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MEASURE AND INTEGRATIOND-MATH, Spring Semester 20
Francesca Da Lio1
May 15, 2020
1Department of Mathematics, ETH Zurich, Ramistrasse 101, 8092 Zurich, Switzerland.
Abstract
The aim of this course is to provide notions of abstract measureand integration which are more general and robust than the classical
notion of Jordan measure and Riemann integral. For a presentationof Jordan measure and Riemann integral we suggest to look at the
lecture notes of Analysis I & II by Michael Struwe(https://people.math.ethz.ch/∼struwe/Skripten).
The concept of measure of a general set in Rn originated from theclassical notion of volume of a cube in Rn as product of the lengthof its sides. Starting from this, by a covering process one can assign
to any subset a nonnegative number which “quantifies its volume”.Such an association leads to the introduction of a set function called
“Lebesgue measure” which is defined over all subsets of Rn. Althoughthe Lebesgue measure was initially developed in Euclidean spaces, the
theory is independent of the geometry of the background space andapplies to abstract spaces as well.
Measure theory is now applied to most areas of mathematics suchas functional analysis, geometry, probability, dynamical systems andother domains of mathematics.
Some References
• Robert G. Bartle. The elements of Integration and Lebesgue Mea-sure, Wiley Classics Library, 1995.
• Piermarco Cannarsa and Teresa D’Aprile. Introduction to mea-sure theory and functional analysis, Vol. 89, Springer, 2015.
• Lawrence Craig Evans and Ronald F. Gariepy. Measure theoryand fine properties of functions, revised edition, CRC Press, 2015.
• Lecture Notes by Prof. Michael Struwe, FS 13,https://people.math.ethz.ch/∼struwe/Skripten/AnalysisIII-FS2013-
12-9-13.pdf
Contents
1 Measure Spaces 1
1.1 Algebras and σ-algebras of sets . . . . . . . . . . . . . 11.1.1 Notation and Preliminaries . . . . . . . . . . . . 1
1.1.2 Algebras and σ-Algebras . . . . . . . . . . . . . 31.2 Measures . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.1 Additive and σ-Additive Functions . . . . . . . 61.2.2 Measures and Measure Spaces . . . . . . . . . . 81.2.3 Construction of Measures . . . . . . . . . . . . 14
1.3 Lebesgue Measure . . . . . . . . . . . . . . . . . . . . . 191.4 Comparison between Lebesgue and Jordan measure . . 25
1.5 An example of a non-measurable set . . . . . . . . . . 281.6 An uncountable Ln-null set: the Cantor Triadic Set . . 30
1.7 Lebesgue-Stieltjes Measure . . . . . . . . . . . . . . . 321.8 Hausdorff Measures . . . . . . . . . . . . . . . . . . . 38
1.8.1 Examples of non-integer Hausdorff-dimension sets 441.9 Radon Measure . . . . . . . . . . . . . . . . . . . . . . 47
2 Measurable Functions 51
2.1 Inverse Image of a Function . . . . . . . . . . . . . . . 512.2 Definition and Basic Properties . . . . . . . . . . . . . 51
2.3 Lusin’s and Egoroff’s Theorems . . . . . . . . . . . . . 572.4 Convergence in Measure . . . . . . . . . . . . . . . . . 61
3 Integration 643.1 Definitions and Basic Properties . . . . . . . . . . . . . 64
3.2 Comparison between Lebesgue and Riemann-Integral . 753.2.1 Review of Riemann Integral . . . . . . . . . . . 75
3.3 Convergence Results . . . . . . . . . . . . . . . . . . . 78
2
3.4 Application: Differentiation of under integral sign . . . 823.5 Absolute Continuity of Integrals . . . . . . . . . . . . . 83
3.6 Vitali’s Theorem . . . . . . . . . . . . . . . . . . . . . 853.7 The Lp(Ω, µ) spaces, 1 ≤ p ≤ ∞ . . . . . . . . . . . . . 91
4 Product Measures and Multiple Integrals 107
4.1 Fubini’s Theorem . . . . . . . . . . . . . . . . . . . . . 1074.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . 118
4.3 Convolution . . . . . . . . . . . . . . . . . . . . . . . . 1204.4 Application:Solution of the Laplace Equation . . . . . . 126
3
Chapter 1
Measure Spaces
In this chapter we develop measure theory from an abstract point of
view in order to construct a large variety of measures on a genericspace X.
In the particular case of X = Rn, a special role is played by Radonmeasures which have important regularity properties.
1.1 Algebras and σ-algebras of sets
1.1.1 Notation and Preliminaries
We denote by X a nonempty set, by P(X) or 2X the set of all subsets
of X and ∅ the empty set. For any subset A of X we denote by Ac itscomplement, i.e.
Ac = x ∈ X |x /∈ A . (1.1.1)
For any A,B ∈ P(X) we set
A \B = A ∩Bc. (1.1.2)
Let (An)n be a sequence in P(X). The De Morgan’s identity holds
( ∞⋃
n=1
An
)c
=
∞⋂
n=1
Acn . (1.1.3)
1
We define1
lim supn→∞
An =∞⋂
n=1
∞⋃
m=n
Am
lim infn→∞
An =
∞⋃
n=1
∞⋂
m=n
Am .
(1.1.4)
If L = lim supn→∞An = lim infn→∞An, then we set L = limn→∞An
and we say that (An)n converges to L.
Remark 1.1.1. Note that:
1. lim infn→∞
An ⊆ lim supn→∞
An.
2. If (An)n is increasing (An ⊆ An+1, n ∈ N), then
limn→∞
An =
∞⋃
n=1
An . (1.1.5)
Proof. Since An is increasing we have
∞⋃
m=1
Am ⊆∞⋃
m=n
Am, ∀n ∈ N .
Therefore
lim supn→∞
An =∞⋂
n=1
∞⋃
m=n
Am =∞⋃
m=1
Am.
On the other hand for every n ∈ N we have∞⋂
m=n= An and
therefore
lim infn→∞
An =
∞⋃
m=1
Am.
1We can “think” about the limit supremum and infimum for sets in the following way:
lim supn→∞
An = x ∈ An, for infinitely many n
lim infn→∞
An = x ∈ An, for all but finitely many n
2
3. If (An)n is decreasing (An ⊇ An+1, n ∈ N), then
limn→∞
An =
∞⋂
n=1
An . (1.1.6)
1.1.2 Algebras and σ-Algebras
Definition 1.1.2. Let A ⊆ P(X), A is called an algebra in XXXif
a) X ∈ A.
b) A,B ∈ A =⇒ A ∪B ∈ A.
c) A ∈ A =⇒ Ac ∈ A.
Remark 1.1.3. If A is an algebra and A,B ∈ A, then A ∩ B andA \ B belong to A. Therefore, the symmetric difference A∆B =
(A \ B) ∪ (B \ A) belongs to A as well. A is also stable under finiteunions and intersections
A1, . . . , An ∈ A =⇒A1 ∪ · · · ∪ An ∈ AA1 ∩ · · · ∩ An ∈ A .
Definition 1.1.4. An algebra E ⊆ P(X) is called a σσσ-algebra iffor any sequence (An)n of elements in E , we have
⋃∞n=1An ∈ E .
Remark 1.1.5. If E is a σ-algebra inX and (An)n ⊆ E , then⋂∞n=1An ∈
E by De Morgan’s identity (1.1.3). Moreover one can show (easy ex-ercise!) that:
lim infn→∞
An ∈ E , lim supn→∞
An ∈ E . (1.1.7)
Exercise 1.1.6.
1. P(X) and A = ∅, X are σ-algebras. P(X) is the largest σ-
algebra in X and A the smallest.
3
2. In X = [0, 1), the class A0 consisting of ∅ and of all finite unionsof half-closed intervals of the form
A =n⋃
i=1
[ai, bi), 0 ≤ ai < bi ≤ ai+1 < bi+1 . . . < bn ≤ 1
is an algebra in [0, 1). Indeed,
Ac = [0, a1) ∪ [b1, a2) ∪ · · · ∪ [bn, 1) ∈ A0,
moreover, A0 is stable under finite unions. It suffices to observethat the union of two (not necessarily disjoint) intervals of the
form [a, b) and [c, d) belongs to A0
3. In an infinite set X, consider the class A.A = A ∈ P(X) |A is finite, or Ac is finite, (1.1.8)
A is an algebra. The only point that needs to be checked is that
A is stable under a finite union. Let A,B ∈ A. If A,B are bothfinite, then so is A ∪ B. In all other cases, (A ∪B)c is finite. Abut not a σ-algebra.
4. In an uncountable set X, consider the class
E = A ∈ P(X) | A is countable or Ac is countable. (1.1.9)
Then E is a σ-algebra which is strictly smaller than P(X) (here“countable” stands for “at most countable”).
5. Let K ⊆ P(X). Verify that the intersection of σ-algebras con-taining K is a σ-algebra and it is the smallest σ-algebra includingK.
Definition 1.1.7. Let K ⊆ P(X). The intersection of all σ-
algebras including K is called the σσσ-algebra generated by Kand will be denoted by σ(K).
4
Example 1.1.8.1. Let (X, d) be a metric space. The σ-algebra generated by all
open sets of X is called the Borel σσσ-algebra of XXX and is denoted byB(X). The elements of B(X) are called Borel sets.
2. Let X = R and I be the set of all half-closed intervals [a, b)
with a ≤ b. We show that σ(I) = B(R).“σ(I) ⊆ B(R)” : this follows from the fact that
[a, b) =∞⋂
n=1
(a− 1
n, b
).
“ B(R) ⊆ σ(I)”: Let V be an open set in R, then it is a countableunion of some family of open intervals. Indeed, let (qk) be the sequence
including all rational numbers of V and denote by Ik the largest openinterval contained in V and containing qk. We clearly have ∪∞
k=0Ik ⊂V , but also the opposite inclusion holds: it suffices to consider, forany x ∈ V , r > 0 such that (x − r, x + r) ⊂ V, and k such that
qk ∈ (x− r, x+ r) to obtain that (x− r, x+ r) ⊂ Ik, by the maximalityof Ik and then x ∈ Ik.
Any open interval (a, b) can be represented as
(a, b) =∞⋃
n=n0
[a+
1
n, b) ( 1
n0< b− a
).
Therefore V ∈ σ(I). An analogous argument proves that B(R) isgenerated by half-closed intervals (a, b] with a ≤ b by open intervals,
by closed intervals and even by open and closed half-lines.
5
1.2 Measures
1.2.1 Additive and σ-Additive Functions
Definition 1.2.1. Let A ⊆ P(X) be an algebra. Let µ: A →[0,+∞] be such that µ(∅) = 0.
i) We say that µµµ is additive if for any finite family
A1, . . . , An ∈ A of mutually disjoint sets, we have
µ( n⋃
k=1
Ak
)=
n∑k=1
µ(Ak). (1.2.1)
ii) We say that µµµ is σσσ-additive if, for any sequence (An)n ⊆ Aof mutually disjoint sets such that
⋃∞k=1Ak ∈ A, we have
µ( ∞⋃
k=1
Ak
)=
∞∑k=1
µ(Ak). (1.2.2)
Remark 1.2.2.
1. Any σ-additive function on A is also additive.
2. If µ is additive on A, it is monotone with respect to the inclusion:
indeed, if A,B ∈ A and B ⊆ A, then µ(A) = µ(B) + µ(A \ B).Therefore, µ(A) ≥ µ(B).
3. Let µ be additive on A and let (An)n ⊆ A be mutually disjointsets such that
⋃∞k=1Ak ∈ A. Then
µ( ∞⋃
k=1
Ak
)≥ µ
( n⋃
k=1
Ak
)=
n∑k=1
µ(Ak) ∀n ∈ N, (1.2.3)
therefore
µ( ∞⋃
k=1
Ak
)≥
∞∑k=1
µ(Ak). (1.2.4)
6
4. Any σ-additive function on µ on A is σσσ-subadditive, that is, forany sequence (An)n in A such that
⋃∞k=1Ak ∈ A
µ( ∞⋃
k=1
Ak
)≤
∞∑k=1
µ(Ak).
Indeed, let (An)n ⊆ A be such that⋃∞
n=1An ∈ A and define
B1 = A1 and Bn = An \ ⋃n−1k=1 Ak ∈ A. Then Bn are mutually
disjoint, µ(Bn) ≤ µ(An) for all n ∈ N and
∞⋃
i=1
Bi =∞⋃
i=1
Ai.
Therefore
µ( ∞⋃
i=1
Ai
)= µ
( ∞⋃
i=1
Bi
)=
∞∑i=1
µ(Bi) ≤∞∑i=1
µ(Ai). (1.2.5)
5. In view of points 3 and 4 an additive function is σ-additive if and only if it is countably subadditive.
Definition 1.2.3. A σ-additive function µ on an algebra A ⊆P(X) is said to be
• finite, if µ(X) < +∞;
• σσσ-finite, if there exists a sequence (An)n ⊆ A such that
∞⋃
n=1
An = X and µ(An) < +∞ ∀n ∈ N.
Exercise 1.2.4. In X = N consider the algebra
A =A ∈ P(X) |A is finite or Ac is finite
. (1.2.6)
• The function µ: A → [0,+∞] defined as
µ(A) =
#A if A is finite
∞ if Ac is finite(1.2.7)
is σ-additive (#A = number of elements of A).
7
• The function ν: A→ [0,+∞] defined as
ν(A) =
∑n∈A
1
2nif A is finite
∞ if Ac is finite
(1.2.8)
is additive but not σ-additive.
1.2.2 Measures and Measure Spaces
Let X be a set.
Definition 1.2.5. A mapping µ: P(X) → [0,+∞] is called a
measure on X if
i) µ(∅) = 0 and
ii) µ(A) ≤∞∑k=1
µ(Ak) whenever A ⊆∞⋃k=1
Ak.
Warning:
Most texts call such a mapping µ an outer measure reservingthe name measure for µ restricted to the collection of µ-measurablesubsets of X (see below).
Remark 1.2.6. The condition ii) in Definition 1.2.5 implies the sub-additivity and the monotonicity.
Definition 1.2.7. Let µ be a measure on X and A ⊆ X, then µrestricted to A written
µ A
is the measure defined by
(µ A)(B) = µ(A ∩B) ∀B ⊆ X. (1.2.9)
8
Definition 1.2.8. (Caratheodory criterion of measurabil-ity)
A ⊆ X is called µµµ-measurable if for all B ⊆ X
µ(B) = µ(B ∩ A) + µ(B \A).µ(B) = µ(B ∩ A) + µ(B \ A).µ(B) = µ(B ∩ A) + µ(B \A). (1.2.10)
Remark 1.2.9.1) If µ(A) = 0, then A is µ-measurable.Indeed, let B ⊆ X. By the subadditivity, we have
µ(B) ≤ µ(B ∩ A) + µ(B \A). (1.2.11)
On the other hand, we have
µ(B ∩ A) + µ(B \ A) ≤ µ(A) + µ(B) = µ(B), (1.2.12)
which yields the claim.
2) By the subadditivity the condition (1.2.10) is equivalent to
µ(B) ≥ µ(B ∩ A) + µ (B \ A).
Theorem 1.2.10. Let µ: P(X) → [0,+∞] be a measure. Then
the setΣ = A ⊆ X: A is µ-measurable
is a σ-algebra.
Proof. We proceed by steps.
i) X ∈ Σ: ∀B ⊆ X, it holds
µ(B ∩X) + µ(B \X) = µ(B) + µ(∅) = µ(B). (1.2.13)
ii) A ∈ Σ =⇒ Ac ∈ Σ: ∀B ∈ X it holds
B ∩ Ac = B \A, B \Ac = B ∩ A. (1.2.14)
Therefore
µ(B ∩Ac) + µ(B \Ac) = µ(B \A) + µ(B ∩A) = µ(B). (1.2.15)
9
iii) Let A1, . . . , Am ∈ Σ. By induction on m we show that⋃m
k=1Ak ∈Σ.
m = 1 is trivial.
(induction step) m− 1 =⇒ m: Let A =m−1⋃k=1
Ak.
By inductive hypothesis, we assume A is µ-measurable. Thus, for∀B ⊆ X, we have
µ(B) = µ(B ∩ A) + µ(B \A). (1.2.16)
Since Am is also µ-measurable, we have
µ(B \ A) = µ((B \A) ∩ Am
)+ µ
((B \A) \ Am
)
= µ((B \A) ∩ Am
)+ µ
(B \ (A ∪ Am)
).
(1.2.17)
Now we observe that
B ∩ (A ∪ Am) = (B ∩ A) ∪ [(B \A) ∩ Am].
Thus by (1.2.17)
µ(B) = µ(B ∩ A) + µ(B \ A)= µ(B ∩ A) + µ
((B \A) ∩ Am
)+ µ
((B \A) \Am
)
≥µ(B ∩ (A ∪ Am)
)+ µ
(B \ (A ∪ Am)
)
(1.2.18)
From Remark 1.2.9 it follows that A ∪ Am =m⋃k=1
Ak ∈ Σ .
iv) Let (Ak)k∈N ⊂ Σ. We show that A =⋃∞
k=1Ak is µ-measurable.We may suppose that Ak ∩ Aℓ = ∅, k 6= ℓ, otherwise we consider
A1 = A1, A2 = A2 \A1, A3 = A3 \ (A1 ∪ A2)
An = An \( n−1⋃
k=1
Ak
).
(1.2.19)
10
We can note that ∞⋃
k=1
Ak =∞⋃
k=1
Ak.
Since each Aℓ is µ-measurable, we have
µ(B ∩
m⋃
k=1
Ak
)= µ
((B ∩
m⋃
k=1
Ak
)∩ Am
)+ µ
((B ∩
m⋃
k=1
Ak
)\Am
)
= µ(B ∩ Am) + µ(B ∩
m−1⋃
k=1
Ak
)(1.2.20)
=m∑k=1
µ(B ∩ Ak).
By using the monotonicity of µ, we get
µ(B) = µ(B ∩
m⋃
k=1
Ak
)+ µ
(B \
m⋃
k=1
Ak
)
≥m∑k=1
µ(B ∩ Ak) + µ(B \ A).(1.2.21)
We let m→ +∞:
µ(B) ≥∞∑k=1
µ(B ∩ Ak) + µ(B \ A)
≥ µ(B ∩ A) + µ(B \ A)
by the σ-subadditivity. Therefore A ∈ Σ, by the Remark 1.2.9
Definition 1.2.11. If µ: P(X) → [0,+∞] is a measure on X,Σ is the σ-algebra of µ-measurable sets, then (X,Σ, µ) is called a
Measure Space.
11
Exercise 1.2.12.
1. Let x ∈ X. Define, for every A ∈ P(X)
δx(A) =
1 x ∈ A0 x /∈ A .
(1.2.22)
Then δx is a measure on P(X), called the Dirac measure atxxx.
2. For every A ∈ P(X) we define
µ(A) =
#A if A is finite+∞ otherwise ;
(1.2.23)
µ is a measure on P(X) and it is called the counting measure.Every A is µ-measurable.
3. µ: P(X) → [0,+∞], µ(∅) = 0, µ(A) = 1, A 6= ∅. In this case,A ⊆ X is µ-measurable ⇐⇒ A = ∅ or A = X.
Theorem 1.2.13. Let (X,Σ, µ) be a Measure Space and let
Ak ∈ Σ, k ∈ N. Then the following conditions hold:
i) Aj ∩ Aℓ = ∅ (j 6= ℓ) =⇒ µ( ∞⋃
k=1
Ak
)=
∞∑k=1
µ(Ak), ( σσσ-
additivity.
ii) A1 ⊂ A2 ⊂ · · · ⊂ Ak ⊂ Ak+1 ⊂ . . . =⇒ µ( ∞⋃
k=1
Ak
)=
limk→+∞
µ(Ak))
iii) A1 ⊃ A2 ⊃ · · · ⊃ Ak ⊃ Ak ⊃ . . . , µ(A1) < +∞. Then
µ( ∞⋂
k=1
Ak
)= lim
k→+∞µ(Ak).
Proof.
12
i) In the proof of Theorem 1.2.10 we showed that ∀B ⊆ X:
µ(B ∩
m⋃
k=1
Ak
)=
m∑k=1
µ(B ∩ Ak) ∀m. (1.2.24)
By taking B = X, we get in particular
µ( m⋃
k=1
Ak
)=
m∑k=1
µ(Ak). (1.2.25)
By using the monotonicity and the σ-subadditivity, we get
µ( ∞⋃
k=1
Ak
)≥ lim
m→+∞µ( m⋃
k=1
Ak
)
= limm→+∞
m∑k=1
µ(Ak)
=∞∑k=1
µ(Ak) ≥ µ( ∞⋃
k=1
Ak
)(1.2.26)
ii) Consider the disjoint union Ak = Ak \Ak−1, A1 = A1:
A1
A2
A3
Fig. 1.1
By using i), we get
µ(A) =∞∑k=1
µ(Ak) = limm→+∞
m∑k=1
µ(Ak)
= limm→+∞
m∑k=2
µ(Ak)− µ(Ak−1) + µ(A1)
= limm→+∞
µ(Am).
(1.2.27)
13
iii) One considers Ak = A1 \ Ak, k ∈ N. We have
∅ = A1 ⊆ A2 ⊂ . . . . (1.2.28)
We observe that ∀k ∈ Nµ(A1) = µ(Ak) + µ(Ak).
Thus
µ(A1)− limn→+∞
µ(Ak) = limk→+∞
µ(Ak)
=ii)µ( ∞⋃
k=1
Ak
)= µ
(A1 −
∞⋂
k=1
Ak
)
= µ(A1) \ µ( ∞⋂
k=1
Ak
).
(1.2.29)
We can conclude.
Remark 1.2.14. If µ(A1) = +∞, then iii) in Theorem 1.2.13
may be false. Consider
µ(A) =
#A if A is finite
+∞ .
Take An := m ∈ N : m ≥ n. We have µ(A1) = +∞ and
µ( ∞⋂
k=1
Ak
)= 0 6= lim
k→+∞µ(Ak).
1.2.3 Construction of Measures
Let X 6= ∅.
Definition 1.2.15. Let K ⊆ P(X). K is called a covering of X
if
i) ∅ ∈ K.
ii) ∃(Kj)j∈N ⊆ K : X =∞⋃j=1
Kj.
14
Example 1.2.16. The open intervals I =∏n
k=1(ak, bk), ak ≤ bk ∈ Rare a covering of X = Rn. Each algebra A ⊆ P(X) is a coveringfor X, since ∅, X ∈ A.
Theorem 1.2.17. Let K be a covering for X, λ: K → [0,+∞]with λ(∅) = 0. Then
µ(A) = inf ∞∑
j=1
λ(Kj) : Kj ∈ K, A ⊆∞⋃j=1
Kj
(1.2.30)
is a measure on X.
Proof of Theorem 1.2.17. We observe that µ ≥ 0, µ(∅) = 0. Weshow that it is σ-subadditive. Let A ⊆ ⋃∞
k=1Ak. The inequality
µ(A) ≤ ∑∞k=1 µ(Ak) is trivial if the right-hand side is infinite. There-
fore assume that∑∞
k=1 µ(Ak) is finite. For any k ∈ N and any ε > 0
there exists (Kj,k)j ⊆ K such that
Ak ⊆∞⋃
j=1
Kj,k and∞∑j=1
λ(Kj,k) ≤ µ(Ak) +ε
2k.
Since A ⊂ ⋃j,kKj,k. By definition of µ, we have
µ(A) ≤ ∑j,k
λ(Kj,k) ≤∞∑k=1
µ(Ak) + ε. (1.2.31)
We let ε→ 0 and the σ-subadditivity follows.
Exercise 1.2.18. X 6= ∅, K = ∅, X, λ: K → [0,+∞], λ(∅) = 0,λ(X) = 1. The measure on X induced by λ is exactly the measure
µ defined by µ(∅) = 0 and µ(A) = 1 if A 6= ∅.
If K = A ⊆ P(X) is an algebra and λ: A → [0,+∞] is additive,a natural question is: is the measure µ defined in Theorem 1.2.17 a
15
σ-additive extension of λ? We will give a necessary condition in orderfor this to hold.
Definition 1.2.19. Let A ⊆ P(X) be an algebra. A mapping λ:
A → [0,+∞] is called a pre-measure if
i) λ(∅) = 0.
ii) λ(A) =∑∞
k=1 λ(Ak) for every A ∈ A such that A =⋃∞
k=1Ak,Ak ∈ A mutually disjoint.
λ is called σσσ-finite if there exists a covering X =⋃∞
k=1 Sk, Sk ∈ Aand λ(Sk) < +∞, ∀k (we may assume without loss of generalitythe Sk’s are mutually disjoint).
Given a pre-measure λ, we obtain as above a measure µ defined by
µ(A) = inf ∞∑
k=1
λ(Ak) : A ⊆∞⋃k=1
Ak, Ak ∈ A. (1.2.32)
Theorem 1.2.20. (Caratheodory-Hahn extension) Let λ: A →[0,+∞] be a pre-measure on X, µ be defined as in (1.2.32). Thenit holds:
i) µ: P(X) → [0,+∞] is a measure.
ii) µ(A) = λ(A) ∀A ∈ A.
iii) ∀A ∈ A is µ-measurable.
Proof of Theorem 1.2.20.
i) It follows from Theorem 1.2.17.
ii) Let A ∈ A. Clearly it holds µ(A) ≤ λ(A).
For the opposite inequality, let A ⊆ ⋃∞k=1Ak, Ak mutually dis-
joint, Ak ∈ A. We set Ak = Ak∩A ∈ A. Clearly, Ak are mutually
16
disjoint and A =⋃∞
k=1 Ak. Since λ is a pre-measure, it follows
λ(A) =∞∑k=1
λ(Ak) ≤∞∑k=1
λ(Ak). (1.2.33)
Thus
λ(A) ≤ inf ∞∑
k=1
λ(Ak) A ⊆∞⋃k=1
Ak, Ak ∈ A
= µ(A).
(1.2.34)
iii) Let A ∈ A and B ⊆ X be arbitrary. For every ε > 0 chooseBk ∈ A, k ∈ N with B ⊆ ⋃∞
k=1Bk and
∞∑k=1
λ(Bk) ≤ µ(B) + ε .
Since A,Bk ∈ A, we have
λ(Bk) = λ(Bk ∩ A) + λ(Bk \A). (1.2.35)
Moreover, we have
B ∩ A ⊆∞⋃
k=1
Bk ∩ A
B \ A ⊆∞⋃
k=1
(Bk \A).(1.2.36)
Thus, by definition of µ, we have
µ(B ∩ A) + µ(B \A) ≤∞∑k=1
λ(Bk ∩ A) +∞∑k=1
λ(Bk \A)
=∞∑k=1
λ(Bk) ≤ µ(B) + ε(1.2.37)
We let ε→ 0 and we conclude the proof.
17
Theorem 1.2.21. (Uniqueness of Charatheodory-Hahnextension)
Let λ: A → [0,+∞] be as in Theorem 1.2.20 and σ-finite, µ be
the measure induced by λ, Σ be the σ-algebra of the µ-measurablesets and let µ: P(X) → [0,+∞] be another measure with µ|A =
λ. Then µ|Σ = µ, namely the Charatheodory-Hahn extension isunique.
Proof. i). Let A ⊆ ∪∞k=1Ak with Ak ∈ A, k ∈ N. From the σ-sub-
additivity of µ it follows
µ(A) ≤ Σ∞k=1µ(Ak) = Σ∞
k=1λ(Ak).
By taking the infimum over all (Ak)k we obtain
µ(A) ≤ µ(A). (1.2.38)
ii) We show that if A ∈ Σ then also the inverse inequality holds. Tothis purpose we consider first of all the case there is S ∈ A such that
A ⊆ S and λ(S) < +∞.By i) we get
µ(S \A) ≤ µ(S \ A) ≤ µ(S) = λ(S) <∞. (1.2.39)
Since A ∈ Σ, S ∈ A ⊂ Σ it follows by the sub-additivity of µ that
µ(A) + µ(S \A) ≤ µ(A) + µ(S \A) = µ(S)
= λ(S) = µ(S) ≤ µ(A) + µ(S \A).(1.2.40)From (1.2.39) and (1.2.40) it follows that µ(A) = µ(A).
Suppose now thatX = ∪∞k=1Sk where (Sk)k∈N are mutually disjoint,
Sk ∈ A, λ(Sk) < ∞, for all k ∈ N. We consider the disjoint union
A = ∪∞k=1Ak, Ak = A ∩ Sk, k ∈ N. For every m it holds
µ(∪mk=1Ak) = µ(∪m
k=1Ak).
By taking the limit m→ +∞ we get
µ(A) ≥ limm→∞
µ(∪mk=1Ak) = lim
m→∞µ(∪m
k=1Ak) = µ(A).
We can conclude the proof.
18
Remark 1.2.22. i) From Theorem 1.2.21 it is not clear if everyA ∈ Σ is also µ-measurable.ii) In general the statement of Theorem 1.2.21 cannot be improved
as the following example shows.
Example 1.2.23. Let X = [0, 1], A = ∅, X and λ : A → [0,+∞],
µ as in the exercise 1.2.18 . We have Σ = ∅, X. We will see inthe next section that Lebesguemeasure L1 is also an extension of λ but
L1([0, 1/2]) = 1/2 6= µ([0, 1/2]) = 1.
1.3 Lebesgue Measure
Lebesgue measure is the extension of “volume” of the so-called ele-
mentary sets.
19
Definition 1.3.1.
i) For a = (a1, . . . , an), b = (b1, . . . , bn) ∈ Rn we set
(a, b) =
n∏i=1
(ai, bi) if ai < bi, 1 ≤ i ≤ n
∅ otherwise.
(1.3.1)
In an analogous way we define the intervals [a, b], (a, b],
[a, b). In case of open (half-open) intervals (ai, bi) we allowai = −∞ or bi = +∞.
ii) The volume of an interval is defined to be
vol((a, b)
)= vol
((a, b]
)= vol
([a, b)
)= vol
([a, b]
)
=
n∏i=1
(bi − ai) ≤ +∞ if ai < bi, 1 ≤ i ≤ n
0 otherwise.
iii) An elementary set is the finite disjoint union of intervalswith volume
vol( m⋃
k=1
Ik
)=
m∑k=1
vol(Ik).
Remark 1.3.2.
i) Let I =m⋃k=1
Ik =n⋃
j=1
Jj where Ik, Jj are intervals with Ik ∩ Iℓ = ∅
if k 6= ℓ and Jh ∩ Ji = ∅, if h 6= i, then
m∑k=1
vol(Ik) =n∑
j=1
vol(Jj).
ii) The class of all elementary sets is an algebra, the vol function
is a pre-measure. In particular, we have the following property:if I is an elementary set such that I =
⋃∞k=1 Ik, Ik elementary
mutually disjoint sets, then
vol(I) =∞∑k=1
vol(Ik). Exercise!
20
In what follows, we shall use the notion of dyadic cubes to obtain abasic decomposition of open sets in Rn. For every k ∈ N let Dk be the
collection of half open cubes
Dk = n∏
i=1
[ai2k,ai+1
2k
), ai ∈ Z
.
In other words, D0 is the collection of cubes with edge length 1 andvertices at points with integer coordinates. Bisecting each edge of a
cube in D0, we obtain from it 2n subcubes of edge length 12. If we
continue bisecting, we obtain finer and finer collections Dk of cubes
such that each cube in Dk has edge length 2−k and is the union of 2n
disjoint cubes in Dk+1.
The cubes of the collectionQ : Q ∈ Dk, k = 0, 1, 2, . . .
are called dyadic cubes.
Remark 1.3.3. Dyadic cubes have the following properties:
a) For every k ≥ 0, Rn =⋃
Q∈DkQ with disjoint union.
b) If Q ∈ Dk and P ∈ Dh with h ≤ k, then Q ⊂ P or P ∩Q = ∅.c) If Q ∈ Dk, then vol(Q) = 2−kn.
Lemma 1.3.4. Every open set in Rn can be written as a countableunion of disjoint dyadic cubes.
Proof. Let ∅ 6= V be an open set in Rn. Let S0 be the collection of allcubes in D0 which lie entirely in V . Let S1 be the set of cubes in D1
which lie in V but which are not subcubes of any cube in S0. Moregenerally, for k ≥ 1, let Sk be the cubes in Dk which lie in V but which
are not subcubes of any cube in S0,S1, . . . ,Sk−1 (see Fig. 1.2).If S = ∪kSk, then S is countable since each Dk is countable and
the cubes in S are non-overlapping by construction. Moreover, since
V is open and the cubes in Dk become arbitrarily small as k → +∞,then by Remark 1.3.3 a) each point of V will eventually lie in a cube
of some Sk. Hence V =⋃
Q∈S Q and the proof is complete.
21
Fig: 1.2
Definition 1.3.5. Lebesguemeasure Ln is the Caratheodory-Hahnextension of the volume defined on the algebra of elementary sets.
Definition 1.3.6. A measure µ on Rn is called Borel, if every
Borel set is µ-measurable.
As a consequence of Lemma 1.3.4 we get that Ln is a Borel measure
and the σ-algebra of the Ln-measurable sets contains the σ-algebra ofthe Borel sets.
In the sequel we are going to list some properties of Lebesguemea-sure.
Theorem 1.3.7. For every A ⊆ Rn it holds
Ln(A) = infA⊆G
Ln(G), G open. (1.3.2)
Proof. “≤”: This inequality follows from the monotonicity: if A ⊆G =⇒ Ln(A) ≤ Ln(G).
“≥”: We may suppose without loss of generality that Ln(A) < +∞(otherwise the inequality is trivial). For ε > 0, choose intervals (Iℓ)ℓ∈Nwith A ⊆ ⋃∞
ℓ=0 Iℓ and
∞∑ℓ=1
vol(Iℓ) ≤ Ln(A) + ε.
22
Since we are assuming that Ln(A) < +∞ we have vol(Iℓ) < +∞for every ℓ > 0.
For every ℓ we choose an open bounded interval Iℓ ⊃ Iℓ such thatvol(Iℓ) ≤ vol(Iℓ) +
ε2ℓ, ℓ ∈ N. We set G =
⋃∞ℓ=1 Iℓ, G is an open set
containing A. Moreover,
Ln(G) ≤∞∑ℓ=1
vol(Iℓ) ≤∞∑ℓ=1
vol(Iℓ) +∞∑ℓ=1
ε
2ℓ
≤ Ln(A) + 2ε.
(1.3.3)
We let ε→ 0 and get the result.
Theorem 1.3.8. Let A ⊆ Rn. The following conditions are equiv-
alent:
i) A is Ln-measurable.
ii) ∀ε > 0 ∃G ⊇ A open: Ln(G \A) < ε.
Proof of Theorem 1.3.8.
i) =⇒ ii):
• Assume first that Ln(A) < ∞. For every ε > 0, choose G ⊇ A,G open with
Ln(G) ≤ Ln(A) + ε.
Since A is Ln-measurable, by choosing B = G in the criterion ofmeasurability, we get
Ln(G) = Ln(G ∩ A) + Ln(G \ A)= Ln(A) + Ln(G \ A).
(1.3.4)
Therefore,Ln(G \A) = Ln(G)− Ln(A) ≤ ε. (1.3.5)
• If Ln(A) = ∞, we set Ak = A ∩ [−k, k]n, k ∈ N, with A =⋃∞
ℓ=1Aℓ.
For every ε > 0 choose Gk ⊇ Ak, Gk open with
Ln(Gk \Ak) <ε
2k, k ∈ N.
23
Then G =⋃∞
ℓ=1 Gk is open, A ⊆ G and
Ln(G \A) ≤∞∑k=1
Ln(Gk \Ak) < ε, since
G \ A =∞⋃
k=1
(Gk \ A) ⊆∞⋃
k=1
(Gk \Ak).
ii) =⇒ i): For ε > 0, choose A ⊆ G, G open with
Ln(G \A) ≤ ε.
Since G is Ln-measurable, we have for every B ⊆ Rn:
Ln(B) = Ln(B ∩G) + Ln(B \G)≥ Ln(B ∩ A) + Ln(B \A)− Ln(G \A)≥ Ln(B ∩ A) + Ln(B \A)− ε.
(1.3.6)
In the second inequality, we have used the relation B \A ⊆ (B \G)∪(G \A). We let ε→ 0 and we get the result.
Corollary 1.3.9. Let A ⊆ Rn is Ln-measurable if and only if itcan be “approximated” from inside and outside: ∀ε > 0 ∃F ⊆A ⊆ G, F closed, G open such that
Ln(G \A) + Ln(A \ F ) ≤ ε. (1.3.7)
Proof. If (1.3.7) holds, then the condition ii) of Theorem 1.3.8 holds
and therefore A is Ln-measurable. Conversely, for ε > 0 choose G ⊇Ac open such that
Ln(G \Ac) < ε (Theorem 1.3.8) .
Set F = Gc ⊆ A. Then F is closed with
Ln(A \ F ) = Ln(A ∩G) = Ln(G \Ac) < ε
and (1.3.7) follows.
24
Corollary 1.3.10. A ⊆ Rn is Ln-measurable iff it holds
∀ε > 0 ∃F ⊆ A ⊆ G, F closed, G open: Ln(G \ F ) < ε.(1.3.8)
Proof. We observe that if F ⊆ A ⊆ G then
G \ F = (G \A) ∪ (A \ F ). (1.3.9)
(1.3.7) =⇒ (1.3.8): From (1.3.9) it follows that
Ln(G \ F ) ≤ Ln(G \A) + Ln(A \ F ).
(1.3.8) =⇒ (1.3.7): We observe that
G \A ⊆ G \ F, A \ F ⊆ G \ F.
ThereforeLn(G \A) + Ln(A \ F ) ≤ 2Ln(G \ F ).
1.4 Comparison between Lebesgue and Jordan mea-
sure
To conclude, we compare Lebesgue measure with Jordan measure.We recall that A ⊆ Rn bounded is Jordan-measurable if µ(A) =
µ(A), where
µ(A) =
∫
Rn
χA dµ := supvol(E): E ⊆ A, E elementary set
µ(A) =
∫
R
χA dµ := infvol(E) : A ⊆ E, E elementary set;
In this case we denote the common value by µ(A).
χA denotes the characteristic function of A:
χA(x) =
1 x ∈ A0 x /∈ A .
25
Theorem 1.4.1. Let A ⊆ Rn be bounded, then
i) µ(A) ≤ Ln(A) ≤ µ(A).
ii) If A is Jordan-measurable, then A is Ln-measurable and
Ln(A) = µ(A).
Proof.
i) It holds
Ln(A) = inf ∞∑
k=1
vol(Ik), A ⊆∞⋃k=1
Ik, Ik intervals
≤ inf m∑
k=1
vol(Ik) A ⊆ E =m⋃k=1
Ik, Ik intervals
= µ(A).
(1.4.10)
Moreover, for every E =⋃m
k=1 Ik ⊆ A, Ik mutually disjoint, itholds
vol(E) = Ln(E) ≤ Ln(A) (1.4.11)
and therefore
µ(A) ≤ Ln(A).
ii) If A is Jordan-measurable, then µ(A) = µ(A) and from i), wehave
µ(A) = µ(A) = µ(A) = Ln(A). (1.4.12)
Now we show that A is Ln-measurable. Assume without loss
of generality that Ln(A) < +∞ (otherwise we consider the setsAk = A ∩ [−k, k]n, k ∈ N which are also Jordan-measurable).
Since A is Jordan-measurable, ∀ε > 0 ∃Iε, Iε elementary setssuch that Iε ⊇ A ⊇ Iε and
vol(Iε)− ε < µ(A) < vol(Iε) + ε.
26
We may assume that G = Iε is open. It follows that
Ln(G \ A) ≤ Ln(Iε \ Iε) = vol(Iε \ Iε)= vol(Iε) \ vol(Iε) < 2ε.
(1.4.13)
Therefore A is Ln-measurable according to Theorem 1.3.8.
Theorem 1.4.2. Lebesgue measure is invariant under isome-
tries namely under Φ: Rn → Rn, x 7→ x0 + Rx, where R is arotation (RtR = Id).
Proof. Exercise!
We present a further property of Lebesgue measure.
Definition 1.4.3. A Borel measure µ on Rn is called Borel reg-
ular if for every A ⊆ Rn there exists B ⊇ A Borel set such that
µ(A) = µ(B).
Corollary 1.4.4. Ln is Borel regular.
Proof. Without loss of generality let Ln(A) <∞ (otherwise B = Rn).Choose Gk open such that A ⊆ Gk and
Ln(Gk) < Ln(A) +1
k, k ∈ N.
In particular Ln(G1) < Ln(A) + 1 < +∞. Without loss of generality
we may assume Gk+1 ⊆ Gk, k ∈ N. Set B =⋂∞
k=1Gk. Then B is Boreland
Ln(B) = limk→+∞
Ln(Gk) = Ln(A).
27
1.5 An example of a non-measurable set
In this section we will construct a subset of R which is not L1 mea-
surable. We will need the axiom of the choice in the following form:
Zermelo’s Axiom:For any family of arbitrary nonempty disjoint sets indexed by a
set A, Eα, α ∈ A, there exists a set consisting of exactly oneelement from each Eα, α ∈ A.
For x, y ∈ [0, 1) we define
x⊕ y =
x+ y, if x+ y < 1
x+ y − 1, if x+ y ≥ 1.(1.5.1)
We observe that if E ⊆ [0, 1) is a L1-measurable set, then E⊕x ⊆ [0, 1)is also L1-measurable and L1(E⊕x) = L1(E) ∀x ∈ [0, 1). Indeed, we
haveE ⊕ x = E1 ∪ E2 (1.5.2)
where
E1 := E ∩ [0, 1− x)⊕ x = E ∩ [0, 1− x) + x
E2 := E ∩ [1− x, 1)⊕ x = E ∩ [1− x, 1) + (x− 1).(1.5.3)
E1, E2 are L1-measurable with
L1(E1) = L1(E ∩ [0, 1− x)
)
L1(E2) = L1(E ∩ [1− x, 1)
)
and E1 ∩ E2 = ∅. Therefore, E ⊕ x is measurable and
L1(E ⊕ x) = L1(E1) + L1(E2)
= L1(E ∩ [0, 1− x)
)+ L1
(E ∩ [1− x, 1)
)
= L1(E).
In [0, 1) we define the following equivalent relation: x, y ∈ [0, 1), x ∼ yif x − y ∈ Q. By the Axiom of Choice, there exists a set P ⊆ [0, 1)
28
such that P consists of exactly one representative point from eachequivalent class. LetQ∩[0, 1) = rjj∈N be an enumeration ofQ∩[0, 1)with r0 = 0, and for j ∈ N we define
Pj = P ⊕ rj. (1.5.4)
Observe the following.
1. (Pj) is a disjoint family.
Let x ∈ Pn ∩ Pm, then x = pn ⊕ rn = pm ⊕ rm, rn, rm ∈ Q.Therefore pn − pm ∈ Q, namely pn, pm are in the same equivalent
class. By construction pn = pm and rn = rm =⇒ Pn ≡ Pm.
2. [0, 1) =+∞⋃j=0
Pj.
Let x ∈ [0, 1) and [x] be its equivalence class. There exists a
unique p ∈ P such that x ∼ p. If x = p, then x ∈ P0 = P ; ifx > p, then x = p + ri = p ⊕ ri for some ri =⇒ x ∈ Pi, if x < p
then x = p + rj − 1 = p ⊕ rj for some rj, therefore x ∈ Pj . Itfollows that
[0, 1) =+∞⋃
j=0
Pj . (1.5.5)
3. If P is Lebesgue measurable, then Pj would be Lebesgue measur-
able as well and
1 = L1([0, 1)
)=
∞∑i=1
L1(Pj) =∞∑i=1
L1(P ).
This is impossible since the right-hand side is = +∞ or = 0.
Since P is not L1-measurable, there exists B ⊆ R such that
L1(B) < L1(B ∩ P ) + L1(B \ P ).
B ∩ P and B \ P are two disjoint sets for which the additivityof the measure does not hold. We observe that L1(P ) > 0 (if
L1(P ) = 0 =⇒ P is L1-measurable). If E ⊆ P is L1-measurable,then L1(E) = 0. Indeed, set Ei = E ⊕ ri, Ei is L1-measurable
29
and L1(Ei) = L1(E). Moreover,⋃∞
i=1Ei = F ⊂ [0, 1), F , L1-measurable with
1 = L1([0, 1]) ≥ L1(F ) =∞∑i=1
L1(Ei)
=∞∑i=1
L1(E)
=⇒ L1(E) = 0
(1.5.6)
Exercise 1.5.1. For every A ⊆ R with L1(A) > 0, there exists B ⊂ A
such that B is not L1-measurable.
Proof. Suppose for instance that A ⊆ (0, 1). Set B = A ∩ P and
Bi = A ∩ Pi. If Bi were L1-measurable, then L1(Bi) = 0 (sinceBi ⊆ Pi), therefore
∑∞i=1L1(Bi) = 0. Since A =
⋃∞i=1Bi, we would
get a contradiction.
Exercise 1.5.2. Show that every countable subset of R has Lebesguemeasure zero.
Proof. Let α ∈ R. Then α ⊆ [α− ε, α+ ε] ∀ε > 0. Therefore
L1α ≤ vol[α− ε, α+ ε] = 2ε ∀ε > 0 (1.5.7)
and this gives L1α = 0.
If A = α1, α2, . . . =∞⋃n=1
αn is a countable set, then
L1(A) ≤ ∑nL1αn = 0. (1.5.8)
1.6 An uncountable Ln-null set: the Cantor Tri-
adic Set
Let X = [0, 1]. For any fixed integer b ≥ 2 and any x ∈ [0, 1] onecan expand x in base b: there exists a sequence of “digits” di(x) ∈0, . . . , b− 1 such that
x =∞∑i=1
di(x) b−i
30
which is also denoted x = 0, d1, d2 . . . . This expansion is not entirelyunique (for instance, in base 10, we have 0.1 = 0.099 = 0.09). How-
ever, the set of those x ∈ X for which there exists more than oneexpansion is countable.
The Cantor set is defined as follows: take b = 3, so the digits are
in 0, 1, 2 and let
C =x ∈ [0, 1] | di(x) ∈ 0, 2 ∀i
be the set of those x which have no digit 1 in their 3-expansion.
Claim 1.6.1. L1(C) = 0, but C is not countable.
Proof of Claim 1.6.1.
• We observe that C =⋂∞
n=1Cn, where
Cn =x ∈ [0, 1] | di(x) 6= 1 ∀i ≤ n
.
Each Cn is a Borel subset of [0, 1]; indeed, we have for instance
C1 =[0, 1
3
]∪[2
3, 1]
C1 =[0, 1
9
]∪[2
9, 13
]∪[2
3, 79
]∪[2
9, 1].
(1.6.1)
More generally, Cn is a Borel subset of [0, 1] which is a disjointunion of 2n intervals of length 3−n. Thus by additivity
L1(Cn) = 2n × 3−n =(2
3
)n
and since Cn+1 ⊆ Cn, L1(C1) < +∞, we have
L1(C) = L1(⋂
n
Cn
)= lim
n→+∞L1(Cn) = 0. (1.6.2)
• C is uncountable.
31
Indeed, the function
f( ∞∑
i=1
di3i
)=
∞∑i=1
di2
1
2i(1.6.3)
maps C onto [0, 1], (see Exercise 1.6.2 below).
Exercise 1.6.2. Show that f in (1.6.3) is surjective but not injec-
tive, (f(x) coincides at the opposing ends of one of the middle thirdremoved. For instance:
7
9= 0.202, 8
9= 0.2200, so
f(7
9
)= 0.101 = f
(8
9
)= 0.11. )
1.7 Lebesgue-Stieltjes Measure
Let F : R→ R be increasing and continuous from the left:
F (x0) = limx→x−
0
F (x) ∀x0 ∈ R.
For a, b ∈ R we set
λF [a, b) =
F (b)− F (a) if a < b
0 otherwise.(1.7.1)
Let ΛF be the induced measure of λF :
ΛF (A) = inf ∞∑
k=1
λF [ak, bk), A ⊆∞⋃k=1
[ak, bk).
ΛF is called the Lebesgue-Stieltjes Measure generated by F . Is ΛF
Borel or even Borel regular?
The following criterion is useful.
Definition 1.7.1. A measure µ on Rn is called metric if∀A,B ⊆ Rn with
dist(A,B) = inf|a− b|, a ∈ A, b ∈ B > 0,
we have
µ(A ∪B) = µ(A) + µ(B). (1.7.2)
32
Theorem 1.7.2. (Caratheodory’s criterion for Borel-measures)
A metric measure µ on Rn is Borel.
Proof of Theorem 1.7.2. It is enough to show that every closed set F
of Rn is µ-measurable, namely ∀B ⊆ Rn we have
µ(B) ≥ µ(B ∩ F ) + µ(B \ F ). (1.7.3)
If µ(B) = ∞ then (1.7.3) is obvious.
Assume instead µ(B) < ∞. For k ≥ 0, define
Fk =x ∈ Rn : d(x, F ) ≤ 1
k
. (1.7.4)
Since µ is metric and
dist(B \ Fk, B ∩ F ) ≥ 1
k> 0,
we have by the monotonicity of µ
µ(B ∩ F ) + µ(B \ Fk) = µ((B ∩ F ) ∪ (B \ Fk)
)
≤µ(B).(1.7.5)
Claim 1.7.3. µ(B \ Fk) → µ(B \ F ) as k → +∞.
Proof of Claim 1.7.3. For k ∈ N we set
Rk = (Fk \ Fk+1) ∩B =x ∈ B :
1
k + 1< d(x, F ) ≤ 1
k
. (1.7.6)
33
Rk are disjoint with
B \ F = B \∞⋂
ℓ=1
Fℓ =
∞⋃
ℓ=1
(B \ Fℓ)
= (B \ F1) ∪∞⋃
ℓ=1
(B \ Fℓ+1) \ (B \ Fℓ)
= (B \ Fk) ∪∞⋃
ℓ=k
(B \ Fℓ+1)− (B \ Fℓ)
= (B \ Fk) ∪∞⋃
ℓ=k
(Fℓ \ Fℓ+1) ∩B
= (B \ Fk) ∪∞⋃
ℓ=k
Rℓ.
Thus
µ(B \ Fk) ≤ µ(B \ F ) ≤ µ(B \ Fk) +∞∑ℓ=k
µ(Rℓ). (1.7.7)
We show that
limk→+∞
∞∑ℓ=k
µ(Rℓ) = 0. (1.7.8)
To this purpose we prove that
∞∑ℓ=1
µ(Rℓ) < +∞. (1.7.9)
We observe that dist(Ri, Rj) > 0, if |i − j| ≥ 2. Since µ is metric, it
follows by induction that
m∑k=1
µ(R2k) = µ( m⋃
k=1
R2k
)≤ µ(B)
m∑k=1
µ(R2k+1) = µ( m⋃
k=1
R2k+1
)≤ µ(B).
(1.7.10)
By adding (1.7.9) and (1.7.10) and letting m → +∞, we prove that∑∞k=1 µ(Rk) < +∞.
34
Theorem 1.7.4. The Lebesgue-Stieltjes measure is Borel regular.
Proof.
Step 1. We first show that ΛF is Borel.According to Theorem 1.7.2 it is enough to show that ΛF is metric.
Let A,B ⊆ R with δ = dist(A,B) > 0. For ε > 0, choose ak, bk ∈ Rwith
A ∪B ⊆∞⋃
k=1
[ak, bk) and∞∑
k=1
λF [ak, bk) < ΛF (A ∪B) + ε.
Up to a further subdivision of the intervals [ak, bk), we can supposethat ∀k ≥ 0 one has |bk − ak| < δ.
Therefore, for every k ∈ N, we have either A ∩ [ak, bk) = ∅ orB ∩ [ak, bk) = ∅. The covering ([ak, bk))k is decomposed in a coveringA of A and a covering B of B. It follows
ΛF (A ∪ B) ≤ ΛF (A) + ΛF (B)
≤ ∑[αk,bk)∈A
ΛF
([ak, bk)
)+
∑[αk,bk)∈B
ΛF
([ak, bk)
)
≤ ∑k
λF([ak, bk)
)≤ ΛF (A ∪B) + ε.
(1.7.11)
We let ε→ 0 and we get the claim.Step 2. We show that ΛF is Borel regular. Let A ⊆ R with
ΛF (A) < +∞ (otherwise we choose B = R). For j ∈ N we choose thesequences (ajk)k, (b
jk)k with
A ⊆∞⋃
k=1
[ajk, bjk) =: Bj
and ∞∑k=1
λF([ajk, b
jk))≤ ΛF (A) +
1
j.
We set B =⋂∞
j=1Bj, B is Borel with A ⊆ B and B ⊆ Bj. It followsthat
ΛF (A) ≤ ΛF (B) ≤ ΛF (Bj)
≤∞∑k=1
λF([ajk, b
jk))≤ ΛF (A) +
1
j.
(1.7.12)
35
We let j → +∞ and get ΛF (A) = ΛF (B) =⇒ ΛF is Borel regular.
The set of half-open intervals is not an algebra. Nevertheless, itholds the following result:
Theorem 1.7.5. For a < b there holds
ΛF
([a, b)
)= λF
([a, b)
)= F (b)− F (a) (1.7.13)
Proof. Let’s consider a, b ∈ R with a < b.
1. By definition we have
ΛF
([a, b)
)≤ λF
([a, b)
)
2. Now we show that
ΛF
([a, b)
)≥ λF
([a, b)
)
Let ([ak, bx))k be a covering of [a, b), namely [a, b) ⊆ ∪k[ak, bx). Since
F is left-continuous for every ε > 0 there δ, δk such that
F (b)− F (b− δ) ≤ ε (1.7.14)
F (ak)− F (ak − δk) ≤ 2−kε, ∀k ≥ 0 (1.7.15)
We observe that
[a, b− δ] ⊆ ∪k≥0(ak − δk, bx).
We can assume without restriction that ak − δk < bk−1 for all k ≥ 1.
Since [a, b− δ] is compact we can extract a finite sub-covering of theopen intervals ((ak − δk, bx))k, i.e.
[a, b− δ] ⊆ ∪mk (ak − δk, bx).
We have
λF ([a, b− δ)) = F (b− δ)− F (a) ≤m∑
k=0
F (bk)− F (ak − δk).
36
Therefore
λF ([a, b)) = F (b)− F (a) = F (b)− F (b− δ) + F (b− δ)− F (a)
≤ ε+
m∑
k=0
F (bk)− F (ak − δk)
= ε+
m∑
k=0
F (bk)− F (ak) +
m∑
k=0
F (ak)− F (ak − δk)(1.7.16)
≤∞∑
k=0
F (bk)− F (ak) + ε+∞∑
k=0
2−kε
=
∞∑
k=0
F (bk)− F (ak) + 4ε . (1.7.17)
By taking in (1.7.16) the infimum over all covering of [a, b) and thenletting ε→ 0 we get
λF ([a, b)) ≤ ΛF ([a, b)),
and we conclude the proof of Theorem 1.7.5.
Exercise 1.7.6.
i) For F (x) = x, ΛF = L1.
ii) For F (x) = 1, x > 0 and F (x) = 0, x ≤ 0, it follows that
ΛF = δ0.
37
1.8 Hausdorff Measures
Hausdorff measures are among the most important measures. They
generalize Lebesgue measure and permit to measure the subsets ofRn of dimension less than n, such as the submanifolds (surfaces and
curves in R3) and the so-called fractal sets (a fractal is a never endingpattern that repeats itself at different scales). They allow us to define
the dimension of every set of Rn even with complicated geometry.
Definition 1.8.1. For s ≥ 0, δ > 0 and ∅ 6= A ⊆ Rn we set
Hsδ(A) = inf
∞∑k=1
rsk, A ⊆∞⋃k=1
B(xk, rk), 0 < rk < δ. (1.8.1)
and we define Hsδ(∅) = 0.
In the definition of Hsδ(A) we require that rk > 0 since we want to
avoid the case 00. Hsδ is a measure on Rn. It is defined in terms of the
diameter of the balls B(xk, rk) := y ∈ Rn: | y − xk | < rk. It holds
Hsδ1(A) ≤ Hs
δ2(A) if δ2 ≤ δ1, (1.8.2)
namely, Hsδ(A) is a non increasing function of δ. Thus there exists the
limit
Hs(A) = limδ↓0
Hsδ(A) = sup
δ>0Hs
δ(A). (1.8.3)
Definition 1.8.2. Hs is called the sss-dimensional Hausdorff
measure on Rn.
Theorem 1.8.3. For s ≥ 0, Hs is a Borel regular measure on
Rn.
Proof of Theorem 1.8.3.
i) HsHsHs is a measure.
Let s ≥ 0. Clearly, Hs(∅) = 0.
38
Let A ⊆ ⋃∞k=1Ak ⊆ Rn. Since for every δ > 0Hs
δ is σ-subadditive,we have the following inequalities
Hsδ(A) ≤
∞∑k=1
Hsδ(Ak) ≤
∞∑k=1
Hs(Ak). (1.8.4)
We let δ → 0 and get the σ-sub-additivity of Hs.
ii) HsHsHs is metric and therefore Borel.
Let A,B ⊆ Rn with δ0 = dist(A,B). We select 0 < δ < δ04 .
Suppose A ∪ B =⋃∞
k=1B(xk, rk), rk < δ. Set
A =B(xk, rk), B(xk, rk) ∩ A 6= ∅
B =B(xk, rk), B(xk, rk) ∩B 6= ∅
.
(1.8.5)
Then
A ⊆⋃
B(xk,rk)∈AB(xk, rk)
B ⊆⋃
B(xk,rk)∈BB(xk, rk) .
(1.8.6)
We observe that if B(xk, rk) ∈ A, then B(xk, rk)∩B 6= 0. Indeed,suppose by contradiction that ∃x ∈ A and y ∈ B such that
|x− xk| < rk, |y − xk| < rk for some k. Then
δ0 = d(A,B) ≤ |x− y| ≤ |x− xk|+ |y − xk|
≤ 2rk ≤ 2δ <δ02,
(1.8.7)
which is a contradiction. Thus
Hsδ(A) +Hs
δ(B) ≤ ∑B(xk,rk)∈A
rsk +∑
B(xk,rk)∈Brsk
≤∞∑k=1
rsk.(1.8.8)
39
Taking the infimum over⋃∞
k=1B(xk, rk) covering A ∪B, we get
Hsδ(A ∪ B) ≥ Hs
δ(A) +Hsδ(B) (1.8.9)
provided δ < δ04 . Let δ → 0 and get
Hs(A ∪ B) ≥ Hs(A) +Hs(B). (1.8.10)
It follows that Hs(A ∪B) = Hs(A) +Hs(B) and we conclude.
iii) HsHsHs is Borel regular.
Let A ⊆ Rn, suppose Hs(A) < +∞ =⇒ Hsδ(A) < +∞ ∀δ > 0.
For δ = 1ℓ , choose
⋃∞k=1B(xk,ℓ, rk,ℓ) ⊇ A with rk,ℓ <
1ℓ and
∞∑k=1
rsk,ℓ ≤ Hs1ℓ
(A) + 1ℓ .
Set Aℓ =⋃∞
k=1B(xk,ℓ, rk,ℓ) and B =⋂∞
ℓ=1Aℓ. B is a Borel set
with A ⊆ B.
For each ℓ, we have
Hs1ℓ
(A) ≤ Hs1ℓ
(B) ≤ Hs1ℓ
(Aℓ) ≤∞∑k=1
rsk,ℓ
≤ Hs1ℓ
(A) +1
ℓ. (1.8.11)
We let ℓ→ +∞ and get
Hs(B) = Hs(A). (1.8.12)
Remark 1.8.4. We observe that H0 is the so-called counting mea-sure. Actually let k be a positive integer and E be a set with kelements and let δ0 > 0 be the minimal distance between the elements
of E. Then every covering of balls of radius 0 < δ < δ0 consists of atleast k sets. On the other hand if E = x1, . . . , xk then ∪k
i=1B(xi, δ)
is a covering of E. It follows that H0δ(E) = k for every 0 < δ < δ0. It
follows that H0(E) = k as well. If E is infinite than by monotonicity
we have H0(E) ≥ k for all k and therefore H0(E) = +∞. FinallyH0(∅) = 0 by definition.
40
In order to define the dimension of an arbitrary set, we need thefollowing
Lemma 1.8.5. Let A ⊆ Rn and 0 ≤ s < t < +∞. It holds
i) Hs(A) < +∞ =⇒ Ht(A) = 0.
ii) Ht(A) > 0 =⇒ Hs(A) = +∞.
Proof.
i) Let Hs(A) < +∞. For δ > 0, A ⊆ ⋃∞k=1B(xk, rk) with rk < δ, it
holds
Htδ(A) ≤
∞∑k=1
rtk =∞∑k=1
rt−sk rsk ≤ δt−s
∞∑k=1
rsk. (1.8.13)
By considering the infimum with respect to⋃∞
k=1B(xk, rk), weget
Htδ(A) ≤ δt−s Hs
δ(A). (1.8.14)
Let δ → 0 and we get Ht(A) = 0.
ii) It follows from i).
Example 1.8.6. Let Q = [−1, 1]n ⊆ Rn. Then it holds
2−nLn(Q) ≤ Hn(Q) ≤ 2−nnn2Ln(Q). (1.8.15)
Proof. Let δ > 0 and k > 0 be such that r = 2−k−1√n < δ. We
decompose Q in subcubes Qℓ of edge length 2−k, 1 ≤ ℓ ≤ 2(k+1)n
41
(−1, 0)
(1, 0)︸ ︷︷ ︸
1
2
(0, 1)
(0, 1)
k = 1
n = 2
ℓ = 16
r =
√n∑
i=1
(2−k
2
)2
= n12 2−(k+1)
r
Fig. 1.3
Each Qℓ is included in a ball with the same center and radius r There-
fore Q ⊆ ⋃2(k+1)n
ℓ=1 B(xℓ, r). Since r = n12 2−(k+1) < δ, we have
Hnδ (Q) ≤
∑1≤ℓ≤2(k+1)n
rn = rn 2(k+1)n
= nn2 2−(k+1)n 2(k+1)n
= nn2 = 2n 2−n n
n2
= Ln(Q)nn2 2−n.
(1.8.16)
42
Conversely, if (B(xk, rk))k with rk < δ is a covering of the ball B(0, 1),then the following estimate holds
wn = Ln(B(0, 1)
)≤ ∑
k
Ln((B(xk, rk)
)
= wn
∑k=1
rnk
where
wn =πn/2
Γ(1 + n2 ), Γ(t) =
∫ +∞
0
xt−1 e−xdx t > 0,
Γ is the Euler function.
Therefore∑
k rnk ≥ 1 and
Hnδ (Q) ≥ Hn
δ
(B(0, 1)
)
= inf∑
k
rnk : B(0, 1) ⊆∞⋃k=1
B(xk, rk), rk < δ
≥ 1 = 2−nLn(Q)
(1.8.17)
Let δ → 0 in (1.8.16) and in (1.8.17) and we get the result.
Remark 1.8.7.
1. Actually, Ln(A) = wn Hn(A) for all Ln-measurable sets A ⊆ Rn.
2. For A ⊆ Rn, if s > n, then it holds Hs(A) = 0. Actually we canwrite Rn = ∪ℓQℓ where Qℓ are dydic cubes. For Example 1.8.6
and Lemma 1.8.5 it follows that Hs(Qℓ) = 0 for every ℓ ∈ N.Therefore Hs(Rn) = 0.
Definition 1.8.8. The Hausdorff dimension of a set A ⊆ Rn
is defined to be
dimH(A) := infs ≥ 0 : Hs(A) = 0.
43
Figure 1.1: Example of fractal in everyday life: the cauliflower
Remark 1.8.9.
1. From Lemma 1.8.5 it follows that
dimH(A) = sups ≥ 0 : Hs(A) = +∞.
2. Observe that dimH(A) ≤ n. Let s = dimH(A). Then Ht(A) = 0
for all t > s and Ht(A) = +∞ for all t < s; Hs(A) may beany number between 0 and ∞, inclusive. Furthermore, dimH(A)need not be an integer. Even if dimH(A) = k is an integer and0 < Hk(A) <∞, A need not be a “k-dimensional surface” in any
sense.
3. If A ⊆ Rn is open, then dimH(A) = n. Indeed, each open setcontains a ball B such that Ln(B) > 0 =⇒ Hn(A) > 0 =⇒dimH(A) ≥ n =⇒ dimH(A) = n.
Sets with non-integer Hausdorff dimension are important and mostsets in nature are fractals with non-integer dimension.
A fractal set is a rough or fragmented geometric set that can be
splitted into parts, each of which is a reduced-size copy of the whole.
1.8.1 Examples of non-integer Hausdorff-dimension sets
1. Triadic Cantor Set
44
Figure 1.2: Approximation of the coastline of Great Britain by polygons
C = limk→+∞
Ck =⋂∞
k=1Ck, Ck closed sets with C0 = [0, 1], Ck is
obtained from Ck−1 by removing the open middle third from each
connected component of Ck−1. In this case:
dimH(C) =ℓn(2)
ℓn(3)=: d 2−(d+1) < Hd(C) < 2−d,
where ℓn(x) = loge(x).
Heuristics:
• ∀s > 0, λ > 0, ∀A ⊆ Rn
Hs(λA) = λsHs(A).
• If there exists d such that Hd(C) ∈ (0,+∞) then, since C isthe union of two copies of C
3, we have
Hd(C) = 2Hd(C
3
)= 2
3dHd(C) =⇒ 2
3d= 1 =⇒
d = log3 2 = ℓn 2
ℓn 3.
2. Cantor Dust
45
PSfrag
y = x− 12
y = −x+ 12
A0 = [0, 1]2
A1 =©1 ∪©2 ∪©3 ∪©4= union of squaresthrough which the two
lines y = x− 12, y = −x+ 1
2 pass
Fig. 1.4
A2 ⊆ A1 is the union of 42 squares of length 4−2 obtained byrepeating the same construction in every sub-squares of A1.
Finally one considers A =⋂∞
k=1Ak. One has
dimH(A) = 1 (1.8.18)
Proof. We show that 12 ≤ H1(A) ≤
√22 .
Each set Ak consists of 4k squares of length 4−k. The 4k balls ofradius 4−k
√22
cover Ak and thus A. Thus
H1δ(A) ≤ H1
δ(Ak) ≤√2
2∀k ≥ k0 : rk0 = 4−k0
√2
2< δ. (1.8.19)
Let Π: R2 → R be the projection Π(x, y) = x. We have Π(A) =[0, 1]. Let A ⊆ ⋃∞
k=1B(zk, rk) with zk = (xk, yk) and rk < δ. We
get
[0, 1] = Π(A) ⊆∞⋃
k=1
Π(B(xk, rk)
)
=∞⋃
k=1
(xk − rk, xk + rk).
(1.8.20)
Thus
1 = L1[0, 1] ≤∞∑k=1
L1(Ik) = 2∞∑k=1
rk =⇒∞∑k=1
rk ≥ 12. (1.8.21)
where Ik = (xk − rk, xk + rk).
46
3. The Koch Curve
Description:
Fig. 1.5
Begin with a straight line. Divide it into 3 equal segments andreplace the middle segment by the two sides of an equilateral tri-
angle of the same length. Now repeat, taking each of the fourresulting segments dividing them into three equal parts and re-
placing each of the middle segments by two sides of an equilateraltriangle. Continue with the construction.
The Koch Curve is the limiting curve obtained by applying thisconstruction on infinite number of times.
The length of the intermediate curve at the nth-iteration of theconstruction is (4
3)n. Therefore the length of the Koch curve is
infinite. Moreover, the length of the Koch curve between any twopoints of the curve is also infinite since there is a copy of the Koch
curve. In this case, the Hausdorff dimension is d = ℓn 4ℓn 3 > 1 (the
curve is made of 4 copies of itself rescaled by the factor 13).
1.9 Radon Measure
Definition 1.9.1. A measure µ on Rn is called a Radon measureif µ is Borel regular and µ(K) <∞, for every compact K ⊆ Rn.
47
Example 1.9.2.
i) Ln is a Radon measure on Rn.
ii) Hs for s < n is not a Radon measure.
iii) If µ is a Radon measure and A ⊆ Rn is µ-measurable, then the
measure µ⌊A defined by
(µ⌊A)(B) := µ(A ∩ B), B ⊆ Rn (1.9.1)
is a Radon measure. Exercise!
For general Radon measures an analogous of Theorem 1.3.7 holds.
Theorem 1.9.3. (Approximation by open and compact sets)
Let µ be a Radon measure on Rn.
i) For every A ⊆ Rn it holds
µ(A) = infµ(G) : A ⊆ G, G open
. (1.9.2)
ii) For every A ⊆ Rn µ-measurable it holds
µ(A) = supµ(F ) : F ⊆ A, F compact
. (1.9.3)
To prove Theorem 1.9.3 we need the following lemma:
Lemma 1.9.4. For every Borel set B ⊆ Rn it holds ∀ε < 0 ∃G ⊇B, G open such that µ(G \B) < ε. (No proof)
Proof of Theorem 1.9.3.
i) If µ(A) = +∞ then (1.9.2) is obvious. Let us suppose µ(A) <+∞.
Assume first that AAA is a Borel set and fix ε > 0. Then byLemma 1.9.4 there exists G open such that G ⊇ A and µ(G\A) <
48
ε. Since µ(G) = µ(G ∩ A) + µ(G \ A), we get µ(G) < µ(A) + εand i) follows.
Now let A be an arbitrary set. Since µ is Borel regular, there
exists a Borel set B ⊇ A such that µ(A) = µ(B). Then
µ(A) = µ(B) = infµ(U), B ⊆ U open≥ infµ(U), A ⊆ U open. (1.9.4)
ii) Let A be a µ-measurable.
• Case 1: µ(A) < +∞.
Set ν = µ⌊A. ν is a Radon measure. We apply i): ∀ε > 0there exists G open such that G ⊇ Rn \A and
ν(G) ≤ ν(Rn \ A) + ε = ε.
Let C = Rn \G, C is closed and C ⊆ A. Moreover,
µ(A \ C) = µ(A ∩ (Rn \ C)
)= ν(Rn \ C)= ν(G) < ε.
(1.9.5)
Thusµ(A) ≤ µ(C) + ε . (1.9.6)
(µ(A) = µ(A∩C)+µ(A\C), since C is measurable). There-
fore
µ(A) = supµ(C): C ⊆ A, C closed. (1.9.7)
• Case 2: µ(A) = +∞.
Define Dk = x : k− 1 ≤ |x| < k. Then A =⋃∞
k=1(Dk ∩A).Since µ is a Radon measure, µ(Dk ∩ A) < +∞. From case 1it follows that ∀k there exists a closed set Ck ⊆ Dk ∩ A, Ck
such that µ(Ck) ≥ µ(Dk ∩ A)− 12k. Now
⋃∞k=1Ck ⊆ A and
limm→+∞
µ( m⋃
k=1
Ck
)= µ
( ∞⋃
k=1
Ck
)=
∞∑k=1
µ(Ck)
≥∞∑k=1
[µ(Dk ∩ A)−
1
2k
]
= µ(A)− 1 = +∞.
49
We observe that⋃m
k=1Ck is closed ∀m, thus, also in the caseµ(A) = +∞, (1.9.7) holds. Finally, we set
Bm = B(0, m) = x ∈ Rn : |x| ≤ m. (1.9.8)
If C is closed, then C ∩ Bm is compact and
µ(C) = limm→+∞
µ(C ∩ Bm).
Hence, for each µ-measurable set A:
supµ(K) : K ⊆ A compact = supµ(C) : C ⊆ A closed.
We conclude the proof.
Theorem 1.9.5. Let µ be a Radon measure on Rn and let A ⊆Rn. The following two conditions are equivalent:
i) A is µ-measurable,
ii) ∀ε > 0 ∃G ⊃ A, G open such that µ(G \ A) < ε.
The proof is the same of that of Theorem 1.3.8 by using Theorem 1.9.3instead of Theorem 1.3.7.
50
Chapter 2
Measurable Functions
2.1 Inverse Image of a Function
Let X, Y be nonempty sets.
For any map ϕ: X → Y and A ∈ P(Y ), we set
ϕ−1(A) = x ∈ X : ϕ(x) ∈ A = ϕ ∈ A, (2.1.1)
ϕ−1 is called the inverse image or preimage of A.
Some Properties
i) ϕ−1(Ac) =(ϕ−1(A)
)c ∀A ∈ P (Y ).
ii) If A,B ∈ P (Y ):
ϕ−1(A ∩B) = ϕ−1(A) ∩ ϕ−1(B).
iii) If Ak ⊆ P (Y ):
ϕ−1( ∞⋃
k=1
Ak
)=
∞⋃
k=1
ϕ−1(Ak).
Consequently, if (Y,F , µ) is ameasure space, then ϕ−1(F) = ϕ−1(A):
A ∈ F is a σ-algebra in X.
2.2 Definition and Basic Properties
Let µ be a measure on Rn, Ω ⊆ Rn µ-measurable.
51
Definition 2.2.1. f : Ω → [−∞,∞] is called µµµ-measurable if
i) f−1+∞, f−1−∞ are µ-measurable.
ii) f−1(U) is µ-measurable for every U ⊆ R open.
Remark 2.2.2. The following two conditions are equivalent to ii) inDefinition 2.2.1.
iii) f−1(B) is µ-measurable for each Borel set B ⊆ R.iv) f−1((−∞, a)) is µ-measurable, ∀a ∈ R.
Remark 2.2.3. We consider R = [−∞,+∞] with the topology gen-erated by the open sets of R and the neighborhoods [−∞, a), (a,+∞],
a ∈ R of ±∞.Then f : Ω → R is µ-measurable if and only if
v) f−1(U) is µ-measurable, ∀U ⊆ R openor if and only ifvi) f−1([−∞, a)) is µ-measurable, ∀a ∈ R.
Proof of vi) =⇒ i) and vi) =⇒ iv):
vi) =⇒ i): We use the representation
f−1(−∞) =⋂
k∈Nf−1
([−∞, k)
)
f−1(+∞) = Ω \⋃
k∈Nf−1
([−∞, k)
).
vi) =⇒ iv): f−1((−∞, a)) = f−1([−∞, a)) \ f−1(−∞.
Exercise 2.2.4. Let f : Ω → R be µ-measurable and g: R → R becontinuous. Then g f is µ-measurable.
52
Theorem 2.2.5. (Properties of Measurable Functions)
i) Let f, g: Ω → R be µ-measurable functions. Then:
f + g, f · g, |f |, sign(f)1 max(f, g), min(f, g) and (if g(x) 6=0) f
gare µ-measurable.
ii) If fk: Ω → R are µ-measurable. Then:
infkfk, sup
kfk lim inf
k→∞fk, lim sup
k→∞fk are µ-measurable.
Proof.
i) In view of Remark 2.2.2, f : Ω → R is µ-measurable iff f−1(−∞, a)
is µ-measurable ∀a ∈ R.
Let f , g: Ω → R be µ-measurable, then
(f + g)−1(−∞, a) =⋃
r+s<ar,s∈Q
f−1((−∞, r)
)∩ g−1
((−∞, s)
). (2.2.1)
Since f 2(−∞, a) = f−1(−∞,√a)\f−1(−∞,−√
a], if a ≥ 0, then
f · g = 1
2
[(f + g)2 − f 2 − g2
](2.2.2)
is µ-measurable as well.
Moreover, it holds
(1
g
)−1((−∞, a)
)=
g−1(1
a, 0
)a < 0
g−1((−∞, 0)
)a = 0
g−1((−∞, 0)
)∪ g−1
((1
a,+∞
))a > 0
(2.2.3)
Therefore1
gand
f
gare µ-measurable.
53
Finally set s+ = maxs, 0, s− = max−s, 0. The maps s→ s+,s→ s− are continuous. Therefore, by Exercise 2.2.4 we have
f+, f−, |f | = f+ + f−
sup(f, g) = f + (g − f)+
inf(f, g) = f − (g − f)−(2.2.4)
are µ-measurable.
The function x 7→ sign(x) is continuous except at the origin. If
U ⊆ R is open, sign−1(U) is either open or of the form V ∪ 0where V is open, so sign(x) is Borel measurable function (in the
sense that the pre-image of a Borel subset of R is a Borel set ofR ). Therefore signf is µ-measurable.
ii) Let fk: Ω → [−∞,+∞] be µ-measurable. Then
infk≥1
f−1k [−∞, a) =
∞⋃
k=1
f−1k [−∞, a)
supk≥1
f−1k [−∞, a) =
∞⋃
ℓ=1
∞⋂
k=1
f−1k
[−∞, a− 1
ℓ
) (2.2.5)
=⇒ infk, fk, supk fk are µ-measurable.
We complete the proof by noting
limk→∞
inf fk = supm≥1
infk≥m
fk
limk→∞
sup fk = infm≥1
infk≥m
fk.
We now discuss the functions that are building blocks for the theory
of integration.Given A ⊆ Rn, the function χA: R
n → R defined by
χA(x) =
1 if x ∈ A
0 if x /∈ A(2.2.6)
54
is called the characteristic function of the set A. It is easily checkedthat the characteristic function χA of A is µ- measurable if and only
if A is measurable.
A simple function is a function of the form
f(x) =+∞∑i=1
di χAi(x), di ∈ R, Ai ⊆ Rn, Ai mutually disjoint.
If Ai are µ-measurable, then f is called a µµµ-measurable simple
function. Equivalently, f : Rn → R is a µ- measurable simple functionif and only if f is µ- measurable and the range of f is an at most
countable subset of R.The following theorem provides a useful way to decompose of a
nonnegative µ-measurable function.
Theorem 2.2.6. Let f : Ω → [0,+∞] be µ-measurable.Then
there are µ-measurable sets Ak ⊆ Ω such that
f =∞∑k=1
1
kχAK
.
Proof of Theorem 2.2.6: We set
A1 = x : f(x) ≥ 1 = f−1[1,+∞], A1 is µ-measurable.
We define by induction the following sets:
Ak =x : f(x) ≥ 1
k+
k−1∑j=1
1
jχAj
k = 2, 3, . . . . (2.2.7)
Claim 2.2.7. f(x) =∞∑k=1
1
kχAk
(x).
Proof of Claim 2.2.7.
“f(x) ≥∞∑k=1
1
kχAk
(x)”:
55
This inequality follows directly from the definition of Ak if supk : x ∈Ak = +∞. Otherwise, consider k0 = maxk: x ∈ Ak, and we use
the fact that x ∈ Ak0. We get
f(x) ≥k0∑
k=1
1
kχAk
(x)
and since x /∈ Ak for ≥ k0 + 1 we can also write
f(x) ≥∞∑
k=1
1
kχAk
(x).
“f(x) ≤∞∑k=1
1
kχAk
(x)”:
We consider different cases.
i) f(x) = +∞ =⇒ x ∈ Ak, ∀k and∞∑k=1
1
kχAk
(x) =∞∑k=1
1
k=
+∞ = f(x).
ii) f(x) = 0 =⇒ x /∈ Ak, ∀k =⇒∞∑k=1
1
kχAk
(x) = 0.
iii) 0 < f(x) < +∞ =⇒ ∀k0 > 0 : x /∈ ⋂k≥k0
Ak, namely ∀k0 >0 : ∃k = k(k0) ≥ k0 such that x /∈ Ak. (Indeed, if ∃k0 andx ∈ ⋂
k≥k0Ak then χAk
(x) = 1 ∀k ≥ k0 and
∞ =∞∑
k=k0
1
kχAk
(x) ≤ f(x) < +∞.
Hence for infinitely many k, we have x /∈ Ak and thus
0 ≤ f(x)−k∑
n=1
1
nχAn
(x) <1
k. (2.2.8)
By letting k → +∞, we get
f(x) ≤∞∑n=1
1
nχAn
(x). (2.2.9)
56
Remark 2.2.8.
1. Set fk =∑k
j=11j χAj
(x). Then fk is an increasing sequence that
converges to f .
If f is bounded, then the convergence is uniform:
supx∈R
|f(x)− fk(x)| → 0 as k → +∞.
2. Through the approximation of f+ and f− we can approximate ev-
ery µ-measurable function f : Ω → [−∞,+∞] by step-functions.
Proposition 2.2.9. Let f : Ω → R be continuous, µ be a Borel
measure. Then f is µ-measurable.
Proof. Let U ⊆ R be an open set. Then f−1(U) is open and thusµ-measurable, since µ is Borel.
Notation:
The expression “µ-a.e.” means “almost everywhere withrespect of the measure µµµ”, that is except possibly on a
set A with µ(A) = 0.
2.3 Lusin’s and Egoroff’s Theorems
Let µ be a Radon measure on Rn.
Theorem 2.3.1. (Egoroff)
Let Ω ⊆ Rn be µ-measurable with µ(Ω) < +∞. Let fk: Ω → R beµ-measurable, ∀k ∈ N, f : Ω → R be µ-measurable and f(x) finiteµ-a.e.. Moreover fk(x) → f(x) as k → +∞ for µ-a.e. x ∈ Ω.
Then it holds: ∀δ > 0 ∃F ⊆ Ω, F compact with µ(Ω\F ) < δ and
supx∈F
|fk(x)− f(x)| → 0 as k → +∞, (2.3.1)
namely (fk)k converges uniformly to f on F .
57
Exercise 2.3.2. Show that the conclusion of Egoroff’s theorem canfail if we drop the assumption that the domain has finite measure.
Solution:
Consider fk(x) = χ[−k,k](x), fk(x) → 1 pointwise as k → +∞ but it
does not converge uniformly outside any bounded set. If there existsF ⊆ R, F compact such that supF |fk(x)− 1| → 0, then ∃N > 0 such
that [−N,N ]c ⊂ F c which is a contradiction.
Proof of Theorem 2.3.1: Let δ > 0. Define
Cij =
∞⋃
k=j
x ∈ Ω : |fk(x)− f(x)| > 1
2i
(i, j = 1, 2, . . . ). (2.3.2)
Then Ci,j+1 ⊆ Ci,j ∀i, j. Since fk(x) → f(x) for ν-a.e x and µ(Ω) <∞,we have
limj→+∞
µ(Cij) = µ( ∞⋂
j=1
Cij
)= 0. (2.3.3)
Hence for every i, ∃N(i) > 0 such that
µ(Ci,N(i)) <δ
2i+1,
we set
A = Ω\∞⋃
i=1
Ci,N(i), µ(Ω\A) = µ( ∞⋃
i=1
Ci,N(i)
)<
∞∑i=1
δ
2i+1=δ
2. (2.3.4)
Moreover, ∀x ∈ A, ∀i, ∀k ≥ N(i)
|fk(x)− f(x)| ≤ 1
2i. (2.3.5)
Thus fk → f is uniformly on A.
Now let F ⊆ A be compact such that µ(A \ F ) < δ2. From (2.3.4)
it follows,
µ(Ω \ F ) ≤ µ(Ω \ A) + µ(A \ F ) ≤ δ/2 + δ/2. (2.3.6)
We can conclude the proof.
58
The type of convergence involved in the conclusion of Egoroff’stheorem is sometimes called almost uniform convergence.
Any continuous function is µ-measurable. The next result gives a
continuity property that characterizes measurable functions.
Theorem 2.3.3. (Lusin’s Theorem)
Let Ω ⊆ Rn be µ-measurable with µ(Ω) < +∞. Let f : Ω → R be
µ-measurable, finite µ-a.e.. Then ∀ε > 0 ∃K ⊆ Ω compact suchthat µ(Ω \K) < ε and f |K is continuous.
Remark 2.3.4. By f |K we mean the restriction of f to the set K.
The conclusion of the theorem states that if f is viewed as a functiondefined only on K, then f is continuous. However, the theorem does
not make the stronger assertion that the function f defined on Ω iscontinuous at the points of K.
Example 2.3.5.
Ω = [0, 1] ⊆ Rf = χ[0,1]\Q
f : [0, 1] → R is not continuous but f |[0,1]\Q: [0, 1] \ Q → R is
continuous.
Example 2.3.5 shows also that we cannot take ε = 0 in Lusin’s theo-
rem.
Proof of Theorem 2.3.3: For each positive integer i, let Bij ⊆ R bedisjoint Borel sets such that R =
⋃∞j=1Bij and diamBij = sup|x−y|,
x, y ∈ Bij <1
i.
Define Aij = f−1(Bij), Aij is µ-measurable and let
Ω =∞⋃
j=1
Aij (Ω = Ω ∪ f−1±∞).
59
Since µ is a Radon measure, there exists Kij ⊆ Aij compact such that
µ(Aij \Kij) <ε
2i+j.
Then
µ(Ω \
∞⋃
j=1
Kij
)= µ
( ∞⋃
j=1
Aij \∞⋃
j=1
Kij
)
≤ µ( ∞⋃
j=1
(Aij \Kij))
≤( ∞∑
j=1
µ(Aij \Kij)
<∞∑j=1
ε
2i+j=ε
2i.
(2.3.7)
Since
limn→+∞
µ(Ω \
n⋃
j=1
Kij
)= µ
(Ω \
∞⋃
j=1
Kij
)<
ε
2i,
∃N(i) > 0 such that
(Ω \
N(i)⋃
j=1
Kij
)<ε
2i.
Set Di =⋃N(i)
j=1 Kij, Di is compact. For each i and j we fix bij ∈ Bij
and define gi(x): Di → R, gi(x) = bij if x ∈ Kij wit j ≤ N(i).
We observe that gi is continuous since it is locally constant (the setsKij j ≤ N(i) are compact, disjoint sets and so there are at positive
distance apart).
Furthermore, |f(x) − gi(x)| < 1i ∀x ∈ Di (diamBij <
1i ). Set
K =⋂∞
i=1Di, K is compact and
µ(Ω \K) = µ( ∞⋃
i=1
(Ω \Di))≤
∞∑i=1
µ(Ω \Di) < ε.
60
Note that
Ω \K = Ω \K ∪ f−1±∞ \K
µ(Ω \K) ≤ µ(Ω \K) ∪ µ(f−1(±∞ \K)
≤ ε+ 0 .
Since |f(x)−gi(x)| < 1i∀x ∈ Di, we see that gi(x) → f(x) is uniformly
on K as i→ +∞. Thus f |K is continuous.
2.4 Convergence in Measure
Let µ be an arbitrary measure on Rn, Ω ⊆ Rn µ-measurable and let
f, fk: Ω → R be µ-measurable and |f(x)| < +∞ µ-a.e..
Definition 2.4.1. The sequence (fk)k converges in measure µ to
f , in short: fkµ→ f as k → ∞ if ∀ε > 0
limk→+∞
µ(x ∈ Ω : |f(x)− fk(x)| > ε) = 0.
Question:
Which is the relation between convergence in measure, point-wise convergence and uniform convergence?
Theorem 2.4.2. Let µ(Ω) < +∞. If fk → f µ-a.e. as k → +∞then fk
µ→ f as k → +∞.
Proof. From Theorem 2.3.1 it follows that for every δ > 0 ∃Fδ ⊆ Ωcompact with µ(Ω \ Fδ) < δ and ∀ε > 0 ∃nε > 0 such that
supx∈Fδ
|fn(x)− f(x)| < ε ∀n ≥ nε.
Therefore, for n ≥ nε
x ∈ Ω |fn(x)− f(x)| > ε ⊆ Ω \ Fδ. (2.4.1)
61
Hence
µx ∈ Ω : |fn(x)− f(x)| > ε ≤ µ(Ω \ Fδ) < δ
Since δ > 0 is arbitrary, we can conclude.
Remark 2.4.3.
1) Theorem 2.4.2 does not hold in general if µ(Ω) = +∞. Take
fk(x) = χ(−k,k)(x), Ω = R, µ = L1. Then fk(x) → f ≡ 1 in R
but fkµ9 1: ∀ε > 0
L1(x ∈ R : |fk(x)− 1| ≥ ε) = L1((−k, k)c
)= ∞ . (2.4.2)
2) The converse of Theorem 2.4.2 does not hold. Take Ω = [0, 1) ⊆R, µ = L1
fk(x) = χ[k−2n
2n, k+1−2n
2n)(x), k ∈ N.
For every k ≥ 1, the index n is chosen so that 2n ≤ k < 2n+1.
We observe that µ(fk(x) > 0) =1
2n<
2
k→ 0 as k → +∞.
Thus fkµ→ f ≡ 0 as k → +∞. But for every k ∈ Ω, n ∈ N,
k ∈ N, we have
fk(x) =
1 if k = [2nx] + 2n
0 for the remaining k ∈ 2n + 1, . . . , 2n+1 − 1,
where [s] = supn ∈ N, n ≤ s. Thus fk cannot converge tof ≡ 0!
However, it holds the following
Theorem 2.4.4. Let fkµ→ f . Then there exists a subsequence
fkn which converges to f µ-a.e. as n→ +∞.
62
Proof. Since fkµ→ f , there exists kn ≥ 1 such that
µ(x ∈ Ω : |fk(x)− f(x)| > 2−n) < 2−n ∀k ≥ kn.
We define
An = x ∈ Ω : |fkn(x)− f(x)| > 2−n (2.4.3)
and for h ≥ 1 fixedEh =
⋃
n≥h
An.
By the σ-subadditivity of the measure, we have
µ(Eh) ≤∞∑n=h
µ(An) ≤∞∑n=h
1
2n= 2−h+1. (2.4.4)
Let x ∈ Ω \ Eh, then x /∈ An ∀n ≥ h. Thus ∀h ∈ N, we have fkn(x)converges to f(x) on Ω \ Eh.
Set E =⋂∞
h=1Eh, since µ(E1) < +∞ and Eh is decreasing, we have
µ(E) = limh→∞
µ(Eh) = 0. (2.4.5)
Since fkn(x) converges to f(x) ∀x ∈ Ω \Eh, ∀h, then fkn(x) convergesto f(x), ∀x ∈ Ω \ E, and we conclude.
63
Chapter 3
Integration
In this chapter we assume that µ is a Radon measure on Rn and
Ω ⊆ Rn is µ-measurable.
3.1 Definitions and Basic Properties
Definition 3.1.1. A function g: Ω → R is called a simple func-
tion if the image of g is at most countable.
Notation:
f+ = max(f, 0), f− = max(−f, 0)
We recall that f = f+ − f−, |f | = f+ + f−.
Definition 3.1.2. (Integral of simple functions)
If g : Ω → [0,+∞] is a nonnegative, simple, µµµ-measurable
function, we define∫
Ω
g dµ ≡ ∑0≤y≤∞
y µ(g−1y). (3.1.1)
We will use the convention that 0 · ∞ = 0.
64
Definition 3.1.3. If g : Ω → [−∞,+∞] is a simple, µ-measurable function, and either
∫Ω g
+dµ < ∞ or∫Ω g
−dµ < ∞,we call g a µµµ-integrable simple function and define
∫g dµ ≡
∫g+ dµ−
∫g− dµ. (3.1.2)
Therefore if g is µ-integrable simple function:∫g dµ =
∑−∞≤y≤∞
yµ(g−1y). (3.1.3)
Definition 3.1.4. Let f : Ω → [−∞,∞]. We define the upper
integral
∫
Ω
f dµ = inf∫
Ω
g dµ : g a µ-integrable simple function,
g ≥ f µ-a.e.∈ R (3.1.4)
and the lower integral
∫
Ω
f dµ = sup∫
Ω
g dµ : g a µ-integrable simple function,
g ≤ f µ-a.e.∈ R. (3.1.5)
Definition 3.1.5. A µ-measurable function f : Ω → R is called
µµµ-integrable if∫Ω f dµ =
∫Ω f dµ, in which case we write
∫
Ω
f dµ ≡∫
Ω
f dµ =
∫
Ω
f dµ. (3.1.6)
65
Exercise 3.1.6. Show that
∫
Ω
f dµ ≤∫
Ω
f dµ. (3.1.7)
Warning:
For us a function is “integrable” provided it hasan integral, even if this integral equals +∞ or −∞(see Evans-Gariepy).
Proposition 3.1.7. Let f : Ω → [0,+∞] µ-measurable. Then fis µ-integrable.
Proof. We may suppose that∫Ω f dµ < +∞.
(Indeed, if∫Ω f dµ = +∞, then
∫Ω f dµ = +∞ as well).
In particular, we have f(x) < +∞ µ-a.e.
• Case 1: Assume µ(Ω) < +∞. For ε > 0 given, we set
Ak = x ∈ Ω : kε ≤ f(x) < (k + 1) ε, k ≥ 0. (3.1.8)
We have ∪k≥0Ak = ∪k≥0f−1([kε, (k + 1)ε)) = Ω. We define the
simple functions
e(x) = ε∞∑k=0
k χAk(x) and
g(x) = ε∞∑k=0
(k + 1)χAk(x).
(3.1.9)
Clearly, we have e(x) ≤ f(x) < g(x) and
66
∫
Ω
e dµ ≤∫
Ω
f dµ ≤∫
Ω
f dµ
≤∫
Ω
g dµ ≤∫
Ω
e dµ+ εµ(Ω).
(3.1.10)
We let ε→ 0 and get the result.
• Case 2: Let Ω ⊆ Rn be a general µ-measurable set. We considerRn =
⋃∞ℓ=1Qℓ, Qℓ disjoint dyadic cubes.
For ε > 0 there are simple functions eℓ, gℓ: Ω ∩Qℓ → R with
eℓ ≤ f ≤ gℓ in Ωℓ = Ω ∩Qℓ
and ∫
Ωℓ
eℓ dµ ≤∫
Ωℓ
gℓ dµ ≤∫
Ωℓ
eℓ dµ+ε
2ℓ. (3.1.11)
Observe that∑∞
ℓ=1 eℓ χΩℓand
∑∞ℓ=1 gℓ χΩℓ
are simple functionswith ∞∑
ℓ=1
eℓ χΩℓ≤ f(x) ≤
∞∑ℓ=1
gℓ χΩℓ.
We sum up in (3.1.11) over ℓ and we get
∞∑ℓ=1
∫
Ω∩Qℓ
eℓ dµ ≤∫
Ω
f dµ ≤∫
Ω
f dµ ≤∞∑ℓ=1
∫
Ω∩Qℓ
eℓ dµ+ ε.
We let ε→ 0 and conclude the proof.
Definition 3.1.8.
i) A function f : Ω → R is µµµ-summable if f is µ-measurable
and ∫
Ω
|f | dµ < +∞.
ii) A function f : Ω → R is locally µµµ-summable in Ω, if f |Kis µ-summable for each compact set K ⊆ Ω.
67
Proposition 3.1.9.
i) If f is µ-summable, then it is µ-integrable.
ii) If f = 0 µ-a.e., then f is µ-integrable and∫Ω f dµ = 0.
Proof of Proposition 3.1.9:
i) We have f = f+ − f−. Since 0 ≤ f± ≤ |f | the functions f± areµ-integrable and
∫Ω f± dµ < +∞. For ε > 0 ∃ simple functions
0 ≤ e± ≤ f± ≤ g± µ-a.e. with
0 ≤∫
Ω
e± dµ ≤∫
Ω
f± dµ ≤∫
Ω
g± dµ
≤∫
Ω
e± dµ+ ε < +∞
(see the proof of Proposition 3.1.7).
Thus e := e+ − g−, g := g+ − e− are simple functions satisfying
e ≤ f ≤ g µ-a.e.
and it holds
∫
Ω
e dµ ≤∫
Ω
f dµ ≤∫
Ω
f dµ ≤∫
Ω
g dµ
≤∫
Ω
e dµ+ 2ε.
We let ε→ 0 and get the result.
ii) It is trivial (choose e = g = 0). In this case we get∫Ω f dµ ≥ 0
and∫Ω f dµ and therefore
∫
Ω
f dµ =
∫
Ω
f dµ = 0.
68
We conclude the proof.
Proposition 3.1.10. Let f : Ω → [0,+∞] be µ-measurable.
i) If∫Ω f dµ = 0 =⇒ f(x) = 0 µ-a.e.
ii) If∫Ω f dµ < +∞ =⇒ f(x) < +∞ µ-a.e.
Proof of Proposition 3.1.10:
i) Assume f is not zero µ-a.e.. Consider the measurable sets
Ak =x : f(x) ≥ 1
k
, k ≥ 1.
Note that Ak ⊆ Ak+1 and⋃
k
Ak = x ∈ Ω : f(x) > 0.
We have
0 < µx ∈ Ω : f(x) > 0 = limk→+∞
µ(Ak). (3.1.12)
Therefore there exists k ≥ 1 such that µ(Ak) > 0. The func-
tions s(x) = 1k χAk
(x) is a simple function such that s(x) ≤ f and∫Ω s(x) dµ = 1
k µ(Ak) > 0 =⇒∫Ω f dµ > 0, which is a contradic-
tion.
ii) Suppose that f(x) = +∞ ∀x ∈ A with µ(A) > 0. Then thesimple function s(x) = +∞χA(x) satisfy 0 ≤ s ≤ f , with∫Ω s(x) dµ = +∞. Hence
∫Ω f dµ = +∞ which is a contradic-
tion.
Proposition 3.1.11. (Monotonicity)
Let f1, f2: Ω → R be µ-integrable with f1 ≥ f2 µ-a.e.. Then∫
Ω
f1 dµ ≥∫
Ω
f2 dµ. (3.1.13)
69
Proof. If g ≥ f1 and f1 ≥ f2 µ-a.e., then g ≥ f2 µ-a.e.. Therefore
∫
Ω
f1 dµ =
∫
Ω
f1 = infg
∫
Ω
g dµ ≥∫
Ω
f2 dµ
=
∫
Ω
f2 dµ,
(3.1.14)
where the infimum is taken over all simple µ-integrable function g ≥ f1µ-a.e..
Corollary 3.1.12. Let f1, f2: Ω → R integrable with f1 = f2µ-a.e. Then ∫
Ω
f1 dµ =
∫
Ω
f2 dµ. (3.1.15)
Proof. It follows directly from Proposition 3.1.11.
Theorem 3.1.13. (Tchebychev Inequality)
Let f : Ω → R be µ-summable. Then for every a > 0 it holds
µ(x ∈ Ω : |f(x)| > a) ≤ 1
a
∫
Ω
|f | dµ. (3.1.16)
Proof. Choose in Proposition 3.1.11 f1 = |f | and f2 = aχx∈Ω:|f(x)|>a.
Corollary 3.1.14. Let f, fk: Ω → R be µ-integrable with
limk→+∞
∫
Ω
|fk − f | dµ = 0.
Then fkµ→ f as k → +∞ and fkn → f µ-a.e. for a subsequence
fkn.
70
Proof. From Theorem 3.1.13 it follows that for ∀ε > 0
µx ∈ Ω : |fk − f | > ε < 1
ε
∫
Ω
|fk − f | → 0 as k → +∞. (3.1.17)
Therefore, fkµ→ f as k → +∞ and the second part is a consequence
of Theorem 2.4.4.
Theorem 3.1.15. Let f, g: Ω → R be µµµ-summable, λ ∈ R.Then f + g, λf are µ-summable and
∫
Ω
(f + g) dµ =
∫
Ω
f dµ+
∫
Ω
g dµ
∫
Ω
λf dµ = λ
∫
Ω
f dµ.
(3.1.18)
Proof.
i) The two relations hold for µ integrable simple functions of the
form
f(x) =∞∑k=1
ak χAkand g(x) =
∞∑k=1
bk χBk(3.1.19)
if both satisfy either∫Ω f
+dµ < +∞ and∫Ω g
+dµ < +∞ or∫Ω f
−dµ < +∞ and∫Ω g
−dµ < +∞. Such conditions are satisfiedfor instance if we assume that both f and g are µ-summable.
We observe that in the representation of f and g in (3.1.19) can
always suppose that Ak = Bk ∀k ∈ N. Otherwise we can consider(Ak∩Bℓ)k,ℓ∈N (we leave all the statements in this part of the proof
as an exercise).
ii) We know that f + g is µ-measurable. For ε > 0 we choose simple
µ-integrable functions fε, fε, gε, g
ε such that
fε ≤ f ≤ f ε and gε ≤ g ≤ gε
∫
Ω
f εdµ−∫
Ω
f dµ < ε and
∫
Ω
f dµ−∫
Ω
fε < ε
∫
Ω
gεdµ−∫
Ω
g dµ < ε and
∫
Ω
g dµ−∫
Ω
gε < ε
71
Thus fε+ gε, fε+ gε are µ-measurable simple functions such that
fε + gε ≤ f + g ≤ f ε + gε (3.1.20)
Moreover since we are assuming that f, g are both µ-summable
we have in particular
∫
Ω
(f ε)−dµ < +∞ and
∫
Ω
(gε)−dµ < +∞,∫
Ω
(fε)+dµ < +∞
∫
Ω
(f ε)+dµ < +∞. It follows that fε + gε,
f ε+ gε are µ-integrable since
∫
Ω
(f ε+ gε)− dµ < +∞ and
∫
Ω
(fε+
gε)+ dµ < +∞ and therefore
∫
Ω
(f ε + gε) dµ =
∫
Ω
f εdµ+
∫
Ω
gεdµ (3.1.21)∫
Ω
(fε + gε) dµ =
∫
Ω
fεdµ+
∫
Ω
gεdµ. (3.1.22)
By combining (3.1.20) and (3.1.21) and (3.1.22) we get
∫
Ω
f + g dµ ≤∫
Ω
(f ε + gε) dµ =
∫
Ω
f εdµ+
∫
Ω
gεdµ
≤∫
Ω
f dµ+
∫
Ω
g dµ+ 2ε
∫
Ω
(f + g) dµ ≥∫
Ω
fε dµ+
∫
Ω
gε dµ
≥∫
Ω
f dµ+
∫
Ω
g dµ− 2ε.
Since ε > 0 is arbitrary, it follows that f + g is µ-integrable with∫
Ω
(f + g) dµ =
∫
Ω
f dµ+
∫
Ω
g dµ.
iii) We apply the result to |f | and |g| and we get∫
Ω
|f + g|dµ ≤∫
Ω
(|f |+ |g|) dµ
=
∫
Ω
|f | dµ+
∫
Ω
|g| fµ < +∞.(3.1.23)
Thus f + g is µ-summable and a) is proved.
72
iv) Let ε > 0, choose simple functions fε ≤ f ≤ f ε such that∫
Ω
f εdµ−∫
Ω
f dµ < ε
∫
Ω
f dµ−∫
Ω
fε dµ < ε.
Thus λ fε, λ fε are simple functions with
λ fε ≤ λf ≤ λf ε if λ ≥ 0
and
λ fε ≥ λf ≥ λf ε if λ < 0.
We consider λ > 0λ > 0λ > 0 (the case λ = 0 is trivial and λ < 0 is similar to
the case λ > 0). It follows
∫
Ω
λf dµ ≤∫
Ω
λf εdµ = λ
∫
Ω
f εdµ
≤ λ
∫
Ω
f dµ+ λε
(3.1.24)
and∫
Ω
(λf) dµ ≥ λ
∫
Ω
fε dµ
≥ λ
∫
Ω
f dµ− λε.
(3.1.25)
Thus λf is µ-integrable with∫
Ω
(λf) dµ = λ
∫
Ω
f dµ. (3.1.26)
Moreover,∫
Ω
|λf | dµ =
∫
Ω
|λ| |f | = |λ|∫
Ω
|f | dµ < +∞. (3.1.27)
Thus λf is µ-summable as well and b) is proved.
73
Corollary 3.1.16. Let f : Ω → R be µ-summable. Then
∣∣∣∫
Ω
f dµ∣∣∣ ≤
∫
Ω
|f | dµ. (3.1.28)
Proof. We have −|f | ≤ f ≤ |f |. Thus
−∫
Ω
|f | dµ ≤∫
Ω
f dµ ≤∫
Ω
|f | dµ (3.1.29)
and we can conclude.
Lemma 3.1.17. Let f : Ω → R be µ-summable and Ω1 ⊆ Ω be
µ-measurable. Then f1 = f |Ω1and fχΩ1
are µ-summable on Ω1
and Ω respectively and∫
Ω1
f1 dµ =
∫
Ω
f χΩ1dµ.
Proof. Exercise!
Corollary 3.1.18. Let f : Ω → R be a µ-summable function and
Ω1 ⊆ Ω with µ(Ω1) = 0. Then∫
Ω1
f dµ = 0. (3.1.30)
Proof. We observe that f χΩ1= 0 µ-a.e.. Therefore
∫Ω f χΩ1
dµ = 0and the result follows from Lemma 3.1.17.
Proposition 3.1.19. Let f : Ω → R be µ-summable, Ω = Ω1∪Ω2,Ω1, Ω2 µ-measurable sets, Ω1 ∩ Ω2 = ∅. Then
∫
Ω
f dµ =
∫
Ω1
f dµ+
∫
Ω2
f dµ. (3.1.31)
74
Proof. We have f = f χΩ1+ f χΩ2
. The result follows from Lemma3.1.17 and Corollary 3.1.18.
3.2 Comparison between Lebesgue and Riemann-
Integral
3.2.1 Review of Riemann Integral
Let I = [a, b] be an interval of R and let P = a = x0, x1, . . . , xn = bbe a partition of I.
If f : I → R is a bounded function and P is a partition of I, we
define the upper and the lower Riemann sums respectively by
S(P, f) =n∑
i=1
(xi − xi−1) supx∈(xi−1,xi]
f(x)
=n∑
i=1
(xi − xi−1)Mi
s(P, f) =n∑
i=1
(xi − xi−1) infx∈(xi−1,xi]
f(x)
=n∑
i=1
(xi − xi−1)mi.
(3.2.1)
Let P be the set of all partitions of I. We define
R∫ b
a
f(x) dx = infS(P, f), P ∈ P
R∫ b
a
f(x) dx = sups(P, f), P ∈ P.(3.2.2)
We say that a bounded function f is Riemann integrable if
R∫ b
a
f(x) dx = R∫ b
a
f(x) dx =:
∫ b
a
f(x) dx.
75
If
ϕ(x) =n∑
i=1
ci χ(xi−1,xi] (3.2.3)
where xii=0,...,n partition of I = [a, b] and c1, . . . , cn ∈ R, its Lebesgueintegral is defined by
∫
[a,b]
ϕ(x) dx =n∑
i=1
ci(xi − xi−1). (3.2.4)
Thus we can write
S(P, f) =
∫
[a,b]
ϕ(x) dx
s(P, f) =
∫
[a,b]
ϕ(x) dx,
(3.2.5)
where
ϕ(x) =n∑
i=1
Mi χ(xi−1,xi] (3.2.6)
ϕ(x) =n∑
i=1
mi χ(xi−1,xi] (3.2.7)
It is clear that ϕ(x) ≤ f(x) ≤ ϕ(x). In view of the above considera-tions we can write
R∫ b
a
f(x) dx = inf∫
[a,b]
ϕ(x) dx, ϕ as in (3.2.3), ϕ ≥ f in I
R∫ b
a
f(x) dx = sup∫
[a,b]
ϕdx, ϕ as in (3.2.3), ϕ ≤ f in I.
(3.2.8)
76
Example 3.2.1. We know that the Dirichlet function χ‖Q∩[0,1] =: fis not Riemann integrable. Indeed, for every partition P of [0, 1],
we have S(P, f) = 1 and s(P, f) = 0. Therefore
R∫ 1
0
f(x) dx = 1, R∫ b
0
f(x) dx = 0. (3.2.9)
Moreover, if we consider the sequence
fn(x) = χr1,...,rn(x),
where rn is an enumeration of Q ∩ (0, 1], we have fn(x) is R-integrable and R
∫0 fn(x) dx = 0 and fn ↑ χ[0,1]∩Q as n→ +∞.
It makes no sense to wonder if
limn→+∞
R∫ 1
0
fn(x) dx = R∫ 1
0
χQ∩[0,1](x) dx.
Unlike the Riemann integral the Lebesgue integral is such that the
characteristic functions of L1-measurable sets are L1-integrable andit behaves well under “passage to the limit” as we will see in thesequel.
Proposition 3.2.2. Let f : [a, b] → R be a bounded R-integrablefunction. Then it is L1-integrable and
R∫ b
a
f(x) dx =
∫
[a,b]
f dL1.
Proof. We observe that every simple function of the type ϕ(x) =∑ni=1 ci χ(xi−1,xi] is aL1-integrable simple function since (xi−1, xi] are
77
L1-measurable. Moreover the following holds:
∫
[a,b]
ϕ(x) dx, ϕ as in (3.2.3), ϕ ≥ f in I
⊆∫
[a,b]
g dL1 : g a µ-integrable simple function, g ≥ f µ-a.e. in I
∫
[a,b]
g dL1 : g a µ-integrable simple function, g ≤ f µ-a.e. in I
⊆∫
[a,b]
ϕdx, ϕ as in (3.2.3), ϕ ≤ f in I
It follows that
R∫ b
a
f(x) dx ≤∫
[a,b]
f dL1 ≤∫
[a,b]
f dL1 ≤ R∫ b
a
f dx.
3.3 Convergence Results
Let µ be a Radon measure on Rn and Ω be µ-measurable.
Theorem 3.3.1. (Fatou’s Lemma)
Let fk: Ω → [0,+∞] be µ-measurable, for all k ∈ N. Then itholds ∫
Ω
lim infk→+∞
fk dµ ≤ lim infk→+∞
∫
Ω
fk dµ. (3.3.1)
Proof. Note that all fk’s are µ-integrable and hence f := lim infk→∞ fkas well. It is sufficient to show that
∫
Ω
g dµ ≤ lim infk→∞
∫
Ω
fk dµ
for every simple function g =∑∞
j=0 aj χAjwith g ≤ f .
78
Without loss of generality we can assume that g ≥ 0, a0 = 0, aj >0 ∀j ≥ 1, Ai ∩ Aj = ∅ (for i 6= j).
We fix 0 < t < 1 and for j ∈ N it holds Aj =∞⋃k=1
Bj,k, where
Bj,k = x ∈ Aj : fℓ(x) > t aj for ℓ ≥ k.
Clearly: Bj,k ⊆ Bj,k+1 and
limk→∞
µ(Bj,k) = µ(Aj). (3.3.2)
Fix J, k ∈ N:∫
Ω
fk dµ ≥J∑
j=1
∫
Aj
fk dµ ≥J∑
j=1
∫
Bj,k
fk dµ
≥ t ·J∑
j=1
aj µ(Bj,k).
• We let first k → +∞:
lim infk→∞
∫
Ω
fk, dµ(3.3.2)≥ t ·
J∑j=1
aj µ(Aj).
• Then we let J → +∞:
lim infk→∞
∫
Ω
fk, dµ ≥ t∞∑j=1
aj µ(Aj) = t ·∫
Ω
g dµ.
• Finally, we let t→ 1 and we conclude the proof.
Example 3.3.2. The condition fk ≥ 0 in Theorem 3.3.1 is neces-
sary. Take µ = Ln, Ω = Rn, fk = − 1kn χBk(0). Then:
lim infk→∞
∫
Ω
fk, dµ = −w1,
as
∫
Rn
fk(x) dx = −Ln(B(0, 1)
)= −w1,
but lim infk→+∞
fk = f ≡ 0.
79
Theorem 3.3.3. (Monotone Convergence Theorem / BeppoLevi’s Theorem)
Let fk: Ω → [0,+∞] be µ-measurable, for all k ≥ 1 be such that
f1 ≤ · · · ≤ fk ≤ fk+1 ≤ . . . . Then it holds∫
Ω
limk→∞
fk dµ = limk→∞
∫
Ω
fk dµ. (3.3.3)
Proof. Since (fk)k is a non decreasing sequence, for every j it holds∫Ω fj dµ ≤
∫Ω limk→∞ fk dµ (j = 1, . . . ) and hence:
limk→∞
∫
Ω
fk dµ ≤∫
Ω
limk→∞
fk dµ.
The opposite inequality follows from Fatou’s Lemma:∫
Ω
limk→∞
fk dµ =
∫
Ω
lim infk→∞
fk dµ
Fatou′s Lemma≤ lim inf
k→∞
∫
Ω
fk dµ = limk→∞
∫
Ω
fk dµ.
Remark 3.3.4. We can also start with Beppo Levi’s Theorem anddeduce Fatou’s Lemma by setting for every k ≥ 1: gk := infℓ≥k fℓ We
have 0 ≤ g1 ≤ gk−1 ≤ gk ≤ fℓ (ℓ ≥ k ≥ 1) and
limk→+∞
gk = lim infk→+∞
fk,
∫gk ≤ inf
ℓ≥k
∫fℓ dµ.
From Beppo Levi’s Theorem it holds∫
Ω
lim infk→∞
fk dµ =
∫
Ω
limk→+∞
gk dµ = limk→+∞
∫
Ω
gk dµ
≤ lim infk→+∞
∫fk dµ.
80
Theorem 3.3.5. (Dominated Convergence / Lebesgue Theorem)
Let g : Ω → [0,∞] be µ-summable and f : Ω → R, fkk : Ω → R
be µ-measurable. Suppose |fk| ≤ g and fk → f µ-a.e. as k →+∞. Then:
limk→+∞
∫
Ω
|fk − f | dµ = 0. (3.3.4)
Moreover:
limk→+∞
∫
Ω
fk dµ =
∫
Ω
f dµ. (3.3.5)
Proof. We observe that
|f | = limk→∞
|fk| ≤ g µ-a.s. .
Thus:
|fk − f | ≤ |fk|+ |f | ≤ 2g (3.3.6)
and hence |fk − f | is µ-summable. Then:∫
Ω
2g dµ =
∫
Ω
lim infk→∞
(2g − |fk − f |) dµ
Fatou′sLemma≤ lim inf
k→∞
∫
Ω
(2g − |fk − f |) dµ
=
∫
Ω
2g dµ− lim supk→∞
∫
Ω
|fk − f | dµ
=⇒ limk→∞
∫
Ω
|fk − f | dµ = 0.
(3.3.7)
The equation (3.3.5) follows from the fact that
∣∣∣∫
Ω
fk dµ−∫
Ω
f dµ∣∣∣ ≤
∫
Ω
|fk − f | dµ.
81
3.4 Application: Differentiation of under integral
sign
Theorem 3.4.1. Let f : R × [0, 1] → R be a given function sat-isfying the following conditions:
i) ∀x: y 7→ f(x, y) is L1-summable on [0, 1].
ii) ∂f
∂xexists and it is bounded in R× [0, 1].
Then the map y 7→ ∂f∂x (x, y) is, for every fixed x ∈ R, L1-
summable on [0, 1] and
d
dx
(∫ 1
0
f(x, y) dy)=
∫ 1
0
∂f
∂x(x, y) dy.
Proof. For x ∈ R, h 6= 0, we have
1
h
(∫ 1
0
f(x+h, y) dy−∫ 1
0
f(x, y) dy)=
(∫ 1
0
1
h
(f(x+h, y)−f(x, y)
)dy
).
(3.4.1)
Given hk → 0 as k → +∞, it holds
ghk(y) =
1
hk
(f(x+ hk, y)− f(x, y)
)→ ∂f
∂x(x, y), as k → +∞.
(3.4.2)
Moreover by the mean value theorem and the assumption ii) we have
|ghk(y)| ≤ sup
R×[0,1]
∣∣∣∂f
∂x(x, y)
∣∣∣≤C. (3.4.3)
∂f
∂x(x, ·) is L1-measurable with respect to y ∈ [0, 1] and L1-summable
since it is limit of L1-measurable function and bounded.
By Dominating Convergence Theorem we have:
∫ 1
0
gk(y) dy →∫ 1
0
∂f
∂x(x, y) dy, as k → +∞.
82
This holds ∀ sequence hk → 0. Hence:
d
dx
∫ 1
0
f(x, y) dy = limh→0
1
h
(∫ 1
0
f(x+ h, y) dy −∫ 1
0
f(x, y) dy)
= limh→0
1
h
∫ 1
0
(f(x+ h, y)− f(x, y)
)dy =
∫ 1
0
∂f
dx(x, y) dy.
3.5 Absolute Continuity of Integrals
Let f : Ω → R be µ-summable. For A ⊆ Ω µ-measurable, we set
ν(A) =
∫
A
f dµ.
According to Corollary 3.1.18 we have
µ(A) = 0 =⇒ ν(A) = 0. (3.5.1)
Exercise 3.5.1. ν is σ-additive, ν is a Radon measure in the case
f ≥ 0 µ-a.e. .
Notation:
ν = µ f .
In particular we have
ν = µ χA = µ A.
Let Σµ and Σν be the σ-algebra respectively of the µ and ν-measurablesets A ⊆ Rn.
Definition 3.5.2. A measure ν such that Σν ⊆ Σµ and with theproperty (3.5.1) is called “absolutely continuous” with respect to
µ and we write ν ≪ µ.
83
Theorem 3.5.3. Let f : Ω → R be µ-summable. Then ∀ε >0 ∃δ > 0 such that
∀A ⊆ Ω µ-measurable with µ(A) < δ =⇒∫
A
|f | dµ < ε.
Proof. We assume by contradiction that there exist ε > 0, and asequence of µ-measurable subsets Ak ⊆ Ω with µ(Ak) < 2−k and
∫
Ak
|f | dµ ≥ ε ∀k ∈ N.
For ℓ ≥ 1 we define
Bℓ :=
∞⋃
k=ℓ
Ak.
The following properties hold
• Bℓ+1 ⊆ Bℓ.
• ∀ℓ: µ(Bℓ) ≤∞∑k=ℓ
µ(Ak) < 21−ℓ.
• µ(B1) < +∞.
•∫
Bℓ
|f | dµ ≥∫
Ak
|f | dµ ≥ ε ∀k ≥ ℓ.
• For A =∞⋂ℓ=1
Bℓ: µ(A) = limℓ→+∞
µ(Bℓ) = 0.
Now observe that gℓ := |f | · χBℓ→ |f | · χA as ℓ → +∞ and |gℓ| ≤ |f |
µ-summable, ∀ℓ.
By Dominated Convergence Theorem it follows
ε ≤ limℓ→+∞
∫
Bℓ
|f | dµ = limℓ→+∞
∫
Ω
gℓ dµ
∫
Ω
limℓ→∞
gℓ dµ =
∫
A
|f | dµ = 0,
We get a contradiction and we can conclude.
84
3.6 Vitali’s Theorem
We formulate in this section a necessary and sufficient condition to
pass to the limit under integral sign.
Let f , fk: Ω → R be µ-summable, k ∈ N.
Definition 3.6.1. The family fk is called uniformly µµµ-
summable if ∀ε > 0 ∃δ > 0: ∀k ∈ N, ∀A ⊆ Ω µ-summable
with µ(A) < δ we have
∫
A
|fk| dµ < ε.
Theorem 3.6.2. (Vitali’s Theorem)
If µ(Ω) < +∞, the following conditions are equivalent:
i) fkµ→ f (as k → +∞) and fk is uniformly µ-summable.
ii) limk→+∞
∫
Ω
|fk − f | dµ = 0.
Remark 3.6.3. The condition µ(Ω) < +∞ is in general necessary:Ω = Rn, µ = Ln, fk =
1kn χB(0,k). We have:
(fk) is uniformly µ-summable, fkµ→ 0 as k → +∞. (3.6.1)
But∫Rn |fk| dµ = wn = Ln(B(0, 1)) > 0 and
∫Rn f dµ = 0.
Proof of Vitali’s Theorem:
ii) =⇒ i):
If∫Ω |fk − f | dµ → 0 as k → +∞, then fk
µ→ f by Corollary 3.1.14.Now we show that (fk is uniformly µ-summable.
For ε > 0, let k0 = k0(ε) > 0 be such that∫
Ω
|fk − f | dµ < ε, for all k ≥ k0.
85
Let δ > 0 be such that ∀A ⊆ Ω µ-measurable, for µ(A) < δ, we have∫
A
|f | dµ < ε (3.6.2)
and
max1≤k∈k0
∫
A
|fk| dµ < ε (3.6.3)
(see Theorem 3.5.3).
Thus for k ≥ k0 and if µ(A) < δ, we have∫
A
|fk| dµ ≤∫
A
|f | dµ+
∫
A
|fk − f | dµ
≤ 2ε .
(3.6.4)
Thus, we have the desired property.
i) =⇒ ii):
Let fkµ→ f as k → +∞ and be uniformly µ-summable. Since fk
µ→ f ,
there exists a sub-sequence fkn → f µ-a.e. as n → +∞ (see Theorem2.4.4).
Given ε > 0, we choose δ > 0 such that for A ⊆ Ω, µ(A) < δ, we
have ∫
A
|f | dµ < ε and
∫
A
|fk| dµ < ε ∀k ≥ 1. (3.6.5)
Correspondent to such a δ > 0 we can find by Egoroff-Theorem a
closed set F ⊆ Ω such that µ(Ω \ F ) < δ and
supx∈F
|fkn(x)− f(x)| → 0 as n→ +∞.
Let n be such that
supx∈F
|fkn(x)− f(x)| < ε
µ(Ω), ∀n ≥ n.
86
Thus for n ≥ n:∫
Ω
|fkn(x)− f(x)| dµ =
∫
Ω\F|fkn − f | dµ+
∫
F
|fkn − f | dµ
≤∫
Ω\F|fkn|+ |f | dµ+
∫
F
supF
|fkn − f | dµ
< 2ε+ε
µ(Ω)µ(F ) < 3ε.
(3.6.6)
Since ε > 0 is arbitrary, we have
limn→+∞
∫
Ω
|fkn(x)− f(x)| dµ = 0. (3.6.7)
Claim 3.6.4. The whole sequence fk satisfies (3.6.7).
Proof of Claim 3.6.4: Suppose that
lim supk→+∞
∫
Ω
|fk(x)− f(x)| dµ > 0.
Then there exists a subsequence fkn such that
limn→+∞
∫
Ω
|fkn(x)− f(x)| dµ = lim supk→+∞
∫
Ω
|fk(x)− f(x)| dµ > 0.
At this point we can repeat the same arguments as above to fkn and
get that ∃fk′n subsequence of fkn such that
limn→+∞
∫
Ω
|fk′n − f | dµ = 0,
which is a contradiction.
Vitali’s Theorem permits to improve Lebesgue’s Theorem.
87
Theorem 3.6.5. Let µ(Ω) < +∞ and fn be a sequence of µ-
measurable functions such that fkµ→ f as k → ∞ . Suppose
that ∀n ∈ N |fn(x)| ≤ |gn(x)| µ-a.e., where (gn)n is a sequence of
µ-summable functions such that
limn→+∞
∫
Ω
|gn(x)− g(x)| dµ = 0
for a µ-summable functions g. Then
limn→+∞
∫
Ω
|fn(x)− f(x)| dµ = 0. (3.6.8)
Proof. Since limn→+∞∫Ω gn(x) − g(x)| dµ = 0, by Vitali’s Theorem,
the sequence gn is uniformly µ-summable, namely ∀ε > 0 ∃δ > 0such that if µ(A) < δ, then
supn∈N
∫
A
|gn(x)| dµ ≤ ε. (3.6.9)
Hence, we have∫
A
|fn(x)| dµ ≤∫
A
|gn(x)| dµ < ε ∀n ∈ N, if µ(A) < δ. (3.6.10)
Thus the sequence (fn)n is also uniformly µ-summable and from Vi-tali’s Theorem, we get the result and we conclude.
Theorem 3.6.6. Let µ(Ω) < +∞ and uk, u: Ω → R be µ-measurable functions such that
∫
Ω
|u|pdµ < +∞,
∫
Ω
|uk|pdµ < +∞∫
Ω
|uk − u|pdµ→ 0 as k → +∞,
where 1 ≤ p < +∞. Then∫
Ω
uk dµ→∫
Ω
u dµ as k → +∞. (3.6.11)
88
Proof. The case p = 1 is trivial. Suppose p > 1.
1) Tchebychev Inequality yields
µx : |uk(x)− u(x)| > ε = µx : |uk(x)− u(x)|p > εp
≤ ε−p
∫
Ω
|uk(x)− u(x)|pdµ→ 0
as k → +∞.
(3.6.12)
Hence ukµ→ u as k → +∞.
ii) (uk)k are uniformly µ-summable.
Indeed for ε > 0 there exists δ > 0, k > 0 such that
µ(A) < δ =⇒∫
A
|u|pdµ < ε and
∫
Ω
|uk − u|pdµ < ε ∀k ≥ k.
We have
|uk|p ≤ (|u|+ |uk − u|)p ≤ 2p−1(|u|p + |uk − u|p). (3.6.13)
Thus, for every A ⊆ Ω µ-measurable we have∫
A
|uk| dµ ≤∫
A
(1 + |uk|p) dµ
≤ µ(A) + 2p−1
∫
A
|u|p + |uk − u|pdµ
≤ δ + 2pε ≤ (1 + 2p) ε
(3.6.14)
if µ(A) < δ < ε and k ≥ k. Thus∫Ω |uk − u| dµ → 0 as k → +∞ by
Vitali’s Theorem and thus∫
Ω
uk dµ→∫
Ω
u dµ as k → +∞. (3.6.15)
Exercise 3.6.7.
89
1. By applying Lebesgue’s Theorem to the counting measure on N,verify that
limn→+∞
n∞∑i=1
sin(2−i
n
)= 1.
2. Let f : [a,+∞) → R be a locally bounded function and there-
fore Riemann integrable. Then f is L1-summable iff f is ab-solutely Riemann integrable in the generalized sense (namely∫ +∞a |f(x)| dµ + limj→±∞
∫ j
a |f(x)| dµ exists and it is finite) andin this case
∫
[a,+∞)
f(x) dL1 = limj→+∞
∫ j
a
f(x) dµ.
Proof. 1. Let gj := |f |χ[a,j]. gj are L1-summable since they areRiemann-integrable. Moreover, gj is an increasing sequence which
converges to |f | pointwise.Beppo-Levi’s Theorem says that that |f | is L1-summable iff
limj→+∞
∫
R
gj dL1 < +∞.
Since ∫
R
gj dL1 =
∫ j
a
|f(x)| dx ∀j ≥ 0
this is equivalent to say that |f | is Riemann integrable in a generalizedsense.
2. Now set fj = fχ[a,j]. fj are R-integrable =⇒ fj are L1-
integrable and fj → f in R as j → +∞, |fj| ≤ |f |. We have justsaid that f is L1-summable iff it is absolutely R-integrable in thegeneralized sense.
By Lebesgue Theorem, we have∫
[a,+∞)
f dL1 = limj→∞
∫
R
fj dL1 = limj→+∞
∫ j
a
f(x) dµ. (3.6.16)
It follows that if f is not absolutely R-integrable in the generalized
sense, then it is not L1-summable. For instance, sinxx is not L1-
summable but there exists limj→+∞∫ j
asinxx dx < +∞.
90
In the same way we can prove that if f is locally R-integrable onX = [a, b) (or X = (a, b], X = (a, b) ), then f is L1-summable on X
iff it is R-absolutely integrable in the generalized sense (namely there
exists finite limε→0
∫ b−ε
a |f(x)| dµ) and in this case:
∫
(a,b)
f dL1 = limε→0
∫ b−ε
a
f(x) dx.
3.7 The Lp(Ω, µ) spaces, 1 ≤ p ≤ ∞We suppose as usual that µ is a Radon-measure and Ω is a µ-measurableset.
Definition 3.7.1. For f : Ω → R, s ≤ p <∞, let
‖f‖Lp(Ω,µ) :=(∫
Ω
|f |pdµ) 1
p ≤ +∞
and
‖f‖L∞(Ω,µ) := µ−ess supx∈Ω
|f(x)|= infC ∈ [0,+∞] : |f | ≤ C µ-a.e. ≤ +∞.
For 1 ≤ p ≤ ∞ we set
Lp(Ω, µ) = f : Ω → R : f µ-measurable, ‖f‖Lp(Ω,µ) < +∞.
Remark 3.7.2. For f ∈ L∞(Ω, µ), we have |f(x)| ≤ ‖f‖L∞ for µ-a.e.
x.
Proof. Consider a sequence ck with ck ↓ ‖f‖L∞ as k → ∞.
Set Ak = x ∈ Ω: |f(x)| > ck k ∈ N. By definition µ(Ak) = 0 ∀kand thus A =
⋃k Ak has measure zero as well. Therefore |f(x)| ≤
‖f‖L∞ ∀x ∈ Ω \A.
91
Remark 3.7.3. Consider the Lebesgue measure L1 on [0, 1] and thespace Lp((0, 1],L1). Let us set for every α ∈ R
ϕα(x) = xα ∀x ∈ (0, 1]. (3.7.1)
Then ϕα ∈ Lp((0, 1]) iff α p+ 1 > 0.
Sol: ϕα ∈ Lp((0, 1]) iff there exists finite the following limit
limδ→0
∫ 1
δ
xpα dx = limδ→0
δpα+1
1 + pα
∣∣∣1
α
= limδ→0
[1− δ1+pα
pα + 1
].
(3.7.2)
The last limit is finite iff αp+ 1 > 0 iff p > − 1α . In this case, we have
∫
[0,1]
xpα dL1 = limδ→0
∫ 1
δ
xpα dL1 =1
1 + αp. (3.7.3)
Exercise 3.7.4. Show that Lp((0, 1),L1) is not an algebra.
We observe that ‖ · ‖p in general is not a norm since ‖ϕ‖p = 0 iff
ϕ(x) = 0 for µ-a.e.
Recall: Let Y be a vector space. A norm on Y is a mapping ‖ · ‖:Y → [0,+∞), y 7→ ‖y‖ such that
i) ‖y‖ = 0 ⇐⇒ y = 0.
ii) ‖αy‖ = |α| ‖y‖ ∀α ∈ R, ∀y ∈ Y .
iii) ‖y1 + y2‖ ≤ ‖y1‖+ ‖y2‖ ∀y1, y2 ∈ Y .
The space Y endowed with the norm ‖ · ‖ is called a normed space.It is also a metric space with the distance given by d(y1, y2) = ‖y1−y2‖y1, y2 ∈ Y .
In order to construct a vector space on which ‖ · ‖p is a norm, letus consider the following equivalence relation on Lp(Ω, µ):
ϕ ∼ ψ ⇐⇒ ϕ = ψ µ-a.e.
92
Let us denote by Lp(Ω, µ) the quotient space Lp(Ω, µ)/∼ = [ϕ]: ϕ ∈Lp(Ω, µ). One can verify that Lp(Ω, µ) is a vector space.
The precise definition of addition of two elements [ϕ1], [ϕ2] ∈Lp(Ω, µ) is the following: let f1, f2 be two “representative” of [ϕ1]and [ϕ2], respectively such that f1, f2 are finite everywhere (such rep-
resentatives exist: exercise). Then [ϕ1] + [ϕ2] = [f1 + f2].We set ‖[ϕ]‖p = ‖ϕ‖p ∀[ϕ] ∈ Lp(Ω, µ).
This definition is independent on the particular element ϕ ∈ [ϕ].Then, since the zero element of Lp(Ω, µ) is the class consisting of allfunctions vanishing µ-a.e., it is clear that ‖[ϕ]‖p = 0 ⇐⇒ [ϕ] = 0.
To simplify the notation we will hereafter identify [ϕ] with ϕ and wewill talk about “Functions in Lp(Ω, µ)” when there is no danger of
confusion with the understanding that we regard equivalent functions(i.e. functions differing only on a set of measure zero) as identical
elements of the space Lp(Ω, µ).
Theorem 3.7.5. The space Lp(Ω, µ) is a complete normed space
for 1 ≤ p ≤ ∞.
Proof. Step 1. We first show that Lp(Ω, µ) is a normed space and in
particular that ‖ · ‖Lp is a norm.i) The positive homogeneity is clear:
‖λf‖Lp =(∫
Ω
‖λf‖pdµ) 1
p
= |λ|(∫
Ω
|f |pdµ) 1
p
.
ii) Triangular InequalityWe need some preliminary results.
Lemma 3.7.6. (Young Inequality) Let 1 < p, q <∞ be such that1p +
1q = 1. Then it holds
ab ≤ ap
p+bq
q∀a, b ≥ 0. (3.7.4)
93
Proof of Lemma 3.7.6: We suppose b > 0. The function a 7→ ab − ap
p
with a ≥ 0 has the maximum at a∗ = b1
p−1 . Thus for all a ≥ 0
ab− ap
p≤ b
pp−1 b− b
pp−1
p
= b1
p−1
(1− 1
p
)=bq
q.
(3.7.5)
The numbers p, q such that 1p +
1q = 1 are called conjugate.
Corollary 3.7.7. (Holder Inequality)
Let 1 ≤ p ≤ q ≤ ∞ be conjugate, f ∈ Lp(Ω, µ), g ∈ Lq(Ω, µ).Then it holds f · g ∈ L1(Ω, µ) and
‖fg‖L1 ≤ ‖f‖Lp ‖g‖Lq. (3.7.6)
Proof of Corollary 3.7.7: We suppose p ≤ q without loss of generality
• Case p = 1, q = ∞|fg| ≤ |f | ‖g‖L∞ µ-a.e.
Thus by monotonicity of the integral, we have∫
Ω
|fg| dµ ≤∫
|f | ‖g‖L∞ dµ
= ‖g‖L∞
∫
Ω
|f | dµ.(3.7.7)
• Case 1 < p ≤ q < +∞The conclusion is trivial if ‖f‖Lp = 0 or ‖g‖Lq = 0. Thus we
assume that ‖f‖Lp > 0 and ‖g‖Lq > 0.
We set f(x) = |f(x)|‖f‖Lp
and g(x) = |g(x)|‖g‖Lq
. We apply Lemma 3.7.6 to
f , g:
f g ≤ f p
p+gq
q. (3.7.8)
94
Integrating over Ω, we get∫
Ω
|f g| dµ ≤ 1
p
∫
Ω
f q dµ+1
q
∫
Ω
gq dµ.
Thus ∫
Ω
|f(x) g(x)| dµ
‖f‖Lp ‖g‖Lq
≤ 1. (3.7.9)
and we conclude.
Corollary 3.7.8.
i) Let f ∈ Lp, g ∈ Lq with 1r = 1
p +1q ≤ 1. Then fg ∈ Lr and
‖fg‖Lr ≤ ‖f‖Lp ‖g‖Lq.
ii) If µ(Ω) <∞, then
Ls(Ω) ⊆ Lr(Ω), 1 ≤ r < s ≤ +∞. (3.7.10)
Proof.
i) The property is clear if p = q = r = +∞. If r <∞, the propertyfollows from Corollary 3.7.7: |f |r ∈ L
pr , |g|r ∈ L
qr and 1
pr
+ 1ar
= 1.
ii) If µ(Ω) < +∞. Set g ≡ 1 ∈ Lrss−r . Observe that 1
r= 1
s+ s−r
rsand
apply i).
95
Exercise 3.7.9. Show that In general the inclusion in ii) is strict inthe sense that there are functions in Lr that does not belong to Ls with
r < s.bf Hint: Take
Ω =(0, 1
2
), f(x) =
1
(− logx)2 x1r
, f ∈ Lr,
but f /∈ Ls, with s > r. Another example is Ω = (0, 1), f(x) =log( 1
x) ∈ Lp ∀p ≥ 1, but f /∈ L∞.
Corollary 3.7.10. (Minkowski Inequality)
Let 1 ≤ p ≤ ∞ and f, g ∈ Lp(Ω, µ). Then f + g ∈ Lp and
‖f + g‖Lp ≤ ‖f‖Lp + ‖g‖Lp.
Proof.
• p = 1: This case follows from the pointwise estimate
|f + g| ≤ |f |+ |g|.
• p = ∞: This case follows from the estimate
|(f + g)(x)| ≤ ‖f‖L∞ + ‖g‖L∞ µ-a.e. x ∈ Ω.
• 1 < p <∞.
From the estimate
|(f + g)(x)|p ≤ 2p−1(|f(x)|p + |g(x)|p
)∀x ∈ Ω
it follows that f + g ∈ Lp and |f + g|p−1 ∈ Lq. By applying the Holder
96
inequality, we get
‖f + g‖pLp =
∫
Ω
|f + g| |f + g|p−1 dµ
≤∫
Ω
|f | · |f + g|p−1 dµ+
∫|g| |f + g|p−1 dµ
≤(∫
Ω
|f |p)1
p(∫
Ω
|f + g|p−1· pp−1
)p−1p
+(∫
Ω
|g|p) 1
p(∫
Ω
|f + g|p−1· pp−1
) p−1p
= (‖f‖p + ‖g‖p)(‖f + g‖p−1
p
).
(3.7.11)
By dividing both sides by ‖f + g‖p−1p , we get
‖f + g‖Lp ≤ ‖f‖Lp + ‖g‖Lp. (3.7.12)
Exercise 3.7.11. (Interpolation Inequality)
Let 1 ≤ p < r < q <∞ and f ∈ Lp ∩ Lq. Then f ∈ Lr and
‖f‖Lr ≤ ‖f‖θLp ‖f‖1−θLq ,
where 1r =
θp +
1−θq , θ ∈ [0, 1].
Exercise 3.7.12. Let µ(Ω) < ∞ and 1 ≤ p < ∞. Show that if f :Ω → R is such that f · g ∈ L1(Ω, µ), ∀g ∈ Lp(Ω, µ), then f ∈ Lq(Ω, µ)
∀q ∈ [1, p′), where p′ is the conjugate of p.Hint: Observe that f ∈ L1(Ω, µ). So by taking g = |f |1/p we
deduce that |f |1+1/p ∈ L1(Ω, µ). Iterate the argument.
We continue with the proof of Theorem 3.7.5.
Lemma 3.7.13. (Lp(Ω, µ) is complete)
Let (fk)k∈N be a Cauchy sequence in Lp(Ω, µ). Then there exists
f ∈ Lp such that ‖fk − f‖Lp → 0 as k → +∞.
97
Proof.
p = ∞: ∀ε > 0 ∃kε ∈ N such that
‖fℓ − fm‖L∞ < ε ∀ℓ,m ≥ kε.
For ℓ,m ∈ N we set
Eℓ,m = x ∈ Ω : |fℓ(x)− fm(x)| > ‖fℓ − fm‖L∞. (3.7.13)
By definition of ess-sup, Eℓ,m has measure zero . We define
E0 =⋃
ℓ,m=1
Eℓ,m. (3.7.14)
We have µ(E0) = 0. Now let x ∈ Ω \ E0, then x /∈ Eℓ,m ∀ℓ,m. Thusif ℓ,m ≥ kε, we have
|fℓ(x)− fx(x)| ≤ ‖fℓ − fm‖∞ < ε. (3.7.15)
This implies that (fk(x))k is a Cauchy sequence in R (∀x ∈ Ω \ E0)and since R is complete, it converges to a real number f(x). Thus if(fk(x))k is a Cauchy sequence in L∞(Ω), then it converges µ-a.e. to a
function f(x).
We consider the inequality
|fℓ(x)− fm(x)| ≤ ε ℓ,m ≥ kε, x ∈ Ω \ E0.
Take the limit as m→ +∞ and we get
|fℓ(x)− f(x)| ≤ ε ∀ℓ ≥ kε ∀x ∈ Ω \ E0. (3.7.16)
Since µ(E0) = 0, we have
‖fℓ(x)− f(x)‖L∞ ≤ ε ∀ℓ ≥ kε (3.7.17)
and
‖f‖L∞ ≤ ‖fℓ − f‖L∞ + ‖fℓ‖L∞ ≤ c if ℓ ≥ kε. (3.7.18)
From (3.7.18) it follows that f ∈ L∞(Ω).
98
p <∞:
We are going to construct a convergent subsequence (fkn)n∈N. Onechooses kn so that
‖fℓ − fm‖Lp <1
2n∀ℓ,m ≥ kn.
(One proceeds as follows: choose k1 such that ‖fℓ−fm‖Lp < 12∀ℓ,m ≥
k1, then choose k2 > k1 such that ‖fℓ − fm‖Lp < 122 ∀ℓ,m ≥ k2, etc.).
We claim that fkn converges in Lp. In order to simplify the notation
we write fn instead of fkn, so that
‖fn+1 − fn‖Lp ≤ 1
2n∀n ≥ 1. (3.7.19)
Let gn(x) =∑n
k=1 |fk+1(x)− fk(x)| so that ‖gn‖Lp ≤ 1.
As a consequence of the monotone convergence theorem, gn(x) con-
verges to g(x) µ-a.e. in Ω where g(x) =∑∞
k=1 |fk+1(x)− fk(x)| ∈ Lp.On the other hand for m ≥ ℓ ≥ 2, we have
|fm(x)− fℓ(x)| ≤ |fm(x)− fm−1(x)|+ · · · + |fℓ+1(x)− fℓ(x)|= gm(x)− gℓ−1(x)
≤ g(x)− gℓ−1(x).
(3.7.20)
It follows that µ-a.e. on Ω, fn(x) is a Cauchy sequence and convergesto a limit f(x). We have µ-a.e. on Ω
|f(x)− fℓ(x)| ≤ 2g(x) ∀ℓ ≥ 2 (3.7.21)
and in particular f ∈ Lp.
Finally we conclude by dominated convergence that ‖fn−f‖Lp → 0
since |fn(x)− f |p → 0 and
|fn − f |p ≤ 2p |g|p ∈ L1.
We can easily see that the entire sequence fnfnfn converges to fff .
99
Indeed, for all n ∈ N and k > kn, we have
‖fk − f‖Lp ≤ ‖fk − fkn‖Lp + ‖fkn − f‖Lp
≤ 1
2n+ ‖fkn − f‖Lp.
(3.7.22)
Thus
lim supk→+∞
‖fk − f‖Lp ≤ limn→+∞
( 1
2n+ ‖fkn − f‖Lp
)= 0. (3.7.23)
The proof of Theorem 3.7.5 is concluded as well.
A Banach space X is called separable if it contains a countabledense subset.
Example 3.7.14.
1. (R, | · |). The countable dense subset is Q.
2. (C0([a, b],R), ‖ · ‖∞). The countable subset is the set of
the polynomials with rational coefficients (Stone-WeierstrassTheorem).
Theorem 3.7.15. Let 1 ≤ p <∞. Then it holds
i) Lp(Ω, µ) is separable.
ii) C0c (Ω) is dense in Lp(Ω, µ).
We recall that C0c (Ω) denotes the space of continuous function f : Ω →
R with supp(f) = x ∈ Ω : f(x) 6= 0 ⊆ Ω and supp f is compact.
Remark 3.7.16. For p = ∞ the conditions of Theorem 3.7.15 donot hold in general.
1. Let Ω = [0, 1], µ = L1. For 0 < t < 1 we consider ft(x) := χ[0,t](x)with ‖ft − fs‖L∞ = 1 if t 6= s.
100
The family (ft)t>0 is more than countable. Suppose there is acountable set (ek)k dense in L∞((0, 1]) with
infk
‖ft − ek‖L∞ < 1
2∀t > 0.
We would have
L∞([0, 1]) ⊆ ∪nB(en, 1/2)
in contrast with the fact no pair of functions of the family (ft)t>0
belongs to the same ball.
2. If Ω ⊂ Rn is compact, then C(Ω) is a closed, proper subset ofL∞(Ω) and therefore C0
c (Ω) L∞(Ω).
Proof of Theorem 3.7.15: i) We suppose Ω = Rn.
• Let E be the countable set defined by
E = N∑
k=1
ak χQk, ak ∈ Q, Qk ⊆ Rn is a cube of a dyadic
decomposition of length 2−ℓ, ℓ ≥ 0.
(3.7.24)
Claim 3.7.17. E is dense in Lp.
Proof of the Claim 3.7.17: Let f ≥ 0. We know that f is the monotonelimit of a sequence of characteristic functions
fk =k∑
j=1
1
jχAj
.
By Lebesgue Theorem ‖fk−f‖Lp → 0. Therefore it is enough to showthe following property:
Claim 3.7.18. χA ∈ E for every µ-measurable A ⊆ Rn withµ(A) <∞.
101
Proof of the Claim 3.7.18: Since µ is a Radon measure, we have
µ(A) = infA⊆G
µ(G) G open. (3.7.25)
Let us choose Gk ⊇ A open Gk+1 ⊆ Gk with µ(Gk) ≤ µ(A)+ 1k, k ∈ N.
Since χGk− χA ∈ 0, 1, it follows that
∫
R
|χGk− χA|p dµ =
∫
R
|χGk\A|p dµ = µ(Gk \A) ≤1
k. (3.7.26)
Thus we may suppose A open. Every open set can be expressed as
countable disjoint union of dyadic cubes Iℓ: A =⋃∞
ℓ=1 Iℓ and µ(A) =∑∞ℓ=1 µ(Iℓ) (Lemma 1.3.4).
In particular:
χ n⋃ℓ=1
Iℓ=
n∑ℓ=1
χIℓ → χA as n→ +∞.
Since µ(A) <∞ (=⇒ χA ∈ Lp), we have by Lebesgue Theorem
‖χA −n∑
ℓ=1
χIℓ ‖Lp → 0 (3.7.27)
as n→ +∞ and clearlyn∑ℓ=i
χIℓ ∈ E ∀n. (3.7.28)
This concludes the proof of Claim 3.7.18 and 3.7.17.
If f is a general function, then we apply the above arguments tof+ and f−.
• C0c (R
n) is dense in Lp(Rn).
Notation:
Truncation operator
Tn: R→ R (n ≥ 1)
Tn(x) =
x if |x| ≤ n
nx
|x| if |x| > n
(Tn is continuous).
102
Claim 3.7.19. Given f ∈ Lp and ε > 0 ∃g ∈ L∞(R) and K ⊆ Rn
compact such that g ≡ 0 on Kc and
‖f − g‖p <ε
2.
Proof of Claim 3.7.19: Let fn = χB(0,n) Tn(f), n ≥ 1.
We have fn → f and by Lebesgue Theorem ‖fn − f‖Lp → 0 asn→ +∞. Let us choose g = fn, n large enough.
Claim 3.7.20. Let g ∈ L∞(Rn) with compact support and δ > 0.
Them there exists g1 ∈ C0c (R
n) such that
‖g − g1‖L1 < δ.
Proof of Claim 3.7.20: We assume g ≥ 0 =⇒ g is the pointwise limit
of simple functions gn and we can pass to the limit under integral signby Lebesgue Theorem, namely ‖gn − g‖L1 → 0 as n→ +∞.
Therefore we can prove Claim 3.7.20 when g = χA, A measurable
and bounded.
We know that ∃F closed and G open such that
F ⊆ A ⊆ G and µ(G \ F ) | < δ
(we may assume that G is compact). By Urysohn Lemma there is g1:Rn → R continuous such that
a) 0 ≤ g1(x) ≤ 1,
b) g1(x) = 0 on Gc,
c) g1(x) = 1 on F ;
and ∫
Rn
|χA − g1| dµ ≤ µ(G \ F ) < δ.
103
We may always suppose without loss of generality that the functiong1 in Claim 3.7.19 satisfies ‖g1‖L∞ ≤ ‖g‖L∞ (otherwise we take g1(x) =
Tn g1 with n = ‖g‖L∞).
We observe that
‖g − g1‖Lp ≤ ‖g − g1‖1p
L1
(2‖g‖L∞
)p−1p .
If δ > 0 is choosen small enough so that (2‖g‖L∞)p−1p δ1/p < ε
2 , then‖g − g1‖Lp < ε
2 . Now we have
‖f − g1‖Lp ≤ ‖f − g‖Lp + ‖g − g1‖Lp ≤ ε
2+ε
2(3.7.29)
and we can conclude.
Theorem 3.7.21. Let f, fk ∈ Lp(Ω, µ), k ∈ N, 1 ≤ p <∞. Then
following two conditions are equivalent:
i) ‖fk − f‖Lp → 0 k → ∞.
ii) fkµ→ f as k → +∞ and ‖fk‖Lp → ‖f‖Lp as k → +∞.
Proof.
i) =⇒ ii):
The fact that fkµ→ f as k → +∞ is a consequence of Tchebychev
inequality as we have already observed several times.
From Minkowski inequality it follows that
| ‖f‖Lp − ‖fk‖Lp| ≤ ‖fk − f‖Lp.
ii) =⇒ i): We observe that
2p−1|fk|p + 2p−1|f |p − |fk − f |p ≥ 0.
104
Let (fkn) be a subsequence of (fk) converging to f as n → +∞. ByFatou’s Lemma, we get
2p∫
Ω
|f |p dµ− lim supn→+∞
∫
Ω
|fkn − f |p dµ =
lim infn→∞
∫2p−1|fkn|p + 2p−1|f |p − |fkn − f |p dµ ≥
∫
Ω
lim infn→∞
(2p−1|fkn|p + 2p−1|f |p − |fkn − f |p) dµ = 2p∫
Ω
|f |p.
=⇒ lim supn→+∞
∫
Ω
|fkn − f |p dµ = 0.
(3.7.30)
It follows that
limn→+∞
∫
Ω
|fkn − f |p dµ = 0.
Claim 3.7.22. ‖fk − f‖Lp → 0 as k → ∞.
Proof of Claim 3.7.22.
Suppose lim supkn→+∞
∫
Ω
|fkn−f | dµ > 0 for some subsequence fkn.
Since fknµ→ f as n → +∞, ∃fk′n → f µ-a.e. Then ‖fk′n − f‖Lp → 0
which contradicts that
lim supk→+∞
∫
Ω
|fkn − f |p dµ > 0.
Remark 3.7.23. For p ≥ 1, the function x 7→ |x|p is convex, therefore∣∣∣x+ y
2
∣∣∣p
≤ 1
2|x|p + 1
2|y|p
|x+ y|p ≤ 2p−1|x|p + 2p−1|y|p ∀x, y ∈ R.(3.7.31)
One can directly prove that ∀a, b ∈ R
|a+ b|p − 2p−1|a|p − 2p−1|b|p ≤ 0.
105
It is enough to consider f(x) = |1 + x|p − 2p−1 − 2p−1|x|p and showthat f(x) ≥ 0.
106
Chapter 4
Product Measures and MultipleIntegrals
4.1 Fubini’s Theorem
Let X, Y be sets.
Definition 4.1.1. Let µ be a measure on X and ν be a measureon Y . We define the measure µ × ν: P (X × Y ) → [0,+∞] bysetting for each S ⊆ X × Y
(µ× ν)(S) = inf ∞∑
i=1
µ(Ai) ν(Bi)
S ⊆ ⋃∞i=1Ai × Bi
Ai ⊆ X µ-measurable
Bi ⊆ Y µ-measurable, i ∈ N.
(4.1.1)
The map µ× ν is called the Product Measure.
Remark 4.1.2. If we define for every A ⊆ X µ-measurable, B ⊆ Yν-measurable
λ(A×B) = µ(A) ν(B). (4.1.2)
Then λ is a pre-measure on the algebra of finite disjoint unions.Thus µ× ν is the Caractheodory extension of λ.
Exercise 4.1.3. X = Rk, Y = Rℓ and µ = Lk, ν = Lℓ =⇒ µ×ν = Ln
n = k + ℓ.
107
Digression
We consider for simplicityR2. Let f : R2 → R be a summable function,
namely a measurable function such that∫
R2
|f(x, y)| dL2 < +∞.
We wonder if (by analogy with Riemann-integral for continuous func-
tions) if the above integral can be splitted in two integrals on R,namely
∫
R2
∫|f(x, y)| dL2 =
∫
R
(∫
R
|f(x, y)| dL1(x))dL1(y),
=
∫
R
(∫
R
|f(x, y)| dL1(y))dL1(x).
In order that the above formula is correct, it is necessary that
i) ∀ y ∈ R: x 7−→ |f(x, y)| L1-measurable.
ii) y 7−→∫
R
|f(x, y)| dL1(x) is L1-measurable and summable.
Example 4.1.4. Let P be a non-measurable subset of R which is
contained in [0, 1] and let E = P × Q. Then E is measurable in R2.Indeed, E ⊆ [0, 1]×Q which is a rectangular of null-measure
L2([0, 1]×Q) = L1[0, 1]×L1(Q)
= 0
=⇒ E is L2-measurable.
Let us consider f(x, y) = χE(x, y) = χP (x)χQ(y), f is L2-measurableand its integral is zero, since f ≡ 0 L2-a.e. Let y ∈ R be fixed. Then
f(x, y) = χP (x) if y ∈ Q and = 0 otherwise.
Thus if y = Q, f(x, y) is not measurable and it has no sense to write∫Rf(x, y) dL1(x). This means that we cannot apparently split the
integral of f over R2 into two integrals. Actually the function f(x, y)
108
is measurable for L1-a.e. y. Therefore we can define for L1-a.e. y ∈ Rthe function y 7−→
∫
R
|f(x, y)| dL1(x) which is identically zero. We
can also re-define the function when y ∈ Q (the value of its integraldoes not change since L1(Q) = 0) in order to obtain a L1-measurable
function which also L1-summable. This example shows that even if thefunction obtained by freezing one of the two variables is not measur-
able, the values of x or y for which we get a non-measurable functionhave measure zero and therefore they can be neglected when we inte-
grate.
Theorem 4.1.5. (Fubini)
Let µ, ν be Radon measures on X = Rk, Y = Rℓ respectively,µ × ν be the product measure on X × Y = Rn, n = ℓ + k. Then
it holds
i) For every A ⊆ X µ-measurable, B ⊆ Y ν-measurable, wehave A× B is µ× ν-measurable and
(µ× ν)(A×B) = µ(A) ν(B). (4.1.3)
109
ii) Let S ⊆ X×Y be (µ×ν)-measurable with (µ×ν)(S) < +∞.
Then the set Sy = x : (x, y) ∈ S is µ-measurable for ν a.e.y and the mapping
y 7−→ µ(Sy) =
∫
X
χSy(x) dµ
is summable with
(µ× ν)(S) =
∫
Y
µ(Sy) dy =
∫
Y
∫
X
χSy(x) dµ dν.
An analogous property holds for Sx = y : (x, y) ∈ S.
iii) µ× ν is a Radon measure.
iv) If f : X × Y → R is µ × ν-summable, then the mappingx 7−→ f(x, y) and y 7−→ f(x, y) are µ-summable for ν-a.e. y
and ν-summable for µ-a.e. x and the mappings
y 7−→∫
X
f(x, y) dµ(x)
x 7−→∫
Y
f(x, y) dν(y)
are respectively ν- and µ-summable and∫
X×Y
f d(µ× ν) =
∫
Y
(∫
X
f(x, y) dµ(x))dν(x)
=
∫
X
(∫
Y
f(x, y) dµ(y))dµ(x).
Proof. We define the set
F =S ⊆ X × Y : S satisfies 1 and 2
(4.1.4)
where
110
1 x 7−→ χS(x, y) is µ-measurable µ-a.e. y
and
2 y 7−→∫
X
χS(x, y) dµ(x)(= µ(Sy)
)is ν-measurable.
For S ∈ F we set
ρ(S) =
∫
Y
(∫
X
χS(x, y) dµ(x))dν(y)
). (4.1.5)
Goal:
To show that every S ⊆ X × Y µ × ν-measurable belongs to F and
f(S) = (µ×ν)(S), in particular for S = A×B. We define the families
P0 = A× B, A ⊆ µ-measurable, B ⊆ Y ν-measurable
P1 = ∞⋃
j=1
Sj, Sj ∈ P0
P2 = ∞⋂
j=1
Rj, Rj ∈ P1
.
(4.1.6)
Claim 4.1.6. P0,P1,P2 ⊆ F .
Proof of the Claim 4.1.6:
i) Clearly, P0 ∈ F . Let S = A×B. We have
Sy =
A if y ∈ B
∅ if y /∈ B .(4.1.7)
Thus Sy is clearly µ-measurable. Moreover, µ(Sy) = µ(A)χB. Thus
y 7−→ µ(Sy) (4.1.8)
is ν-measurable as well and ρ(S) = µ(A)µ(B).
111
ii) P1 ⊆ F . We first observe that for A1 ×B1, A2 ×B2 ∈ P0 it holds
(A1 ×B1) ∩ (A2 ×B2) = (A1 ∩ A2)× (B1 ∩ B2) ∈ P0 (4.1.9)
and
(A1×B1)\(A2×B2) =[(A1\A2)×B1
]∪[(A1∩A2)×B1\B2
]. (4.1.10)
A2 ×B2
(A1 × B1) \ (A2 × B2)
Fig. 4.1
Let now S =⋃∞
j=1 Sj. We define
S1 = S1, S2 = S2 \ S1, Sn = Sn \n−1⋃
k=1
Sk. (4.1.11)
Therefore each member of P1 is a countable disjoint union of membersof P0. If S =
⋃∞j=1 Sj, Sj ∈ P0 mutually disjoint, then the map
x 7−→ χS(x, y) =∞∑0χSj
(x, y) = limk→+∞
k∑j
χSj(x, y)
is measurable since it is the limit of µ-measurable functions.
We have AJ : y 7−→∫X χSj
(x, y) dµ(x) is ν-measurable. Thus
y 7−→∫
X
χS(x, y) dµ(x) =
∫
X
limk→+∞
k∑0
χSj(x, y) dµ(x)
=ւ
limk→∞
k∑0
∫
X
χSj(x, y) dµ(x)
by Beppo-Levi Theorem
(4.1.12)
112
is ν-measurable. Moreover,
ρ(S) =
∫
Y
(∫
X
χS(x, y) dµ)dν
=
∫
Y
∞∑j
∫
X
χSj(x, y) dµ dν
=∞∑j=0
∫
Y
(∫
X
χSj(x, y) dµ
)dν
=∞∑j=0
ρ(Sj).
(4.1.13)
Above we have applied two times Beppo Levi-s Theorem.
Claim 4.1.7. For R =⋂∞
j=0Rj, Rj ∈ P1 and ρ(Rj) < +∞ ∀j,then
ρ(R) = limk→+∞
ρ( k⋂
0Rj
). (4.1.14)
Proof of Claim 4.1.7: Let R =⋂∞
j=0Rj, Rj ∈ P1, ρ(Rj) < +∞ ∀j.We may assume that Rj+1 ⊆ Rj (otherwise we consider the family
(Rj)j:
R1 = R1, R2 = R1 ∩ R2
Rj = Rj+1 ∩ Rj.(4.1.15)
Since χRj(x, y) ≤ χRj
(x, y) we apply twice the Lebesgue Theorem andget
ρ(R) =
∫
Y
(∫
X
χR(x, y) dµ)dν
=
∫
Y
(∫
X
limyχRj
(x, y) dµ)dν
=
∫
Y
limj→+∞
(∫
X
χRj(x, y) dµ
)dν
= limj→+∞
∫
Y
(∫
X
χRj(x, y) dµ) dν
= limj→+∞
ρ(Rj).
(4.1.16)
113
Therefore (4.1.14) holds.
Claim 4.1.8. For S ⊆ X × Y , we have
(µ× ν)(S) = infρ(R) |S ⊆ R, R ∈ P1
. (4.1.17)
Proof of Claim 4.1.8: First we observe that if S ⊆ R =⋃∞
i=1Ai × Bi,
then
ρ(R) ≤↓
n∑i=1
ρ(Ai × Bi) =∞∑i=1
µ(Ai)µ(Bi)
Claim 4.1.6
(4.1.18)
Thusinfρ(R) : S ⊆ R ∈ P1 ≤ (µ× ν)(S1). (4.1.19)
Moreover, there exists a disjoint sequence A′j × B′
j in P0 such that
R =∞⋃
j=0
(A′j × B′
j).
Thus
ρ(R) =∞∑j=0
ρ(A′j × B′
j) =∞∑j=0
µ(A′i) ν(B
′j)
≥ (µ× ν)(S).
(4.1.20)
We prove i) of Theorem 4.1.5.
Let A×B ∈ P0. By definition of µ×ν and that ρ(A×B) = µ(A) ν(B),
we have
(µ× ν)(A×B) ≤ µ(A) ν(B) = ρ(A×B) ≤ ρ(R)
∀R ∈ P1 with A×B ⊆ R. Claim 4.1.8 implies that
(µ× ν)(A× B) = ρ(A×B) = µ(A) ν(B).
114
Claim 4.1.9. A× B is µ× ν-measurable.
Proof of Claim 4.1.9. Take T ⊆ X × Y , R ∈ P1 with T ⊆ R. ThenR\ (A×B), R∩ (A×B) are disjoint and belong to P1. Consequently,
(µ× ν)(T \ (A×B)
)+ (µ× ν)
(T ∩ (A× B)
)≤
ρ(R \ (A+ B) + ρ(R ∩ (A× B)
)= ρ(R).
According to Claim 4.1.8, we get
(µ× ν)(T \ (A×B)
)+ (µ× ν)
(T ∩ (A×B)
)≤ (µ× ν)(T ). (4.1.21)
=⇒ A× B is µ× ν-measurable.
Claim 4.1.10. For each S ⊆ X × Y there is a set R ⊆ P2 such
that S ⊆ R andρ(R) = (µ× ν)(S).
Proof of Claim 4.1.10: If (µ × ν)(S) = +∞, then take R = X × Y .Otherwise, for j ∈ N, we choose S ⊆ Rj ∈ P1 with (µ × ν)(S) ≤ρ(Rj) ≤ (µ× ν)(S) + 1
j .
We may suppose Rj+1 ⊂ Rj. We set R =⋂
j Rj ⊇ S. Since byClaim 4.1.7 limj→∞ ρ(Rj) = ρ(R), we get the property, since R ∈ P2.
We prove ii) of Theorem 4.1.5.
Let S ⊆ X × Y be µ × ν-measurable and (µ × ν)(S) < +∞. Wechoose R ∈ P2 according to Claim 4.1.10 with S ⊆ R and
(µ× ν)(S) = ρ(R).
We have two cases.
a) (µ × ν)(S) = 0 =⇒ ρ(R) = 0 =⇒∫
Y
µ(Ry) dν = 0 =⇒ µ(Ry) =
0 ν-a.s. y.
Now we have
0 ≤ XS(x, y) ≤ XR(x, y),
115
and in particular0 ≤ χSy
(x) ≤ χRy(x).
Since µ(Ry) = 0 =⇒ µ(Sy) = 0. Thus Sy is µ-measurable. More-over,
y 7−→∫
y
µ(Sy) dν = 0 (4.1.22)
and thus it is ν-measurable. In particular, ρ(S) = 0.
b) (µ× ν)(S) = ρ(R) > 0. Consider R \ S:
(µ× ν)(R \ S) = (µ× ν)(R)− (µ× ν)(S)
= ρ(R)− (µ× ν)(S) = 0.
We remark that every member of P2 is µ × ν-measurable and ifR ∈ P2, (µ× ν)(R) <∞, then (µ× ν)(R) = ρ(R).
Since (µ×ν)(R\S) = 0, then by a) ρ(R\S) = 0. This means that
µ(R \ S)y = µ(Ry \ Sy) = 0.
Since Ry − Sy is µ-measurable (having null measure), then Sy is µ-measurable as well and µ(Ry) = µ(Sy),
(µ× ν)(S) = ρ(R) =
∫
Y
µ(Ry) dy
=
∫
Y
µ(Sy) dy = ρ(S).
iii) No proof.
Proof of iv) of Theorem 4.1.5. If f = χS < +∞, then iv) follows fromii). If f ≥ 0 we write
f =∞∑j=1
1
jχAj
(x, y), (4.1.23)
with Aj(µ× ν)-measurable.
116
Because of ii), the linearity of integrals, the property holds for thepartial sums
fk =k∑
j=1
1
jχAj
. (4.1.24)
By passing to the limit as k → +∞ and applying Beppo-Levi Theo-rem, we get the property for f∫
Y
(∫
X
f(x, y) dµ)dν =
∫
Y
∫
X
( ∞∑j=1
1
jXj(x) dµ(x)
)dν(y)
=∞∑j=1
1
j
∫
Y
(∫
X
χAjdµ(x)
)dν(y)
=∞∑j=1
1
j(µ× ν)(Aj) =
∫
X×Y
f(x, y) dµ× ν.
(4.1.25)
For general f , consider f = f+ − f−.
Remark 4.1.11. A set S ⊆ X × Y that satisfies 1 and 2 is
not necessarily (µ × ν)-measurable: take X = Y = R, µ = ν = L1,
A ⊆ [0, 1] not measurable and set
S = (x, y) : |x− χA(y)| <1
2; y ∈ [0, 1]. (4.1.26)
S is not (µ × ν)-measurable, but ∀y ∈ [0, 1], Sy is µ-measurable andµ(Sy) = 1.
• If f : X×Y → R+ is µ×ν-measurable, then f is µ×ν-summableif one of the two iterated integrals exists finite (Tonelli Theorem).
• It may happen that the two iterated integrals exist without fbeing µ× ν-measurable (take f = χS, S as above) or without f
being µ× ν-summable (take f(x, y) = sin(y)x with (x, y) ∈ (0, 1]×
[0, 2π]. In this case we have∫ 2π
0 f(x, y)dy = 0 for all x ∈ (0, 1]but the function is not summable).
117
4.2 Applications
1. Show that f(x, y) = x2−y2
(x2+y2)2is not Lebesgue summable in [0, 1]×
[0, 1], namely
∫
[0,1]×[0,1]
∣∣∣x2 − y
(x2 + y2)2
∣∣∣ dx dy = +∞.
1st Proof. We apply Fubini’s Theorem: if f ∈ L1(X × Y ) then for µ
a.e. x ∈ X, f(x, ·) ∈ L1(Y ) and S∫Y f(x, y) dν ∈ L1(X), for ν-a.e.
y ∈ Y , f(·, y) ∈ L1(X) and∫X f(x, y) dµ ∈ L1(Y ). Moreover,
∫
X×Y
f(x, y) d(µ× ν) =
∫
X
(∫
Y
f(x, y) dν)dµ
=
∫
Y
(∫
X
f(x, y) dµ)dν .
(4.2.1)
If f ∈ L1([0, 1]× [0, 1]) then the two iterated integrals should be equal.We compute separately the two iterated integrals:
∫ 1
0
x2 − y2
(x2 + y2)2dy =
∫ 1
0
x2 + y2 − 2y2
(x2 + y2)2dy
=
∫ 1
0
1
x2 + y2dy +
∫ 1
0
−2y2
(x2 + y2)2dy
=
∫ 1
0
1
x2 + y2dy +
y
x2 + y2
∣∣∣1
0−
∫ 1
0
1
x2 + y2dy
=1
x2 + 1
=⇒∫ 1
0
1
x2 + 1dx = arc tg x
∣∣∣1
0=
π
4.
On the other hand:∫ 1
0
∫ 1
0
x2 − y2
(x2 + y2)2dx dy = −π
4.
118
2nd Proof. One computes directly
∫
[0,1]
∣∣∣ x2 − y2
(x2 + y2)2
∣∣∣ d(µ× ν) =
∫ 1
0
[ ∫ y
0
y2 − x2
(x2 + y2)2︸ ︷︷ ︸1
dx+
∫ 1
y
x2 − y2
(x2 + y2)2︸ ︷︷ ︸2
]dy
∫ y
0
y2 − x2
(x2 + y2)2dx =
∫ y
0
1
(x2 + y2)dx−
∫ y
0
2x2
(x2 + y2)2dx
=
∫ y
0
1
(x2 + y2)dx+
x
y2 + x2
∣∣∣y
0−∫ y
0
1
(x2 + y2)2dx
=y
2y2=
1
2y∫ 1
y
x2 − y2
(x2 + y2)2dx = −
∫ 1
y
1
(x2 + y2)dx+
∫ 1
y
2x2
(x2 + y2)dx
= −∫ 1
y
1
(x2 + y2)dx− x
x2 + y2
∣∣∣1
y+
∫ 1
y
1
x2 + y2dx
=1
1 + y2+
y
2y2= − 1
1 + y2+
1
2y.
Therefore∫ 1
0
[∫ y
0
y2 − x2
(x2 + y2)2dx+
∫ 1
y
x2 − y2
(x2 + y2)2
]dy =
∫ 1
0
1
ydy +
∫ 1
0
− 1
1 + y2dy
= log |y|∣∣∣1
0− arc tg y
∣∣∣1
0= +∞.
2. Gaussian Integral
We want to compute ∫ +∞
−∞e−x2
.
There is no elementary indefinite integral for∫e−x2
dx but the definite
119
integral∫ +∞−∞ e−x2
dx can be evaluated:∫
R2
e−(x2+y2)dx dy =
∫
R
(∫
R
e−(x2+y2) dx)dy
=(∫ +∞
−∞e−x2
dx)(∫ +∞
−∞e−y2dy
)
=(∫ +∞
−∞e−x2
dx)2
.
On the other hand:∫
R2
e−(x2+y2)dx dy =
∫ 2π
0
∫ +∞
0
e−r2r dx dσ
= 2π
∫ +∞
0
re−r2dr =↓
s=−r2
π
∫ ∞
0
e−sds = π
Therefore∫ +∞−∞ e−x2
=√π.
4.3 Convolution
We suppose µ = Ln is the n-dimensional Lebesgue measure. For
Ω ⊆ Rn, 1 ≤ p ≤ ∞ we will write for simplicity Lp(Ω) or Lp insteadof Lp(Ω,Ln).
Lemma 4.3.1. Let f : Rn → R be Ln-measurable. Then the
functionF (x, y) = f(x− y) : Rn × Rn → R
is L2n-measurable.
Proof. With the substitution z := x− y we obtain
F−1([−∞, a))= (x, y) : f(x− y) < a = (x, x− z) : f(z) < a
= T (Rn × z : f(z) < a)
where T : R2n → R2n (x, z) 7→ (x, x− z).
120
T is Lipschitz continuous andRn×z : f(z) < a is L2n-measurable.The claim follows from the following lemma:
Lemma 4.3.2. (no proof)
Let T : Rn → Rn be Lipschitz continuous with
|T (x)− T (y)| ≤ L |x− y| ∀x, y ∈ Rn
and let A ⊆ Rn be Ln-measurable. Then T (A) is Ln-measurable.
Theorem 4.3.3. Let f, g ∈ L1(Rn). Then the function f ⋆ g:Rn → R defined by
(f ⋆ g)(x) =
∫
Rn
f(x− y) g(y) dy =
∫
Rn
f(z) g(x− z) dz (4.3.1)
is Ln-measurable and f ⋆ g ∈ L1(Rn) with
‖f ⋆ g‖L1 ≤ ‖f‖L1 ‖g‖L1.
The function f ⋆ g is called the convolution of f and g. We observethat f ⋆ g = g ⋆ f (commutativity).
Proof of Theorem 4.3.3. By Lemma 4.3.1 the function F (x, y) =f(x− y) g(y) is L2n-measurable.
i) We suppose f, g ≥ 0. Then F is L2n-integrable. By Tonelli’s
Theorem, F is L2n-summable if and only if one of the iterated
121
integrals exist finite
‖f ⋆ g‖L1 =
∫
Rn
(∫
Rn
f(x− y) g(y) dy)dx
=
∫
Rn
(∫
Rn
f(x− y) g(y) dx)dy
=
∫
Rn
g(y)(∫
Rn
f(x− y) dx)dy
= ‖f‖L1 ‖g‖L1 < +∞ .
ii) For general f, g it follows from i) that
|F (x, y)| = |f(x− y)| |g(y)| ∈ L1(R2n).
Thus by Fubini’s Theorem |f | ⋆ |g| ∈ L1(Rn). We have
‖f ⋆ g‖L1 =∫
Rn
∣∣∣∫
Rn
f(x− y) g(y) dy∣∣∣ dx ≤
∫
Rn
∫
Rn
|f(x− y)| |g(y)| dy dx =
∫
Rn
|f | ⋆ |g|(x) dx = ‖ |f | ⋆ |g| ‖L1 = ‖f‖L1 ‖f‖L2.
(4.3.2)
Corollary 4.3.4. Let 1 ≤ p ≤ +∞, 1 ≤ q ≤ +∞.
i) f ∈ L1, g ∈ Lp =⇒ f ⋆ g ∈ Lp and ‖f ⋆ g‖Lp ≤ ‖f‖L1 ‖g‖Lp.
ii) f ∈ Lp, g ∈ Lq, 1 ≤ 1p +
1q = 1 + 1
r . Then f ⋆ g ∈ Lr and
‖f ⋆ g‖Lr ≤ ‖f‖Lp ‖g‖Lr.
Proof of Corollary 4.3.4.
122
i) The case p = 1 follows from Theorem 4.3.3. For 1 < p < +∞, let1 < q < ∞ be the conjugate of p (1p +
1q = 1):
∫
Rn
|f ⋆ g|pdx =
∫
Rn
∣∣∣∫
Rn
f(x− y) g(y) dy∣∣∣p
dx
≤∫
Rn
∣∣∣∫
Rn
f(x− y)|1q |f(x− y)| 1p |g(y)| dy|pdx.
From the case p = 1, it follows that for a.e. x ∈ Rn, y 7−→|f(x− y)| |g(y)|p is µ-summable and thus
y 7−→ |f(x− y)| 1p |g(y)| ∈ Lp(Rn). (4.3.3)
Moreover, y 7−→ |f(x − y)|1q ∈ Lq. From Holder inequality itfollows
|f(x− y)| |g(y)| = |f(x− y)|1q |f(x− y)| 1p |g(y)| ∈ L1(Rn). (4.3.4)
Thus
∫Rn |f ⋆ g|pdx ≤
∫
Rn
(∫
Rn
|f(x− y)| dy) p
q(∫
Rn
|f(x− y)| |g(y)|p)dx
≤ ‖f‖pq
L1 ‖ |f | ⋆ |g|p‖L1
≤ ‖f‖pq
L1 ‖ |f |L1 ‖ |g|p‖L1
≤ ‖f‖pq+1
L1 ‖|g‖pLp
=⇒ ‖f ⋆ g‖Lp ≤ ‖f‖L1 ‖g‖Lp.(4.3.5)
The case p = ∞ follows form Holder inequality.
ii) We assume without loss of generality that p ≤ q. Since 1 + 1r =
1p+ 1
q≤ 1 + 1
q, we get p ≤ q ≤ r.
• The case r = ∞ follows from Holder inequality.
123
• The case p = 1 follows from i) (q = r).
Thus 1 < p ≤ q < r <∞. We write∫
Rn
|f(x− y)| |g(y)| dy =∫
Rn
|f(x− y)|1− pr |g(y)|1− q
r |f(x− y)|pr |g(y)| qrdy.(4.3.6)
We have
|f(x− y)|1− pr ∈ Ls, s =
p
1− pr
=pr
r − p
|g(y)|1− qr ∈ Lt, t =
q
1− qr
=rq
r − q
|f(x− y)|pr |g(y)| qr ∈ Lr
(4.3.7)
(observe that |f |p ∈ L1, |g|q ∈ L1 and from Theorem 4.3.3 |f |p ⋆|g|q ∈ L1). We have
1
r+
1
s+
1
t=
1
r+(1p− 1
r
)+(1q− 1
r
)= 1. (4.3.8)
We apply the generalized Holder inequality and get
|f ⋆ g|r ≤ ‖ |f |1− pr‖rLs‖ |g|1− q
r‖rLt|f |p ⋆ |g|q
=(∫
|f |pdx)r ( r−p
rp)(∫
|g|q)r( r−q
rq)
|f |p ⋆ |g|q
= ‖f‖r(1−pr)
Lp ‖g‖r(1−qr)
Lq |f |p ⋆ |g|q
(4.3.9)
Therefore:
‖f ⋆ g‖Lr ≤(‖f‖r(1−
pr)
Lp ‖f‖pLq ‖g‖r(1−qr)
Lq ‖g‖qLq
)1r
= ‖f‖Lp ‖g‖Lq .
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Other properties of convolutions
Definition 4.3.5. Let 1 ≤ p ≤ ∞. We say that a function f :
Rn → R belongs to Lploc(R
n) if f χK ∈ Lp(Rn) for every compactK ⊆ Rn.
Note that if f ∈ Lploc(R
n) then f ∈ L1loc(R
n).
Proposition 4.3.6. Let f ∈ Cc(Rn) and g ∈ L1
loc(Rn). Then
(f ⋆ g)(x) is well defined for every x ∈ Rn and moreover f ⋆ g ∈C(Rn).
Proof. Note that for every x ∈ Rn the function y 7−→ f(x − y) g(y)
is Ln-summable on Rn and therefore (f ⋆ g)(x) is defined for everyx ∈ Rn.
Let xn → x and let K be a fixed compact set in Rn such thatxn− supp f ⊆ K ∀n. Therefore, we have f(xn− y) = 0 ∀n and y /∈ K(y /∈ K =⇒ y /∈ xn − supp f =⇒ y − xn /∈ −supp f).
We deduce from the uniform modulus of continuity of f that
|f(xn − y)− f(x− y)| ≤ wf(|xn − x|) χk(y)
where wf is the continuity of f :
|(f ⋆ g)(xn)− (f ⋆ g)(y)|
=∣∣∣∫
Rn
(f(xn − y)− f(x− y)
)g(y) dy
∣∣∣
≤∫
Rn
wf(|xn − x|)χK(y) |g(y)| dy
≤ wf(|xn − n|) ‖g‖L1(K) → 0 as n → +∞.
125
Proposition 4.3.7. (no proof)
Let f ∈ Ckc (R
n) k ≥ 1 and g ∈ L1loc(R
n). Then f ⋆ g ∈ Ck(Rn)and
Dα(f ⋆ g) = Dαf ⋆ g ∀|α| ≤ k.
In particular, if f ∈ C∞c (Rn) and g ∈ L1
esc(Rn), then f ⋆ g ∈
C∞(Rn).
4.4 Application:
Solution of the Laplace Equation
For u ∈ C2(Rn) let
∆u =n∑
i=1
∂2u
∂x2i= div(∇u).
Given f ∈ C∞c (Rn) we look for a solution u ∈ C∞(Rn) of the Laplace
equation
−∆u = f in Rn.
We consider first the problem to determine a function u ∈ C∞c (Rn)
from its Laplacean. We recall the Gauss Theorem.
Theorem 4.4.1. Let Ω ⊆ Rn be a smooth bounded open set,ϕ, ψ ∈ C2(R). Then it holds
∫
Ω
ϕ∆ψ − ψ∆ϕ =
∫
∂Ω
ϕϕ∂ψ
∂n− ψ∂ϕ
∂ndσ (4.4.1)
where ~n is the outward normal vector to ∂Ω and
∂ϕ
∂n= ∇ϕ · ~n.
126
We choose ψ = u ∈ C∞c (Rn) and Ω open set containing supp(u) =
x : u(x) 6= 0. For every ϕ ∈ C2(Rn) it follows from Theorem 4.4.1
−∫
Ω
u∆ϕdx = −∫
Ω
ϕ∆u dx.
If we succeed in finding ϕ such that
−∆ϕ = δ0
(δ0 is the Delta-Dirac distribution with δ0(u) = u(0) ∀u ∈ C∞c (Rn)),
then we expect a representation of the form
u(n) = ϕ ⋆ (−∆u)
=
∫
Rn
ϕ(x− y)(−∆u)(y) dy .
n = 1 ϕ(x) = −1
2|x| ∀x ∈ R.
In this case, we have
∆ϕ(x) = ϕ1(x) = 0, x 6= 0.
127
Moreover,
ϕ ⋆ (−∆u)(x0) = −∫ +∞
∞u′′(x) ϕ(x0 − x) dx
= − limε→0
[ ∫ x0−ε
−∞u′′(x)ϕ(x0 − x) dx+
∫ +∞
x0,+ε
u′′(x)ϕ(x0 − x) dx]
= − limε→0
[u′(x)
∣∣x0−ε
−∞ + ϕ(x0 − x)∣∣+∞n0+ε
]
= − limε→0
[ ∫ x0−ε
−∞u′(x)ϕ−1(x0 − x) du+
∫ +∞
x0+ε
u′(x)ϕ(x0 − ε) dx]
= − limε→0
[[u(x)ϕ−1(x0 − x)
] ∣∣∣x0−ε
−∞+ u(x)ϕ′(x0 − x)
∣∣∣+∞
x0+ε
+
∫ x0−ε
−∞u(x) dx+
∫ +∞
x0+ε
u(x) dx]
= − limε→0
[u(x0 − ε)ϕ′(ε)− u(x0 + ε)ϕ′(−ε)
]
= u(x0) limε→0
[ϕ′(−ε)− ϕ′(ε)]
= u(x0).(4.4.2)
n = 2 ϕ(x) =1
2πlog
( 1
|x|).
n = 3
ϕ(x) =1
(n− 2)αn−1 |n|n−2:= Cn |n|2−n
dn−1 = vol(Sn−1). ϕ is called the fundamental solution of the Laplacean.
Theorem 4.4.2. Let f ∈ C∞c (Rn). Then u = ϕ ⋆ f ⊂ C∞(Rn)
and solves −∆u = f .
Proof. See for instance Evan’s book: Partial Differential Equations.
128
