Constructing measures - Forsiden · measure we know what we want the measure of a cylinder set to be. The art is to extend such \pre-measures" to full-blown measures. We shall use

Chapter 6

Constructing measures

So far we have been taking measures for granted; except for a few, almosttrivial examples we have not shown that they exist. In this chapter we shalldevelop a powerful technique for constructing measures with the propertieswe want, e.g. the Lebesgue measures and the coin tossing measure describedin Section 5.1.

When we construct a measure, we usually start by knowing how we wantit to behave on a family of simple sets: For the Lebesgue measure on R, wewant the intervals (a, b) to have measure b − a, and for the coin tossingmeasure we know what we want the measure of a cylinder set to be. Theart is to extend such “pre-measures” to full-blown measures.

We shall use a three step procedure to construct measures. We startwith a collection R of subsets of our space X and a function ρ : R → R+.The idea is that the sets R in R are the sets we “know” the size ρ(R) of – if,e.g., we want to construct the Lebesgue measure, R could be the collectionof all open intervals, and ρ would then be given by ρ((a, b)) = b− a. FromR and ρ, we first construct an “outer measure” µ∗ which assigns a “size”µ∗(A) to all subsets A of X. The problem with µ∗ is that it usually fails tobe countably additive; i.e. the crucial equality

µ∗(⋃n∈N

An) =∞∑n=1

µ∗(An)

doesn’t hold for all disjoint sequences {An} of subsets of X. The second stepin our procedure is therefore to identify a σ-algebra A such that countableadditivity holds when the disjoint sets An belong to A. The restriction of µ∗to A will then be our desired measure µ. The final step in the procedure isto check that µ really is an extension of ρ, i.e. that R ⊂ A and µ(R) = ρ(R)for all R ∈ R. This is not always the case, but requires special properties ofR and ρ.

We begin by constructing outer measures.

1

2 CHAPTER 6. CONSTRUCTING MEASURES

6.1 Outer measure

To construct outer measures, we don’t need to require much of R and ρ:

Definition 6.1.1 Assume that X is a nonempty set and that R is a collec-tion of subsets of X such that ∅, X ∈ R. Throughout this section, we assumethat ρ : R → R+ is a function such that ρ(∅) = 0.

Assume that B is a subset of X. A covering of X is a countable collectionC = {Cn}n∈N of sets from R such that

B ⊂⋃n∈N

Cn

We define the size of the covering to be

|C| =∞∑n=1

ρ(Cn)

We are now ready to define the outer measure µ∗ generated by R and ρ: Forall B ⊂ X, we set

µ∗(B) = inf{|C| : C is a covering of B}

We see why µ∗ is called an outer measure; it is obtained by approximatingsets from the outside by unions of sets in R.

The essential properties of outer measures are not hard to establish:

Proposition 6.1.2 The outer measure µ∗ satisfies:

(i) µ∗(∅) = 0.

(ii) (Monotonicity) If B ⊂ C, then µ∗(B) ≤ µ∗(C).

(iii) (Countable subadditivity) If {Bn}n∈N is a sequence of subsets of Rd,then

µ∗(∞⋃n=1

Bn) ≤∞∑n=1

µ∗(Bn)

Proof: (i) Since C = {∅, ∅, ∅, . . .} is a covering of ∅ and ρ(∅) = 0, we getµ∗(∅) = 0.

(ii) Since any covering of C is a covering of B, we have µ∗(B) ≤ µ∗(C).

6.1. OUTER MEASURE 3

(iii) If µ∗(Bn) = ∞ for some n ∈ N, there is nothing to prove, and wemay hence assume that µ∗(Bn) 0 be given. For eachn ∈ N, we can find a covering C(n)1 , C

(n)2 , . . . of Bn such that

∞∑k=1

ρ(C(n)k ) < µ

∗(Bn) +�

2n

The collection {C(n)k }k,n∈N of all sets in all the coverings is a countablecovering of

⋃∞n=1Bn, and∑

k,n∈Nρ(C

(n)k ) =

∞∑n=1

( ∞∑k=1

ρ(C(n)k )

)≤∞∑n=1

(µ∗(Bn) +

�

2n

)=

∞∑n=1

µ∗(Bn) + �

(if you feel unsure about these manipulations, take a look at Exercise 5).This means that

µ∗(⋃n∈N

Bn) ≤∞∑n=1

µ∗(Bn) + �

and since � is an arbitrary, positive number, we must have

µ∗(⋃n∈N

Bn) ≤∞∑n=1

µ∗(Bn)

2

Remark: Note that property (iii) in the proposition above also holds forfinite sums; i.e.

µ∗(N⋃n=1

Bn) ≤N∑n=1

µ∗(Bn)

(to see this, just apply (iii) to the sequence B1, B2, . . . , BN , ∅, ∅, . . .). In par-ticular, we always have µ∗(A ∪B) ≤ µ∗(A) + µ∗(B).

We have now completed the first part of our program: We have con-structed the outer measure and described its fundamental properties. Thenext step is to define the measurable sets, to prove that they form a σ-algebra, and show that µ∗ is a measure when we restrict it to this σ-algebra.

Exercises for Section 6.1

1. Show that µ∗(R) ≤ ρ(R) for all R ∈ R.

2. Let X = {1, 2}, R = {∅, {1}, {1, 2}}, and define ρ : R → R by ρ(∅) = 0,ρ({1}) = 2, ρ({1, 2}) = 1. Show that µ∗({1}) < ρ({1}).

3. Assume that X = R, and let R consist of ∅, X, plus all open intervals (a, b),where a, b ∈ R. Define ρ : R → R by ρ(∅) = 0, ρ(R) =∞, ρ((a, b)) = b− a.


a) Show that if I = [c, d] is a closed and bounded interval, and C = {Cn}is a covering of I, then there is a finite number Ci1 , Ci2 , . . . , Cin of setsfrom C that covers I (i.e., such that I ⊂ Ci1 ∪ Ci2 ∪ . . . ∪ Cin). (Hint:Compactness.)

b) Show that µ∗([c, d]) = ρ([c, d]) = d − c for all closed and boundedintervals.

4. Assume that R is a σ-algebra and that ρ is a measure on R. Let (X, R̄, µ̄)be the completion of (X,R, µ). Show that µ∗(A) = µ̄(A) for all A ∈ R̄.

5. Let {an,k}n,k∈N be a collection of nonnegative, real numbers, and let A bethe supremum over all finite sums of distinct elements in this collection, i.e.

A = sup{I∑i=1

ani,ki : I ∈ N and all pairs (n1, k1), . . . , (nI , kI) are different}

a) Assume that {bm}m∈N is a sequence which contains each element in theset {an,k}n,k∈N exactly ones. Show that

∑∞m=1 bm = A.

b) Show that∑∞n=1 (

∑∞k=1 an,k) = A.

c) Comment on the proof of Proposition 6.1.2(iii).

6.2 Measurable sets

The definition of a measurable set is not difficult, but it is rather myste-rious in the sense that it is not easy to see why it captures the essence ofmeasurability.

Definition 6.2.1 A subset E of X is called measurable if

µ∗(A ∩ E) + µ∗(A ∩ Ec) = µ∗(A)

for all A ⊂ X. The collection of all measurable sets is denoted by M.

This approach to measurability was introduced by the Greek mathematicianConstantin Carathéodory (1873-1950) and replaced more cumbersome (buteasier to understand) earlier approaches (see Exercise 6.3.4 for one such).As already mentioned, it is not at all easy to explain why it captures theintuitive notion of measurability. The best explanation I can offer, is that thereason why some sets are impossible to measure (and hence non-measurable),is that they have very irregular boundaries. The definition above says thata set is measurable if we can use it to cut any other set in two parts withoutintroducing any further irregularities – hence all parts of its boundary mustbe reasonably regular. I admit that this explanation is rather vague, and abetter argument may simply be to show that the definition works. So let usget started.

6.2. MEASURABLE SETS 5

Let us first of all make a very simple observation. Since A = (A ∩ E) ∪(A∩Ec), subadditivity (recall Proposition 6.1.2(iii)) tells us that we alwayshave

µ∗(A ∩ E) + µ∗(A ∩ Ec) ≥ µ∗(A)

Hence to prove that a set is measurable, we only need to prove that

µ∗(A ∩ E) + µ∗(A ∩ Ec) ≤ µ∗(A)

Our first result is easy.

Lemma 6.2.2 If E has outer measure 0, then E is measurable. In partic-ular, ∅ ∈ M.

Proof: If E has outer measure 0, so has A ∩ E since A ∩ E ⊂ E. Hence

µ∗(A ∩ E) + µ∗(A ∩ Ec) = µ∗(A ∩ Ec) ≤ µ∗(A)

for all A ⊂ X. 2

Next we have:

Proposition 6.2.3 M satisfies:1

(i) ∅ ∈ M.

(ii) If E ∈M, then Ec ∈M.

(iii) If E1, E2, . . . , En ∈M, then E1 ∪ E2 ∪ . . . ∪ En ∈M.

(iv) If E1, E2, . . . , En ∈M, then E1 ∩ E2 ∩ . . . ∩ En ∈M.

Proof: We have already proved (i), and (ii) is obvious from the definitionof measurable sets. Since E1 ∪ E2 ∪ . . . ∪ En = (Ec1 ∩ Ec2 ∩ . . . ∩ Ecn)c byDe Morgan’s laws, (iii) follows from (ii) and (iv). Hence it only remains toprove (iv).

To prove (iv) is suffices to prove that if E1, E2 ∈M, then E1 ∩E2 ∈Mas we can then add more sets by induction. If we first use the measurabilityof E1, we see that for any set A ⊂ Rd

µ∗(A) = µ∗(A ∩ E1) + µ∗(A ∩ Ec1)

Using the measurability of E2, we get

µ∗(A ∩ E1) = µ∗((A ∩ E1) ∩ E2) + µ∗((A ∩ E1) ∩ Ec2)1As you probably know from Chapter 5, a family of sets satisfying (i)-(iii) is usually

called an algebra. As (iv) is a consequence of (ii) and (iii) using one of De Morgan’s laws,the proposition simply states that M is an algebra, but in the present context (iv) is inmany ways the most important property.


Combining these two expressions, we have

µ∗(A) = µ∗((A ∩ (E1 ∩ E2)) + µ∗((A ∩ E1) ∩ Ec2) + µ∗(A ∩ Ec1)

Observe that (draw a picture!)

(A ∩ E1 ∩ Ec2) ∪ (A ∩ Ec1) = A ∩ (E1 ∩ E2)c

and hence

µ∗(A ∩ E1 ∩ Ec2) + µ∗(A ∩ Ec1) ≥ µ∗(A ∩ (E1 ∩ E2)c)

Putting this into the expression for µ∗(A) above, we get

µ∗(A) ≥ µ∗((A ∩ (E1 ∩ E2)) + µ∗(A ∩ (E1 ∩ E2)c)

which means that E1 ∩ E2 ∈M. 2

We would like to extend parts (iii) and (iv) in the proposition above tocountable unions and intersection. For this we need the following lemma:

Lemma 6.2.4 If E1, E2, . . . , En is a disjoint collection of measurable sets,then

µ∗(A∩ (E1 ∪E2 ∪ . . .∪En)) = µ∗(A∩E1) + µ∗(A∩E2) + . . .+ µ∗(A∩En)

Proof: It suffices to prove the lemma for two sets E1 and E2 as we can thenextend it by induction. Using the measurability of E1, we see that

µ∗(A ∩ (E1 ∪E2)) = µ∗((A ∩ (E1 ∪E2)) ∩E1) + µ∗(A ∩ (E1 ∪E2)) ∩Ec1) =

= µ∗(A ∩ E1) + µ∗(A ∩ E2)

2

We can now prove that M is closed under countable unions.

Lemma 6.2.5 If An ∈M for each n ∈ N, then⋃n∈NAn ∈M.

Proof: Define a new sequence {En} of sets by E1 = A1 and

En = An ∩ (E1 ∪ E2 ∪ . . . ∪ En−1)c

for n > 1, and note that En ∈M sinceM is an algebra. The sets {En} aredisjoint and have the same union as {An} (make a drawing!), and hence itsuffices to prove that

⋃n∈NEn ∈M, i.e.

µ∗(A) ≥ µ∗(A ∩

∞⋃n=1

En)

+ µ∗(A ∩

( ∞⋃n=1

En)c)

6.2. MEASURABLE SETS 7

Since⋃Nn=1En ∈M for all N ∈ N, we have:

µ∗(A) = µ∗(A ∩

N⋃n=1

En)

+ µ∗(A ∩

( N⋃n=1

En)c) ≥

≥N∑n=1

µ∗(A ∩ En) + µ∗(A ∩

( ∞⋃n=1

En)c)

where we in the last step have used the lemma above plus the observationthat

(⋃∞n=1En

)c ⊂ (⋃Nn=1En)c. Since this inequality holds for all N ∈ N,we get

µ∗(A) ≥∞∑n=1

µ∗(A ∩ En) + µ∗(A ∩

( ∞⋃n=1

En)c)

By sublinearity, we have∑∞

n=1 µ∗(A ∩ En) ≥ µ∗(

⋃∞n=1(A ∩ En)) = µ∗(A ∩⋃∞

n=1(En)), and hence

µ∗(A) ≥ µ∗(A ∩

∞⋃n=1

En)

+ µ∗(A ∩

( ∞⋃n=1

En)c)

2

Lemmas 6.2.3 and 6.2.5 tell us that M is a σ-algebra. We may restrictµ∗ to M to get a function

µ :M→ R+

defined by2

µ(A) = µ∗(A) for all A ∈M

We can now complete the second part of our program:

Theorem 6.2.6 M is a σ-algebra, and µ is a complete measure on M.

Proof: We already know that M is a σ-algebra, and if we prove that µ isa measure, the completeness will follow from Lemma 6.2.2. As we alreadyknow that µ(∅) = µ∗(∅) = 0, we only need to prove that

µ(

∞⋃n=1

En) =

∞∑n=1

µ(En) (6.2.1)

for all disjoint sequences {En} of sets from M2Since µ is just the restriction of µ∗ to a smaller domain, it may seem a luxury to intro-

duce a new symbol for it, but in some situations it is important to be able to distinguishquickly between µ and µ∗.


By Proposition 6.1.2(iii), we already know that

µ(∞⋃n=1

En) ≤∞∑n=1

µ(En)

To get the opposite inequality, we use Lemma 6.2.4 with A = X to see that

N∑n=1

µ(En) = µ(N⋃n=1

En) ≤ µ(∞⋃n=1

En)

Since this holds for all N ∈ N, we must have∞∑n=1

µ(En) ≤ µ(∞⋃n=1

En)

Hence we have both inequalities, and (6.2.1) is proved. 2

We have now completed the second part of our program — we haveshown how we can turn an outer measure µ∗ into a measure µ by restrictingit to the measurable sets. This is in an interesting result in itself, but we stillhave some work to do – we need to compare the measure µ to the originalfunction ρ.


1. Explain in detail why 6.2.3(iii) follows from (ii) and (iv).

2. Carry out the induction step in the proof of Proposition 6.2.3(iv).

3. Explain the equality (A∩E1 ∩Ec2)∪ (A∩Ec1) = A∩ (E1 ∩E2)c in the proofof Lemma 6.2.3.

4. Carry out the induction step in the proof of Lemma 6.2.4.

5. Explain why the sets En in the proof of Lemma 6.2.5 are disjoint and havethe same union as the sets An. Explain in detail why the sets En belong toM.

6.3 Carathéodory’s Theorem

In the generality we have have been working so far, there isn’t much that canbe said about the relationship between the original set function ρ and themeasure µ generated via the outer measure construction. Since R, ∅, ∅, . . . isa covering of R, we always have µ∗(R) ≤ ρ(R) for all R ∈ R, but if we wantequality instead of inequality, we need to impose conditions on R and ρ:

Definition 6.3.1 R is called an algebra of sets on X if the following con-ditions are satisfied:

6.3. CARATHÉODORY’S THEOREM 9

(i) ∅ ∈ R

(ii) If R ∈ R, then Rc ∈ R.

(iii) If R,S ∈ R, then R ∪ S ∈ R

If R is an algebra of sets, the function ρ : R → R+ is called a premeasureif the following conditions are satisfied:

(iv) ρ(∅) = 0

(v) If {Rn}n∈N is a disjoint sequence of sets in R whose union happens tobelong to R, then3

ρ(⋃n∈N

Rn) =∞∑n=1

ρ(Rn)

We are now ready for the main result of this section.

Theorem 6.3.2 (Carathéodory’s Extension Theorem) Assume that Ris an algebra and that ρ is a premeasure on R. Then there exists a com-plete measure µ extending ρ, i.e., a measure µ defined on a σ-algebra Mcontaining R such that µ(R) = ρ(R) for all R ∈ R.

Proof: We know that the outer measure construction generates a σ-algebraM of measurable sets and a complete measure µ on M, and it suffices toshow that all sets in R are measurable and that µ(R) = ρ(R) for all R ∈ R.

Let us first prove that any set R ∈ R is measurable, i.e. that

µ∗(A) ≥ µ∗(A ∩R) + µ∗(A ∩Rc)

for any set A ⊂ X. Given � > 0, we can find a covering C = {Cn}n∈N of Asuch that

∑∞n=1 ρ(Cn) < µ

∗(A) + �. Note that {Cn ∩R} and {Cn ∩Rc} arecoverings of A ∩R and A ∩Rc, respectively, and hence

µ∗(A) + � >

∞∑n=1

ρ(Cn) =

∞∑n=1

(ρ(Cn ∩R) + ρ(Cn ∩Rc)

)=

=

∞∑n=1

ρ(Cn ∩R) +∞∑n=1

ρ(Cn ∩Rc) ≥ µ∗(A ∩R) + µ(A ∩Rc)

Since � < 0 is arbitrary, this means that µ∗(A) ≥ µ∗(A ∩ R) + µ∗(A ∩ Rc),and hence R is measurable.

It remains to prove that µ(R) = µ∗(R) = ρ(R) for all R ∈ R. As wealready know that µ∗(R) ≤ ρ(R), it suffices to prove the opposite inequality.

3In general, there is no reason why such a union should belong to R, but it does happen,e.g. when all but a finite number of the sets Rn equals ∅


For any � > 0, there is a covering C = {Cn}n∈N of R such that∑∞

n=1 ρ(Cn) <µ∗(R) + �. The sets C ′n = R ∩ (Cn \

⋃n−1k=1 Ck) are disjoint elements of R

whose union is R, and hence by condition (v) in the definition above

ρ(R) =∞∑n=1

ρ(C ′n) ≤∞∑n=1

ρ(Cn) < µ∗(R) + �

Since � > 0 is arbitrary, this means that ρ(R) ≤ µ∗(R), and since we alreadyhave the opposite inequality, we have proved that µ∗(R) = ρ(R). 2

In general, there is more than one measure extending a given premeasureρ, but for most spaces that occur in practice, there isn’t too much freedom(but see Exercise 2 for an extreme case):

Proposition 6.3.3 Let ρ be a premeasure on an algebra R and let (X,M, µ)be the measure space obtained by the outer measure construction. Assumethat B is a σ-algebra containing A and that ν is a measure on B such thatν(R) = ρ(R) for all R ∈ R. Then ν(A) ≤ µ(A) for all A ∈ M ∩ B, withequality whenever µ(A)


(i) If R,S ∈ R, then R ∩ S ∈ R.

(ii) If R ∈ R, then Rc is a a disjoint union of sets in R.

Starting with a semi-algebra, it is not hard to build an algebra:

Lemma 6.3.5 Assume that R is a semi-algebra on a set X and let A consistof ∅ plus all finite, disjoint unions of sets in R. Then A is the algebragenerated by R.

Proof: As all sets in A clearly have to be in any algebra containing R, weonly need to show that A is an algebra, and for this it suffices to show thatA is closed under complements and finite intersections.

Let us start with the intersections. Assume that A,B are two nonemptysets in A, i.e. that

A = R1 ∪R2 ∪ . . . ∪Rn

B = Q1 ∪Q2 ∪ . . . ∪Qm

are disjoint unions of sets from R. Then

A ∩B =⋃i,j

(Ri ∩Qj)

is clearly a disjoint, finite union of sets in R, and hence A ∩ B ∈ A. Byinduction, we see that any finite intersection of sets in A is in A.

Turning to complements, we first observe that X = ∅c belongs to A.To see this, pick a nonempty set R ∈ R. Since R is a semi-algebra, Rc =R1 ∪ . . . ∪Rn for disjoint sets R1, . . . , Rn ∈ R, and hence

X = R ∪R1 ∪ . . . ∪Rn

which proves that X ∈ A. Assume next that A = R1 ∪ R2 ∪ . . . ∪ Rn is adisjoint union of sets in R. Then by one of De Morgan’s laws

Ac = Rc1 ∩Rc2 ∩ . . . ∩Rcn

Since R is a semi-algebra, each Rci is a disjoint union of sets in R and hencebelongs to A. Since we have already proved that A is closed under finiteintersections, Ac ∈ A. 2

Here is the promised extension of Carathéodory’s theorem to semi-algebras:

Theorem 6.3.6 (Carathéodory’s Theorem for Semi-Algebras) Let Rbe a semi-algebra of subsets of X and assume that λ : R → R+ is a functionsuch that


(i) if a set R ∈ R is a disjoint, finite union R =⋃ni=1Ri of sets in R,

then

λ(R) =n∑i=1

λ(Ri)

(ii) If a set R ∈ R is a disjoint, countable union R =⋃i∈NRi of sets in

R, then

λ(R) =∞∑i=1

λ(Ri)

Then λ has an extension to a complete measure on a σ-algebra containingR.4

Before we turn to the proof of the theorem, it is convenient to prove thefollowing lemma.

Lemma 6.3.7 Assume that R and λ are as in the theorem above.

(i) Assume that a set A ⊂ X can be written as disjoint, finite unions ofsets in R in two different ways: A =

⋃ni=1Ri and A =

⋃mj=1 Sj. Then

n∑i=1

λ(Ri) =m∑j=1

λ(Sj)

(ii) Assume that a set A ⊂ X can be written as disjoint unions of sets inR in a finite and a countable way: A =

⋃ni=1Ri and A =

⋃j∈N Sj.

Thenn∑i=1

λ(Ri) =∞∑j=1

λ(Sj)

Proof: We only prove (ii). The proof of (i) is similar, the only difference be-ing that we in one place use condition (i) of the theorem instead of condition(ii).

To prove (ii), observe that since R is a semi-algebra, the intersectionsRi ∩ Sj belong to R, and hence by condition (ii) in the theorem

λ(Ri) =∞∑j=1

λ(Ri ∩ Sj)

Similarly by condition (i) in the theorem

λ(Sj) =

n∑i=1

λ(Ri ∩ Sj)

4If ∅ ∈ R and λ(∅) = 0, (i) is just a special case of (ii), but since we haven’t assumedthat ∅ ∈ R, we have to treat finite and infinite unions separately.


This means that

n∑i=1

λ(Ri) =n∑i=1

∞∑j=1

λ(Ri ∩ Sj) =∞∑j=1

n∑i=1

λ(Ri ∩ Sj) =∞∑j=1

λ(Sj)

which is what we wanted to prove. 2

Proof of Carathéodory’s theorem for semi-algebras: Let A be the algebragenerated by R. The plan is to extend λ to a premeasure ρ on A, and thenuse the version of Carathéodory’s theorem that we have already proved.

Since any non-empty set A in A is a finite, disjoint union A =⋃ni=1Ri

of sets in R, we define ρ : A → R+ by putting ρ(∅) = 0 and

ρ(A) =n∑i=1

λ(Ri)

(ρ is well-defined as part (i) of the lemma tells us that if A =⋃mj=1 Sj is

another way of writing A as a disjoint union of sets in R, then∑n

i=1 λ(Ri) =∑mj=1 λ(Sj)). To use Carathéodory’s Extension Theorem 6.3.2, we need to

show that ρ is a premeasure, i.e. that

ρ(A) =

∞∑n=1

ρ(An)

whenever A and A1, A2, . . . are in A and A is the disjoint union of A1, A2, . . ..Since A and the An’s are in A, they can be written as finite, disjoint unionsof sets in R:

A =

M⋃j=i

Rj

and

An =

Nn⋃k=1

Sn,k

Since A =⋃n∈NAn =

⋃n,k Sn,k, part (ii) of the lemma tells us that

M∑j=1

λ(Rj) =∑n,k

λ(Sn,k)

The rest is easy: By definition of ρ

ρ(A) =

M∑j=1

λ(Rj) =∑n,k

λ(Sn,k) =

∞∑n=1

Nn∑k=1

λ(Sn,k) =

∞∑n=1

ρ(An)


We have now proved that ρ is a premeasure on the algebra A, and bythe original Carathéodory’s theorem 6.3.2, there is a complete measure ona σ-algebra containing A which extends ρ (and hence λ). 2

We now have the machinery we need to construct measures, and in thenext section we shall use it to construct Lebesgue measure on R.


1. Prove part (i) of Lemma 6.3.7.

2. A measure space (X,A, µ) is called σ-finite if there there is a disjoint family{An}n∈N of sets in A such that X =

⋃n∈NAn and µ(An)

6.4. LEBESGUE MEASURE ON R 15

b) Show that if E is measurable, then E is ∗-measurable. (Hint: UseDefinition 6.2.1 with A = X.)

c) Show that if E is ∗-measurable, then for every � > 0 there are measur-able sets D,F such that D ⊂ E ⊂ F and

µ∗(D) > µ∗(E)− �2

and µ∗(F ) < µ∗(E) +�

2

d) Show that µ∗(F \D) < � and explain that µ∗(F \E) < � and µ∗(E\D) <�.

e) Explain that for any set A ⊂ X

µ∗(A ∩ F ) = µ∗(A ∩D) + µ∗(A ∩ (F \D)) < µ∗(A ∩D) + �

and

µ∗(A ∩Dc) = µ∗(A ∩ F c) + µ∗(A ∩ (F \D)) < µ∗(A ∩ F c) + �

and use this to show that µ∗(A∩D) > µ∗(A∩E)− � and µ∗(A∩F c) >µ∗(A ∩ Ec)− �.

f) Explain that for any set A ⊂ X

µ∗(A) ≥ µ∗(A ∩ (F c ∪D)) =

= µ∗(A ∩ F c) + µ∗(A ∩D) ≥ µ∗(A ∩ Ec) + µ∗(A ∩ E)− 2�

and use it to show that if E is ∗-measurable, then E is measurable. Thenotions of measurable and ∗-measurable hence coincide when µ∗(X) isfinite.

6.4 Lebesgue measure on R

In this section we shall use the theory in the previous section to constructthe Lebesgue measure on R. Essentially the same argument can be used toconstruct Lebesgue measure on Rd for d > 1, but as the geometric consid-erations become a little more complicated, we shall restrict ourselves to theone dimensional case. In Section 6.6 we shall see how we can obtain higherdimensional Lebesgue measure by a different method.

Recall that the one dimensional Lebesgue measure is a generalization ofthe notion of length: We know how long intervals are and want to extend thisnotion of size to a full-blown measure. For technical reasons, it is convenientto work with half-open intervals (a, b].

Definition 6.4.1 In this section, R consists of the following subsets of R:

(i) All finite, half-open intervals (a, b], where a, b ∈ R, a < b.

(ii) All infinite intervals of the form (−∞, b] and (a,∞) where a, b ∈ R.


The advantage of working with half-open intervals become clear when wecheck what happens when we take intersections and complements:

Proposition 6.4.2 R is a semi-algebra of sets.

Proof: I leave it to the reader to check the various cases. The crucial obser-vation is that the complement of a half-open interval is either a half-openinterval or the union of two half-open intervals. 2

We define λ : R → R+ simply by letting λ(R) be the length of the intervalR. To use Carathéodory’s theorem for semi-algebras, we first need to checkthat the main condition is satisfied.

Lemma 6.4.3 If a set R ∈ R is a disjoint, countable union R =⋃i∈NRi

of sets in R, then

λ(R) =

∞∑i=1

λ(Ri)

The same holds for finite unions: If R ∈ R is a disjoint, finite union R =R1 ∪R2 ∪ . . . ∪Rn of sets in R, then λ(R) = λ(R1) + λ(R2) + . . . λ(Rn).

Proof: The case for finite unions is easy and left to the reader. For thecountable case, note that for any finite N , the nonoverlapping intervalsR1, R2, . . . , RN can be ordered from left to right, and obviously make up apart of R in a such a way that λ(R) ≥

∑Nn=1 λ(Rn). Since this holds for all

finite N , we must have λ(R) ≥∑∞

n=1 λ(Rn).The opposite inequality is more subtle, and we need to use a compactness

argument. Let us first assume that R is a finite interval (a, b]. Given an� > 0, we extend each interval Rn = (an, bn] to an open interval R̂n =(an, bn +

�2n ). These open intervals cover the compact interval [a + �, b],

and by Theorem 2.6.6 there is a finite subcover R̂n1 , R̂n2 , . . . , R̂nk . Sincethese finite intervals cover an interval of length b − a − �, we clearly have∑k

j=1 λ(R̂nj ) ≥ (b − a − �). But since λ(R̂n) = λ(Rn) +�2n , this means

that∑k

j=1 λ(Rnj ) ≥ (b − a − 2�). Consequently,∑∞

n=1 λ(Rn) ≥ b − a −2� = λ(R)− 2�, and since � is an arbitrary, positive number, we must have∑∞

n=1 λ(Rn) ≥ λ(R).It remains to see what happens if R is an interval of infinite length,

i.e. when λ(R) = ∞. For each n ∈ N, the intervals {Ri ∩ (−n, n]}i∈Nare disjoint and have union R ∩ (−n, n]. By what we have already proved,λ(R ∩ (−n, n]) =

∑∞i=1 λ(Ri ∩ (−n, n]) ≤

∑∞i=1 λ(Ri). Since limn→∞ λ(R ∩

(−n, n]) =∞, we see that∑∞

i=1 λ(Ri) =∞ = λ(R). 2

We have reached our goal: The lemma above tells us that we can applyCarathéodory’s theorem for semi-algebras to the semi-algebra R and thefunction λ. The resulting measure µ is the Lebesgue measure on R. Let ussum up the essential points.


Theorem 6.4.4 The Lebesgue measure µ is the unique, completed Borelmeasure on R such that µ(I) = |I| for all intervals I.

Proof: By construction, the µ-measure of all half-open intervals is equalto their length, and by continuity of measure (Proposition 5.1.5), the samemust hold for open and closed intervals. By Proposition 6.3.3, if ν is anyother completed, Borel measure with the same property, µ(A) = ν(A) forall Borel sets such that µ(A) < ∞. But the same must hold for a generalBorel set A as we can slice it up into pieces of finite measure:

µ(A) =∑n∈Z

µ(A ∩ (n, n+ 1]

)=∑n∈Z

ν(A ∩ (n, n+ 1]

)= ν(A)

2

The next result tells us that measurable sets can be approximated arbi-trarily well by open and closed set.

Proposition 6.4.5 Assume that A ⊂ R is a measurable set. For each � > 0,there is an open set G ⊃ A such that µ(G \A) < �, and a closed set F ⊂ Asuch that µ(A \ F ) < �.

Proof: We begin with the open sets. Assume first that A has finite measure.Then for every � > 0 there is a covering {Cn} of A by half-open rectanglesCn = (an, bn] such that

∞∑n=1

|Cn| < µ(A) +�

2

If we replace the half-open intervals Cn = (an, bn] by the open intervalsBn = (an, bn +

�2n+1

), we get an open set G =⋃∞n=1Bn containing A with

µ(G) ≤∞∑n=1

µ(Bn) =

∞∑n=1

(|Cn|+

�

2n+1

)< µ(A) + �

and henceµ(G \A) = µ(G)− µ(A) < �

by Lemma 5.1.4c).If µ(A) is infinite, we slice A into pieces of finite measure An = A ∩

(n, n+1] for all n ∈ Z, and use what we have already proved to find an openset Gn such that An ⊂ Gn and µ(Gn \ An) < �2n+2 . Then G =

⋃n∈ZGn is

an open set which contains A, and since G \A ⊂⋃n∈Z(Gn \An), we get

µ(G \A) ≤∑n∈Z

µ(Gn \An) <∑n∈Z

�

2n+2< �,


proving the statement about approximation by open sets.To prove the statement about closed sets, just note that if we apply

the first part of the theorem to Ac, we get an open set G ⊃ Ac such thatµ(G \ Ac) < �. This means that F = Gc is a closed set such that F ⊂ A,and since A \ F = G \Ac, we have µ(A \ F ) < �. 2

Another important property of the Lebesgue measure is that it is trans-lation invariant — if we move a set a distance to the left or to the right, itkeeps its measure. To formulate this mathematically, let E be a subset of Rand a a real number, and write

E + a = {e+ a | e ∈ E}

for the set we obtain by moving all points in E a distance a.

Proposition 6.4.6 If E ⊂ R is measurable, so is E + a for all a ∈ R, andµ(E + a) = µ(E).

Proof: We shall leave this to the reader (see Exercise 7). The key observa-tion is that µ∗(E + a) = µ∗(E) holds for outer measure since intervals keeptheir length when we translate them. 2

One of the reasons why we had to develop the rather complicated ma-chinery of σ-algebras, is that we can not in general expect to define a measureon all subsets of our measure space X — some sets are just so complicatedthat they are nonmeasurable. We shall now take a look at such a set. Beforewe begin, we need to modify the notion of translation so that it works onthe interval [0, 1). If x, y ∈ [0, 1), we first define

x.

+ y =

x+ y if x+ y ∈ [0, 1)

x+ y − 1 otherwise

If E ⊂ [0, 1) and y ∈ [0, 1), let

E.

+ y = {e.

+ y | e ∈ E}

Note that E.

+ y is the set obtained by first translating E by y and thenmoving the part that sticks out to the right of [0, 1) one unit to left so thatit fills up the empty part of [0, 1). It follows from translation invariance thatE

.+ y is measurable if E is, and that µ(E) = µ(E

.+ y).

Example 1: A nonmeasurable set. We start by introducing an equiva-lence relation ∼ on the interval [0, 1):

x ∼ y ⇐⇒ x− y is rational


Next, we let E be a set obtained by picking one element from each equiva-lence class.5 We shall work with the sets E

.+ q for all rational numbers q

in the interval [0, 1), i.e., for all q ∈ Q̂ = Q ∩ [0, 1).First observe that if q1 6= q2, then (E

.+ q1) ∩ (E

.+ q2) = ∅. If not, we

could write the common element x as both x = e1.

+ q1 and x = e2.

+ q2for some e1, e2 ∈ E. The equality e1

.+ q1 = e2

.+ q2, implies that e1 − e2

is rational, and by definition of E this is only possible if e1 = e2. Henceq1 = q2, contradicting the assumption that q1 6= q2.

The next observation is that [0, 1) =⋃q∈Q̂(E

.+ q). To see this, pick an

arbitrary x ∈ [0, 1). By definition, there is an e in E that belongs to thesame equivalence class as x, i.e. such that q = x− e is rational. If q ∈ [0, 1),then x ∈ E

.+ q, if q < 0 (the only other possibility), we have x ∈ E

.+ (q+1)

(check this!). In either case, x ∈⋃q∈Q̂(E

.+ q).

Assume for contradiction that E is Lebesgue measurable. Then, as al-ready observed, E

.+ q is Lebesgue measurable with µ(E

.+ q) = µ(E) for

q ∈ Q̂. But by countable additivity

µ([0, 1)) =∑q∈Q̂

µ(E.

+ q)

and since µ([0, 1)) = 1, this is impossible — a sum of countable many, equal,nonnegative numbers is either ∞ (if the numbers are positive) or 0 (if thenumbers are 0).

Note that this argument works not only for the Lebesgue measure, butfor any (non-zero) translation invariant measure on R. This means that it isimpossible to find a translation invariant measure on R that makes all setsmeasurable. ♣

The existence of nonmeasurable sets is not a surprise – there is no reasonto expect that all sets should be measurable – but it is a nuisance whichcomplicates many arguments. As we have seen in this chapter, the hardpart is often to find the right class of measurable sets and prove that it is aσ-algebra.


1. Complete the proof of Proposition 6.4.2.

2. Prove the “finite case” of Lemma 6.4.3.

3. In the proof of Lemma 6.4.3, explain why∑∞n=1 |In| ≥ b− a− 2� = |I| − 2�.

4. Explain that A \ F = G \Ac at the end of the proof of Proposition 6.4.5.5Here we are using a principle from set theory called the Axiom of Choice which allows

us to make a new set by picking one element from each set in an infinite family. It hasbeen proved that it is not possible to construct a nonmeasurable subset of R without usinga principle of this kind.


5. A subset of R is called a Gδ-set if it is the intersection of countably manyopen sets, and it is called a Fσ-set if it is union of countably many closed set.

a) Explain why all Gδ- and Fσ-sets are measurable.b) Show that if A ⊂ R is measurable, there is a Gδ-set G such that A ⊂ G

and µ(G \A) = 0.c) Show that if A ⊂ R is measurable, there is a Fσ-set F such that F ⊂ A

and µ(A \ F ) = 0.

6. Assume that A ⊂ R is a measurable set with finite measure. Show that forevery � > 0, there is a compact set K ⊂ A such that µ(A \K) < �.

7. Prove Proposition 6.4.6 by:

a) Showing that if E ⊂ R and a ∈ R, then µ∗(E + a) = µ(E).b) Showing that if E ⊂ R is measurable, then so is E + a for any a ∈ R.c) Explaining how to obtain the proposition from a) and b).

8. Check that the equivalence relation in Example 1 really is an equivalencerelation.

9. If A is a subset of R and r is a positive, real number, let

rA = {ra | a ∈ A}

Show that if A is measurable, then so is rA and µ(rA) = rµ(A).

10. Use Proposition 6.4.6 to prove that that if E ⊂ [0, 1) is Lebesgue measurable,then E

.+ y is Lebesgue measurable with µ(E

.+ y) = µ(E) or all y ∈ [0, 1).

6.5 The coin tossing measure

As a second example of how to construct measures, we shall construct thenatural probability measure on the space of infinite sequences of unbiasedcoin tosses (recall Example 8 of Section 5.1). Be aware that this section israther sketchy – it is more like a structured sequence of exercises than anordinary section. The results here will not be used in the sequel.

Let us recall the setting. The underlying space Ω consists of all infinitesequences

ω = (ω1, ω2, . . . , ωn, . . .)

where each ωn is either H (for heads) or T (for tails). If a = a1, a2, . . . , anis a finite sequence of H’s and T’s, we let

Ca = {ω ∈ Ω |ω1 = a1, ω2 = a2, . . . , ωn = an}

and call it the cylinder set generated by a. We call n the length of Ca. LetR be the collection of all cylinder sets (of all possible lengths).

Lemma 6.5.1 R is a semi-algebra.

6.5. THE COIN TOSSING MEASURE 21

Proof: The intersection of two cylinder sets Ca and Cb is either equal to oneof them (if one of the sequences a, b is an extension of the other) or empty.The complement of a cylinder set is the union of all other cylinder sets ofthe same length. 2

We define a function λ : R → [0, 1] by putting

λ(Ca) =1

2n

where n is the length of Ca. (There are 2n cylinder sets of length n and theycorrespond to 2n equally probable events).

We first check that λ behaves the right way under finite unions:

Lemma 6.5.2 If A1, A2, . . . , AN are disjoint sets in R whose union⋃Nn=1An

belongs to R, then

λ(

N⋃n=1

An) =

N∑n=1

λ(An)

Proof: Left to the reader (see Exercise 1). 2

To use Carathéodory’s theorem for semi-algebras we must extend the resultabove to countable unions, i.e., we need to show that if {An} is a disjointsequence of cylinder sets whose union is a cylinder set, then

λ(⋃n∈N

An) =∞∑i=1

λ(An)

The next lemma tells us that this condition is trivially satisfied becausethe situation never occurs – a cylinder set is never a disjoint, countableunion of infinitely many cylinder sets! As this is the difficult part of theconstruction, I spell out the details. The argument is actually a compactnessargument in disguise, and corresponds to Lemma 6.4.3 in the constructionof the Lebesgue measure.

Before we begin, we need some notation and terminology. For eachk ∈ N, we write ω ∼k ω̂ if ωi = ω̂i for i = 1, 2, . . . , k, i.e., if the first k cointosses in the sequences ω and ω̂ are equal. We say that a subset A of Ω isdetermined at k ∈ N if whenever ω ∼k ω̂ then either both ω and ω̂ belong toA or none of them do (intuitively, this means that you can decide whether ωbelongs to A by looking at the k first coin tosses). A cylinder set of length kis obviously determined at time k. We say that a set A is finally determinedif it is determined at some k ∈ N.

Lemma 6.5.3 A cylinder set A is never an infinite, countable, disjointunion of cylinder sets.


Proof: Assume for contradiction that {An}n∈N is a disjoint sequence ofcylinder sets whose union A =

⋃n∈NAn is also a cylinder set. Since the

An’s are disjoint and nonempty, the set BN = A \⋃Nn=1An is nonempty for

all N ∈ N. Note that the sets BN are finitely determined since A and theAn’s are.

Since the sets BN are decreasing and nonempty, there must either bestrings ω = (ω1, ω2, . . .) starting with ω1 = H in all the sets BN or stringsstarting with ω1 = T in all the sets BN (or both, in which case we justchoose H). Call the appropriate symbol ω̂1. Arguing in the same way, wesee that there must either be strings starting with (ω̂1, H) or strings start-ing with (ω̂1, T ) in all the sets BN (or both, in which case we just chooseH). Call the appropriate symbol ω̂2. Continuing in this way, we get aninfinite sequence ω̂ = (ω̂1, ω̂2, ω̂3, . . .) such that for each k ∈ N, there is asequence starting with (ω̂1, ω̂2, ω̂3, . . . , ω̂k) in each BN . Since BN is finitelydetermined, this means that ω̂ ∈ BN for all N (just choose k so large thatBN is determined at k). But this implies that ω̂ ∈ A \

⋃n∈NAn, which is a

contradiction since A =⋃n∈NAn by assumption. 2

We are now ready for the main result:

Theorem 6.5.4 There is a complete measure P on Ω such that P (A) =λ(A) for all finitely determined sets A. The measure is unique on the σ-algebra of measurable sets.

Proof: The three lemmas above give us exactly what we need to applyCarathéodory’s theorem for semi-algebras. The uniqueness follows fromProposition 6.3.3. The details are left to the reader. 2


1. Prove Lemma 6.5.2 (Hint: If the cylinder sets A1, A2, . . . , AN have lengthK1,K2, . . . ,KN , respectively, then this is really a statement about finitesequences of length K = max{K1,K2, . . . ,KN}.)

2. Fill in the details in the proof of Theorem 6.5.4.

3. Let {Bn}n∈N be a decreasing sequence of nonempty, finitely determined sets.Show that

⋂n∈NBn 6= ∅.

4. Let E = {ω ∈ Ω |ω contains infinitely many H’s}. Show that E is notfinitely determined.

5. Let H = 1, T = 0. Show that

E = {ω ∈ Ω | limn→∞

1

n

n∑i=1

ωi =1

2}

is not finitely determined.

6.5. THE COIN TOSSING MEASURE 23

6. Show that a set is in the algebra generated by R if and only if it is finitelydetermined. Show that if E is a measurable set, then there is for any � > 0a finitely determined set D such that P (E 4D) < �

7. (You should do Exercise 6 before you attempt this one.) Assume that I is anonempty subset of N. Define an equivalence relation ∼I on Ω by

ω ∼I ω̂ ⇐⇒ ωi = ω̂i for all i ∈ I

We say that a set B ⊂ Ω is I-determined if whenever ω ∼I ω̂, then eitherω and ω̂ are both in B or both in Bc (intuitively, this means that we candetermine whether ω belongs to B by looking at the coin tosses ωi wherei ∈ I).

a) Let A be the σ-algebra of all measurable sets, and define

AI = {A ∈ A |A is I-determined}

Show that AI is a σ-algebra.

b) Assume that A ∈ AI and that C is a finitely determined set such thatA ⊂ C. Show that there is a finitely determined Ĉ ∈ AI such thatA ⊂ Ĉ ⊂ C.

c) Assume that A ∈ AI . Show that for any � > 0, there is a finitelydetermined B ∈ AI such that P (A4B) < �.

d) Assume that I, J ⊂ N are disjoint. Show that if B ∈ AI and D ∈ AJare finitely generated, then P (B∩D) = P (B)P (D). In the language ofprobability theory, B and D are independent events. (Hint: This is justfinite combinatorics and has nothing to do with measures. Note thatfinitely determined sets are in the algebra generated by R, and hencetheir measures are given directly in terms of λ.)

e) We still assume that I, J ⊂ N are disjoint. Show that if A ∈ AI andC ∈ AJ , then P (A ∩ C) = P (A)P (C). (Hint: Combine c) and d).)

f) Let In = {n, n + 1, n + 2, . . .}. A set E ⊂ Ω is called a tail event ifE ∈ In for all n ∈ N. Show that the sets E in Exercises 4 and 5 are tailevents.

g) Assume that E is a tail event and that A is finitely generated. Showthat P (A ∩ E) = P (A)P (E).

h) Assume that E is a tail event, and let En be finitely determined setssuch that P (E 4 En) < 1n (such sets exist by Exercise 6). Show thatP (E ∩ En)→ P (E) as n→∞.

i) Show that on the other hand P (E ∩ En) = P (En)P (E) → P (E)2.Conclude that P (E) = P (E)2, which means that P (E) = 0 or P (E) =1. We have proved Kolmogorov’s 0-1-law : A tail event can only haveprobability 0 or 1.


6.6 Product measures

In calculus you learned how to compute double integrals by iterated inte-gration: If R = [a, b]× [c, d] is a rectangle in the plane, then∫∫

Rf(x, y) dxdy =

∫ dc

[∫ baf(x, y) dx

]dy =

∫ ba

[∫ dcf(x, y) dy

]dx

There is a similar result in measure theory that we are now going to lookat. Starting with two measure spaces (X,A, µ) and (Y,B, ν), we shall firstconstruct a product measure space (X × Y,A ⊗ B, µ × ν) and then provethat∫

f d(µ× ν) =∫ [∫

f(x, y) dµ(x)

]dν(y) =

∫ [∫f(x, y) dν(y)

]dµ(x)

(I am putting in x’s and y’s to indicate which variables we are integrat-ing with respect to). The guiding light in the construction of the productmeasure µ× ν is that we want it to satisfy the natural product rule

µ× ν(A×B) = µ(A)ν(B)

for all A ∈ A and all B ∈ B (think of the formula for the area of an ordinaryrectangle).

As usual, we shall apply Carathéodory’s theorem for semi-algebras. Ifwe define measurable rectangles to be subsets of X×Y the form R = A×B,where A ∈ A and B ∈ B, we first observe that the class of all such sets forma semi-algebra.

Lemma 6.6.1 The collection R of measurable rectangles is a semi-algebra.

Proof: Observe first that since

(R1 × S1) ∩ (R2 × S2) = (R1 ∩R2)× (S1 ∩ S2)

the intersection of two measurable rectangles is a measurable rectangle.Moreover, since

(A×B)c = (Ac ×Bc) ∪ (Ac ×B) ∪ (A×Bc) (6.6.1)

the complement of measurable rectangle is a finite, disjoint union of mea-surable rectangles, and hence R is a semi-algebra. 2

The next step is to define a function λ : R → R+ by

λ(A×B) = µ(A)ν(B)

6.6. PRODUCT MEASURES 25

To use Carathéodory’s theorem for semi-algebras, we need to show that if

A×B =⋃n∈N

(Cn ×Dn)

is a disjoint union, then λ(A×B) =∑∞

n=1 λ(Cn ×Dn), or in other words

µ(A)ν(B) =∞∑n=1

µ(Cn)ν(Dn) (6.6.2)

(note that in this case, ∅ ∈ R and there is no need to treat finite sums sepa-rately). Observe that since µ(C) =

∫1C(x)dµ(x) and ν(D) =

∫1D(y)dν(y),

we have

µ(C)ν(D) =

∫1C(x)dµ(x)

∫1D(y)dν(x) =

=

∫ [∫1C(x)1D(y)dµ(x)

]dν(x) =

∫ [∫1C×D(x, y)dµ(x)

]dν(x)

for any two sets C ∈ A and D ∈ B. If A×B =⋃n∈N(Cn×Dn) is a disjoint

union, the Monotone Convergence Theorem 5.5.6 thus tells us that

∞∑n=1

µ(Cn)ν(Dn) = limN→∞

N∑n=1

µ(Cn)ν(Dn) =

= limN→∞

N∑n=1

∫ [∫1Cn×Dn(x, y)dµ(x)

]dν(x)

= limN→∞

∫ [∫ N∑n=1

1Cn×Dn(x, y)dµ(x)

]dν(x)

=

∫ [limN→∞

∫ N∑n=1

1Cn×Dn(x, y)dµ(x)

]dν(x)

=

∫ [∫limN→∞

N∑n=1

1Cn×Dn(x, y)dµ(x)

]dν(x)

=

∫ [∫ ∞∑n=1

1Cn×Dn(x, y)dµ(x)

]dν(x)

=

∫ [∫1C×D(x, y)dµ(x)

]dν(x) = µ(C)ν(D)

which proves equation (6.6.2).We are now ready to prove the main theorem. Remember that a measure

space is σ-finite if it is a countable union of sets of finite measure.


Theorem 6.6.2 Assume that (X,A, µ) and (Y,B, ν) are two measure spacesand let A⊗B be the σ-algebra generated by the measurable rectangles A×B,A ∈ A, B ∈ B. Then there exists a measure µ× ν on A⊗ B such that

µ× ν(A×B) = µ(A)ν(B) for all A ∈ A, B ∈ B

If µ and ν are σ-finite, this measure is unique and is called the productmeasure of µ and ν.

Proof: The existence follows from Theorem 6.3.6 as formula (6.6.2) guaran-tees that conditions (i) and (ii) hold (as ∅ = ∅ × ∅ ∈ R, we need not treatfinite and countable sums separately in this case). The uniqueness followsfrom Proposition 6.3.3 (see also Exercise 6.3.1c)). 2

Product measures can be used to construct Lebesgue measure in higherdimension. If µ is Lebesgue measure on R, the completion of the productµ×µ is the Lebesgue measure on R2. To get Lebesgue measure on R3, takea new product (µ× µ)× µ and complete it. Continuing in this way, we getLebesgue measure in all dimensions.


1. Check that formula (R1 × S1) ∩ (R2 × S2) = (R1 ∩ R2) × (S1 ∩ S2) in theproof of Lemma 6.6.1 is correct.

2. Check that formula (6.6.1) is correct and that the union is disjoint.

3. Assume that µ is the counting measure on N. Show that µ × µ is countingmeasure on N2.

4. Show that any open set in Rd is a countable union of open boxes of the form

(a1, b1)× (a2, b2)× . . .× (ad, bd)

where a1 < b1, a2 < b2, . . . , ad < bd (this can be used to show that theLebesgue measure on Rd is a completed Borel measure).

5. In this problem we shall generalize Proposition 6.4.5 from R to R2. Let µ bethe Lebesgue integral on R and let λ = µ× µ.

a) Show that if D,E are open sets in R, then D × E is open in R2.b) Assume that A × B is a measurable rectangle with µ(A), µ(B) < ∞.

Show that for any � > 0 there are open sets D,E ⊂ R such thatA×B ⊂ D × E and λ(E ×D)− λ(A×B) < �.

c) Assume that Z ⊂ R2 is measurable with λ(Z) 0, there is an open setG ⊂ R2 such that Z ⊂ G and λ(G)−λ(Z) < �.Explain why this mean that λ(G \ Z) < �.

d) Assume that Z ⊂ R2 is measurable. Show that for any � > 0, there isan open set G ⊂ R2 such that Z ⊂ G and λ(G \ Z) < �.

e) Assume that Z ⊂ R2 is measurable. Show that for any � > 0, there isa closed set F ⊂ R2 such that that Z ⊃ F and λ(Z \ F ) < �.

6.7. FUBINI’S THEOREM 27

6.7 Fubini’s Theorem

In this section we shall see how we can integrate with respect to a productmeasure; i.e. we shall prove the formulas

∫f d(µ× ν) =

∫ [∫f(x, y) dµ(x)

]dν(y) =

∫ [∫f(x, y) dν(y)

]dµ(x) (6.7.1)

mentioned in the previous section. As one might expect, these formulas donot hold for all measurable functions, and part of the challenge is to findthe right conditions. We shall prove two theorems; one (Tonelli’s Theorem)which only works for nonnegative functions, but doesn’t need any additionalconditions; and one (Fubini’s Theorem) which works for functions takingboth signs, but which has integrability conditions that might be difficult tocheck. Often the two theorems are used in combination – we use Tonelli’sTheorem to show that the conditions for Fubini’s Theorem are satisfied.

We have to begin with some technicalities. For formula (6.7.1) to makesense, the functions x 7→ f(x, y) and y 7→ f(x, y) we get by fixing one ofthe variables in the function f(x, y), have to be measurable. To simplifynotation, we write fy(x) for the function x 7→ f(x, y) and fx(y) for thefunction y 7→ f(x, y). Similarly for subsets of X × Y :

Ey = {x ∈ X | (x, y) ∈ E} and Ex = {y ∈ Y | (x, y) ∈ E}

These sets and functions are called sections of f and E, respectively (makea drawing).

Lemma 6.7.1 Assume that (X,A, µ) and (Y,B, ν) are two measure spaces,and let A⊗ B be the product σ-algebra.

(i) For any E ∈ A ⊗ B, we have Ey ∈ A and Ex ∈ B for all y ∈ Y andx ∈ X.

(ii) For any A⊗B-measurable f : X × Y → R, the sections fy and fx areA- and B-measurable, respectively, for all y ∈ Y and x ∈ X.

Proof: I only prove the lemma for the y-sections, and leave the x-sectionsto the readers.

(i) Since A⊗B is the smallest σ-algebra containing the measurable rect-angles, it clearly suffices to show that

C = {E ⊂ X × Y |Ey ∈ A}

is a σ-algebra containing the measurable rectangles. That the measurablerectangles are in C, follows from the observation

(A×B)y =

A if y ∈ B

∅ if y /∈ B


To show that C is closed under complements, just note that if E ∈ C, thenEy ∈ A, and hence (Ec)y = (Ey)c ∈ A (check this!) which means thatEc ∈ C. Similarly for countable unions: If for each n ∈ N, En ∈ C, then(En)

y ∈ A for all n, and hence⋃n∈N(En)

y = (⋃n∈NEn)

y ∈ A (check this!)which means that

⋃n∈NEn ∈ C.

(ii) We need to check that (fy)−1(I) ∈ A for all intervals of the form[−∞, r). But this follows from (i) and the measurability of f since

(fy)−1(I) = (f−1(I))y

(check this!) 2

There is another measurability problem in formula (6.7.1): We need toknow that the integrated sections

y 7→∫f(x, y) dµ(x) =

∫fy(x) dµ(x)

and

x 7→∫f(x, y) dν(y) =

∫fx(y) dν(y)

are measurable. This is a more complicated question, and we shall needa quite useful and subtle result known as the Monotone Class Theorem.A monotone class of subsets of a set Z is a collection M of subsets of Zthat is closed under increasing countable unions and decreasing countableinteresections. More precisely:

(i) If E1 ⊂ E2 ⊂ . . . ⊂ En ⊂ . . . are in M, then⋃n∈N ∈M

(ii) If E1 ⊃ E2 ⊃ . . . ⊃ En ⊃ . . . are in M, then⋂n∈N ∈M

All σ-algebras are monotone classes, but a monotone class need not be a σ-algebra. If R is a collection of subsets of Z, there is (by the usual argument)a smallest monotone class containing R. It is called the monotone classgenerated by R.

Theorem 6.7.2 (Monotone Class Theorem) Assume that Z is a nonemptyset and that A is an algebra of subsets of Z. Then σ-algebra and the mono-tone class generated by A coincide.

Proof: Let C be the σ-algebra and M the monotone class generated by A.Since all σ-algebras are monotone classes, we must have M⊂ C.

To prove the opposite inclusion, we show thatM is a σ-algebra. Observethat it suffices to prove that M is an algebra as closure under countableunions will then take care of itself: If {En} is a sequence from M, thesets Fn = E1 ∪ E2 ∪ . . . ∪ En are in M since M is an algebra, and hence


⋃n∈NEn =

⋃n∈N Fn ∈ M since M is closed under increasing countable

unions.

To prove that M is an algebra, we use a trick. For each M ∈M define

M(M) = {F ∈M|F \M,F \M,F ∩M ∈M}

It is not hard to check that sinceM is a monotone class, so isM(M). Notealso that by symmetry, N ∈M(M)⇐⇒M ∈M(N).

Our aim is to prove thatM(M) =M for all M ∈M. This would meanthat the intersection and difference between any two sets in M are in M,and since Z ∈ M (because Z ∈ A ⊂ M), we may conclude that M is analgebra.

To show that M(M) = M for all M ∈ M, pick a set A ∈ A. SinceA is an algebra, we have A ⊂ M(A). Since M is the smallest monotoneclass containing A, this means that M(A) =M, and hence M ∈M(A) forall M ∈ M. By symmetry, A ∈ M(M) for all M ∈ M. Since A was anarbitrary element in A, we have A ⊂M(M) for all M ∈ M, and using theminimality of M again, we see that M(M) =M for all M ∈M. 2

The advantage of the Monotone Class Theorem is that it is often mucheasier to prove that families are closed under monotone unions and intersec-tions than under arbitrary unions and intersections, especially when there’sa measure involved in the definition. The next lemma is a typical case. Notethe σ-finiteness condition; Exercise 9 shows that the result does not alwayshold without it.

Lemma 6.7.3 Let (X,A, µ) and (Y,B, ν) be two σ-finite measure spaces,and assume that E ⊂ X × Y is A ⊗ B-measurable. Then the functionsx 7→ ν(Ex) and y 7→ µ(Ey) are A- and B-measurable, respectively, and∫

ν(Ex) dµ(x) =

∫µ(Ey) dν(y) = µ× ν(E)

Proof: We shall prove the part about the x-sections Ex and leave the (sim-ilar) proof for y-sections to the readers. We shall first carry out the prooffor finite measure spaces, i.e. we assume that µ(X), ν(Y )


To show that any measurable rectangle E = A × B belongs to C, justobserve that

ν(Ex) =

ν(B) if x ∈ A

0 if x /∈ A

and that∫ν(Ex) dµ(x) =

∫A ν(B) dµ = µ(A)ν(B) = µ× ν(E).

A set F in the algebra generated by the measurable rectangles, is adisjoint union F =

⋃ni=1Ei of measurable rectangles, and since ν(Fx) =∑n

i=1 ν((Ei)x), the function x 7→ ν(Fx) is A-measurable (a sum of measur-able functions is measurable) and

∫ν(Fx) dµ(x) =

∫ ∑ni=1 ν((Ei)x)dµ(x) =∑n

i=1 µ× ν(Ei) = µ× ν(F ). Hence F ∈ C.To show that C is monotone class, assume that {En} is an increasing

sequence of sets in C. Let E =⋃n∈NEn, and note that Ex =

⋃∞n=1(En)x.

By continuity of measure, ν(Ex) = limn→∞ ν((En)x), and hence x 7→ ν(Ex)is measurable as the limit of a sequence of measurable functions. Moreover,by the Monotone Convergence Theorem and continuity of measures,∫

ν(Ex) dµ(x) =

∫limn→∞

ν((En)x) dµ(x)) = limn→∞

∫ν((En)x) dµ(x)) =

= limn→∞

µ× ν(En) = µ× ν(E)

This means that E ∈ C.We must also check monotone intersections. Assume that {En} is a

decreasing sequence of sets in C. Let E =⋂n∈NEn, and note that Ex =⋂∞

n=1(En)x. By continuity of measure (here we are using that ν is a finitemeasure), ν(Ex) = limn→∞ ν((En)x), and hence x 7→ ν(Ex) is measurable.Moreover, using the Dominated Convergenge Theorem (since the measurespace is finite, we can use the function that is constant 1 as the dominatingfunction), we see that∫

ν(Ex) dµ(x) =

∫limn→∞

ν((En)x) dµ(x)) = limn→∞

∫ν((En)x) dµ(x)) =

= limn→∞

µ× ν(En) = µ× ν(E)

and hence E ∈ C.Since this shows that C is a monotone class and hence a σ-algebra con-

tainingA×B, we have proved the lemma for finite measure spaces. To extendit to σ-finite spaces, let {Xn} and {Yn} be increasing sequence of subsetsof X and Y of finite measure such that X =

⋃n∈NXn and Y =

⋃n∈N Yn.

If E is a A⊗ B-measurable subset of X × Y , it follows from what we havealready proved that x 7→ ν((E ∩ (Xn × Yn)x) is measurable and∫

ν((E ∩ (Xn × Yn)x) dµ(x) = µ× ν(E ∩ (Xn × Yn))


The lemma for σ-finite spaces now follows from the Monotone ConvergenceTheorem and continuity of measures. 2

Remark: The proof above illustrates a useful technique. To prove that allsets in the σ-algebra F generated by a family R satisfies a certain propertyP , we prove

(i) All sets in R has property P .

(ii) The sets with property P form a σ-algebra G.

Then F ⊂ G (since F is the smallest σ-algebra containing R), and hence allsets in F has property P .

We are now ready to prove the first of our main theorems.

Theorem 6.7.4 (Tonelli’s Theorem) Let (X,A, µ) and (Y,B, ν) be twoσ-finite measure spaces, and assume that f : X ×Y → R+ is a nonnegative,A ⊗ B-measurable function. Then the functions x 7→

∫f(x, y) dν(y) and

y 7→∫f(x, y)dµ(x) are A- and B-measurable, respectively, and∫

f d(µ× ν) =∫ [∫

f(x, y) dν(y)

]dµ(x) =

∫ [∫f(x, y) dµ(x)

]dν(y)

Proof: We prove the first equality and leave the second to the reader. Noticefirst that if f = 1E is an indicator function, then by the lemma∫

1E d(µ× ν) = µ× ν(E) =∫ν(Ex) dµ(x) =

∫ [∫1E(x, y) dν(y)

]dµ(x)

which proves the theorem for indicator functions. By linearity, it also holdsfor nonnegative, simple functions.

For the general case, let {fn} be an increasing sequence of nonnegativesimple functions converging pointwise to f . The functions x 7→

∫fn(x, y) dν(y)

increase to x 7→∫f(x, y) dν(y) by the Monotone Convergence Theorem.

Hence the latter function is measurable (as the limit of a sequence of mea-surable functions), and using the Monotone Convergence Theorem again,we get ∫

f(x, y) d(µ× ν) = limn→∞

∫fn(x, y) d(µ× ν) =

= limn→∞

∫ [∫fn(x, y) dν(y)

]dµ(x) =

∫ [limn→∞

∫fn(x, y) dν(y)

]dµ(x) =

=

∫ [∫limn→∞

fn(x, y) dν(y)

]dµ(x) =

∫ [∫f(x, y) dν(y)

]dµ(x)

2

Fubini’s Theorem is now an easy application of Tonelli’s Theorem.


Theorem 6.7.5 (Fubini’s Theorem) Let (X,A, µ) and (Y,B, ν) be twoσ-finite measure spaces, and assume that f : X×Y → R is µ×ν-integrable.Then the functions fx and f

y are integrable for almost all x and y, and theintegrated functions x 7→

∫f(x, y) dν(y) and y 7→

∫f(x, y)dµ(x) are µ- and

ν-integrable, respectively. Moreover,∫f d(µ× ν) =

∫ [∫f(x, y) dν(y)

]dµ(x) =

∫ [∫f(x, y) dµ(x)

]dν(y)

Proof: Since f is integrable, it splits as the difference f = f+ − f− betweentwo nonnegative, integrable functions. Applying Tonelli’s Theorem to |f |,we get ∫ [∫

|f(x, y)| dν(y)]dµ(x) =

∫ [∫|f(x, y)| dµ(x)

]dν(y) =

=

∫|f | d(µ× ν)


Theorem 6.7.6 Let (X,A, µ) and (Y,B, ν) be two complete, σ-finite mea-sure spaces, and assume that f : X × Y → R is µ× ν-measurable, whereµ× ν is the completion of the product measure. Then the functions fxand fy are measurable for almost all x and y, and the integrated functionsx 7→


∫f(x, y)dµ(x) are measurable as well. More-

over

(i) (Tonelli’s Theorem for Completed Measures) If f is nonnega-tive,∫f dµ× ν =

∫ [∫f(x, y) dν(y)

]dµ(x) =

∫ [∫f(x, y) dµ(x)

]dν(y)

(ii) (Fubini’s Theorem for Completed Measures) If f is integrable,the functions fx and f

y are integrable for almost all x and y, and theintegrated functions x 7→


∫f(x, y)dµ(x) are

µ- and ν-integrable, respectively. Moreover,∫f dµ× ν =

∫ [∫f(x, y) dν(y)

]dµ(x) =

∫ [∫f(x, y) dµ(x)

]dν(y)


1. Show that (Ec)y = (Ey)c (here c i referring to complements).

2. Show that (⋃n∈NEn)

y =⋃n∈NE

yn.

3. Show that (fy)−1(I) = (f−1(I))y.

4. Show thatM = {M ⊂ R | 0 ∈M}

is a monotone class, but not a σ-algebra.

5. Show that the setsM(M) in the proof of the Monotone Class Theorem reallyare monotone classes.

6. In this problem µ Lebesgue measure on R, while ν is counting measure onN. Let λ = µ× ν be the product measure and let

f : R× N→ R

be given by

f(x, n) =1

1 + (2nx)2

Compute∫f dλ. Remember that

∫1

1+u2 du = arctanu+ C.

7. Let f : [0, 1] × [0, 1] → R be defined by f(x, y) = x−y(x+y)3 (the expressiondoesn’t make sense for x = y = 0, and you may give the function whatevervalue you want at that point). Show by computing the integrals that∫ 1

0

[∫ 10

f(x, y) dx

]dy = −1

2


and ∫ 10

[∫ 10

f(x, y) dy

]dx =

1

2

(you may want to use that f(x, y) = 1x2+y2 −2y

(x+y)3 in the first integral and

argue by symmetry in the second one). Let µ be the Lebesgue measure on[0, 1] and λ = µ× µ. Is f integrable with respect to λ?

8. Define f : [0,∞)× [0,∞) by f(x, y) = xe−x2(1+y2).

a) Show by performing the integrations that∫ ∞0

[∫ ∞0

f(x, y) dx

]dy =

π

4

b) Use Tonelli’s theorem to show that∫∞0

[∫∞0f(x, y) dy

]dx = π4

c) Make the substitution u = xy in the inner integral of∫ ∞0

[∫ ∞0

f(x, y) dy

]dx =

∫ ∞0

[∫ ∞0

xe−x2(1+y2)) dy

]dx

and show that∫∞0

[∫∞0f(x, y) dy

]dx =

(∫∞0e−u

2

du)2

.

d) Conclude that∫∞0e−u

2

du =√π2 .

9. Let X = Y = [0, 1], let µ be the Lebesgue measure on X (this is just therestriction of the Lebesgue measure on R to [0, 1]) and let ν be the countingmeasure. Let E = {(x, y) ∈ X × Y |x = y}, and show that

∫ν(Ex) dµ(x),∫

µ(Ey) dν(y), and µ× ν(E) are all different (compare lemma 6.7.3).

10. Assume that X = Y = N and that µ = ν is the counting measure. Letf : X × Y → R be defined by

f(x, y) =

1 if x = y

−1 if x = y + 1

0 otherwise

Show that∫f dµ×ν =∞, but that the iterated integrals

∫ [∫f(x, y) dµ(x)

]dν(u)

and∫ [∫

f(x, y) dν(y)]dµ(x) are both finite, but unequal.

11. In this exercise, we shall sketch the proof of Theorem 6.7.6, and we assumethat (X,A, µ) and (Y,B, ν) are as in that theorem.

a) Assume that E ∈ A⊗ B and that µ× ν(E) = 0. Show that µ(Ey) = 0for ν-almost all y and that ν(Ex) = 0 for µ-almost all x.

b) Assume that N is an µ × ν-null set, i.e. there is an E ∈ A ⊗ B suchthat N ⊂ E and µ × ν(E) = 0. Show that for ν-almost all y, Ny isµ-measurable and µ(Ny) = 0. Show also that for µ-almost all x, Nx isν-measurable and ν(Nx) = 0. (Here you need to use that the originalmeasure spaces are complete).


c) Assume that D ⊂ X × Y is in the completion of A⊗B with respect toµ × ν. Show that for ν-almost all y, Dy is µ-measurable, and that forµ-almost all x, Dx is ν-measurable (use that by Theorem 5.2.5 D canbe written as a disjoint union D = E ∪N , where E ∈ A ⊗ B and N isa null set).

d) Let D be as above. Show that the functions x 7→ ν(Dx) and y 7→µ(Dy) are A- and B-measurable, respectively (define ν(Dx) and µ(Dy)arbitrarily on the sets of measure zero where Dx and D

y fail to bemeasurable), and show that∫

ν(Dx) dµ(x) =

∫µ(Dy) dν(y) = µ× ν(D)

e) Prove Theorem 6.7.6 (this is just checking that the arguments that gotus from Lemma 6.7.3 to Theorems 6.7.4 and 6.7.5, still works in thenew setting).

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