshev’s inequality then gives

λ(E) ≤ 2

ε‖gn − f‖1 +

2

ε‖gn − f‖1.

Taking n→∞ shows λ(E) = 0, and the result follows.

Corollary 4.1.5 (The Lebesgue density theorem). Suppose E ⊂ Rd is a Borelset, then for almost every x ∈ E we have

limr→0

λ(B(r, x) ∩ E)

λ(B(r, x))= 1. (4.2)

Proof. This follows immediately from the Lebesgue differentiation theorem byconsidering the characteristic function f = 1E .

Points where (4.2) holds are called points of density for the set E. In lightof the triangle inequality for integration, the following gives a slight improvementon Lebesgue’s differentiation theorem.

Theorem 4.1.6. Let f ∈ L1loc(Rd), then for almost every x ∈ Rd we have

limr→0

1

λ(B(r, x))

∫B(r,x)

|f(y)− f(x)| dλ(y). (4.3)

Proof. For each rational number q let Zq denote the set of points where theformula

limr→0

1

λ(B(r, x))

∫B(r,x)

|f(y)− q| dλ(y) = |f(x)− q|

does not hold. Since x 7→ |f(x)−q| is locally integrable it follows from Lebesgue’sdifferentiation theorem that λ(Zq) = 0 for every q ∈ Q. If we set Z = ∪q∈QZqthen we have λ(Z) = 0. For any x ∈ Rd, q ∈ Q, and r > 0 we have

1

λ(B(r, x))

∫|f(y)− f(x)| dλ(y) ≤ 1

λ(B(r, x))

∫|f(y)− q| dλ(y) + |q − f(x)|.

Therefore if x 6∈ Z we have

lim supr→0

1

λ(B(r, x))

∫|f(y)− f(x)| dλ(y) ≤ 2|f(x)− q|,

for every q ∈ Q. Since Q is dense in R the result then follows.

4.1. THE LEBESGUE DIFFERENTIATION THEOREM 101

Any point x ∈ Rd which satisfies (4.3) is called a Lebesgue point of f .

Theorem 4.1.7. Let µ be a complex Borel measure on Rd, let µ = µac + µs beits Lebesgue decomposition so that µac λ and µs ⊥ λ, and let f = dµac

dλ be theRadon-Nikodym derivative. Then for λ-almost every x ∈ Rd we have

limr→0

µ(B(r, x))

λ(B(r, x))= f(x).

Proof. In light of Lebesgue’s density theorem it is enough to show that when

µ ⊥ λ we have limr→0µ(B(r,x))λ(B(r,x)) = 0 for λ-almost every x ∈ Rd. Note also that

since∣∣∣µ(B(r,x))λ(B(r,x))

∣∣∣ ≤ |µ|(B(r,x))λ(B(r,x)) it is enough to consider the case when µ is a finite

positive measure.Let A be a Borel set so that µ(A) = λ(Ac) = 0. Fix r > 0 and set

F =

x ∈ A ∩ [−r, r] | lim sup

r→0

µ(B(r, x))

λ(B(r, x))> 1/r

.

Fix ε > 0, and take O an open set so that A ⊂ O and µ(O) < ε. Let V consist

of all closed balls B ⊂ O so that µ(B)λ(B) > r. Then V is a Vitali cover of F ∩ A

and hence by Vitali’s covering theorem there is a pairwise disjoint subcollectionG so that λ(F \ (∪C∈GC)) = 0.

Therefore, λ(F ) ≤∑C∈G λ(C) ≤ r

∑C∈G µ(C) ≤ rµ(O) < rε. Since ε > 0

was arbitrary we conclude that λ(F ) = 0, and since r > 0 was arbitrary theresult follows.

4.1.3 Exercises

A net of measurable sets Sαα∈I is said to shrink regularly to x if

1. the diameter of Sα tends to 0, and

2. there exists K > 0 so that for all α ∈ I, if B is the smallest ball withcenter x containing Sα, then λ(B) ≤ Kλ(Sα).

Exercise 4.1.8. If Sαα∈I shrinks regularly to x, and if x is a Lebesgue pointof f ∈ L1

loc(Rd), then

limα→∞

1

λ(Sα)

∫Sα

|f(y)− f(x)| dλ(y) = 0.

Exercise 4.1.9 (The Lebesgue density topology on R). For a Lebesgue mea-

surable set A ⊂ R, set D(A) = x ∈ R | limx∈I,diam(I)→0λ(A∩I)λ(I) = 1. We

define a topology on R by letting the open sets be all measurable sets A suchthat A ⊂ D(A). Give a description of meager sets and use this to show that Rwith the Lebesgue density topology is a Baire space.


4.2 Functions of bounded variation

Lemma 4.2.1. Suppose f : R → R is monotone, then f is continuous exceptat countably many points.

Proof. Let D ⊂ R denote the set of points where f is discontinuous. As f ismonotone it can only have jump discontinuities, so that for each x ∈ D wehave limh→0− f(x) < limh→0+ f(x). We let Dn = x ∈ R | limh→0− f(x) <limh→0+ f(x) + 1/n. If F ⊂ Dn ∩ [a, b] is finite then we have f(b) − f(a) ≥∑x∈F limh→0+ f(x) − limh→0− f(x) > |F |/n. Therefore |Dn ∩ [a, b]| < ∞ for

each n ≥ 0 and a, b ∈ Q. We then have that D is countable.

Suppose f is a complex function which is defined in a neighborhood of apoint x0 ∈ R, the Dini numbers associated to f at x0 are

• D1f(x0) = lim suph→0+f(x0+h)−f(x0)

h .

• D2f(x0) = lim infh→0+f(x0+h)−f(x0)

h .

• D3f(x0) = lim suph→0−f(x0+h)−f(x0)

h .

• D4f(x0) = lim infh→0+f(x0+h)−f(x0)

h .

We say that f is differentiable at x0 if D1f(x0) = D2f(x0) = D3f(x0) =D4f(x0) and in this case this common value is the derivative f ′(x0).

Theorem 4.2.2. Suppose f : [a, b] → R is monotone increasing, then f isalmost everywhere (with respect to Lebesgue measure) differentiable, f ′ is mea-surable, and we have ∫ b

a

f ′ dλ ≤ f(b)− f(a).

Proof. We obviously have 0 ≤ D2f ≤ D1f and 0 ≤ D4f ≤ D3f , thus it isenough to show that at almost every point we have D1 ≤ D4 and D3 ≤ D2. Wewill show that D1 ≤ D4 almost everywhere. A similar argument will apply forD3 ≤ D2.

It is enough to show that for all r, s ∈ Q with r > s > 0 the set

A = Ar,s = x ∈ (a, b) | D1f(x) > r > s > D4f(x)

has measure zero.Fix ε > 0 and take O ⊂ (a, b) open so that A ⊂ O and λ(O \A) ≤ ε. Let V

denote the collection of intervals [x−h, x] with h > 0 so that [x−h, x] ⊂ O and

f(x− h)− f(x)

−h< s. (4.4)

As D4f(x) < s for x ∈ A it follows that V gives a Vitali covering of A. ByVitali’s covering theorem there exists [xi − hi, xi]i ⊂ V pairwise disjoint so

4.2. FUNCTIONS OF BOUNDED VARIATION 103

that λ((∪i[xi − hi, xi]) \A) = 0. Since (4.4) holds we have∑i

f(xi)− f(xi − hi) < s∑i

hi = sλ(∪i[xi − hi, xi]) ≤ sλ(O) ≤ sλ(A) + sε.

(4.5)Set B = A ∩ (∪i(xi − hi, xi)) and note that λ(B) = λ(A). Let W be the

collection of intervals [y, y + k] so that [y, y + k] ⊂ (xi − hi, xi) for some i, andsuch that

f(y + k)− f(y)

k> r. (4.6)

As D1(x) > r for all x ∈ B ⊂ A it follows that W is a Vitali covering of B.Again by Vitali’s covering theorem there exists [yj , yj + kj ]j ⊂ W pairwisedisjoint so that λ((∪j [yj , yj + kj ]) \B) = 0. Since (4.6) holds we have∑

j

f(yj + kj)− f(yj) > r∑j

kj = rλ(∪j [yj , yj + kj ]) ≥ rλ(A). (4.7)

If Ji denotes the collection of j’s so that [yj , yj + kj ] ⊂ [xi − hi, xi] then asf is monotone increasing we have∑

j∈Ji

f(yj + kj)− f(yj) ≤ f(xi)− f(xi − hi).

Summing over all i ∈ I and using (4.5) and (4.7) then gives

rλ(A) ≤∑j∈J

f(yj + kj)− f(yj) ≤∑i∈I

f(xi)− f(xi − hi) ≤ sλ(A) + sε.

As ε > 0 was arbitrary this then shows rλ(A) ≤ sλ(A). Since r > s we concludethat λ(A) = 0.

We have therefore established that the derivative f ′ exists almost every-where. We let D ⊂ (a, b) denote the set of points where f ′ exists. Then D is

clearly measureable and if we set fk : D → [0,∞) as fk(x) = f(x+k−1)−f(x)k−1 ,

then fk are measurable and fk → f ′ pointwise which shows that f ′ is measur-able. If a < c < d < b and f is continuous at c and d then by Fatou’s lemmawe have ∫ d

c

f ′ dλ ≤ lim infk→∞

∫ d

c

fk dλ

= lim infk→∞

(k

∫ d+k−1

d

f dλ− k∫ c

c−k−1

f dλ

)= f(d)− f(c) ≤ f(b)− f(a).

By Lemma 4.2.1 we may take limits c→ a and d→ b, which then gives∫ b

a

f ′ dλ ≤ f(b)− f(a).


If f : R → C and Γ = x0, x1, . . . , xn with x0 < x1 < · · · < xn, thenconsider the sum

SΓ(f) =

n∑i=1

|f(xi)− f(xi−1)|.

The variation of f is then defined as

V (f) = supΓSΓ(f)

If V (f) <∞ then we say that f is a function of bounded variation. If I ⊂ Ris an interval then we set VI(f) = supΓ⊂I SΓ(f), and say that f is a function ofbounded variation on I.

For example, if f : R→ R is monotone then f is of bounded variation on anybounded interval [a, b] and we have V[a,b](f) = |f(b)−f(a)|, if we moreover havelimx→−∞ |f(x)| <∞ and limx→∞ |f(x)| <∞ then f is of bounded variation onall of R. Another example is given by f = 1x in which case we have V (f) = 2.Recall that a function is Lipschitz continuous with Lipschitz constant C if forall x, y ∈ R we have

|f(x)− f(y)| ≤ C|x− y|.

It is easy to see that a Lipschitz function is of bounded variation on any boundedinterval [a, b], and we have

V[a,b](f) ≤ C(b− a).

Proposition 4.2.3. If f, g are functions of bounded variation on an interval I,then f and g are bounded, and both f + g and fg are of bounded variation onI. Moreover, we have

VI(f + g) ≤ VI(f) + VI(g); VI(fg) ≤ ‖f‖∞VI(g) + ‖g‖∞VI(f).

Also, if f is of bounded variation and ‖1/f‖∞ < ∞, then 1/f is also ofbounded variation and we have

VI(1/f) ≤ ‖1/f‖2∞VI(f).

Proof. If f(xn) → ∞, then taking Γn = x1, . . . , xn (unordered) its easy tosee that we have SΓn(f) → ∞. Therefore functions of bounded variation arebounded. Using the triangle inequality it is easy to see that for any finitepartition Γ we have

SΓ(f + g) ≤ SΓ(f) + SΓ(g);

SΓ(fg) ≤ ‖f‖∞SΓ(g) + ‖g‖∞SΓ(f);

SΓ(1/f) ≤ ‖1/f‖2∞SΓ(f).

Taking suprema then gives the result.

4.2. FUNCTIONS OF BOUNDED VARIATION 105

Note that of course any scalar multiple of function of bounded variationis again of bounded variation, and we also have VI(f) = VI(f). Therefore itfollows that f is of bounded variation on an interval I if and only if its real andimaginary parts are of bounded variation on I.

If Γ = x0, . . . , xn is a partition and f is real valued we set

PΓ(f) =

n∑i=1

(f(xi)− f(xi−1))+;

NΓ(f) =

n∑i=1

(f(xi)− f(xi−1))−.

Note thatPΓ(f) +NΓ(f) = SΓ(f), (4.8)

whilePΓ(f)−NΓ(f) = f(b)− f(a). (4.9)

The positive (resp. negative) variation of f on an interval I is given by

P (f) = supΓ⊂I

PΓ(f) (resp. N(f) = supΓ⊂I

NΓ(f) ).

Proposition 4.2.4. If any of the three of P (f), N(f), V (f) are finite then allthree must be finite, and in this case we have

P (f) +N(f) = V (f); P (f)−N(f) = f(b)− f(a).

Proof. From (4.8) we see that P (f), N(f) ≤ V (f), so that if V (f) is finitethen so is P (f) and N(f). Also, if either of P (f) or N(f) is finite then from(4.9) we see that they both must be finite, and from (4.8) we see that V (f) ≤P (f) + N(f) so that V (f) is also finite. If we take a sequence of partitonsΓ1n, so that PΓ1

n(f) → P (f), and Γ2

n, so that NΓ2n(f) → N(f) then setting

Γn = Γ1n ∪ Γ2

n we see that PΓn(f)→ P (f) and NΓn(f)→ N(f). From (4.9) wethen deduce that P (f) − N(f) = f(b) − f(a), and from (4.8) we deduce thatP (f) +N(f) ≤ V (f).

Theorem 4.2.5 (Jordan decomposition). If f : [a, b] → R is of bounded vari-ation then there exist monotone increasing functions f1, f2 : [a, b] → [0,∞) sothat f = f1 − f2.

Proof. Since f is of bounded variation the restriction of f to [a, x] is also ofbounded variation for any a < x ≤ b. We set f1(x) = P[a,x](f) and set f2(x) =N[a,x](f) − c. The functions f1 and f2 are increasing and from the previousproposition we have f1(x)− f2(x) = f(x).

The previous theorem can also be easily extended to unbounded intervals.

Corollary 4.2.6. Let f : R → C be a function of bounded variation, then thederivative f ′ exists almost everywhere, and is integrable.

Proof. This follows from the Jordan decomposition and Theorem 4.2.2.


4.2.1 Exercises

Exercise 4.2.7. There is a continuous function f : R → R which is not ofbounded variation on any interval of positive length.

Exercise 4.2.8. There is a function of bounded variation f : R → R which isnot monotone on any interval of positive length.

4.3 Absolutely continuous and singular functions

Let I be an interval in the real line. A function f : I → C is absolutelycontinuous if for all ε > 0 there exists δ > 0 so that whenever (ai, bi)i is apairwise disjoint collection of subintervals of I which satisfy

∑i(bi− ai) < δ we

have∑i |f(bi)− f(ai)| < ε. A function f : I → C is singular if it its derivative

exists and equals 0 almost everywhere.

Lemma 4.3.1. Suppose g : [a, b] → C is integrable, then f(x) =∫ xag dλ is

absolutely continuous.

Proof. Since |g|λ λ this follows easily from Proposition 2.7.1.

Lemma 4.3.2. If f is absolutely continuous then f is of bounded variation onany compact interval.

Proof. Since f is absolutely continuous there exists δ so that if (ai, bi)i is apairwise disjoint collection of subintervals of I which satisfy

∑i(bi − ai) < δ

we have∑i |f(bi) − f(ai)| < 1. Then we have V[a,b](f) ≤ 1 for any interval

[a, b] ⊂ I which satisfies b − a < δ. If we take a1 < b1 = a2 < · · · < bN sothat I = ∪Ni=1[ai, bi] and such that bi − ai < δ then we must have VI(f) =∑Ni=1 V[ai,bi](f) ≤ N .

Corollary 4.3.3. If f is absolutely continuous then f is differentiable almosteverywhere.

Lemma 4.3.4. If f : I → C is absolutely continuous and singular then f isconstant.

Proof. Take a, b ∈ I, with a < b. Let E = x ∈ (a, b) ∈ I | f ′(x) = 0, andfix ε > 0. Since f is singular we have λ(E) = b − a. Since f is absolutelycontinuous on I there exists δ > 0 so that whenever (ai, bi)i is a pairwisedisjoint collection of subintervals of I which satisfy

∑i(bi − ai) < δ then we

have∑i |f(bi)− f(ai)| < ε.

Let V denote the collections of intervals [a0, b0] ⊂ (a, b) so that

|f(b0)− f(a0)| < (b0 − a0)ε/(b− a). (4.10)

Then V is a Vitali covering of E and hence by Vitali’s covering theorem thereexists a finite pairwise disjoint collection of intervals [ai, bi]Ni=1 ⊂ V so that

λ(E \ (∪i[ai, bi])) < δ.

4.3. ABSOLUTELY CONTINUOUS AND SINGULAR FUNCTIONS 107

Rearranging we may assume that a1 < b1 ≤ a2 < b2 ≤ · · · < bn. From (4.10)we have

N∑i=1

|f(bi)− f(ai)| ≤ εN∑i=1

(bi − ai)/(b− a) ≤ ε. (4.11)

If we set b0 = a and aN+1 = b we then have

N∑i=0

(bi − ai+1) = λ(∪Ni=0(bi, ai+1)) = λ([a, b] \ (∪i[xi, xi + hi])) < δ.

HenceN∑i=0

|f(bi)− f(ai+1)| < ε. (4.12)

Combining (4.11) and (4.12) with the triangle inequality then gives

|f(b)− f(a)| ≤N∑i=1

|f(bi)− f(ai)|+N∑i=0

|f(bi)− f(ai+1)| < 2ε.

As ε > 0 was arbitrary we then have f(b) = f(a), and as a < b was arbitrary itthen follows that f is constant.

Theorem 4.3.5. Let f : [a, b]→ C be a function of bounded variation, then

fac(x) =

∫ x

a

f ′ dλ

is absolutely continuous and f−fac is singular. Moreover, if f = gac+gs wheregac is absolutely continuous and gs is singular then there exists a constant c ∈ Cso that gac = fac + c and gs = fs − c.

Proof. Since f is of bounded variation f ′ exists almost everywhere and is inte-grable, therefore fac is well defined. Moreover, by Lebesgue’s differentiation the-orem we have that fac is differentiable almost everywhere and we have f ′ac = f ′

almost everywhere, hence f − fac is singular.f = gac + gs where gac is absolutely continuous and gs is singular, then

f ′ac = f ′ = g′ac almost everywhere and hence fac−gac is an absolutely continuousfunction which is also singular, hence constant by the previous lemma.

Corollary 4.3.6. A function f : [a, b]→ C is absolutely continuous if and onlyif f ′ exists almost everywhere, is integrable, and we have

f(x)− f(a) =

∫ x

a

f ′ dλ

for each x ∈ [a, b].


4.3.1 Exercises

Exercise 4.3.7. Every absolutely continuous function is uniformly continuous.

Exercise 4.3.8. The Cantor function is monotone, uniformly continuous, butnot absolutely continuous.

Exercise 4.3.9. Every Lipschitz continuous function is absolutely continuous.

Exercise 4.3.10. The sum and product of two absolutely continuous functionon a compact interval remains absolutely continuous.

Exercise 4.3.11. Let µ be a complex Borel measure on R and set f(x) =µ((−∞, x]). Then f is of bounded variation. Moreover, f is absolutely contin-uous if and only if µ λ, and f is singular if and only if µ ⊥ λ.

Chapter 5

Lp spaces

Suppose (X,µ) is a measure space, and 0 < p <∞. We denote by Lp(X,µ) thecollection of all measurable functions f ∈ M(X,µ) such that |f |p ∈ L1(X,µ).We identify two functions if they agree almost everywhere. Given f ∈ Lp(X,µ)we set

‖f‖p =

(∫|f |p dµ

)1/p

.

We will almost exclusively be interested in the case when p ≥ 1. When p ≥ 1we will show that Lp(X,µ) is a vector space, and ‖ · ‖p gives a complete normon Lp(X,µ).

Throughout this chapter we will use the convention 1∞ = 0.

5.1 Holder’s and Minkowski’s inequalities

Lemma 5.1.1 (Young’s inequality). Suppose 0 ≤ a, b < ∞, and 1 ≤ p, q ≤ ∞with 1

p + 1q = 1, then

ab ≤ ap

p+bq

q.

Proof. Set t = 1p , so that 1− t = 1

q . As logarithm is concave we have

log(tap + (1− t)bq) ≥ t log(ap) + (1− t) log(bq) = log(a) + log(b) = log(ab).

Exponentiating then gives the inequality.

Theorem 5.1.2 (Holder’s inequality). Let (X,µ) be a measure space and sup-pose 1 ≤ p, q ≤ ∞ with 1

p + 1q = 1, then for f ∈ Lp(X,µ) and g ∈ Lq(X,µ) we

have fg ∈ L1(X,µ) and

‖fg‖1 ≤ ‖f‖p‖g‖q.

109

110 CHAPTER 5. LP SPACES

Proof. We assume that neither f nor g is essentially 0 since otherwise the in-

equality is trivial. Applying Young’s inequality to a = |f(x)|‖f‖p and b = |g(x)|

‖g‖qgives

|f(x)g(x)|‖f‖p‖g‖q

≤ |f(x)|p

p‖f‖pp+|g(x)|q

q‖g‖qq.

Since ‖f‖pp =∫|f |p dµ and ‖g‖qq =

∫|g|q dµ, integrating the right hand side

gives 1p + 1

q = 1. It then follows that fg ∈ L1(X,µ) and integrating the lefthand side gives

‖fg‖1‖f‖p‖g‖q

≤ 1,

so that Holder’s inequality follows.

Theorem 5.1.3. Let (X,µ) be a measure space and suppose 1 ≤ p, q ≤ ∞ withp <∞ such that 1

p + 1q = 1, then for any f ∈ Lp(X,µ) we have

‖f‖p = sup‖g‖q=1

∫|fg| dµ.

Proof. Set L = sup‖g‖q=1

∫|fg| dµ. Then by Holder’s inequality we have L ≤

‖f‖p. If f is essentially 0 then the result is trivial so we assume that f is notessentially 0. Set

g0 = |f |p−1sgnf.

If q =∞ then ‖g0‖∞ = 1, otherwise ‖g0‖qq =∫|f |(p−1)q dµ =

∫|f |p dµ = ‖f‖pp,

so that ‖g0‖q = ‖f‖p−1p . If we set g = g0

‖f‖p−1p

then ‖g‖q = 1. Hence,

L ≥∫fg dµ =

∫|f |p dµ‖f‖p−1

p

= ‖f‖p.

Note that by Lemma 2.7.7, the previous theorem holds in the case p =∞ ifand only if (X,µ) is semifinite. In the σ-finite case we also have the inequalityeven if f 6∈ Lp(X,µ):

Theorem 5.1.4 (Minkowski’s inequality). Let (X,µ) be a measure space andsuppose 1 ≤ p ≤ ∞, then Lp(X,µ) is a vector space and ‖ · ‖p gives a norm onLp(X,µ), i.e., for f, g ∈ Lp(X,µ) we have f + g ∈ Lp(X,µ) and

‖f + g‖p ≤ ‖f‖p + ‖g‖p.

Proof. This is just the pointwise triangle inequality for p = ∞ so we consideronly the case when p < ∞. Note first that from convexity of the functiont 7→ tp we have the pointwise inequality |f + g|p ≤ 2p−1(|f |p + |g|p), so thatf + g ∈ Lp(X,µ).

5.1. HOLDER’S AND MINKOWSKI’S INEQUALITIES 111

By the previous theorem if we take 1 < q ≤ ∞ so that 1p + 1

q = 1 then wehave

‖f + g‖p = sup‖h‖q=1

∫|(f + g)h| dµ

≤ sup‖h‖q=1

∫|fh| dµ+ sup

‖k‖q=1

∫|gk| dµ = ‖f‖p + ‖g‖p.

Theorem 5.1.5. Let (X,µ) be a measure space and 1 ≤ p ≤ ∞. Then Lp(X,µ)is a vector space and ‖ · ‖p gives a complete norm on Lp(X,µ).

Proof. We have already shown this for p = ∞, and so we may assume p < ∞.It’s enough to show that every absolutely convergent series in Lp(X,µ) ac-tually converges in Lp(X,µ). Suppose fn∞n=1 ⊂ Lp(X,µ) such that A =∑∞n=1 ‖fn‖p < ∞. Set Gn =

∑nk=1 |fk|, and G =

∑∞k=1 |fk|. Then by

Minkowski’s inequality we have ‖Gn‖p ≤∑nk=1 ‖fk‖p ≤ A. By the mono-

tone convergence theorem we then have∫Gp dµ = limn→∞

∫Gpk dµ ≤ Ap. In

particular, we have that the series∑∞n=1 fn converges almost everywhere to a

function F such that |F | ≤ G.Then |F−

∑nk=1 fk|p ≤ (2G)p ∈ L1(X,µ), and by the dominated convergence

theorem we have ∥∥∥∥∥F −n∑k=1

fk

∥∥∥∥∥p

p

=

∫ ∣∣∣∣∣F −n∑k=1

fk

∣∣∣∣∣p

dµ→ 0.

Therefore the series∑∞n=1 fn converges to F in Lp(X,µ).

The next proposition extends Theorem 5.1.3 to the case when f 6∈ Lp(X,µ).

Proposition 5.1.6. Let (X,µ) be a σ-finite measure space and suppose 1 ≤p, q ≤ ∞ with q <∞ such that 1

p + 1q = 1. If f 6∈ Lp(X,µ) then we have

sup‖g‖q≤1

∫|fg| dµ =∞.

Proof. We leave the case p =∞ to the reader. For 1 ≤ p <∞ we first considerthe case when (X,µ) is finite, so that if a < ∞ and E = x ∈ X | |f(x)| ≤ athen 1Ef ∈ Lp(X,µ) and hence 1Ecf 6∈ Lp(X,µ). Therefore, we may constructa pairwise disjoint sequence of measurable sets En of the form En = x ∈ X |a < |f(x)| ≤ b so that 1Enf ∈ Lp(X,µ) for each n and ‖1Enf‖p > 4n.

By Theorem 5.1.3 there then exists gn ∈ Lq(X,µ) with ‖gn‖q ≤ 1 so that∫En|fgn| dµ ≥ 4n. We set g =

∑∞n=1 2−ngn1En , so that by Minkowski’s in-

equality we have ‖g‖q ≤ 1. For each n ≥ 1 we then have∫|fg| dµ ≥

∫En

2−n|fgn| dµ ≥ 2n.


Hence,∫|fg| dµ =∞.

Now suppose (X,µ) is σ-finite and suppose X = ∪nXn where Xnn isan increasing sequence of finite measure subsets. If f 6∈ Lp(Xn, µ) for somen then the result follows from the finite case above. Otherwise we have thatf ∈ Lp(Xn, µ) for each n and ‖1Xnf‖p →∞. By Theorem 5.1.3 there then existsa sequence gn ∈ Lq(Xn, µ) with ‖gn‖q so that

∫|fgn| dµ→∞, completing the

proof.

5.1.1 Minkowski’s integral inequality

Minkowski’s inequality shows that the Lp-norm of a sum is dominated by thesum of their Lp-norms. Generalizing from sums to integrals gives the following:

Theorem 5.1.7 (Minkowski’s integral inequality). Suppose (X,M, µ) and (Y,N , ν)are σ-finite measure spaces, 1 ≤ p < ∞, and F : X × Y → [0,∞) is M⊗N -measurable. Then(∫

Y

(∫X

F (x, y) dµ(x)

)pdν(y)

)1/p

≤∫X

(∫Y

F (x, y)p dν(y)

)1/p

dµ(x).

Proof. If g ∈ Lq(Y, ν) then by Tonelli’s theorem and Holder’s inequality we have∫ (∫F (x, y) dµ(x)

)|g(y)| dν(y) =

∫∫F (x, y)|g(y)| dν(y)dµ(x)

≤ ‖g‖q∫ (∫

F (x, y)p dν(y)

)1/p

dµ(x).

Proposition 5.1.6 and Theorem 5.1.3 then gives the result.

5.1.2 Exercises

Exercise 5.1.8. Suppose (X,µ) is a finite measure space, then for 1 ≤ p ≤ q ≤∞ we have Lq(X,µ) ⊂ Lp(X,µ).

Exercise 5.1.9. Let (X,µ) be a measure space. If 0 < p < q < r ≤ ∞ thenLq(X,µ) ⊂ Lp(X,µ) + Lr(X,µ).

Exercise 5.1.10. Let (X,µ) be a measure space. If 0 < p < q < r ≤ ∞ thenLp(X,µ)∩Lr(X,µ) ⊂ Lq(X,µ) and if λ ∈ [0, 1] satisfies 1

q = λ 1p +(1−λ) 1

q then

‖f‖q ≤ ‖f‖λp‖f‖1−λr .

Hint: Apply Holder’s inequality with |f |λq ∈ Lp/λq(X,µ) and |f |(1−λ)q ∈Lr/(1−λ)q(X,µ).

Exercise 5.1.11. If X is any set and 0 < p < q ≤ ∞ then `p(X) ⊂ `q(X) and‖f‖q ≤ ‖f‖p.

5.2. THE DUAL OF LP -SPACES 113

Exercise 5.1.12. If (X,µ) is a measure space with µ(X) = 1, and 0 < p < q ≤∞ then Lq(X,µ) ⊂ Lp(X,µ) and ‖f‖p ≤ ‖f‖q.

Exercise 5.1.13. Suppose (X,µ) is a measure space and f ∈ Lp(X,µ) ∩L∞(X,µ) for some p < ∞ (hence f ∈ Lq(X,µ) for q ≥ p), then ‖f‖∞ =limq→∞ ‖f‖q.

Exercise 5.1.14 (Chebyshev’s inequality). Let (X,µ) be a measure space, 0 <p <∞, and f ∈ Lp(X,µ). Then for any α > 0 we have

µ(x ∈ X | |f(x)| > α) ≤(‖f‖pα

)p.

Exercise 5.1.15. Let (X,µ) be a measure space and suppose 1 ≤ p, q, r ≤ ∞such that 1

p + 1q = 1

r . If f ∈ Lp(X,µ) and g ∈ Lq(X,µ) then fg ∈ Lr(X,µ) and

‖fg‖r ≤ ‖f‖p‖g‖q.

Exercise 5.1.16. Let (X,µ) be a semifinite measure space, 1 ≤ p ≤ ∞, andg ∈ L∞(X,µ). The operator Mg : Lp(X,µ) → Lp(X,µ) given by Mg(f) = gfsatisfies ‖Mg‖B(Lp(X,µ)) = ‖g‖∞.

5.2 The dual of Lp-spaces

Lemma 5.2.1. Let (X,µ) be a finite measure space and suppose 1 ≤ p, q ≤ ∞such that 1

p + 1q = 1. If g ∈ L1(X,µ) then

‖g‖p = supf∈L∞(X,µ),‖f‖q≤1

∣∣∣∣∫ fg dµ

∣∣∣∣ .Proof. Replacing f with |f | sgn g we see that

supf∈L∞(X,µ),‖f‖q≤1

∣∣∣∣∫ fg dµ

∣∣∣∣ = supf∈L∞(X,µ),‖f‖q≤1

∫|fg| dµ.

If f ∈ Lq(X,µ), then setting Ek = x ∈ X | |f |(x) ≤ k we have 1Ekf ∈L∞(X,µ) with ‖1Ekf‖q ≤ ‖f‖q and by the monotone convergence theorem wehave

limk→∞

∫|1Ekfg| dµ =

∫|fg| dµ.

Therefore,

supf∈L∞(X,µ),‖f‖q≤1

∫|fg| dµ = sup

f∈Lq(X,µ),‖f‖q≤1

∫|fg| dµ,

and the result then follows from Proposition 5.1.6.


Theorem 5.2.2. Let (X,µ) be a measure space and suppose 1 < p, q < ∞such that 1

p + 1q = 1. For each g ∈ Lq(X,µ) let Ξg ∈ Lp(X,µ)∗ be given

by Ξg(f) =∫fg dµ. Then Ξ defines an isometric isomorphism from Lq(X,µ)

onto Lp(X,µ)∗.

Proof. First, note that by Theorem 5.1.3 we have that Ξ is a well defined andisometric map, thus we only need to show that it is surjective.

Suppose ϕ ∈ Lp(X,µ)∗. We first consider the case when µ is finite so thatall simple functions are in Lp(X,µ). For each measurable set E ⊂ X we setν(E) = ϕ(1E). If En∞n=1 is a pairwise disjoint sequence of measurable setsand E = ∪∞n=1En then we have 1E =

∑∞n=1 1En where the series converges in

Lp-norm since ∥∥∥∥∥∞∑n=K

1En

∥∥∥∥∥p

p

= µ (∪∞n=KEn)→ 0.

Since ϕ is continuous we then have

ν(E) =

∞∑n=1

ϕ(1En) =

∞∑n=1

ν(En).

Therefore ν describes a complex measure. Also, if µ(E) = 0 then 1E = 0 inLp(X,µ) and hence ν(E) = ϕ(0) = 0, so that µ is absolutely continuous withrespect to µ. By the Radon-Nikodym theorem there then exists g ∈ L1(X,µ)so that ν(E) =

∫Eg dµ for any measurable set E, and hence for any simple

function f ∈ Lp(X,µ) we have

ϕ(f) =

∫fg dµ. (5.1)

If f ∈ L∞(X,µ) then taking simple functions fn ∈ L∞(X,µ) with ‖f−fn‖∞we have

∫fng dµ →

∫fg dµ. Moreover, since ‖f − fn‖p ≤ µ(X)‖f − fn‖∞ we

also have ϕ(fn) → ϕ(f). Thus, for any f ∈ L∞(X,µ) we have ϕ(f) =∫fg dµ

and hence

supf∈L∞(X,µ),‖f‖p≤1

∣∣∣∣∫ fg dµ

∣∣∣∣ = supf∈L∞(X,µ),‖f‖p≤1

|ϕ(f)| ≤ ‖ϕ‖.

By Lemma 5.2.1 we then have g ∈ Lp(X,µ) with ‖g‖p ≤ ‖ϕ‖. Since L∞(X,µ)is dense in Lp(X,µ), applying Holder’s inequality to approximate f in Lp(X,µ)we see that equation (5.1) holds for all f ∈ Lp(X,µ).

We now consider the case when µ is σ-finite, so that we may write X =∪nn=1Fn where Fn are increasing measurable sets of finite measure. ConsideringLp(Fn, µ) as a subset of Lp(X,µ) we may then restrict ϕ and from the argumentabove it then follows that there exists gn ∈ Lq(Fn, µ) so that ϕ(f) =

∫fg dµ

for all f ∈ Lp(Fn, µ). It’s easy to check that for m ≥ n we have gn = gm|Fn ,µ-almost everywhere. Thus, we may essentially define a function g : X → C byletting g|Fn = gn. Since ‖gn‖q ≤ ‖ϕ‖ it follows from the monotone convergence

5.2. THE DUAL OF LP -SPACES 115

theorem that g ∈ Lq(X,µ). If f ∈ Lp(Fn, µ) then we have ϕ(f) =∫fg dµ and

since ∪∞n=1Lp(Fn, µ) is dense in Lp(X,µ) it then follows that ϕ(f) =

∫fg dµ

for all f ∈ Lp(X,µ).Finally, we consider the general case. From above, for each σ-finite set E ⊂

X there exists an essentially unique function gE ∈ Lq(E,µ) so that ‖gE‖q ≤ ‖ϕ‖and ϕ(f) =

∫fgE dµ for any f ∈ Lp(E,µ). We let M ≤ ‖ϕ‖ be the supremum

of ‖gE‖q as E varies over all σ-finite subsets of X, and we take a sequenceEn so that ‖gEn‖q → M . Set F = ∪∞n=1En. Then F is σ-finite, and we have‖gF ‖q = M . If E ⊂ F c is any σ-finite set then M ≥ ‖gF + gE‖q ≥ ‖gF ‖q = M ,and hence it follows that gE is essentially 0. In particular, if f ∈ Lp(X,µ)vanishes on F it then follows that ϕ(f) = 0 =

∫fgF dµ. Thus, we then see that

for general f ∈ Lp(X,µ) we have ϕ(f) =∫fgF dµ.

5.2.1 Exercises

Exercise 5.2.3. Define ϕn ∈ `∞(N)∗ by ϕn(f) = 1n

∑nk=1 f(n), show that if ϕ

is a weak∗ cluster point of ϕnn then ϕ 6∈ `1(N).

If (X,M) is a measurable space then a (complex) finitely additive mea-sure on (X,M) is a function m :M→ C, such that there exists K > 0 so thatwhenever E1, . . . , En ∈M are disjoint we have

1. m(∪nk=1Ek) =∑nk=1m(Ek);

2.∑nk=1 |m(Ek)| ≤ K.

If µ is a measure on (X,M) then we say that m is absolutely continuouswith respect to µ if m(E) = 0 whenever µ(E) = 0.

Exercise 5.2.4. Let (X,M, µ) be a measure space and suppose that m is afinitely additive measure on (X,M) which is absolutely continuous with respectto µ. There exists a unique continuous linear functional ϕ ∈ L∞(X,µ)∗ so thatϕ(1E) = m(E) for all E ∈ M. Moreover, every continuous linear functional onL∞(X,µ) arrises in this way.


Chapter 6

Functional analysis

6.1 Topological vector spaces

Suppose K = C, or K = R. A topological K-vector space consists of aK-vector space X, which is also a Hausdorff topological space such that vectoraddition and scalar mulitiplication give continuous maps X × X → X andK×X → X respectively.

Examples of topological vector spaces that we have already encounteredinclude normed spaces, as well as the duals of normed spaces enowed with theweak∗-topology. As was the case for normed spaces, if X is a topological vectorspace then we may consider the space X∗ consisting of all continuous linearfunctions into K. The weak∗-topology on X∗ is the coarsest topology suchthat the evaluation maps X∗ 3 ϕ 7→ ϕ(x) ∈ K are continuous for each x ∈ X.The weak-topology on X is the coarsest topology such that the evaluationmaps X 3 x 7→ ϕ(x) ∈ K are still continuous for each ϕ ∈ X∗.

If X is a topological vector space then a net xii∈I is Cauchy if for anyneighborhood U of 0 there exists α ∈ I so that xi − xj ∈ U whenever i, j ≥ α.We say that X is complete if every Cauchy net converges to a point in X. Ametric d is translation invariant if d(x+ z, y+ z) = d(x, y) for all x, y, z ∈ X.If d is a translation invariant metric on X which is compatible with the topologythen the notions of Cauchy and completeness are the same as those notions ford. In particular, if X has two translation invariant metrics, both of which givethe topology, then X is complete with respect to one if and only if X is completewith respect to the other.

6.1.1 Locally convex spaces

Suppose X is a vector space over K and F is a family of seminorms on X. Wesay that F separates points if ρ(x) = 0 for all ρ ∈ F only when x = 0. If Fseparates points then the coarsest topology for which every ρ ∈ F is continuousgives a Hausdorff topological vector space structure to X. In this case we seethat a net xii is Cauchy in X if and only if for each ρ ∈ F the net is Cauchy

117

118 CHAPTER 6. FUNCTIONAL ANALYSIS

with respect to (X, ρ). We also have that xi → x in X if and only if ρ(x−xi)→ 0for each ρ ∈ F .

We now wish to find a topological characterization of those spaces whosetopology arises from seminorms. If X is a topological vector space and C ⊂ X,then C is convex if for all x, y ∈ C, and 0 ≤ t ≤ 1 we have tx + (1− t)y ∈ C.The set C is balanced if for all x ∈ C and λ ∈ K with |λ| ≤ 1 we have λx ∈ C.The set C is absorbing if ∪t≥0tC = X.

If ρ is a semi-norm on X then the unit ball x ∈ X | ρ(x) < 1 is a convex,balanced, absorbing, open set. In general, if C is a an absorbing set we definethe Minkowski functional ρC : X → [0,∞) by

ρC(x) = inft ≥ 0 | x ∈ tC.

Proposition 6.1.1. If X is a topological vector space and C is a convex, bal-anced, and open set, then the Minkowski functional is a continuous semi-normon X, such that C = x ∈ X | ρC(x) < 1.

Proof. First note that since C is open if x ∈ X then limn→∞ 1nx = 0 and hence

1nx ∈ C for some n. Therefore C is also absorbing and so we have ρC(x) < ∞for all x ∈ X. If t > 0 then it is clear that ρC(tx) = tρC(x). Also if α ∈ C with|α| = 1 then as C is balanced we have that ρC(tαx) = ρC(tx) = tρC(x).

If 0 < a, b < ∞ and x ∈ aC, y ∈ bC, then as C is convex we have aa+bx +

ba+by ∈ C, and hence x+ y ∈ (a+ b)C.

If x, y ∈ X, t = ρC(x), s = ρC(y), then for every ε > 0 we have x ∈ (t+ ε)Cand y ∈ (s + ε)C. Therefore we see that x + y ∈ (s + t + 2ε)C. Since ε > 0 isarbitrary we then have ρC(x+ y) ≤ s+ t = ρC(x) + ρC(y).

Thus, it remains to show C = x ∈ X | ρC(x) < 1. If ρC(x) < 1 then sinceC is absorbing it following that x ∈ C. Conversely, if x ∈ C, then as C is openwe have tx ∈ C for t sufficiently close to 1, therefore ρC(x) < 1.

Theorem 6.1.2. Let X be a topological vector space. Then there is a familyof seminorms on X which generates the topology if and only if there exists aneighborhood base at 0 consisting of convex, balanced, absorbing sets.

Proof. If F is a family of seminorms which generate the topology on X then wehave a neighborhood base at 0 consisting of the convex balanced and absorbingsets x ∈ X | ρ(x) < a for ρ ∈ F and a > 0. Conversely, if Cαα is a familyof open convex balanced and absorbing sets which give a neighborhood base at0 then by the previous proposition the seminorms ρCα are continuous andsatisfy x ∈ X | ρCα(x) < 1 = Cα, hence the family generates the topology onX.

A topological vector space X is locally convex if it satisfies one of thehypotheses of the previous theorem.

Proposition 6.1.3. Let X be a locally convex topological vector space, then thefollowing conditions are equivalent:

6.1. TOPOLOGICAL VECTOR SPACES 119

1. The topology on X is given by a translation invariant metric.

2. X is metrizable.

3. X is first countable.

4. There exists a countable family of semi-norms on X which generate thetopology on X.

5. There exists a countable neighborhood base at 0 consisting of convex, bal-anced, and absorbing sets.

Proof. The equivalence between the last two conditions above follow as in The-orem 6.1.2. Also, every metric space is first countable.

If ρnn is a countable family of seminorms which generate the topologythen consider the translation invariant metric

d(x, y) =∑n

2−nρn(x− y)

1 + ρn(x− y).

A net converges with respect to this metric if and only if it converges withrespect to each seminorm ρn, thus the metric d describes the topology on X.

If X is first countable then there exists a countable neighborhood base Unnat 0. Since X is locally convex, for each n there exists a convex, balanced,and absorbing set Cn such that Cn ⊂ Un. We then have that Cnn givesa neighborhood base at 0 which consists of convex, balanced, and absorbingsets.

A Frechet space is a locally convex topological vector space which is hasa complete translation invariant metric.

6.1.2 The open mapping and closed graph theorems

Lemma 6.1.4. Let X be a locally convex topological vector space, Y a Frechetspace, and T : X → Y a continuous surjective linear map. Then for anyneighborhood G of 0 in X we have that T (G) is a neighborhood of 0 in Y .

Proof. Let G be a convex, balanced, absorbing neighborhood of 0 in X. ThenX = ∪n≥1nG and as T is surjective we have ∪n≥1nT (G) = ∪n≥1T (nG) =T (X) = Y .

Since Y is a Frechet space it satisfies the Baire property, and so cannot bea countable union of nowhere dense sets. Hence for some n we must have thatnT (G) contains a non-empty open set O. We then have that U = O − O isan open neighborhood of 0 in Y and U ⊂ nT (G) − nT (G) ⊂ 2nT (G). ThusV = 1

2nU is an open neighborhood of 0 and V ⊂ T (G).

Theorem 6.1.5 (The open mapping theorem). Let X and Y be Frechet spacesand T : X → Y a continuous surjective linear map, then T is an open map. Inparticular, if T is a bijection then T is a homeomorphism.


Proof. Fix compatible translation invariant metrics dX and dY on X and Yrespectively. The map T is an open map if neighborhoods of a point x aremapped to neighborhoods of Tx, and by translation it suffices to consider thecase x = 0. Fix 0 < r ≤ 1, then by Lemma 6.1.4 we have that T (BX(r, 0)) is aneighborhood of 0, and thus it suffices to show that T (BX(r, 0)) ⊂ T (BX(2r, 0)).

Fix y ∈ T (BX(r, 0)). For n ≥ 0 we inductively define a sequence xn, so that

xn ∈ BX(r2−n, 0),

andy − T (x0 + x1 + · · ·+ xn) ∈ BY (2−n, 0) ∩ T (BX(r2−n, 0)).

Indeed, y ∈ T (BX(r, 0)) and by Lemma 6.1.4 T (BX(r2−1, 0)) is a neighborhoodof 0, therefore there exists x0 ∈ BX(r, 0) so that

y − Tx0 ∈ BY (1, 0) ∩ T (BX(r2−1, 0)).

Now suppose that x0, . . . , xn−1 have been chosen. Since

y − T (x0 + x1 + · · ·+ xn−1) ∈ T (BX(r2−n, 0))

and since T (BX(r2−n−1, 0)) is a neighborhood of 0 there exists xn ∈ BX(r2−n, 0)such that

y − T (x0 + x1 + · · ·+ xn) ∈ BY (2−n, 0) ∩ T (BX(r2−n−1, 0)).

Then ∑nk=0 xkn≥1 gives a Cauchy sequence and we have y = T (

∑∞n=0 xn).

Since∑∞n=0 xn ∈ B(2r, 0) we then have that y ∈ T (B(2r, 0)).

Corollary 6.1.6. Let X and Y be Frechet spaces and suppose T : X → Y isa continuous linear bijective map, then T gives an isomorphism of topologicalvector spaces.

Theorem 6.1.7 (The closed graph theorem). Let X and Y be Frechet spacesand T : X → Y a linear map such that the graph of T

G(T ) = (x, Tx) | x ∈ X ⊂ X × Y

is closed in the product topology, then T is continuous.

Proof. X × Y is also a Frechet space and hence so is G(T ), being a closedsubspace. The projection map pX : G(T ) → X is continuous bijective andhence by the open mapping theorem has continuous inverse. Since pY is alsocontinuous it then follows that T = pY p−1

X is continuous.

Theorem 6.1.8 (The Banach-Steinhaus uniform boundedness principle). LetX be a Banach space and Y a normed vector space. Suppose F ⊂ B(X,Y ) issuch that for all x ∈ X one has

supT∈F‖T (x)‖Y <∞,

thensupT∈F‖T‖B(X,Y ) <∞.

6.1. TOPOLOGICAL VECTOR SPACES 121

Proof. For each n ∈ N let

Xn = x ∈ X | supT∈F‖T (x)‖Y ≤ n.

Then Xn is closed and ∪∞n=1Xn = X. By the Baire category theorem thereexists n so that Xn has non-empty interior, i.e., there exists x0 ∈ Xn and ε > 0so that B(ε, x0) ⊂ Xn.

Let y ∈ X with ‖y‖ ≤ 1, and suppose T ∈ F . Then

‖T (y)‖Y = ε−1‖T (x0 + εy)− T (x0)‖Y≤ ε−1(‖T (x0 + εy)‖Y + ‖T (x0)‖Y ) ≤ 2ε−1n.

Therefore supT∈F ‖T‖B(X,Y ) ≤ 2ε−1n <∞.

6.1.3 Exercises

Exercise 6.1.9. Let X be a σ-compact, locally compact Hausdorff space. Con-sider C(X) endowed with the topology of uniform convergence on compact sets.Then C(X) is a Frechet space.

Exercise 6.1.10. Let X be a vector space over K and suppose ‖ · ‖1 and ‖ · ‖2are two complete norms on X such that there exists C > 0 with ‖x‖1 ≤ C‖x‖2for all x ∈ X. Show that there exists C ′ > 0 so that ‖x‖2 ≤ C ′‖x‖1 for allx ∈ X.

Exercise 6.1.11. The vector space Cm(R) of all m-times continuously differen-tiable functions is a Frechet space with the semi-norms ‖f‖k,n = sup|f (k)(x)| |x ∈ [−n, n], for 0 ≤ k ≤ m.

Exercise 6.1.12. The vector space C∞(R) of all infinitely differentiable func-tions f : R→ C is a Frechet space with the semi-norms ‖f‖k,n = sup|f (k)(x)| |x ∈ [−n, n], for 0 ≤ k <∞.

Exercise 6.1.13. If Xii∈I is a family of locally convex topological vectorspaces, then

∏i∈I Xi with the product topology and coordinatewise operations

is again a locally convex topological vector space. Moreover, if I is countableand each Xi is a Frechet space then so is

∏i∈I Xi.

Exercise 6.1.14. If X is a topological vector space, then X∗ with the weak∗-topology, and X with the weak-topology are also topological vector spaces.

Exercise 6.1.15. Let X be a topological vector space and suppose ϕ ∈ X∗ isnot the zero functional, then ϕ is an open map, i.e., if G ⊂ X is open then ϕ(G)is also open.

Exercise 6.1.16. If X is a topological vector space, then (X,wk)∗ = X∗.

Exercise 6.1.17. Let X be a set. The pairing between c0X and `1X given by〈f, g〉 =

∑x∈X f(x)g(x) gives an isomorphism `1X ∼= c0X

∗.


Exercise 6.1.18. Let (X,µ) be a σ-finite measure space, and let fnn∈N ⊂L1(X,µ) be a uniformly bounded sequence of non-negative functions. Thenfn → 0 weakly if and only if

∫fn dµ→ 0.

Exercise 6.1.19. Consider L∞([0, 1]) = L1([0, 1])∗. Then spanei2πntn∈Z isweak∗-dense in L∞([0, 1]). Hint: Use Lusin’s theorem and the Stone-Weierstrasstheorem.

Exercise 6.1.20. The sequence ei2πntn∈N ⊂ L∞([0, 1]) converges to 0 in theweak∗-topology.

Exercise 6.1.21. A topological vector space X is locally convex if and onlyif it is isomorphic (as topological vector spaces) to a subspace of a product ofBanach spaces.

Exercise 6.1.22. If X is a Frechet space then a subspace Y ⊂ X is a Frechetspace if and only if Y is closed.

Exercise 6.1.23. If X is a Frechet space then there exists a complete metricd on X which is compatible with the topology and is translation invariant, i.e.,d(x+z, y+z) = d(x, y) for all x, y, z ∈ X. Hint: First find a translation invariantmetric which is compatible with the topology and then use the previous exerciseto show that this metric is complete.

Exercise 6.1.24. Let X be a locally convex topological vector space whosetopology is defined by a family of seminorms F . If ϕ ∈ X∗ then there existρ1, . . . , ρn ∈ F and K > 0 so that |ϕ(x)| ≤ K

∑ni=1 ρi(x), for all x ∈ X.

6.2 The Hahn-Banach theorem

Let X be a real vector space. A function f : X → R is a sublinear functionalif

1. f(tx) = tf(x), for all t > 0 and x ∈ X;

2. f(x+ y) ≤ f(x) + f(y) for all x, y ∈ X.

Note that any seminorm is a sublinear functional. Further examples are givenby the following lemma whose proof we leave to the reader.

Lemma 6.2.1. Let X be a real vector space and let C ⊂ X be a convex andabsorbing set. Then the Minkowski functional

ρC(x) = inft > 0 | x ∈ tC

is a sublinear funcitonal, and C = x ∈ X | ρC(x) < 1.

6.2. THE HAHN-BANACH THEOREM 123

Theorem 6.2.2 (The Hahn-Banach theorem I). Let X be a real vector spacewith a sublinear functional f : X → R. Suppose Y ⊂ X is a subspace, andϕ : Y → R is a linear functional such that

ϕ(y) ≤ f(y), y ∈ Y.

Then there exists a linear functional ψ : X → R such that ψ(y) = ϕ(y) fory ∈ Y , and such that

ψ(x) ≤ f(x), x ∈ X.

Proof. We let F denote the set of all linear functionals ψ : Z → K such thatY ⊂ Z ⊂ X, ψ|Y = ϕ and ψ(z) ≤ f(z) for all z ∈ Z. If ψ1, ψ2 ∈ F withψ1 : Z1 → K and ψ2 : Z2 → K then we write ψ1 ≺ ψ2 if Z1 ⊂ Z2 and ψ2|Z1

= ψ1.This then gives a partial ordering on F . If ψαα∈I is an increasing chain inF with ψα : Zα → K then by setting Z = ∪αZα and defining ψ : Z → K byψ(x) = ψα(x) for x ∈ Zα we have that ψ ∈ F is well defined and ψα ≺ ψ foreach α. Thus every chain has an upper bound and hence by Zorn’s lemma thereexists a maximal element ψ ∈ F .

To finish the theorem it then suffices to show that any maximal element in Fmust have the entirety of X in its domain. Suppose that ψ ∈ F with ψ : Z → Rsuch that X 6= Z. Take x0 ∈ X \ Z and set Z = z + αx0 | α ∈ R. If t ∈ Rthen we may define a linear function ψ : Z → R by ψ(z + αx0) = ψ(z) + αt. Inorder for ψ to belong to F we need to be able to choose t so that for all z ∈ Zwe have

ψ(z) + αt ≤ f(z + αx0).

Equivalently, for α > 0 we need

−f( zα− x0

)+ ψ

( zα

)≤ t ≤ f

( zα

+ x0

)− ψ

( zα

),

for all z ∈ Z.Note that for z1, z2 ∈ Z we have

ψ(z2) + ψ(z1) = ψ(z2 + z1) ≤ f(z2 + z1) ≤ f(z2 + x0) + f(z1 − x0).

Therefore,−f(z1 − x0) + ψ(z1) ≤ f(z2 + x0)− ψ(z2).

If we set c1 = supz1∈Z−f(z1 − x0) + ψ(z1) and we set c2 = infz2∈Zf(z2 +x0) − ψ(z2) then we have shown that c1 ≤ c2. Taking t so that c1 ≤ t ≤ c2then gives the extension ψ ∈ F , showing that ψ was not maximal.

Theorem 6.2.3 (The Hahn-Banach theorem II). Let X be a vector space overK and ρ a semi-norm on X. Suppose Y ⊂ X is a subspace, and ϕ : Y → K isa linear functional such that

|ϕ(y)| ≤ ρ(y), y ∈ Y.

Then there exists a linear functional ψ : X → K such that ψ(y) = ϕ(y) fory ∈ Y , and such that

|ψ(x)| ≤ ρ(x), x ∈ X.


Proof. We first consider the case K = R. Since ρ is a sublinear functional andϕ(y) ≤ |ϕ(y)| ≤ ρ(y) for y ∈ Y it follows from the previous theorem that thereexists a linear functional ψ : X → R so that ψ|Y = ϕ and ψ(x) ≤ ρ(x) forx ∈ X. Considering −x then shows that −ψ(x) = ψ(−x) ≤ ρ(−x) = ρ(x) andhence |ψ(x)| ≤ ρ(x) for all x ∈ X.

We now consider the case K = C. We have |Re (ϕ(x))| ≤ |ϕ(y)| ≤ ρ(y) andhence viewing X as a real vector space it follows from above that there existsa R-linear functional ψ0 : X → R so that ψ0(y) = Re (ϕ(y)) for y ∈ Y , and|ψ0(x)| ≤ ρ(x) for x ∈ X. Set ψ : X → C by ψ(x) = ψ0(x) − iψ0(ix). Then ψis R-linear and we also have ψ(ix) = iψ(x), hence ψ is C-linear.

Moreover, for y ∈ Y we have ψ(y) = Re (ϕ(y))−iRe (ϕ(iy)) = ϕ(y). Finally,if x ∈ X choose θ so that eiθψ(x) = |ψ(x)|. Then

|ψ(x)| = ψ(eiθx) = Re (ψ(eiθx)) = ψ0(eiθx) ≤ ρ(eiθx) = ρ(x).

Corollary 6.2.4. Let X be a normed space, then the map ι : X → X∗∗ givenby ι(x)(ϕ) = ϕ(x) is isometric.

Proof. Fix x ∈ X and define ϕ : Kx → K by ϕ(αx) = α‖x‖. Then ϕ is linearand we have |ϕ(αx)| = ‖αx‖. By the Hahn-Banach theorem there then exists alinear functional ψ : X → K so that ψ(x) = ‖x‖, and |ψ(z)| ≤ ‖z‖ for all z ∈ X,i.e., ‖ψ‖ ≤ 1. Therefore we see ‖ι(x)‖ ≥ |ι(x)(ψ)| = |ψ(x)| = ‖x‖. The reverseinequality is trivial.

Lemma 6.2.5. Let X be a vector space over K and suppose ϕ,ϕ1, . . . , ϕn arelinear functionals on X such that ∩nk=1 ker(ϕk) ⊂ ker(ϕ). Then we have ϕ ∈spanϕ1, . . . , ϕn.

Proof. We may assume that ϕ1, . . . , ϕn is linearly independent. Set L =∩nk=1 ker(ϕk). Then ϕ,ϕ1, . . . , ϕn all give well defined linear functionals onX/L which has dimension at most n. Since ϕ1, . . . , ϕn ⊂ (X/L)∗ is a set ofn linearly independent vectors it follows that ϕ1, . . . , ϕn also spans (X/L)∗.Therefore, ϕ ∈ spanϕ1, . . . , ϕn.

Proposition 6.2.6. If X is a locally convex topological vector space, then themap Ξ : X → (X∗,wk∗)∗ given by Ξx(ϕ) = ϕ(x) is bijective.

Proof. If x ∈ X, x 6= 0 then there exists a continuous semi-norm ρ so thatρ(x) > 0. If we consider ϕ : Kx → K given by ϕ(αx) = αρ(x), then as inCorollary 6.2.4 we may apply the Hahn-Banach theorem to produce a linearfunctional ψ : X → K so that ψ(x) = ρ(x) and |ψ(z)| ≤ ρ(z) for all z ∈ X.Hence ψ is continuous and Ξ(x)(ψ) = ψ(x) = ρ(z) 6= 0. Therefore Ξ is injective.

If ζ ∈ (X∗,wk∗)∗ then there exit K > 0 and x1, . . . , xn ∈ X so that|ζ(ϕ)| ≤ K

∑ni=1 |ϕ(xi)|, for all ϕ ∈ X∗. In particular we have ker(ζ) ⊂

∩ni=1 ker(Ξ(xi)), and it follows from the previous lemma that ζ = Ξ(x) forsome x ∈ spanx1, . . . , xn.


6.2.1 Separating convex sets

If X is a topological vector space over K and A,B ⊂ X, then A and B areseparated if there exists a continuous linear functional ϕ ∈ X∗, and α ∈ R sothat

Re (ϕ(a)) ≤ α ≤ Re (ϕ(b)), a ∈ A, b ∈ B (6.1)

If the inequalities in Equation (6.1) may be taken to be strict then we say thatA and B are strictly separated. Note, that if A and B are (strictly) separatedthen the convex sets they generate are also (strictly) separated.

Lemma 6.2.7. Let X be a topological vector space over K. If G ⊂ X is convexopen, and x0 6∈ G, then G and x are separated.

Proof. We first consider the case K = R and we assume that G is nonempty.Taking g ∈ G and replacing G with G − g and x0 with x0 − g it is enough toconsider the case when 0 ∈ G. Since G is open and contains 0 we have thatG is absorbing and hence by Lemma 6.2.1 the Minkowski functional ρG(y) =inft > 0 | y ∈ tG is sublinear and satisfies G = y ∈ X | ρG(y) < 1. It isalso easy to see that ρG is continuous.

Define ϕ : Rx0 → R by ϕ(αx0) = αρG(x0) ≤ ρG(αx0). By the Hahn-Banachtheorem there then exists a linear functional ψ : X → R so that ψ(αx0) =αρG(x0) and ψ(z) ≤ ρG(z) for all z ∈ X. If zα → z then we have ρG(zα−z)→ 0and hence lim supα→∞ ψ(zα−z) ≤ 0. Since we also have ρG(z−zα)→ 0 we alsoobtain lim infα→∞ ψ(zα − z) ≥ 0 and hence limα→∞ ψ(zα − z) = 0. Thereforeψ is continuous. Finally, for x ∈ G we have

ψ(x) ≤ ρG(x) < 1 ≤ ρG(x0) = ψ(x0).

We now consider the case K = C. Treating X as a R-vector space it followsfrom above that there exists a continuous R-linear functional ψ and α ∈ R sothat for all x ∈ G we have ψ(x) ≤ α ≤ ψ(x0). Setting ψ(x) = ψ(x) − iψ(ix)then gives the result.

Proposition 6.2.8. Let X be a topological vector space over K. If G,H ⊂ X aredisjoint convex sets such that G is open, then G and H are separated. Moreover,if H is also open then G and H are strictly separated.

Proof. Consider the set G −H = x − y | x ∈ G, y ∈ H. Since G and H areconvex it follows that G−H is also convex. Moreover, as G and H are disjointwe have that 0 6∈ G − H. Writing G − H = ∪y∈H(G − y) we see that G − His open. Thus, by Lemma 6.2.7 we can separate G − H and 0 by some linearfunctional ϕ, i.e., replacing ϕ with −ϕ if needed, we have 0 ≤ Re (ϕ(x− y)) forall x ∈ G and y ∈ H. This then shows that Re (ϕ(y)) ≤ Re (ϕ(x)) for all x ∈ Gand y ∈ H, and hence G and H are separated.

If H is also open then by Exercise 6.1.15 we have that Re (ϕ(G)) andRe (ϕ(H)) are disjoint open sets in R showing that G and H are strictly sepa-rated.


Theorem 6.2.9 (The Hahn-Banach separation theorem). Let X be a locallyconvex topological vector space over K. If K,F ⊂ X are disjoint closed convexsets and K is compact, then K and F are strictly separated.

Proof. Since F is closed and X is locally convex for each k ∈ K there existsa balanced convex open neighborhood Uk of 0 so that (k + Uk) ∩ F = ∅. Thefamily k + 1

2Ukk∈K gives an open over of K and so by compactness thereexists a finite subcover ki + 1

2Ukini=1. Set U = ∩ni=1

14Uki . Then K + U and

F + U are convex open sets.We claim that K + U and F + U are disjoint. Indeed if k + u = f + v with

k ∈ K and u, v ∈ U then we have k ∈ (ki + 12Uki) for some i and so

f = k + u− v ∈(ki +

1

2Uki

)+

1

4U +

1

4U ⊂ (ki + Uki) ⊂ F c.

Hence (K + U) ∩ (F + U) = ∅ and thus these sets are strictly separated byProposition 6.2.8. Since K ⊂ K+U and F ⊂ F +U it then follows that K andF are strictly separated.

Note that the hypothesis that K is compact is necessary. Indeed, even in R2

the closed sets (t, 0) | t > 0 and (t, t−1) | t > 0 cannot be strictly separated.

Corollary 6.2.10. Let X be a K-vector space and suppose that T1 and T2 aretopologies on X giving the structure of a locally convex topological vector spaceand such that (X, T1) and (X, T2) have the same continuous linear functionals.Then a convex set C ⊂ X is closed in the T1-topology if and only if C is closedin the T2-topology.

Proof. Suppose C ⊂ X is convex and closed in the T1-topology. For each pointx 6∈ C it follows from the Hahn-Banach separation theorem that x and C arestrictly separated. Thus, there exists a T1-continuous linear functional ϕx andα ∈ R so that Re (ϕx(y)) ≤ α for all y ∈ C and Re (ϕx(x)) > α. We thenhave C = ∩x∈Xy ∈ X | Re (ϕx(y)) ≤ α. Since the two topologies have thesame continuous linear functionals it then follows that C is also closed in theT2-topology. The converse also holds by symmetry.

Corollary 6.2.11. Let X be a locally convex topological vector space and sup-pose C ⊂ X is a convex set, then C is closed if and only if C is weakly closed.In particular, a subspace Y ⊂ X is closed if and only if it is weakly closed.

Proof. Since X with its given topology and X with the weak topology have thesame continuous linear functionals this follows from the previous corollary.

Recall from Corollary 6.2.4 that for a normed space X we have a naturalisometric embedding ι : X → X∗∗ given by ι(x)(ϕ) = ϕ(x). We will thereforeidentify X as a subspace of X∗∗.

Proposition 6.2.12. Let X be a normed space, then the unit ball of X isweak∗-dense in the unit ball of X∗∗. In particular, X is weak∗-dense in X∗∗.


Proof. Let B be the unit ball of X and let C be the weak∗-closure of B in X∗∗.By Proposition 6.2.6 any weak∗-continuous linear functional on X∗∗ is of theform η 7→ η(ϕ) for some ϕ ∈ X∗. Therefore by the Hahn-Banach separationtheorem if ζ ∈ X∗∗ is not in C then there exists ϕ ∈ X∗ and α ∈ R so that forany x ∈ B we have Re (ϕ(x)) < α < Re (ζ(ϕ)). Since 0 ∈ B we have 0 < α.

Since eiθB = B it then follows that |ϕ(x)| < α for each x ∈ B and hence‖ϕ‖ ≤ α. Therefore α < Re (ζ(ϕ)) ≤ |ζ(ϕ))| ≤ ‖ζ‖‖ϕ‖ ≤ α‖ζ‖, and weconclude that ‖ζ‖ > 1. By contraposition it then follows that C agrees with theunit ball in X∗∗.

A Banach space X is reflexive if X∗∗ = X. For example, Theorem 5.2.2shows that if (X,µ) is a measure space and 1 < p < ∞, then Lp(X,µ) isreflexive.

Theorem 6.2.13. Let X be a Banach space, then the following conditions areequivalent:

1. X is reflexive.

2. The weak and weak∗ topologies on X∗ agree.

3. X∗ is reflexive.

4. The closed unit ball in X is weakly compact.

Proof. By Proposition 6.2.12 the unit ball of X is weak∗-dense in the unit ballof X∗∗ in the weak∗-topology, which agrees with the weak topology on X.Therefore if the unit ball of X is weakly compact we must have that the unitball of X is equal to the unit ball of X∗∗ and hence it follows that X = X∗∗.This then shows that (4) =⇒ (1). While the Banach-Alaoglu theorem gives(1) =⇒ (4).

We also clearly have (3) =⇒ (2). And (2) together with the Banach-Alaoglutheorem shows that the closed unit ball of X∗ is weakly compact which thenshows that X∗ is reflexive from the implication (4) =⇒ (1) applied to X∗.Thus we see that (2) and (3) are also equivalent.

To complete the theorem it then suffices to show (3) =⇒ (1) as (1) =⇒(3) would then follow by considering X∗∗. Suppose therefore that (3) (andhence also (2)) holds. If ζ ∈ X∗∗ then ζ is continuous with respect to theweak topology on X∗. Since the weak and weak∗ topologies agree we then havethat ζ is continuous with respect to the weak∗ topology and hence ζ ∈ X byProposition 6.2.6. This then shows that X is reflexive.

6.2.2 The Krein-Milman theorem

If X is a K-vector space and C ⊂ X is a nonempty convex set, then a point k ∈ Cis an extreme point if it cannot be written as a non-trivial convex combinationof elements in C, i.e., if x, y ∈ C x, y 6= k and t ∈ [0, 1] then tx+ (1− t)y 6= k.More generally, we say that a nonempty subset K ⊂ C is an extreme subset if


whenever x, y ∈ C \K and t ∈ [0, 1] then we have tx+ (1− t)y 6∈ K. We denoteby ext(C) the set of extreme points of C. If F ⊂ X we let co(F ) denote thesmallest closed convex set which contains F .

Lemma 6.2.14. Let X be a vector space over K and suppose C ⊂ X is anonempty convex set, ϕ is a linear functional and α ∈ R so that K = C ∩ x ∈X | Re (ϕ(x)) ≤ α is nonempty. Then K is an extreme subset of C.

Proof. Suppose x, y ∈ C \K and t ∈ [0, 1] then ϕ(tx+ (1− t)y) = tϕ(x) + (1−t)ϕ(y) > tα+ (1− t)α = α.

Theorem 6.2.15 (The Krein-Milman theorem). Suppose K is a compact con-vex subset of a locally convex topological vector space over K, then K = co(extK).

Proof. We consider the family F consisting of compact convex extreme subsetsof C which we consider ordered by decreasing inclusion. If Kαα is a chainof elements in F then set K = ∩αKα. Then K is convex and by compactnesswe have that K is nonempty. It is also easy to see that K is an extremesubset. Thus any chain has an upper bound and by Zorn’s lemma there thenexists a nonempty convex extreme set K which has no proper subset with thisproperty. If x, y ∈ K with x 6= y then by the Hahn-Banach theorem there existsa continuous linear functional ϕ so that Re (ϕ(x)) < Re (ϕ(y)). We would thenhave that K ∩ z ∈ X | Re (ϕ(z)) ≤ Re (ϕ(x)) is a non-empty convex compactextreme subset which does not contain y contradicting minimality of K. Thuswe conclude that K consists of a single point. and so ext(C) 6= ∅.

Let B = co(ext(C)) and suppose B 6= C so that there exists x ∈ C \ B.As B is a closed convex set from the Hahn-Banach separation theorem thereexists a continuous linear functional ϕ and α ∈ R so that Re (ϕ(x)) ≤ α andRe (ϕ(y)) > α for all y ∈ B. We then have that C ∩ x ∈ X | Re (ϕ(x)) ≤ α isa nonempty convex compact extreme subset which is disjoint from B. Howeverthe argument above shows that C ∩ x ∈ X | Re (ϕ(x)) ≤ α must contain anextreme point and hence cannot be disjoint from B. Therefore we conclude thatB = C.

6.2.3 Exercises

Exercise 6.2.16. Let X be a Banach space and suppose xn∞n=1 ⊂ X is asequence which converges weakly, then xn∞n=1 is bounded.

Exercise 6.2.17. Let (X,µ) be a σ-finite measure space. Then L1(X,µ) isreflexive if and only if L1(X,µ) is finite dimensional.

Exercise 6.2.18. Let X be a Frechet space, then X∗ is a Frechet space in theweak∗-topology if and only if X is isomorphic to a separable Banach space. Hint:If ρnn∈N is a family of seminorms which give the topology on X, first showthat An = ϕ ∈ X∗ | |ϕ(x)| ≤ n

∑nk=1 ρk(x) for all x ∈ X is weak∗-compact

by the Banach-Alaoglu theorem, then show that X = ∪nAn and use the Baireproperty.

6.3. HILBERT SPACE 129

Exercise 6.2.19 (The Markov-Kakutani fixed point theorem). Let K ⊂ X bea non-empty compact convex subset of a locally convex space X, and suppose Sis a non-empty family of pairwise commuting continuous maps which are affine,i.e., T (tk1 + (1 − t)k2) = tT (k1) + (1 − t)T (k2) whenever T ∈ S, k1, k2 ∈ Kand t ∈ [0, 1]. Then there is a point in K which is a common fixed points forall maps in S. Hint: First consider the case when S consists of a single map T ,take k0 ∈ K and consider the sequence 1

n

∑n−1i=0 T

ik0.

6.3 Hilbert space

6.3.1 Inner product spaces

Let H be a vector space over K, where K = R, or K = C. An inner producton H is a map (ξ, η) 7→ 〈ξ, η〉 from H×H → K so that

1. 〈ξ, ξ〉 ∈ (0,∞) for all nonzero ξ ∈ H.

2. 〈αξ + η, ζ〉 = α〈ξ, ζ〉+ 〈η, ζ〉, for α ∈ K, ξ, η, ζ ∈ H.

3. 〈ξ, η〉 = 〈η, ξ〉, for ξ, η ∈ H.

Observe that from the last two conditions above we also have 〈ξ, αη + ζ〉 =α〈ξ, η〉+ 〈ξ, ζ〉 for α ∈ K, and ξ, η, ζ ∈ H. Given ξ ∈ H we define

‖ξ‖ =√〈ξ, ξ〉.

An inner product space is a vector space together with an inner product onthat space. As an example, suppose (X,µ) is a measure space, and considerL2(X,µ) = f ∈ M(X,µ) | |f |2 ∈ L1(X,µ) where we identity two functionswhich agree almost everywhere. From the inequalities

|ab| ≤ |a|2 + |b|2, |a+ b|2 ≤ 2(|a|2 + |b|2), a, b ∈ C,

we deduce that for f, g ∈ L2(X,µ) we have gf ∈ L1(X,µ), and f+g ∈ L2(X,µ).Therefore, L2(X,µ) is a vector space and we obtain an inner product by setting

〈f, g〉 =

∫gf dµ.

Proposition 6.3.1 (The Cauchy-Schwarz inequality). Let H be an inner prod-uct space, then for all ξ, η ∈ H we have

|〈ξ, η〉| ≤ ‖ξ‖‖η‖.

Proof. For any λ ∈ K, ξ0, η0 ∈ H we have

‖ξ0‖2 + 2Re (λ〈ξ0, η0〉) + |λ|2‖η0‖2 = ‖ξ0 + λη0‖2 ≥ 0.

Taking λ so that |λ| = 1 and λ〈ξ0, η0〉 ≤ 0 gives

‖ξ0‖2 + ‖η0‖2 ≥ 2|〈ξ0, η0〉|.


Setting ξ0 = ‖η‖ξ and η0 = ‖ξ‖η gives

2‖ξ‖2‖η‖2 ≥ 2‖ξ‖‖η‖|〈ξ, η〉|,

and the inequality follows.

Proposition 6.3.2. Let H be an inner product space, then the map H 3 ξ 7→‖ξ‖ defines a norm on H.

Proof. The only nontrivial thing to check is the triangle inequality. Supposeξ, η ∈ H, then from the Cauchy-Schwarz inequality we have

‖ξ + η‖2 = ‖ξ‖2 + 2Re 〈ξ, η〉+ ‖η‖2 ≤ ‖ξ‖2 + 2‖ξ‖‖η‖+ ‖η‖2 = (‖ξ‖+ ‖η‖)2.

Lemma 6.3.3. Let H be an inner product space, then the inner product isjointly continuous with respect to the topology induced by the norm.

Proof. Suppose ξn → ξ and ηn → η, then ηnn∈N and the Cauchy-Schwarzinequality gives

|〈ξn, ηn〉 − 〈ξ, η〉| ≤ |〈ξn − ξ, ηn〉|+ |〈ξ, ηn − η〉|≤ ‖ξn − ξ‖‖ηn‖+ ‖ξ‖‖ηn − η‖ → 0.

A Hilbert space is an inner product space which is complete with re-spect to the given norm. For example, if (X,µ) is any measure space then byTheorem 5.1.5 we have that L2(X,µ) is a Hilbert space with inner product〈f, g〉 =

∫gf dµ.

Proposition 6.3.4 (The parallelogram identity). Let H be an inner productspace, then for ξ, η ∈ H we have

‖ξ + η‖2 + ‖ξ − η‖2 = 2(‖ξ‖2 + ‖η‖2).

Proof. Just add the formulas ‖ξ ± η‖2 = ‖ξ‖2 ± 2Re 〈ξ, η〉+ ‖η‖2.

The next proposition shows that the norm completely determines the innerproduct.

Proposition 6.3.5 (The polarization identity). Let H be a complex inner-product space and suppose µ is a measure on the circle T so that µ(T) = 1, and∫λ dµ(λ) =

∫λ2 dµ(λ) = 0. Then for ξ, η ∈ H we have∫

λ‖ξ + λη‖2 dµ(λ) = 〈ξ, η〉


Proof. We may compute directly∫λ‖ξ + λη‖2 dµ(λ) =

∫λ‖ξ‖2 dµ(λ) +

∫〈ξ, η〉 dµ(λ)

+

∫λ2〈η, ξ〉 dµ(λ) +

∫λ‖η‖2 dµ(λ)

= 〈ξ, η〉.

The most commonly used case of the polarization identity is when µ =14

(δ1 + δi + δ−1 + δ−i

), in which case we obtain the formula

〈ξ, η〉 =1

4

3∑k=0

ik‖ξ + ikη‖2.

Another case is if we take Lebesgue measure on the interval [0, 1] and we considerthe corresponding Lebesgue measure on the circle, which is the push-forwardunder the map t 7→ e2πit.

Corollary 6.3.6. Let H and K be two complex inner product spaces. A linearmap U : H → K is isometric if and only if 〈Uξ, Uη〉 = 〈ξ, η〉 for all ξ, η ∈ H.

Proof. If U : H → K is isometric, and ξ, η ∈ H, then by the polarization identitywe have

〈Uξ, Uη〉 =1

4

3∑k=0

ik‖U(ξ + ikη)‖2 =1

4

3∑k=0

ik‖(ξ + ikη)‖2 = 〈ξ, η〉.

The previous result is also valid for real inner product spaces, and we leaveit as an exercise.

6.3.2 Orthogonal subspaces and the Riesz representationtheorem

Given an inner product space H, two vectors ξ, η ∈ H are orthogonal if〈ξ, η〉 = 0. A set ξii∈I is orthogonal if 〈ξi, ξj〉 = 0 for i 6= j, and ξii∈I isorthonormal if it is orthogonal and we also have ‖ξi‖ = 1 for all i ∈ I.

If A ⊂ H we set

A⊥ = ξ ∈ H | 〈ξ, η〉 = 0 for all η ∈ A.

Theorem 6.3.7. Let H be a Hilbert space, K ⊂ H a nonempty closed convexsubset, and η0 ∈ H. Then there exists a unique element ξ0 ∈ K with minimaldistance to η0.


Proof. By considering K = K − η0 it suffices to consider the case when η0 = 0.Set d = inf‖ξ‖ | ξ ∈ K, and choose a sequence ξn ∈ K such that ‖ξn‖ → d.Then for n,m ∈ N we have

d2 ≤∥∥∥∥1

2ξn +

1

2ξm

∥∥∥∥2

=1

4‖ξn‖2 +

1

2Re 〈ξn, ξm〉+

1

4‖ξm‖2.

Hence, limn,m→∞Re 〈ξn, ξm〉 = d2.

Therefore,

limn,m→∞

‖ξn − ξm‖2 = limn,m→∞

‖ξn‖2 − 2Re 〈ξn, ξm〉+ ‖ξm‖2

= d2 − 2d2 + d2 = 0.

Hence ξnn∈N is Cauchy and converges to vector ξ0 ∈ K, which satisfies ‖ξ0‖ =d. Since the sequence ξnn∈N was chosen arbitrary it follows that ξ0 must beunique.

Let H be a Hilbert space and suppose K ⊂ H is a closed subspace. If ξ ∈ Hwe let PK(ξ) denote the unique vector in K with minimal distance to ξ and wecall the map PK the orthogonal projection from H onto K.

Proposition 6.3.8. Let H be a Hilbert space, K ⊂ H a closed subspace and fixξ ∈ H. Then PK(ξ) is the unique vector in K such that ξ − PK(ξ) ∈ K⊥.

Proof. As PK(ξ) minimizes the distance to ξ it follows that for all a > 0 andη ∈ K we have

‖ξ−PK(ξ)‖2 ≤ ‖ξ−PK(ξ)−aη‖2 = ‖ξ−PK(ξ)‖2−2aRe 〈ξ−PK(ξ), η〉+a2‖η‖2.

Rearranging and dividing by a then gives

Re 〈ξ − PK(ξ), η〉 ≤ a‖η‖2.

As a > 0 was arbitrary this then shows that Re 〈ξ−PK(ξ), η〉 ≤ 0, and replacingη with ikη for k = 0, 1, 2, 3 then shows that 〈ξ − PK(ξ), η〉 = 0. Thereforeξ − PK(ξ) ∈ K⊥.

Conversely, suppose that η ∈ K is such that ξ − η ∈ K⊥. Then for ζ ∈ K wehave 〈ξ − η, ζ〉 = 0, hence

‖ξ − η − ζ‖2 = ‖ξ − η‖2 + ‖ζ‖2 ≥ ‖ξ − η‖2.

It then follows that η − PK(ξ).

Corollary 6.3.9. Let H be a Hilbert space, K ⊂ H a closed subspace, then PKis a linear map and ‖PK‖ ≤ 1.


Proof. If ξ, η ∈ H, α ∈ C and ζ ∈ K then by the previous proposition we have

〈αξ + η − αPK(ξ)− PK(η), ζ〉 = α〈ξ − PK(ξ), ζ〉+ 〈η − PK(η), ζ〉 = 0.

Hence, again by the previous proposition we have αPK(ξ)+PK(η) = PK(αξ+η),which shows that PK is linear.

Also, since PK(ξ) and ξ−PK(ξ) are orthogonal we have ‖PK(ξ)‖2 ≤ ‖PK(ξ)‖2+‖ξ − PK(ξ)‖2 = ‖ξ‖2 so that PK is a contraction.

Theorem 6.3.10 (Riesz representation theorem). Let H be a Hilbert space,and for each η ∈ H consider the map Ξη ∈ H∗ given by Ξη(ξ) = 〈ξ, η〉. ThenΞ : H → H∗ gives an isometric anti-linear surjection.

Proof. Clearly Ξ is anti-linear. By the Cauchy-Schwarz inequality we have|Ξη(ξ)| = |〈ξ, η〉| ≤ ‖ξ‖‖η‖ which shows that Ξ is a contraction. MoreoverΞη(η) = ‖η‖2 which then shows that Ξ is isometric.

Suppose now that we have ϕ ∈ H∗, and assume that ϕ 6= 0. Then ker(ϕ) isa proper closed subspace. Take η ∈ H so that ϕ(η) = 1 and by replacing η withη − Pker(ϕ)(η) we assume that η ∈ ker(ϕ)⊥.

If ξ ∈ H, then ξ − ϕ(ξ)η ∈ ker(ϕ) and hence is orthogonal to η. Thus,

0 = 〈ξ − ϕ(ξ)η, η〉 = 〈ξ, η〉 − ϕ(ξ)‖η‖2.

Thus, ϕ(ξ) = 〈ξ, η‖η‖−2〉 for all ξ ∈ H which shows that Ξ is surjective.

6.3.3 Orthonormal bases and dimension

Note that if ξ1, . . . , ξn is a finite orthonormal set then expanding the innerproduct gives ∥∥∥∥∥

n∑i=1

αiξi

∥∥∥∥∥2

=∑i∈I|αi|2.

We will use this equality throughout this section.

Proposition 6.3.11 (Bessel’s inequality). Let H be a Hilbert space and supposeξii∈I is an orthonormal set, then for any η ∈ H we have∑

i∈I|〈η, ξi〉|2 ≤ ‖η‖2.

In particular, i ∈ I | 〈η, ξi〉 6= 0 is countable.

Proof. If ξ1, . . . , ξn is an orthonormal set and K denotes it’s span, then forη ∈ H set η0 =

∑ni=1〈η0, ξi〉ξi. For 1 ≤ j ≤ n we then have

〈η − η0, ξj〉 = 〈η, ξj〉 −n∑i=1

〈η0, ξi〉〈ξi, ξj〉 = 〈η, ξj〉 − 〈η, ξj〉 = 0.

Thus, by Proposition 6.3.8 we have PK(η) = η0 =∑ni=1〈η, ξi〉ξi.

In particular, we have∑ni=1 |〈η, ξi〉|2 = ‖PK(η)‖2 ≤ ‖η‖2. Thus, Bessel’s

inequality holds for finite sets and the general case then follows easily.


If H is a Hilbert space then an orthonormal set ξii∈I is an orthonormalbasis if 0 is the only vector which is orthogonal to every ξi. For example, if Xis a set then the family of Dirac functions δxx∈X forms an orthonormal basis.

Proposition 6.3.12 (Parseval’s identity). Suppose H is a Hilbert space withorthonormal basis ξii∈I , then for η ∈ H we have

‖η‖2 =∑i∈I|〈η, ξi〉|2,

and η =∑i∈I〈η, ξi〉ξi, where the sum converges absolutely in H.

Proof. Given η ∈ H set η0 =∑i∈I〈η, ξi〉ξi and note that this sum converges

absolutely in H by Bessel’s inequality.For j ∈ I we have 〈η − η0, ξj〉 = 〈η, ξj〉 − 〈η, ξj〉 = 0. Since ξii∈I is an

orthonormal basis we then have η = η0 =∑i∈I〈η, ξi〉ξi. By approximating η

by finite sums it then follows that

‖η‖2 =∑i∈I|〈η, ξi〉|2.

Theorem 6.3.13. Every Hilbert space has an orthonormal basis. Moreover,any two orthonormal bases have the same cardinality.

Proof. Let H be a Hilbert space. By Zorn’s lemma it follows easily that H hasa maximal (with respect to inclusion) orthonormal set ξii∈I . Suppose η ∈ Hsuch that 〈η, ξi〉 = 0 for all i ∈ I. If η 6= 0 then the set η‖η‖−1 ∪ ξii∈Iwould be an orthonormal set which is strictly larger and hence would contradictZorn’s lemma. Thus, we must have η = 0 and hence ξii∈I is an orthonormalbasis.

To see that any two bases have the same cardinality we consider separatelythe finite and infinite cases. In the finite case an orthonormal basis is also analgebraic basis and this is then a standard fact from abstract linear algebrawhich we will not present here.

Suppose therefore that ξii∈I and ηjj∈J are two infinite orthonormalbases. By Bessel’s inequality to each i ∈ I the set of j’s such that 〈ξi, ηj〉 6= 0is a non-empty countable set. Since every ηj is not orthogonal to some ξi itfollows that there exists a surjective map from I ×N onto J . Since I is infinitewe have |I| = |I×N| ≥ |J |. By symmetry we also have that |J | ≥ |I| and hence|I| = |J |.

If H is a Hilbert space then the dimension of H is the cardinality of anyorthonormal basis, and denoted by dimH. If H and K are Hilbert spaces thena unitary operator is a surjective linear isometry U : H → K.

Theorem 6.3.14. If H is a Hilbert space, and X is a set then dimH = |X| ifand only if there exists a unitary operator U : `2(X)→ H.


Proof. If U : `2(X) → H is a unitary operator then Uδxx∈X gives an or-thonormal basis with cardinality |X|.

Conversely, if H has an orthonormal basis ξii∈I with cardinality |I| = |X|,then there exists a bijection θ : X → I. We define an operator U : `2X → H,by setting U(f) =

∑x∈X f(x)ξθ(x). It’s then an easy calculation to see that U

is a unitary operator.

6.3.4 Exercises

Exercise 6.3.15. If we consider N with the counting measure, then `2(N) is aHilbert space.

Exercise 6.3.16. Let X be a real normed space with norm ‖ · ‖2. Then ‖ · ‖2comes from an inner product if and only if the parallelogram identity ‖ξ+η‖2 +‖ξ − η‖2 = 2(‖ξ‖2 + ‖η‖2) holds.

Exercise 6.3.17. Let H and K be two real inner product spaces. A linear mapU : H → K is isometric if and only if 〈Uξ, Uη〉 = 〈ξ, η〉 for all ξ, η ∈ H.

Exercise 6.3.18 (The Banach-Saks theorem). Let H be a Hilbert space andξnn∈N ⊂ H a uniformly bounded sequence, then there exists a subsequence

ξnkk so that the Cesaro means 1K

∑Kk=1 ξnk converges in H. Hint: Using the

Banach-Alaoglu theorem you may assume that ξn has a weak limit.

Exercise 6.3.19. Let H be an inner product space, and A ⊂ H, then A⊥ is aclosed subspace.

Exercise 6.3.20. IfH is a Hilbert space and A ⊂ H is a subspace then (A⊥)⊥ =A. This does not hold for general inner product spaces.

Exercise 6.3.21. Let H, and K be Hilbert spaces and suppose T : H → K isa bounded linear operator, then there exists a unique bounded linear operatorT ∗ : K → H so that for all ξ ∈ H and η ∈ K we have

〈Tξ, η〉 = 〈ξ, T ∗η〉.

The operator T ∗ : K → H is called the adjoint of the operator T .

Exercise 6.3.22. Let H and K be Hilbert spaces.

1. A bounded linear operator P ∈ B(H) is an orthogonal projection operatorif and only if P = P ∗ and P 2 = P .

2. A bouned operator U ∈ B(H,K) is isometric if and only if U∗U = id.

Exercise 6.3.23. Let H and K be Hilbert spaces, and suppose T ∈ B(H,K).Then ker(T ) = Range(T ∗)⊥.

A linear operator V ∈ B(H,K) is a partial isometry if V ∗V is an orthog-onal projection.


Exercise 6.3.24. Suppose V ∈ B(H,K) is a partial isometry.

1. V ∗V is the orthogonal projection onto ker(V )⊥.

2. Range(V ) is closed and V V ∗ is the orthogonal projection onto Range(V ).In particular, V ∗ is also a partial isometry.

Exercise 6.3.25. Let M be a closed subspace of L2([0, 1], λ) such that M iscontained in C([0, 1]).

1. There exists C > 0 such that ‖f‖∞ ≤ C‖f‖L2 for all f ∈M .

2. For each x ∈ [0, 1] there exists gx ∈ M so that f(x) = 〈f, gx〉 for allf ∈M . Moreover, ‖gx‖L2 ≤ C.

3. dimM ≤ C2. Hint: If fii is an orthonormal sequence in M then∑i |fi(x)|2 ≤ C2 for all x ∈ [0, 1].

Exercise 6.3.26 (Von Neumann’s mean ergodic theorem). Let U be a unitaryoperator on a Hilbert space H, set K = ξ ∈ H | Uξ = ξ, and let P denote the

orthogonal projection onto K. If Sn = 1n

∑n−1k=0 U

k then for all ξ ∈ U we haveSnξ → Pξ. Hint: Use Exercise 6.3.23 applied to the operator 1− U .

Exercise 6.3.27. Let (X,µ) be a finite measure space and fix f ∈ L∞(X,µ).Set an =

∫|f |n dµ.

1. The sequence an+1

anis non-decreasing as n increases. Hint: Apply the

Cauchy-Schwarz inequality for ξ = |f |n/2 and η = |f |(n+2)/2.

2. The sequence an+1

anconverges to ‖f‖∞. Hint: Show that the series

∑∞n=1 b

n|f |n

converges absolutely in L1(X,µ) if b < ‖f‖−1∞ , and diverges in L1(X,µ) if

b > ‖f‖−1∞ .

Index

AAlaoglu, Leonidas (1914-1981), 74Alexandrov, Pavel (1896-1982), 88,

89Arzela, Cesare (1847-1912), 74Ascoli, Giulio (1843-1896), 74

BBaire, Rene-Louis (1874-1932), 84Banach, Stefan (1892-1945), 22, 74,

120, 123, 135Bendixson, Ivar Otto (1861-1935),

87Benyamini, Yoav (1943- ), 89Bernstein, Felix (1878-1956), 8Bessel, Friedrich (1784-1846), 133Bolzano, Bernhard (1781-1848), 20Borel, Emile (1871-1956), 20, 89,

92Brouwer, L. E. J. (1881-1966), 86

CCantor, Georg (1845-1918), 8, 9,

15, 86, 87Caratheodory, Constantin

(1873-1950), 33, 35Cauchy, Augustin-Louis

(1789-1857), 18, 129Cech, Eduard (1893-1960), 80, 83Cesaro, Ernesto (1859-1906), 135Chebyshev, Pafnuty (1821-1894),

113

Dde Bruijn, Nicolaas Govert

(1918-2012), 76Dini, Ulisse (1845-1918), 102

EEgorov, Dmitri (1869-1931), 49Erdos, Paul (1913-1996), 76

FFatou, Pierre (1878-1929), 48Frechet, Maurice (1878-1973), 119Fubini, Guido (1879-1943), 50

GGottschalk, Walter (1918-2004), 76

HHahn, Hans (1879-1934), 56, 123Hausdorff, Felix (1868-1942), 12, 88Heine, Eduard (1821-1881), 20Hilbert, David (1862-1943), 81, 129Holder, Otto Ludwig (1859-1937),

109

JJordan, Camille (1838-1922), 57,

105

KKakutani, Shizuo (1911-2004), 129Konig, Denes (1884-1944), 16Kuratowski, Kazimierz

(1896-1980), 83, 93

LLebesgue, Henri (1875-1941), 26,

38, 89, 98, 100

137

138 INDEX

Lipschitz, Rudolf (1832-1903), 17Lusin, Nikolai (1883-1950), 41, 88,

92

MMarkov, Andrey Jr. (1903-1979),

129Minkowski, Hermann (1864-1909),

110, 118Moore, Robert Lee (1882-1974), 66

NNikodym, Otto M. (1887-1974), 59

PParseval, Marc-Antoine

(1755-1836), 134

RRadon, Johann (1887-1956), 59Riesz, Frigyes (1880-1956), 133

SSaks, Stanis law (1897-1942), 135Schroder, Ernst (1841-1902), 8Schwarz, Hermann (1843-1921),

129Sierpinski, Wac law (1882-1969), 89Sorgenfrey, Robert (1915-1995), 66Steinhaus, Hugo (1887-1972), 120

Stieltjes, Thomas Joannes(1856-1894), 38

Stone, Marshall Harvey(1903-1989), 77, 80

Suslin, Mikhail Yakovlevich(1894-1919), 88, 92

TTietze, Heinrich Franz Friedrich

(1880-1964), 70Tikhonov, Andrey Nikolayevich

(1906-1993), 70, 73Tonelli, Leonida (1885-1946), 50

UUrysohn, Pavel (1898-1924), 70,

82, 89

VVitali, Giuseppe (1875-1932), 25,

97Von Neumann, John (1903-1957),

14, 94, 95, 136

WWeierstrass, Karl (1815-1897), 20,

77

ZZariski, Oscar (1899-1986), 66Zorn, Max August (1906-1993), 12

Bibliography

[hg] Andrey Gogolev (http://mathoverflow.net/users/2029/andrey gogolev), If fis infinitely differentiable then f coincides with a polynomial, MathOverflow,URL:http://mathoverflow.net/q/34067 (version: 2011-11-01).

139

Documents

Real Analysis - Vanderbilt University · negative integers (including zero), the integers, the rational numbers, the real numbers, and the complex numbers. If Ais a collection of