introduction to measure theory and lebesgue integration

1

INTRODUCTION TO MEASURE THEORYAND LEBESGUE INTEGRATION

Eduard EMELYANOV

Ankara — TURKEY

2007

2

FOREWORD

This book grew out of a one-semester course for graduate students that theauthor have taught at the Middle East Technical University of Ankara in 2004-06. It is devoted mainly to the measure theory and integration. They form thebase for many areas of mathematics, for instance, the probability theory, and atleast the large part of the base of the functional analysis, and operator theory.

Under measure we understand a σ-additive function with values in R+ ∪ ∞defined on a σ-algebra. We give the extension of this basic notion to σ-additivevector-valued measures, in particular, to signed and complex measures and referfor more delicate topics of the theory of vector-valued measures to the excellentbook of Diestel and Uhl [3]. The σ-additivity plays a crucial role in our book. Wedo not discuss here the theory of finitely-additive functions defined on algebrasof sets which are sometimes referred as finitely-additive measures.

There are two key points in the measure theory. The Caratheodory extensiontheorem and construction of the Lebesgue integral. Other results are more orless technical. Nevertheless, we can also emphasize the importance of the Jor-dan decomposition of signed measure, theorems about convergence for Lebesgueintegral, Cantor sets, the Radon – Nikodym theorem, the theory of Lp-spaces,the Liapounoff convexity theorem, and the Riesz representation theorem.

We provide with proofs only basic results, and leave the proofs of the others tothe reader, who can also find them in many standard graduate books on themeasure theory like [1], [4], and [5]. Exercises play an important role in thisbook. We expect that a potential reader will try to solve at least half of them.Exercises marked with ∗ are more difficult, and sometimes are too difficult forthe first reading. In the case if someone will use this book as a basic text-bookfor the analysis graduate course, the author recommends to study subsectionsmarked with , omitting the material of subsections marked with !.

I am indebted to many. I thank Professor Safak Alpay for reading the manuscriptand offering many valuable suggestions. I thank to our students of 2004-2006METU graduate real analysis classes, especially Tolga Karayayla, who mademany corrections. Finally, I thank to my wife Svetlana Gorokhova for helpingin preparing of the manuscript.

3

Contents

Introduction

Chapter 1

1.1. σ-Algebras, Measurable Functions, Measures, the Egoroff Theorem, Ex-haustion Argument

1.2. Vector-valued, Signed and Complex Measures, Variation of a Vector-valuedMeasure, Operations with Measures, the Jordan Decomposition Theorem, Ba-nach Space of Signed Measures of Bounded Variation

1.3. Construction of the Lebesgue Integral, the Monotone Convergence Theo-rem, the Dominated Convergence Theorem,

Chapter 2

2.1. The Caratheodory Theorem, Lebesgue Measure on R, Lebesgue – StieltjesMeasures, the Product of Measure Spaces, the Fubini Theorem

2.2. Lebesgue Measure on Rn, Lebesgue Integral in Rn, the Lusin Theorem,Cantor Sets

Chapter 3

3.1. The Radon – Nikodym Theorem, Continuity of a Measure with Respectto another Measure, the Hahn Decomposition Theorem

3.2. Holder’s and Minkowski’s Inequalities, Completeness, Lp-Spaces, Duals

3.3. The Liapounoff Convexity Theorem

Chapter 4

4.1. Vector Spaces of Functions on Rn, Convolutions

4.2. Radon Measures, the Riesz Representation Theorem

Bibliography

Index

Chapter 1

1.1 σ-Algebras, Measurable Functions, Measures

Here we define the notion of σ-algebra which plays the key role in the measuretheory. We study basic properties of σ-algebras and measurable functions. Inthe end of this section we define and study the notion of measure.

1.1.1. σ-Algebras. The following notion is the principal in the measuretheory.

Definition 1.1.1 A collection A of subsets of a set X is called a σ-algebra if

(a) X ∈ A;

(b) if A ∈ A then X \ A ∈ A;

(c) given a sequence (Ak)k ⊆ A, we have⋃k

Ak ∈ A.

It follows from this definition that the empty set ∅ belongs to A since X ∈ Aand ∅ = X \X. Further, given a sequence (Ak)k ⊆ A, we have⋂

k

Ak = X \⋃k

(X \ Ak) ∈ A.

Finally, the difference A \B := A ∩ (X \B) and the symmetric difference

A4B := (A \B) ∪ (B \ A)

both belong to A.

Any σ-algebra of subsets of a set X has at least two elements: ∅ and X itself.One of the main and mostly obvious examples of a σ-algebra is P(X) is the setof all subsets of X. The following simple proposition shows us how to constructnew σ-algebras from a given family of σ-algebras.

5

6 CHAPTER 1.

Proposition 1.1.1 Let Ωαα∈A be a nonempty family of σ-algebras in P(X),then Ω = ∩αΩα is also a σ-algebra. 2

This proposition leads to the following definition.

Definition 1.1.2 Let G ⊆ P(X), then the set of all σ-algebras containing Gis nonempty since it contains P(X). Hence we may talk about the minimal σ-algebra containing G. This σ-algebra is called the σ-algebra generated by G.

An important special case of this notion is the following.

Definition 1.1.3 Let X be a topological space and let G be the family of allopen subsets of X. The σ-algebra generated by G is called the Borel algebraof X and denoted by B(X).

1.1.2. Measurable functions. Let f be a real-valued function defined ona set X. We suppose that some σ-algebra Ω ⊆ P(X) is fixed.

Definition 1.1.4 We say that f is measurable, if f−1([a, b]) ∈ Ω for any realsa < b.

The following three propositions are obvious.

Proposition 1.1.2 Let f : X → R be a function. Then the following conditionsare equivalent:

(a) f is measurable;

(b) f−1([0, b)) ∈ Ω for any real b;

(c) f−1((b,∞)) ∈ Ω for any real b;

(d) f−1(B) ∈ Ω for any B ∈ B(R). 2

Proposition 1.1.3 Let f and g be measurable functions, then

(a) α · f + β · g is measurable for any α, β ∈ R;

(b) functions maxf, g and f · g are measurable.

In particular, functions f+ := maxf, 0, f− := (−f)+, and |f | := f+ + f− aremeasurable. 2

1.1. σ-ALGEBRAS, MEASURABLE FUNCTIONS, MEASURES 7

Proposition 1.1.4 Let (fn)∞n=1 be a sequence of measurable functions, then

lim supn→∞

fn and lim infn→∞

fn

are measurable. 2

1.1.3. Measures. The main definition here is the following.

Definition 1.1.5 Let A be a σ-algebra. A function µ : A → R ∪ ∞ is calleda measure, if:

(a) µ(∅) = 0;

(b) µ(A) ≥ 0 for all A ∈ A; and

(c) µ(⋃k Ak) =

∑k µ(Ak) for any sequence (Ak)k of pairwise disjoint sets from

A, that is Ai ∩ Aj = ∅ for i 6= j.

The axiom (c) is called σ-additivity of the measure µ. As usual, we will alsoassume that any measure under consideration satisfies the following axiom:

(d) for any subset A ∈ A with µ(A) =∞, there exists B ∈ A such that B ⊆ Aand 0 < µ(B) <∞.

This axiom allows to avoid a pathological case that for some A, with µ(A) = 0it holds either µ(B) = 0 or µ(B) =∞ for any B ⊆ A, B ∈ A.

We will use often the following simple proposition, the proof of which is left tothe reader.

Proposition 1.1.5 Let µ be a measure on a σ-algebra A, An ∈ A, and An → A.Then A ∈ A and µ(A) = lim

n→∞µ(An). In particular, if (Bn)∞n=1 is a decreasing

sequence of elements of A such that⋂∞n=1Bn = ∅, then µ(Bn)→ 0. 2

1.1.4. Measure spaces. We consider a fixed but arbitrary σ-algebra witha measure.

Definition 1.1.6 If A is a σ-algebra of subsets of X and µ is a measure on A,then the triple (X,A, µ) is called a measure space. The sets belonging to Aare called measurable sets because the measure is defined for them.

8 CHAPTER 1.

Now we present several examples of measure spaces.

Example 1.1.1 Let X = x1, . . . , xN be a finite set, A be the σ-algebra of allsubsets of X and a measure is defined by setting to each xi ∈ X a nonnegativenumber, say pi. This follows that the measure of a subset xα1 , . . . , xαk

⊆ X isjust pα1 + · · · + pαk

. If pi = 1 for all i, then the measure is called a countingmeasure because it counts the number of elements in a set.

Example 1.1.2 If X is a topological space, then the most natural σ-algebra ofsubsets of X is the Borel algebra B(X). Any measure defined on a Borel algebrais called Borel measure. In Section 2.1, we will prove that on the Borel σ-algebra B(R) there exists a unique measure such that µ([a, b]) = b − a for anyinterval [a, b] ⊆ R.

Whenever we consider spaces X = [a, b], X = R, or X = Rn, we usually assumethat the measure under consideration is the Borel measure.

As presented in Definition 1.1.6, the notion of measure space is extremely gen-eral. In almost all applications, the following specific class of measure spaces isadequate.

Definition 1.1.7 A measure space (X,A, µ) is called σ-finite if there is a se-quence (Ak)

∞k=1, Ak ∈ A, satisfying

X =∞⋃k=1

Ak and µ(Ak) <∞ for all k.

Obviously, any σ-finite measure satisfies the axiom (d). Moreover, a measure µis σ-finite iff (= if and only if) there exists an increasing sequence (Xn)∞n=1 ofsubsets of finite measure such that X =

⋃∞n=1Xn. If X = R, then Ak may be

chosen as intervals [−k, k]. In Rd, they may be chosen as balls of radius k, etc.

Definition 1.1.8 A measure space (X,A, µ) is called finite if µ(X) < ∞. Inparticular, if µ(X) = 1, then the measure space is said to be probabilistic, andµ is said to be a probability. A measure µ is called separable if there exists acountable family R ⊆ A such that for any A ∈ A, µ(A) <∞, and for any ε > 0there exists B ∈ R satisfying µ(A4B) < ε. A measure µ is called complete ifthere holds:

[ µ(A) = 0 & B ⊆ A ] ⇒ B ∈ A .


Every Borel measure on R (or on [0, 1], or on Rn, etc.) possesses a unique com-pletion, which is called a Lebesgue measure.

1.1.5. Convergence of functions and the Egoroff theorem. For mea-surable functions, there are several natural types of convergence. Some of themare given by the following definition.

Definition 1.1.9 Let (fn)∞n=1 be a sequence of R-valued functions defined on X.We say that:

(a) fn → f pointwise, if fn(x)→ f(x) for all x ∈ X;

(b) fn → f almost everywhere (a.e.), if fn(x)→ f(x) for all x ∈ X excepta set of measure 0;

(c) fn → f uniformly, if for any ε > 0 there is n(ε) such that

sup|fn(x)− f(x)| : x ∈ X ≤ ε

for all n ≥ n(ε).

Theorem 1.1.1 (Egoroff’s theorem) Suppose that µ(X) < ∞, fn∞n=1 andf are measurable functions on X such that fn → f a.e. Then, for every ε > 0,there exists E ⊆ X such that µ(E) < ε and fn → f uniformly on Ec = X \ E.

Proof: Without loss of generality, we may assume that fn → f everywhere onX and (by replacing fn with fn − f) that f ≡ 0. For k, n ∈ N, let

En(k) :=∞⋃m=n

x : |fm(x)| ≥ k−1.

Then, for a fixed k, En(k) decreases as n increases, and⋂∞n=1En(k) = ∅. Since

µ(X) <∞, we conclude that µ(En(k))→ 0 as n→∞. Given ε > 0 and k ∈ N,choose nk such that

µ(Enk(k)) < ε · 2−k,

and set

E :=∞⋃k=1

Enk(k).

10 CHAPTER 1.

Then µ(E) < ε, and we have

|fn(x)| < k−1 (∀n > nk, x 6∈ E).

Thus fn → 0 uniformly on X \ E. 2

1.1.6. Exhaustion argument.Let (X,Σ, µ) be a σ-finite measure space. Given a sequence (Un)∞n=1 ⊆ Σ, a setA ∈ Σ is called (Un)n-bounded if there exists n such that A ⊆ Un µ-almosteverywhere.

Theorem 1.1.2 (Exhaustion theorem) Let (Yn)∞n=1 ⊆ Σ be a sequence satis-fying Yn ↑ X and µ(Yn) <∞ for all n. Let P be some property of (Yn)n-boundedmeasurable sets, such that A ∈ P iff B ∈ P for all B, µ(A4B) = 0. Supposethat any (Yn)n-bounded set A, µ(A) > 0, has a subset B ∈ Σ, µ(B) > 0 with theproperty P. Moreover, assume that either(a) A1 ∪ A2 ∈ P for every A1, A2 ∈ P, or(b) ∪nBn ∈ P for every at most countable family (Bn)n of pairwise disjointsets possessing the property P.Then there exists a sequence (Xn)∞n=1 ⊆ Σ such that Xn ↑ X, and P 3 Xn ⊆ Ynfor all n. Moreover, there exists a pairwise disjoint sequence (An)∞n=1 ⊆ Σ such

that∞⋃n=1

An = X and An ∈ P for all n.

Proof: Let A be a (Yn)n-bounded set with µ(A) > 0. Denote

PA := B ∈ P : B ⊆ A & m(A) := supµ(B) : B ∈ PA .

I(a) Suppose P satisfies (a). Then there exists a sequence (Fn)∞n=1 ⊆ PA suchthat m(A) = lim

n→∞µ(Fn), We may assume, that Fn ↑ . By Proposition 1.1.5, the

set F = ∪∞n=1Fn satisfies µ(F ) = m(A). We show that µ(A) = m(A). If notthen µ(A \ F ) > 0. The set A \ F has a subset of positive measure F0 ∈ P .Then Fn ∪ F0 ∈ PA and µ(Fn ∪ F0) > m(A) for a sufficiently large n, whichcontradicts to the definition of m(A). Therefore, µ(A) = m(A).

Now we apply this for A = Yn. Thus, there exists a sequence (X ′n)n ⊆ Σ suchthat X ′n ⊆ Yn, X ′n ∈ P , and µ(Yn \X ′n) < n−1 for all n. By (a), we may assumethat X ′n ↑. The set X ′0 = ∪∞n=1X

′n satisfies Yn\X ′0 ⊆ Yn\X ′n, so µ(Yn\X ′0) < n−1


for all n. Then µ(Yn \X ′0) = 0, and µ((∪∞n=1Yn)\X ′0) = 0, or µ(X \X ′0) = 0. LetXn := (X ′n∪(X \X ′0))∩Yn, then the sequence (Xn)n has the required properties.

The desired pairwise disjoint sequence (An)∞n=1 is given recurrently by A1 = X1

and Ak+1 = Xk+1 \ ∪ki=1Ai.

I(b) Suppose P satisfies (b). Let FA be the family of all pairwise disjointfamilies of elements of PA of nonzero measure. Then FA is ordered by inclusionand, obviously, satisfies the conditions of the Zorn lemma. Therefore, we havea maximal element in FA, say ∆. Then ∆ is at most countable family, say∆ = Dnn. By (b), its union D := ∪nDn is an element of PA as well. If D isa proper subset of A, then µ(A \ D) > 0. The set A \ D has a subset F ∈ Pof the positive measure. Then ∆1 := ∆ ∪ F is an element of FA which isstrictly greater then ∆. The obtained contradiction, shows that A ∈ P for every(Yn)n-bounded set A. So, we may take Xn := Yn for each n.

Now we apply this for A = Zm := Ym \ ∪m−1k=1 Yk. Let Zm = ∪nDm

n be a pairwisedisjoint union, where Dm

n ∈ P for all n,m. The family Dmn n,m is an at most

countable disjoint decomposition of X, say Dmn n,m = (An)∞n=1. The sequence

(An)∞n=1 satisfies the required properties. 2

1.1.7. Exercises to Section 1.1.

Exercise 1.1.1 Prove that the cardinality of any σ-algebra is either finite or uncountable.

Exercise 1.1.2 Let f and g be measurable R-valued functions. Show that f · g is measurable.

Exercise 1.1.3 For each X ⊆ N and for each k ∈ N define:

nkX := card(X ∩ 1, ..., k) .

Let A be a family of all X ⊆ N such that the number µ(X) = limk→∞

k−1nkX exists. Is A a

σ-algebra? Is µ a measure on A?

Exercise 1.1.4 Prove Proposition 1.1.5.

Exercise 1.1.5 ∗ Let (Ω,Σ, µ) be a measure space. Show that the set

R(µ) := µ(E) : E ∈ Σ

is, in general, neither closed nor convex in R∪∞. Is R(µ) closed if, additionally, µ is finiteor non-atomic (see Exercise 1.1.9 for the definition)?

12 CHAPTER 1.

Exercise 1.1.6 Show that each separable measure which satisfies Axiom 1.1.3 d) is σ-finite.

Exercise 1.1.7 ∗ Give an example of a family (Ωα)α∈[0,2π] of σ-subalgebras of the Borelalgebra B([0, 1]2) such that for any α, β ∈ [0, 2π], α 6= β:

Ωα ∩ Ωβ = ∅, [0, 1]2

and(∀α)[α ∈ [0, π]](∀a)[0 ≤ a ≤ 1](∃B ∈ Ωα)[µ(B) = a].

Exercise 1.1.8 ∗∗ Construct a sequence (Σn)∞n=1 of σ-subalgebras of the Borel algebra B([0, 1]2)such that, for any n < m,

Σm ⊆ Σn & Σn 6= Σmand

(∀n)(∀a)[0 ≤ a ≤ 1](∃B ∈ Σn)[µ(B) = a].

Exercise 1.1.9 ∗∗ Let (Ω,Σ, µ) be a measure space equipped with a probabilistic measure µsuch that µ has no atom (non-atomic measure), i.e., for any A ∈ Σ, µ(A) > 0 implies thatthere exists A0 ⊆ A such that 0 < µ(A0) < µ(A).(a) Show that, for any 0 ≤ α ≤ 1, there exists B ∈ Σ such that µ(B) = α.(b) Give an example of a measure ν on (N,P(N)) such that ν(N) = 1 and, for any 0 ≤ α ≤ 1,there exists B ∈ P(N) such that µ(B) = α.

Exercise 1.1.10 (The Borel – Cantelli lemma) Let (X,Σ, µ) be a measure space. Showthat if Ann ⊆ Σ and

∑∞n=1 µ(An) <∞ then µ(lim sup

n→∞An) = 0.

Exercise 1.1.11 Show that Egoroff’s theorem may fail if µ(X) =∞.

Exercise 1.1.12 ∗∗∗ Let (X,Ω, µ) be a measure space and µ(X) < ∞. Say that A ∼ B ifµ(A4B) = 0.(a) Show that ∼ is an equivalence relation on Ω;(b) Show that the function ρ : (Ω/ ∼)2 → R:

ρ([A], [B]) := µ(A4B),

is a metric, and (Ω/ ∼, ρ) is a complete metric space.

Exercise 1.1.13 ∗∗∗ Let (Ω/ ∼, ρ) be the metric space from Exercise 1.1.12.(a) Show that (Ω/ ∼, ρ) is compact, if (X,Ω, µ) is a purely atomic measure space.(b) Show that (Ω/ ∼, ρ) is not compact, if (X,Ω, µ) is not purely atomic.Here, the measure space (X,Ω, µ) is called purely atomic if for any A ∈ Ω, µ(A) <∞, thereexists a sequence (an)∞n=1 of elements of A such that µ(A) =

∑n µ(an).

Exercise 1.1.14 ∗ Show that in Proposition 1.1.4 pointwise limits lim supn→∞


fn

may be replaced by a.e.-limits lim supn→∞


fn.

1.2. VECTOR-VALUED MEASURES 13

1.2 Vector-valued Measures

In this section, we extend the notion of a measure. Then we study the basicoperations with signed measures and present the Jordan decomposition theo-rem. We refer for further delicate results about vector-valued measures to [3]and about spaces of signed measures to [7] and [12].

1.2.1. Vector-valued, signed and complex measures. Let Σ be aσ-algebra of subsets of a set X, and let E = (E, ‖ · ‖) be a Banach space.

Definition 1.2.1 A function µ : Σ → E ∪ ∞ is called a vector-valuedmeasure (or E-valued measure) if

(a) µ(∅) = 0;(b) µ(

⋃k Ak) =

∑k µ(Ak) for any pairwise disjoint sequence (Ak)k ⊆ Σ;

(c) for any A ∈ Σ, µ(A) =∞, there exists B ∈ Σ such that B ⊆ A and

0 < ‖µ(B)‖ <∞.

Example 1.2.1 Take Σ = P(N), and c0 is the Banach space of all convergentC-valued sequences with a fixed element (αn)n ∈ c0. Define for any A ⊆ N:

ψ(A) := (βn)n,

where βn = αn if n ∈ A and βn = 0 if n 6∈ A. Then ψ is a c0-valued measure onP(N).

Example 1.2.2 Let X be a set and let Ω be a σ-algebra in P(X). Then for anyfamily µkmk=1 of finite measures on Ω and for any family wkmk=1 of vectors ofRn, the Rn-valued measure Ψ on Ω is defined by the formula

Ψ(E) :=m∑k=1

µk(E) · wk (E ∈ Ω).

Example 1.2.3 Let X be a set and let Ω be a σ-algebra in P(X). Then forany family µkmk=1 of finite measures on Ω, for any family Akmk=1 of pairwisedisjoint sets in Ω, and for any family wkmk=1 of vectors of Rn, the Rn-valuedmeasure Φ on Ω is defined by the formula

Φ(E) :=m∑k=1

µk(E ∩ Ak) · wk (E ∈ Ω).

14 CHAPTER 1.

From Definition 1.2.1, it is quite easy to see (and we leave this as an exercise tothe reader) that if µ is an E-valued measure, then for two linearly independentvectors x, y ∈ E, ‖x‖ = ‖y‖ = 1, it is impossible to find sequences (An)∞n=1 and(Bn)∞n=1 of elements of Σ such that

‖µ(An)‖ → ∞ , limn→∞

µ(An)

‖µ(An)‖= x , and ‖µ(Bn)‖ → ∞ , lim

n→∞

µ(An)

‖µ(Bn)‖= y.

This can be expressed as: any vector-valued measure cannot be infinite in twodifferent directions.

The next two definitions are special cases of Definition 1.2.1.

Definition 1.2.2 A function µ : Σ→ R ∪ ∞ is called a signed measure ifµ(∅) = 0, µ(

⋃k Ak) =

∑k µ(Ak) for any sequence (Ak)k of pairwise disjoint sets

from Σ, and, for any A ∈ Σ, µ(A) = ∞, there exists B ∈ Σ such that B ⊆ Aand 0 < |µ(B)| <∞.

Definition 1.2.3 A function µ : Σ→ C ∪ ∞ is called a complex measureif µ(∅) = 0, µ(

⋃k Ak) =

∑k µ(Ak) for any sequence (Ak)k of pairwise disjoint

sets from Σ, and, for any A ∈ Σ, µ(A) = ∞, there exists B ∈ Σ such thatB ⊆ A and 0 < |µ(B)| <∞.

There are many interesting classes of vector-valued measures. In this book weconsider only a few of them, in particular, the classes given by the followingdefinition.

Definition 1.2.4 An E-valued measure µ is called finite if µ(A) ∈ E for everymeasurable A. An E-valued measure µ is called σ-finite if there is a sequence

(Ak)k, Ak ∈ Σ, satisfying X =∞⋃k=1

Ak and µ|Akis finite measure for all k ∈ N.

An E-valued measure µ is called separable if there exists a countable familyR ⊆ Σ such that, for any A ∈ Σ of a finite measure and for any ε > 0, thereexists B ∈ R satisfying ‖µ(A4B)‖ < ε.

The definitions of σ-finite, finite, and separable signed or complex measure areobvious.


1.2.2. The variation of a vector-valued measure. Now we show how inthe natural way construct a new measure from a given vector-valued measure.Let µ be a E-valued measure on (X,Σ). We define a function

|µ| : Σ→ R+ ∪ ∞

by the following formula:

|µ|(A) := supn=j∑n=1

‖µ(An)‖ : An ∈ Σ,

n=j⋃n=1

An = A,Ak ∩ Ap = ∅ if k 6= p.

(1.2.1)This function is called the variation of µ. It is an exercise for the reader toshow that |µ| is additive and therefore monotone.

Theorem 1.2.1 Let µ be an E-valued measure on (X,Σ). Then |µ| is a mea-sure.

Proof: Suppose that µ is an E-valued measure. We have to show only theσ-additivity of the function |µ| : Σ→ R+ ∪ ∞.

Let (An)∞n=1 be a pairwise disjoint sequence of elements of Σ and A :=⋃nAn. If

|µ|(A) =∞, then there is nothing to proof. So we may assume that |µ|(A) <∞.

Let π be a finite partition of A into pairwise disjoint members of Σ. Then∑E∈π

‖µ(E)‖ =∑E∈π

‖µ(E ∩ A)‖ =∑E∈π

‖∑n

µ(E ∩ An)‖ ≤

∑E∈π

∑n

‖µ(E ∩ An)‖ =∑n

∑E∈π

‖µ(E ∩ An)‖ ≤∑n

|µ|(An).

Since it holds for any partition π, we obtain the inequality

|µ|(⋃n

An) ≤∑n

|µ|(An). (1.2.2)

Since |µ| is additive and monotone, then for each n:

i=n∑i=1

|µ|(Ai) = |µ|(n⋃i=1

Ai) ≤ |µ|(⋃n

An),

16 CHAPTER 1.

and by taking the limit:

∞∑i=1

|µ|(Ai) ≤ |µ|(⋃n

An). (1.2.3)

Combining (1.2.2) and (1.2.3) give us σ-additivity of |µ|. 2

It is also easy to see that µ is a σ-finite or finite E-valued measure iff |µ| is.

Definition 1.2.5 An E-valued measure µ is called a vector-valued measure ofbounded variation if |µ| is finite.

The vector-valued measure constructed in Example 1.2.1 is of bounded variationif αn = 2−n; and, of unbounded variation if αn = 1/n.

1.2.3. Operations with vector-valued measures. Let Σ be a σ-algebraof subsets of a non-empty set X and let E be a Banach space. The set Mb =ME

b (X,Σ) of all E-valued measures of bounded variation on Σ is a vector spacewith respect to the natural operations of addition and scalar multiplication, i.e.,

(µ1 + µ2)(A) = µ1(A) + µ2(A), (αµ)(A) = αµ(A) (1.2.4)

for every A ∈ Σ.

Theorem 1.2.2 Mb is a Banach space with respect to the norm ‖ · ‖b:

‖µ‖b := |µ|(X) (∀µ ∈Mb)

which is called the total variation of a measure µ.

Proof: By (1.2.1) and (1.2.4), the map ‖ · ‖b : Mb → R+ satisfies all axioms ofnorm. To show that Mb is complete under the norm ‖ · ‖b, it is enough to showthat if

∞∑n=1

‖µn‖b <∞ (1.2.5)

for some sequence (µn)∞n=1 of E-valued measures, then

µ :=∞∑n=1

µn ∈Mb. (1.2.6)


For any A ∈ Σ, the series∞∑n=1

µn(A)

is convergent in E, because E is a Banach space, and there hold (1.2.5) and

‖µn(A)‖ ≤ ‖µn‖b (∀n).

So µ in (1.2.6) is well defined. We leave to the reader to show that µ ∈ Mb asan exercise. 2

In general, if µ1 and µ2 are not finite E-valued measures, the formula (1.2.4) doesnot define the sum of them. Thus, in general, the set of all E-valued measuresis not a vector space.

Now we are going to study the important case of finite signed measures. Let Mb

be the Banach space of all finite signed measures on (X,Σ). First, we define apartial ordering in Mb saying that µ1 ≤ µ2 whenever

µ1(A) ≤ µ2(A) (∀A ∈ Σ), (1.2.7)

then Mb becomes a partially ordered vector space. We show that Mb is avector lattice, which means, by the definition, that for every µ, ν ∈ Mb thereexists supµ, ν ∈Mb.

For this purpose, we denote for µ ∈Mb the number

‖µ‖ := sup|µ(A)| : A ∈ Σ.

From|(µ1 + µ2)(A)| ≤ |µ1(A)|+ |µ2(A)| ≤ ‖µ1‖+ ‖µ2‖

holding for µ1 and µ2 in Mb and arbitrary A ∈ Σ, it follows that

‖µ1 + µ2‖ ≤ ‖µ1‖+ ‖µ2‖.

Furthermore, it is evident that ‖αµ‖ = |α| · ‖µ‖ for µ ∈Mb and a real α. Thus,‖ · ‖ is, clearly, a norm in Mb (it is equivalent to ‖ · ‖b in Theorem 1.2.2).

We prove now that supµ1, µ2 exists in Mb for all µ1 and µ2 in Mb. For anyA ∈ Σ, let the number ν(A) be defined by

ν(A) = supµ1(B) + µ2(A \B) : A ⊇ B ∈ Σ.

18 CHAPTER 1.

It is not difficult to see that ν is a signed measure on Σ. Since the supremumon the right is less than or equal to ‖µ1‖+ ‖µ2‖, we get

sup|ν(A)| : A ∈ Σ ≤ ‖µ1‖+ ‖µ2‖.

It follows that ν is bounded and ‖ν‖ ≤ ‖µ1‖+ ‖µ2‖. Having shown that ν is anelement of Mb, we prove now that ν = supµ1, µ2. From the definition of ν, itis clear that ν(A) ≥ µ1(A) and ν(A) ≥ µ2(A) for every A ∈ Σ, so ν is an upperbound of µ1 and µ2. Let τ be another upper bound. Then, for any A ∈ Σ andA ⊇ B ∈ Σ, we have

τ(A) = τ(B) + τ(A \B) ≥ µ1(B) + µ2(A \B),

soτ(A) ≥ supµ1(B) + µ2(A \B) : A ⊇ B ∈ Σ = ν(A).

This shows that ν = supµ1, µ2 in Mb. Note now that infµ1, µ2 exists aswell, because

infµ1, µ2 = − sup−µ1,−µ2.

Hence Mb is a vector lattice.

As a direct consequence of this we have the following theorem in the case offinite signed measures.

Theorem 1.2.3 (The Jordan decomposition theorem) If µ is a signed mea-sure then there exist unique positive measures ν and ξ such that µ = ν − ξ and,for any positive measures ν ′ and ξ′ such that µ = ν ′ − ξ′, the inequalities ν ≤ ν ′

and ξ ≤ ξ′ hold.

Proof: In the case when µ is finite, it is enough to take ν := µ+ = sup0, µand ξ := µ− = − inf0, µ. If µ is not finite (or even not σ-finite) the proof ismore complicated. We will not treat it in this book, and refer the reader for theproof to [5] and [4]. 2

We list main properties of the Banach space Mb = (Mb, ‖ · ‖b) of all signedmeasures of bounded variation on (X,Σ). Proofs are easy and left to the reader.

(i) The norm ‖ · ‖b on Mb satisfies

|µ| ≤ |ν| ⇒ ‖µ‖b ≤ ‖ν‖b (∀µ, ν ∈Mb),


where |µ| := supµ,−µ. In other words, Mb is a Banach lattice.

(ii) The norm ‖ · ‖b is additive on the positive cone M+b of Mb:

µ ≥ 0, ν ≥ 0⇒ ‖µ+ ν‖b = ‖µ‖b + ‖ν‖b.

(iii) Mb is Dedekind complete in the following sense. For any order boundednonempty family µαα∈A, the supremum supα∈A µα exists and can be evaluatedby the formula

supα∈A

µα(B) = supk∑j=1

µαj(Bj) : Bj ∈ Σ, αj ∈ A,

j=k⋃j=1

Bj = B,Bi∩Bj = ∅ if i 6= j

(1.2.8)for any B ∈ Σ .


Exercise 1.2.1 ∗∗ Prove the property of vector-valued measures given after Definition 1.2.1

Exercise 1.2.2 ∗ Let µ be a E-valued measure on (X,Σ). Show that

|µ|(A) := sup∞∑n=1

‖µ(An)‖ : An ∈ Σ,∞⋃n=1

An = A,Ak ∩Ap = ∅ if k 6= p.

Exercise 1.2.3 Let µ be a E-valued measure on (X,Σ). Show that |µ| is additive andmonotone in the following sense:

|µ|(A ∪B) = |µ|(A) + |µ|(B)

for any A,B ∈ Σ, A ∩B = ∅, and|µ|(A) ≤ |µ|(B)

for any A,B ∈ Σ, A ⊆ B.

Exercise 1.2.4 ∗∗ Show that a signed measure µ is finite iff its variation |µ| is finite. Showthat a signed measure µ is σ-finite iff its variation |µ| is σ-finite.

Exercise 1.2.5 Show that the vector-valued measure constructed in Example 1.2.1 is of un-bounded variation if αn = 1/n; and, of bounded variation if αn = 2−n.

20 CHAPTER 1.

Exercise 1.2.6 Let µ be a E-valued measure. Show that |µ| is a separable iff µ is separable.

Exercise 1.2.7 Show that the function µ from Σ to E, which is defined in the proof of The-orem 1.2.2 by

µ(A) =∞∑n=1

µn(A) (∀A ∈ Σ),

is an E-valued measure of bounded variation.

Exercise 1.2.8 ∗∗∗ Let µ be a non-atomic signed measure of bounded variation on Σ (i.e.for any A ∈ Σ, µ(A) 6= 0, there exists B ∈ Σ, B ⊆ A, such that 0 < |µ(B)| < |µ(A)|). UsingExercise 1.1.9, show that, for any α,

infµ(A) : A ∈ Σ ≤ α ≤ supµ(A) : A ∈ Σ ,

there is Aα ∈ Σ such that µ(Aα) = α.

Exercise 1.2.9 Let µ be a signed non-atomic measure on (Ω,Σ). Show that the set

R(µ) := µ(E) : E ∈ Σ.

is closed in R ∪ ∞.

Exercise 1.2.10 ∗ Let Mb be the Banach space of all finite signed measures on (X,Σ). Showthat Mb is a partially ordered Banach space with respect to the order given in (1.2.7).

Exercise 1.2.11 Show that Mb satisfies 1.2.4(i).

Exercise 1.2.12 Show that Mb satisfies 1.2.4(ii).

Exercise 1.2.13 ∗∗ Show that (1.2.8) defines the supremum supα∈A µα.

Exercise 1.2.14 Show that the Rn-valued measure Ψ : Ω→ Rn defined in Example 1.2.2 hasa bounded variation and

|Ψ|(X) ≤m∑k=1

µk(X) · ‖wk‖.

Exercise 1.2.15 ∗∗ Show that the Rn-valued measure Φ : Ω → Rn defined in Example 1.2.3has a bounded variation and

|Φ|(X) ≤m∑k=1

µk(Ak) · ‖wk‖.

Show that the rangeR(Φ) := Φ(E) : E ∈ Ω

of Φ is a compact subset of Rn if all µk are non-atomic.

Exercise 1.2.16 ∗∗ Take the Rn-valued measure Φ as in Exercise 1.2.15 and assume that anyµk, which was used in the construction of Φ, is non-atomic on Ak. Show that R(Φ) is convexin Rn.

1.3. THE LEBESGUE INTEGRAL 21

1.3 The Lebesgue Integral

Historically, many approaches to integration have been used (see the book [2]of S.B. Chae for nice historical remarks). No doubt that the Lebesgue integra-tion is the most successful approach. In the following, we fix a σ-finite measurespace (X,A, µ). To avoid unnecessary details, we consider only the case of R-valued functions and leave obvious generalizations to C- or Rn-valued cases tothe reader.

1.3.1. The construction of the Lebesgue integral. Recall that if acertain property involving the points of measure space is true, except a subsethaving measure zero, then we say that this property is true almost everywhere(abbreviated as a.e.). We denote

f+(x) = max(0, f(x)),

f−(x) = max(0,−f(x)),

and|f |(x) = f+(x) + f−(x).

Let Ai ∈ A, i = 1, . . . , n, be such that µ(Ai) <∞ for all i, and Ai ∩ Aj = ∅ forall i 6= j. The function

g(x) =n∑i=1

λiχAi(x), λi ∈ R,

is called a simple function. The Lebesgue integral of a simple function g(x)is defined as ∫

X

g dµ :=n∑i=1

λi · µ(Ai).

It is a simple exercise to show that the Lebesgue integral of a simple function iswell defined. We leave this to the reader.

Definition 1.3.1 Suppose that µ is finite. Let f : X → R be an arbitrarynonnegative bounded measurable function and let (gn)∞n=1 be a sequence of simplefunctions which converges uniformly to f . Then the Lebesgue integral of f is∫

X

f dµ := limn→∞

∫X

gn dµ.

22 CHAPTER 1.

It can be easily shown that the limit in Definition 1.3.1 exists and does notdepend on the choice of a sequence (gn)∞n=1, and moreover, the sequence (gn)∞n=1

can be chosen such that 0 ≤ gn ≤ f for all n.

Definition 1.3.2 Let f : X → R be an arbitrary nonnegative bounded measur-able function. Then the Lebesgue integral of f is∫

X

f dµ := sup ∫A

f dµ : A ∈ A , µ(A) <∞

Remark that the existence of the supremum in Definition 1.3.2 is guarantied by1.1.3(d).

Definition 1.3.3 Let f : X → R be a nonnegative unbounded measurable. Put

fM(x) =

f(x) if 0 ≤ f(x) ≤M,M if M < f(x).

Then the Lebesgue integral of f is∫X

f dµ := limM→∞

∫X

fM dµ.

Definition 1.3.4 Let f : X → R be a measurable function. Then the Lebesgueintegral of f is defined by∫

X

f dµ :=

∫X

f+dµ−∫X

f−dµ.

If both of these terms are finite then the function f is called integrable. In thiscase we write f ∈ L1.

1.3.2. Elementary properties of the Lebesgue integral. We will usethe following notation. For any A ∈ A:∫

A

f dµ :=

∫X

f · χA dµ.

Let us give several obvious properties of Lebesgue integral.


Lemma 1.3.1 Let f : X → R be an arbitrary nonnegative measurable functionthen∫

X

f dµ = sup∫X

φ dµ : φ is a simple function such that 0 ≤ φ ≤ f.

Proof: Firstly consider the case of bounded f and finite µ. Then by thedefinition of

∫Xf dµ there exists a sequence (see the remark after Definition

1.3.1) (gn)∞n=1 of simple functions such that 0 ≤ gn ≤ f for all n, and (gn)converges uniformly to f , and satisfies∫

X

f dµ = limn→∞

∫X

gn dµ = supn

∫X

gn dµ .

Now the using of Definitions 1.3.2 and 1.3.3 completes the proof. 2

L1. If f, g : X → R are measurable, g is integrable, and |f(x)| ≤ g(x), then fis integrable and ∣∣∣∣∫

X

f dµ

∣∣∣∣ ≤ ∫X

g dµ.

L2.∫X|f | dµ = 0 if and only if f(x) = 0 a.e.

L3. If f1, f2 : X → R are integrable then, for λ1, λ2 ∈ R, the linear combinationλ1f1 + λ2f2 is integrable and∫

X

[λ1f1 + λ2f2] dµ = λ1

∫X

f1 dµ+ λ2

∫X

f2 dµ.

If f is integrable function, we write f ∈ L1 = L1(X,A, µ).

L4. Let f ∈ L1, then the formula

ν(A) :=

∫A

f dµ

defines a signed measure on the σ-algebra A.

1.3.3. Convergence. If (fn)∞n=1 is a sequence of integrable functions on aset X, the statement “fn → f as n→∞” can be taken in many different senses,for example, for pointwise or uniform convergence (cf. the end of Section 1.1).One of them is L1-convergence.

24 CHAPTER 1.

Definition 1.3.5 We say that a sequence (fn)∞n=1 of integrable functions L1-converges to f (or converges in L1) if∫

|fn − f | dµ→ 0 as n→∞.

Of course, uniform convergence implies pointwise convergence, which in turnimplies a.e.-convergence, but these modes of convergence do not imply L1-con-vergence. The reader should keep in mind the following examples:

(i) fn = n−1χ(0,n);(ii) fn = χ(n,n+1);(iii) fn = nχ[0,1/n];(iv) f1 = χ[0,1], f2 = χ[0,1/2], f3 = χ[1/2,1], f4 = χ[0,1/4], f5 = χ[1/4,1/2], and, ingeneral, fn = χ[j/2k,(j+1)/2k], where n = 2k + j with 0 ≤ j < 2k.

In (i), (ii), and (iii), fn → 0 uniformly, pointwise, and a.e., respectively, butfn 6→ 0 in L1 (in fact,

∫|fn| dµ =

∫fn dµ = 1 for all n). In (iv), fn → 0

in L1 since∫|fn| dµ = 2−k for 2k ≤ n < 2k+1, but fn(x) does not converge

for any x ∈ [0, 1], since there are infinitely many n for which fn(x) = 0 andinfinitely many, for which fn(x) = 1. On the other hand, if fn → f a.e. and|fn| ≤ g ∈ L1 for all n, then fn → f in L1. This will be clear from the domi-nated convergence theorem (see Theorem 1.3.2 below) since |fn−f | ≤ 2g. Also,we will see below that if fn → f in L1 then some subsequence converges to f a.e.

1.3.4. The monotone convergence theorem. Now we state one of themost important results about convergence.

Theorem 1.3.1 (The monotone convergence theorem) If (fn)∞n=1 is a se-quence in L+

1 such that fj ≤ fj+1 for all j and f = supn fn then∫X

f dµ = limn→∞

∫X

fn dµ.

Proof: The limit of the increasing sequence (∫Xfn dµ)∞n=1 (finite or infinite)

exists. Moreover by 1.3.2.(L1),∫Xfn dµ ≤

∫Xf dµ for all n, so

limn→∞

∫X

fn dµ ≤∫X

f dµ.


To establish the reverse inequality, fix α ∈ (0, 1), let φ be a simple function with0 ≤ φ ≤ f , and let En = x : fn(x) ≥ αφ(x). Then (En)∞n=1 is an increasingsequence of measurable sets whose union is X, and we have∫

X

fn dµ ≥∫En

fn dµ ≥ α

∫En

φ dµ (∀n ≥ 1).

By 1.3.2.(L4) and by Proposition 1.1.5, limn→∞

∫Enφ dµ =

∫Xφ dµ, and hence

limn→∞

∫X

fn dµ ≥ α

∫X

φ dµ.

Since this is true for all α, 0 < α < 1, it remains true for α = 1:

limn→∞

∫X

fn dµ ≥∫X

φ dµ .

Using Lemma 1.3.1, we may take the supremum over all simple functions φ,0 ≤ φ ≤ f . Thus

limn→∞

∫X

fn dµ ≥∫X

f dµ. 2

Proofs of the following two corollaries of Theorem 1.3.1 are left to the reader.

Corollary 1.3.1 If (fn)∞n=1 is a sequence in L+1 and f =

∑∞n=1 fn pointwise

then ∫f dµ =

∑n

∫fn dµ. 2

Corollary 1.3.2 If (fn)∞n=1 is a sequence in L+1 , f ∈ L+

1 , and fn ↑ f a.e., then∫fn dµ ↑

∫f dµ. 2

1.3.5. The Fatou lemma. The condition in the previous subsection thatthe sequence (fn)∞n=1 is increasing can be omitted in the following way.

Lemma 1.3.2 (Fatou’s lemma) If (fn)∞n=1 is any sequence in L+1 then∫

lim infn→∞

fn dµ ≤ lim infn→∞

∫fn dµ .

26 CHAPTER 1.

Proof: For each k ≥ 1, we have

infn≥k

fn ≤ fj (∀j ≥ k),

hence ∫infn≥k

fn dµ ≤∫fj dµ (∀j ≥ k),

hence ∫infn≥k

fn dµ ≤ infj≥k

∫fj dµ.

Now let k →∞ and apply Theorem 1.3.1:∫lim inf

n→∞fn dµ = lim

k→∞

∫infn≥k

fn dµ ≤ lim infn→∞

∫fn dµ . 2

Corollary 1.3.3 If (fn)∞n=1 is a sequence in L+1 , f ∈ L+

1 , and fn → f a.e., then∫f dµ ≤ lim inf

n→∞

∫fn dµ. 2

1.3.6. The dominated convergence theorem. Now we present thefollowing convergence theorem, which has many applications in PDEs, functionalanalysis, operator theory, etc.

Theorem 1.3.2 (The dominated convergence theorem) Let f and g bemeasurable, let fn be measurable for any n such that

|fn(x)| ≤ g(x) a.e.,

and fn → f a.e. If g is integrable then f and fn are also integrable and

limn→∞

∫fn dµ =

∫f dµ.

Proof: f is measurable by Exercise 1.1.14 and, since |f | ≤ g a.e., we havef ∈ L1. We have that g + fn ≥ 0 a.e. and g − fn ≥ 0 so, by Fatou’s lemma,∫

g dµ+

∫f dµ ≤ lim inf

n→∞

∫(g + fn) dµ =

∫g dµ+ lim inf

n→∞

∫fn dµ,

1.3. THE LEBESGUE INTEGRAL 27∫g dµ−

∫f dµ ≤ lim inf

n→∞

∫(g − fn) dµ =

∫g dµ− lim sup

n→∞

∫fn dµ.

Therefore

lim infn→∞

∫fn dµ ≥

∫f dµ ≥ lim sup

n→∞

∫fn dµ,

and the result follows. 2

1.3.7.! Convergence in measure. Another mode of convergence, that isfrequently useful, is convergence in measure. We say that a sequence (fn)∞n=1 ofmeasurable functions on (X,M, µ) is Cauchy in measure if, for every ε > 0,

µ(x : |fn(x)− fm(x)| ≥ ε)→ 0 as m,n→∞,

and that (fn)∞n=1 converges in measure to f if, for every ε > 0,

µ(x : |fn(x)− f(x)| ≥ ε)→ 0 as n→∞.

For example, the sequences (i), (iii), and (iv) above converge to zero in measure,but (ii) is not Cauchy in measure.

Proposition 1.3.1 If fn → f in L1 then fn → f in measure.

Proof: Let En,ε = x : |fn(x)− f(x)| ≥ ε. Then∫|fn − f | dµ ≥

∫En,ε

|fn − f | dµ ≥ εµ(En,ε),

so µ(En,ε) ≤ ε−1∫|fn − f | dµ→ 0. 2

The converse of Proposition 1.3.1 is false, as Examples 1.3.3.(i) and 1.3.3.(iii)show.

Theorem 1.3.3 Suppose that (fn)∞n=1 is Cauchy in measure. Then there is ameasurable function f such that fn → f in measure, and there is a subsequence(fnj

)j that converges to f a.e. Moreover, if fn → g in measure then g = f a.e.

28 CHAPTER 1.

Proof: We can choose a subsequence (gj)j = (fnj)j of (fn)∞n=1 such that if

Ej = x : |gj(x)− gj+1(x)| ≥ 2−j

then µ(Ej) ≤ 2−j. If Fk =⋃∞j=k Ej then µ(Fk) ≤

∑∞k=1 2−j = 21−k, and if

x 6∈ Fk we have for i ≥ j ≥ k

|gj(x)− gi(x)| ≤i−1∑l=j

|gl+1(x)− gl(x)| ≤i−1∑l=j

2−l ≤ 21−j. (1.3.1)

Thus (gj)j is pointwise Cauchy on F ck . Let

F =∞⋂k=1

Fk = lim supjEj.

Then µ(F ) = 0, and if we set f(x) = lim gj(x) for x 6∈ F , and f(x) = 0 forx ∈ F , then f is measurable and gj → f a.e. By (1.3.1), we have that

|gj(x)− f(x)| ≤ 21−j

for x 6∈ Fk and j ≥ k. Since µ(Fk) → 0 as k → ∞, it follows that gj → f inmeasure, because

x : |fn(x)−f(x)| ≥ ε ⊆ x : |fn(x)−gj(x)| ≥ 1

2ε∪x : |gj(x)−f(x)| ≥ 1

2ε,

and the sets on the right both have small measure when n and j are large.Likewise, if fn → g in measure

x : |f(x)− g(x)| ≥ ε ⊆ x : |f(x)− fn(x)| ≥ 1

2ε ∪ x : |fn(x)− g(x)| ≥ 1

2ε

for all n, hence µ(x : |f(x)− g(x)| ≥ ε) = 0 for all ε > 0, and f = g a.e. 2

Theorem 1.3.4 Let fn → f in L1, then there is a subsequence (fnk)k such that

fnk→ f a.e.

Proof: LetEn,ε = x : |fn(x)− f(x)| ≥ ε.


Then ∫|fn − f | dµ ≥

∫En,ε

|fn − f | dµ ≥ εµ(En,ε),

so µ(En,ε) → 0. Then, by Theorem 1.3.3, there is a subsequence (fnk)k such

that fnk→ f a.e. 2


Exercise 1.3.1 Check that the limit in Definition 1.3.1 exists and does not depend on thechoice of a sequence (gn)∞n=1.

Exercise 1.3.2 Prove the properties L1− L4 of the Lebesgue integral from 1.3.2.

Exercise 1.3.3 Investigate simple properties of sequences (fn)∞n=1 in 1.3.3.(i)-(iv).

Exercise 1.3.4 Let [a, b] be a compact interval in R and µ be a standard Borel measure on[a, b]. Show that the Lebesgue integral from any continuous function f : [a, b] → R coincideswith the Riemann integral from f .

Exercise 1.3.5 Show that the condition “µ(X) < ∞” in Egoroff’s theorem can be replacedby “|fn| ≤ g for all n, where g ∈ L1”.

Exercise 1.3.6 Prove the dominated convergence theorem by using of the Egoroff theorem.

Exercise 1.3.7 ∗ Let fn → f in measure and gn → g in measure. Show that

fn · gn → f · g

in measure if µ(X) <∞, but not necessarily if µ(X) =∞.

Exercise 1.3.8 Prove Corollary 1.3.1.


Exercise 1.3.10 Let fn → f in measure and fn ≥ 0 for all n. Show that∫f dµ ≤ lim inf

n→∞

∫fn dµ.

30 CHAPTER 1.


Exercise 1.3.12 ∗∗ Let χAn→ f in measure. Show that f is a.e. equal to the characteristic

function of a measurable set.

Exercise 1.3.13 Fix the Lebesgue measure on R.

(a) Given a nonempty set Ξ ⊆ L+1 (R), show that

inf∫

ξ(t) dt : ξ ∈ Ξ≤∫f(t) dt

for every f ∈ co(Ξ), where co(Ξ) is the convex hull of Ξ in the vector space L1(R), whichis, by the definition, the intersection of all convex sets containing Ξ.

(b) Construct an example of a nonempty set Θ ⊆ L+1 (R) such that

inf∫

ξ(t) dt : ξ ∈ Θ<

∫f(t) dt

for every f ∈ co(Θ).

(c) Show that a nonempty set ∆ ⊆ L+1 (R) satisfying

inf∫

ξ(t) dt : ξ ∈ ∆<

∫f(t) dt

for every f ∈ co(∆) can not be finite.

Exercise 1.3.14 Let f ∈ L1(R) and F (t) :=∫ t−∞ f(s)ds. Show that F is continuous. Com-

pare with Exercise 1.3.2.

Chapter 2

2.1 The Extension of Measure

In this section we present the Caratheodory extension theorem which is the mostimportant result in the measure theory. By using of this theorem, we can con-struct almost all important measures, in particular, the Lebesgue measure andLebesgue – Stieltjes Measures on R. Then we consider the product of measurespaces and prove the Fubini theorem.

2.1.1. Outer measures. Let X be a nonempty set. An outer measure(or submeasure) on X is a function

ξ : P(X)→ [0,∞]

that satisfies:

(a) ξ(∅) = 0;

(b) ξ(A) ≤ ξ(B) if A ⊆ B;

(c) ξ

(∞⋃j=1

Aj

)≤∞∑j=1

ξ(Aj) for all sequences (Aj)j of subsets of X.

The common way to obtain an outer measure is to start with a family G of“elementary sets” on which a notion of measure (= length, mass, charge, orvolume) is defined (such as rectangles or cubes in Rn) and then approximatearbitrary sets from the outside by countable unions of members of G.

31

32 CHAPTER 2.

Proposition 2.1.1 Let G ⊆ P(X) and let ρ : G → [0,∞] be such that ∅ ∈ G,X ∈ G, and ρ(∅) = 0. For any A ⊆ X, define

ξ(A) = ρ∗(A) = inf

∞∑j=1

ρ(Gj) : Gj ∈ G and A ⊆∞⋃j=1

Gj

. (2.1.1)

Then ξ is an outer measure.

Proof: For any A ⊆ X, we have A ⊆⋃∞j=1X, so ξ is well defined. Obviously

ξ(∅) = 0. To prove countable subadditivity, suppose Aj∞j=1 ⊆ P(X) and ε > 0.For each j, there exists Gk

j∞k=1 ⊆ G such that Aj ⊆⋃∞k=1G

kj and

∞∑k=1

ρ(Gkj ) ≤ ξ(Aj) + ε2−j.

Then if A =⋃∞j=1Aj, we have

A ⊆∞⋃

j,k=1

Gkj &

∑j,k

ρ(Gkj ) ≤

∑j

ξ(Aj) + ε,

whence ξ(A) ≤∑

j=1 ξ(Aj) + ε. Since ε > 0 is arbitrary, we have done. 2

2.1.2. The Caratheodory theorem. The principle step that leads fromouter measures to measures is following. Let ξ be an outer measure on X.

Definition 2.1.1 A set A ⊆ X is called ξ-measurable if

ξ(B) = ξ(B ∩ A) + ξ(B ∩ (X \ A)) (∀B ⊆ X). (2.1.2)

Of course, the inequality

ξ(B) ≤ ξ(B ∩ A) + ξ(B ∩ (X \ A))

holds for any A and B. So, to prove that A is ξ-measurable, it suffices to provethe reverse inequality, which is trivial if ξ(B) = ∞. Thus, we see that A isξ-measurable iff

ξ(B) ≥ ξ(B ∩ A) + ξ(B ∩ (X \ A)) (∀B ⊆ X, ξ(B) <∞). (2.1.3)

2.1. THE EXTENSION OF MEASURE 33

Theorem 2.1.1 (Caratheodory’s theorem) Let ξ be an outer measure onX. Then the family Σ of all ξ-measurable sets is a σ-algebra, and the restrictionof ξ to Σ is a complete measure.

Proof: First, we observe that Σ is closed under complements, since the def-inition of ξ-measurability of A is symmetric in A and Ac := X \ A. Next, ifA,B ∈ Σ and E ⊆ X,

ξ(E) = ξ(E ∩ A) + ξ(E ∩ Ac)

= ξ(E ∩ A ∩B) + ξ(E ∩ A ∩Bc) + ξ(E ∩ Ac ∩B) + ξ(E ∩ Ac ∩Bc).

But (A ∪B) = (A ∩B) ∪ (A ∩Bc) ∪ (Ac ∩B) so, by subadditivity,

ξ(E ∩ A ∩B) + ξ(E ∩ A ∩Bc) + ξ(E ∩ Ac ∩B) ≥ ξ(E ∩ (A ∪B)),

and henceξ(E) ≥ ξ(E ∩ (A ∪B)) + ξ(E ∩ (A ∪B)c).

It follows that A ∪ B ∈ Σ, so Σ is an algebra. Moreover, if A,B ∈ Σ andA ∩B = ∅,

ξ(A ∪B) = ξ((A ∪B) ∩ A) + ξ((A ∪B) ∩ Ac) = ξ(A) + ξ(B),

so ξ is finitely additive on Σ.

To show that Σ is a σ-algebra, it suffices to show that Σ is closed under countabledisjoint unions. If (Aj)

∞j=1 is a sequence of disjoint sets in Σ, set

Bn =n⋃j=1

Aj & B =∞⋃j=1

Aj.

Then, for any E ⊆ X,

ξ(E ∩Bn) = ξ(E ∩Bn ∩ An) + ξ(E ∩Bn ∩ Acn)

= ξ(E ∩ An) + ξ(E ∩Bn−1),

so a simple induction shows that ξ(E ∩Bn) =∑n

j=1 ξ(E ∩ Aj). Therefore

ξ(E) = ξ(E ∩Bn) + ξ(E ∩Bcn) ≥

n∑j=1

ξ(E ∩ Aj) + ξ(E ∩Bc)

34 CHAPTER 2.

and, letting n→∞, we obtain

ξ(E) ≥∞∑j=1

ξ(E ∩ Aj) + ξ(E ∩Bc) ≥ ξ

(∞⋃j=1

(E ∩ Aj)

)+ ξ(E ∩Bc)

= ξ(E ∩B) + ξ(E ∩Bc) ≥ ξ(E).

Thus the inequalities in this last calculation become equalities. It follows B ∈ Σ.Taking E = B we have ξ(B) =

∑∞1 ξ(Aj), so ξ is σ-additive on Σ. Finally, if

ξ(A) = 0 then we have for any E ⊆ X

ξ(E) ≤ ξ(E ∩ A) + ξ(E ∩ Ac) = ξ(E ∩ Ac) ≤ ξ(E),

so A ∈ Σ. Therefore ξ(E ∩ A) = 0 and ξ|Σ is a complete measure. 2

Combination of Proposition 2.1.1 and Theorem 2.1.1 gives the following corollarywhich is also called Caratheodory’s theorem.

Corollary 2.1.1 Let G ⊆ P(X) be such that ∅ ∈ G, X ∈ G, and let ρ : G →[0,∞] satisfy ρ(∅) = 0. Then the family Σ of all ρ∗-measurable sets (where ρ∗

is given by (2.1.1)) is a σ-algebra, and the restriction of ρ∗ to Σ is a completemeasure. 2

2.1.3. Premeasures. Our main definition in this subsection is following.

Definition 2.1.2 Let A be an algebra of subsets of X, i.e. A contains ∅ andX, and A is closed under finite intersections and complements. A functionζ : A → [0,∞] is called a premeasure if ζ(∅) = 0 and

ζ(∞⋃j=1

Aj) =∞∑j=1

ζ(Aj)

for any disjoint sequence (Aj)j of elements of A such that⋃∞j=1Aj ∈ A.

Theorem 2.1.2 If ζ is a premeasure on an algebra A ⊆ P(X) and

ζ∗ : P(X)→ [0,∞]

is given by (2.1.1) then ζ∗|A = ζ and every A ∈ A is ζ∗-measurable.


Proof: First, ζ is obviously additive and hence monotone on A, i.e.

A ⊆ B ⇒ ζ(A) ≤ ζ(B) (∀A,B ∈ A).

Suppose that A ∈ A, An ∈ A, and A ⊆⋃∞n=1 An. Then

Bn := A ∩ (An \n−1⋃j=1

Aj) ∈ A,

since A is an algebra. The sequence (Bn)∞n=1 is disjoint. This implies

ζ(A) =∞∑n=1

ζ(Bn) ≤∞∑n=1

ζ(An),

and therefore ζ(A) ≤ ζ∗(A) (here we used the monotonity of ζ). The reverseinequality ζ∗(A) ≤ ζ(A) follows from the definition of ζ∗.

To show that every A ∈ A is ζ∗-measurable, take A ∈ A, Y ⊆ X, and ε > 0.Then, by (2.1.1), there exists a sequence (An)∞n=1 in A such that

Y ⊆∞⋃n=1

An &∞∑n=1

ζ(An) ≤ ζ∗(Y ) + ε.

Since ζ is additive on A, Y ∩A ⊆⋃∞n=1(An ∩A), and Y ∩ (X \A) ⊆

⋃∞n=1(An ∩

(X \ A)) then

ζ∗(Y ) + ε ≥∞∑n=1

ζ(An∩A) +∞∑n=1

ζ(An∩ (X \A)) ≥ ζ∗(Y ∩A) + ζ∗(Y ∩ (X \A)).

Since ε > 0 is arbitrary and ζ∗ is an outer measure,

ζ∗(Y ) ≥ ζ∗(Y ∩ A) + ζ∗(Y ∩ (X \ A)) ≥ ζ∗(Y ).

Y ⊆ X is arbitrary, so A is ζ∗-measurable. 2

Theorem 2.1.3 If ζ is a premeasure on an algebra A ⊆ P(X) and A is theσ-algebra generated by A, then there exists a measure ζ on A such that ζ|A = ζ.Moreover, if ζ is σ-finite, then ζ is unique extension of ζ.

36 CHAPTER 2.

Proof: By Caratheodory’s theorem and Theorem 2.1.2, such an extension ζ∗

exists. We leave to the reader to show that measure ζ on A, which extends ζfrom A, is unique if the premeasure ζ is σ-finite. 2

2.1.4. The Lebesgue and Lebesgue – Stieltjes measure on R. Themost important application of Caratheodory’s theorem is the construction ofthe Lebesgue measure on R. Take G as the set of all intervals [a, b], wherea, b ∈ R∪−∞,+∞ and [a, b] = ∅ if a > b. Define the function ρ : G → R∪∞by

ρ([a, b]) := b− a (∀a ≤ b) & ρ([a, b]) = 0 (∀a > b).

The function ρ has the obvious extension (which we denote also by ρ) to thealgebra A generated by all finite or infinite intervals, and this extension is apremeasure on A.

The σ-algebra Σ given by Corollary 2.1.1 is called the the Lebesgue σ-algebrain R, and the restriction of ρ∗ to Σ = Σ(R) is called the Lebesgue measureon R and is denoted by µ. By Theorem 2.1.3, µ is the unique extension of ρ.

By the construction, B(R) ⊆ Σ(R). Hence the Lebesgue measure is a Borelmeasure. It can be shown that B(R) 6= Σ(R) (see Exercise 2.1.8) and that theLebesgue measure can be obtained also as the completion of any Borel measureω such that

ω([a, b]) = b− a (∀a ≤ b).

The notion of the Lebesgue measure on R has the following very natural gener-alization. Suppose that µ is a σ-finite Borel measure on R, and let

F (x) := µ((−∞, x]) (∀x ∈ R). (2.1.4)

Then F is increasing and right continuous (see Exercise 2.1.9). Moreover, ifb > a, (−∞, b] = (−∞, a] ∪ (a, b], so

µ((a, b]) = F (b)− F (a).

Our procedure used above can be to turn this process around and construct ameasure µ starting from an increasing, right-continuous function F . The specialcase F (x) = x will yield the usual Lebesgue measure. As building blocks we canuse the left-open, right-closed intervals in R i.e. sets of the form (a, b] or (a,∞)or ∅, where −∞ ≤ a < b < ∞. We call such sets h-intervals. The family Aof all finite disjoint unions of h-intervals is an algebra, moreover, the σ-algebragenerated by A is the Borel algebra B(R). We omit the direct and routine proofof the following elementary lemma.


Lemma 2.1.1 Given an increasing and right continuous function F : R → R,if

(aj, bj] (j = 1, . . . , n)

are disjoint h-intervals, let

µ0

(n⋃1

(aj, bj]

)=

n∑1

[F (bj)− F (aj)] ,

and let µ0(∅) = 0. Then µ0 is a premeasure. 2

Theorem 2.1.4 If F : R → R is any increasing, right continuous function,there is a unique Borel measure µF on R such that

µF ((a, b]) = F (b)− F (a) (∀a, b).

If G is another such function then µF = µG iff F −G is constant.

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

F (x) =

µ((0, x]) if x > 0,0 if x = 0,−µ((x, 0]) if x < 0,

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

Proof: Each F induces a premeasure on B(R) by Lemma 2.1.1. It is clearthat F and G induce the same premeasure iff F −G is constant, and that thesepremeasures are σ-finite (since R =

⋃∞−∞(j, j + 1]). The first two assertions

follow now from Exercise 2.1.11. As for the last one, the monotonicity of µimplies the monotonicity of F , and the continuity of µ from above and frombelow implies the right continuity of F for x ≥ 0 and x < 0. It is evident thatµ = µF on A, and hence µ = µF on B(R) (accordingly to Exercise 2.1.11). 2

Lebesgue – Stieltjes measures possess some important and useful regularity prop-erties. Let us fix a complete Lebesgue – Stieltjes measure µ on R associated to anincreasing, right continuous function F . We denote by Σµ the Lebesgue algebracorrespondent to µ. Thus, for any E ∈ Σµ,

µ(E) = inf

∞∑j=1

[F (bj)− F (aj)] : E ⊆∞⋃j=1

(aj, bj]

38 CHAPTER 2.

= inf

∞∑j=1

µ((aj, bj]) : E ⊆∞⋃j=1

(aj, bj]

.

Since B(R) ⊆ Σµ, we may replace in the second formula for µ(E) h-intervals byopen intervals, namely.

Lemma 2.1.2 For any E ∈ Σµ,

µ(E) = inf

∞∑j=1

µ((aj, bj)) : E ⊆∞⋃j=1

(aj, bj)

. 2

Theorem 2.1.5 If E ∈ Σµ then

µ(E) = infµ(U) : U ⊇ E and U is open= supµ(K) : K ⊆ E and K is compact.

Proof: By Lemma 2.1.2, for any ε > 0, there exist intervals (aj, bj) such that

E ⊆∞⋃j=1

(aj, bj) & µ(E) ≤∞∑j=1

µ((aj, bj)) + ε.

If U =⋃∞j=1(aj, bj) then U is open, E ⊆ U , and µ(U) ≤ µ(E) + ε. On the other

hand, µ(U) ≥ µ(E) whenever E ⊆ U so the first equality is valid.

For the second one, suppose first that E is bounded. If E is closed then E iscompact and the equality is obvious. Otherwise, given ε > 0, we can choose anopen U , E \ E ⊆ U , such that µ(U) ≤ µ(E \ E) + ε. Let K = E \ U . Then Kis compact, K ⊆ E, and

µ(K) = µ(E)− µ(E ∩ U) = µ(E)− [µ(U)− µ(U \ E)]

≥ µ(E)− µ(U) + µ(E \ E) ≥ µ(E)− ε.If E is unbounded, let Ej = E ∩ (j, j + 1]. By the preceding argument, forany ε > 0, there exist a compact Kj ⊆ Ej with µ(Kj) ≥ µ(Ej) − ε2−j. LetHn =

⋃j=nj=−nKj. Then Hn is compact, Hn ⊆ E, and

µ(Hn) ≥ µ(

j=n⋃j=−n

Ej)− ε.

Since µ(E) = limn→∞

µ(⋃j=nj=−nEj), the result follows. 2

The proofs of the following two theorems are left to the reader as exercises.


Theorem 2.1.6 If E ⊆ R, the following are equivalent:(a) E ∈ Σµ;(b) E = V \N1, where V is a Gδ–set and µ(N1) = 0;(c) E = H ∪N2, where H is an Fσ–set and µ(N2) = 0. 2

Theorem 2.1.7 If E ∈ Σµ and µ(E) <∞ then, for every ε > 0, there is a setA that is a finite union of open intervals such that µ(E4A) < ε.2

2.1.5. Product measures. Let (Xα,Σα, µα)α∈∆ be a nonempty family ofmeasure spaces. In this subsection we show how to construct a product-measureon

∏α∈∆

Xα. First we define the family Ω of blocks:

A(Aα1 , Aα2 , . . . , Aαn) := Aα1×Aα2×· · ·×Aαn×∏

α∈∆; α 6=αk: k=1,...,n

Xα (Aαk∈ Σαk

);

and define a function: µ : Ω→ R+ ∪ ∞:

µ(A(Aα1 , Aα2 , . . . , Aαn)) := µα1(Aα1)·µα2(Aα2)·. . .·µαn(Aαn)·∏

α∈∆; α 6=αk:k=1,...,n

µα(Xα).

This function possesses an extension (by additivity) on the algebra A generatedby Ω. It is an exercise to show that µ is a premeasure on A.

Definition 2.1.3 The measure µ on the σ-algebra Σ generated by A accordinglyto Theorem 2.1.3 is called the product measure of µαα∈∆, and the triple

(∏α∈∆

Xα,Σ, µ).

is called the product of measure spaces (Xα,Σα, µα). We denote the σ-algebra Σ by

⊗α∈∆ Σα, and the measure µ by

⊗α∈∆ µα.

This construction is essentially non-trivial only in the following two cases. Thefirst one is when card(∆) <∞. The second one is when every µα is a probabilitymeasure (i.e. µα(Xα) = 1 for all α).

There are many interesting and important results about the product of measurespaces. The mostly used one is the Fubini theorem about changing the order

40 CHAPTER 2.

of integration. Let us use the the following notations. If E ⊆ X1 × X2 andx1 ∈ X1, x2 ∈ X2, we define

Ex1 := x ∈ X2 : (x1, x) ∈ E & Ex2 := x ∈ X1 : (x, x2) ∈ E .

Also, if f : X1×X2 → R is a function, we define fx1 : X2 → R and fx2 : X1 → Rby

fx1(x) := f(x1, x) & fx2(x) := f(x, x2) .

In this notations, for example, (χE)x1 = χEx1and (χE)x2 = χEx2 .

Theorem 2.1.8 (Fubini’s theorem) Let µ1, µ2 be σ-finite measures on (X1,Σ1)and (X2,Σ2),

(X1 ×X2, Σ1 ⊗ Σ2, µ1 ⊗ µ2) = (X1, Σ1, µ1)× (X2, Σ2, µ2),

and let f ∈ L1(X1×X2, Σ1⊗Σ2, µ1⊗ µ2). Then fx1 ∈ L1(X2, Σ2, µ2) µ1-a.e.,and fx2 ∈ L1(X1, Σ1, µ1) µ2-a.e., and∫

X1×X2

f d(µ1 ⊗ µ2) =

∫X2

[∫X1

fx2 dµ1

]dµ2 =

∫X1

[∫X2

fx1 dµ2

]dµ1 .

To prove this theorem, we need several lemmas.

Lemma 2.1.3 Let (X1,Σ1, µ1) and (X2,Σ2, µ2) be measure spaces, E ∈ Σ1⊗Σ2,and let f be a Σ1 ⊗ Σ2-measurable function on X1 ×X2, then:

(a) Ex1 ∈ Σ2 for all x1 ∈ X1 and Ex2 ∈ Σ1 for all x2 ∈ X2;

(b) fx1 is Σ2-measurable and fx2 is Σ1-measurable for all x1 ∈ X1 and x2 ∈ X2.

Proof: Denote by A the collection of all A ⊆ X1 ×X2 such that

Ax1 ∈ Σ2 & Ax2 ∈ Σ1 (∀x1 ∈ X1 , x2 ∈ X2) .

The family A contains all rectangles. Thus, since

[∞⋃n=1

An]x1 =∞⋃n=1

[An]x1 , [∞⋃n=1

Bn]x2 =∞⋃n=1

[Bn]x2 (2.1.5)

and

[X1 ×X2 \ A]x1 = X2 \ Ax1 , [X1 ×X2 \ A]x2 = X1 \ Ax2 , (2.1.6)

A is a σ-algebra. So Σ1 ⊗ Σ2 ⊆ A, and (a) is proved.

Now the part (b) follows from (a) due to

f−1x1

(A) = [f−1(A)]x1 & [fx2 ]−1(A) = [f−1(A)]x2 (∀A ⊆ R) . 2 (2.1.7)


Definition 2.1.4 A family M ⊆ P(X) is called a monotone class if M isclosed under countable increasing unions and countable decreasing intersections.

The proof of the following lemma is an elementary exercise and we leave it tothe reader.

Lemma 2.1.4 If A ⊆ P(X) is an algebra then the monotone class generatedby A coincides with the σ-algebra generated by A. 2

Lemma 2.1.5 Let (X1,Σ1, µ1) and (X2,Σ2, µ2) be measure spaces, E ∈ Σ1⊗Σ2.Then the functions x1 → µ2(Ex1) and x2 → µ1(Ex2) are measurable on (X1,Σ1)and (X2,Σ2), and

µ1 ⊗ µ2(E) =

∫X2

µ1(Ex2) dµ2 =

∫X1

µ2(Ex1) dµ1 . (2.1.8)

Proof: First we consider the case when µ1 and µ2 are finite. Let A bethe family of all E ∈ Σ1 ⊗ Σ2 for which (2.1.8) is true. If E = A × B, thenµ1(Ex2) = µ1(A)χB(x2) and µ2(Ex1) = µ2(B)χA(x1), so E ∈ A. By additivity,it follows that finite disjoint unions of rectangles are in A so, by Lemma 2.1.4,it will suffice to show that A is a monotone class. If (En)∞n=1 is an increasingsequence in A and E =

⋃∞n=1En, then the function fn(x2) = µ1((En)x2) are

measurable and increase pointwise to f(y) = µ1(Ex2). Hence f is measurableand, by the monotone convergence theorem,∫

X2

µ1(Ex2) dµ2 = limn→∞

∫X2

µ1((En)x2) dµ2 = limn→∞

µ1 × µ2(En) = µ1 × µ2(E).

Likewise µ1 × µ2(E) =∫X1µ2(Ex) dµ1, so E ∈ A. Similarly, if (En)∞n=1 is a

decreasing sequence in A and E =⋂∞n=1En, the function x2 → µ1((E1)x2) is in

L1(µ2) because µ1((E1)x2) ≤ µ1(X1) < ∞ and µ2(X2) < ∞, so the dominatedconvergence theorem can be applied to show that E ∈ A. Thus, A is a monotoneclass, and the proof is complete for the case of finite measure spaces.

Finally, if µ1 and µ2 are σ-finite, we can write X1 × X2 as the union of anincreasing sequence (Xj

1×Xj2)∞n=1 of rectangles of finite measure. If E ∈ Σ1⊗Σ2,

the preceding argument applies to E ∩ (Xj1 ×X

j2) for each j gives us

µ1 × µ2(E ∩ (Xj1 ×X

j2)) =

∫Xj

2

µ1(Ex2 ∩Xj1) dµ2 =

∫Xj

1

µ2(Ex1 ∩Xj2) dµ1 .

The application of the monotone convergence theorem then yields the desiredresult. 2

42 CHAPTER 2.

Lemma 2.1.6 (Tonelli’s theorem) Let (X1,Σ1, µ1) and (X2,Σ2, µ2) be mea-sure spaces, and f : X1×X2 → R+ be a Σ1⊗Σ2-measurable function. Then thefunctions

fµ2(x1) :=

∫X2

fx1 dµ2 & fµ1(x2) :=

∫X1

fx2 dµ1

are Σ1-measurable and Σ2-measurable, respectively, and∫X1×X2

f dµ1 ⊗ µ2 =

∫X2

[∫X1

fx2 dµ1

]dµ2 =

∫X1

[∫X2

fx1 dµ2

]dµ1 . (2.1.9)

Proof: In the case when f is a characteristic function, the statement ofthis lemma follows from Lemma 2.1.5. Therefore, by linearity, it holds alsofor nonnegative simple functions. If a nonnegative measurable function f isarbitrary, there exists a sequence of nonnegative simple functions which increasepointwise to f , say (fn)∞n=1. By the monotone convergence theorem,∫

X1

fµ2 dµ1 = limn→∞

∫X1

fnµ2dµ1 = lim

n→∞

∫X1×X2

fn dµ1 ⊗ µ2,

and ∫X2

fµ1 dµ2 = limn→∞

∫X2

fnµ1dµ2 = lim

n→∞

∫X1×X2

fn dµ1 ⊗ µ2,

where

fnµ2(x1) :=

∫X2

[fn]x1 dµ2 , fnµ1

(x2) :=

∫X1

[fn]x2 dµ1 .

This proves (2.1.9) and the lemma. 2

Proof of Theorem 2.1.8: Since an R-valued function f is Lebesgue inte-grable iff its positive f+ and negative f− parts are integrable, it is sufficient toprove the theorem only for nonnegative function f ∈ L1(X1 ×X2, Σ, µ1 ⊗ µ2).But this was exactly done in Lemma 2.1.6. 2


Exercise 2.1.1 Show that if (aα, bα) : α ∈ G is a finite or countable family of intervals inR such that [0, 1] ⊆

⋃α∈G(aα, bα) then

∑α∈G |aα − bα| > 1.


Exercise 2.1.2 ∗ Show that the length of intervals in R is a premeasure on the subalgebra inP(R) generated by all intervals.

Exercise 2.1.3 ∗ Give an example of an outer measure on P(N) which is not a premeasureon any subalgebra A ⊆ P(N) of infinite cardinality.

Exercise 2.1.4 ∗∗ Give an example of an algebra A in P(N) which is not a σ-algebra, andan example of a function ρ : A → [0, 1], such that ρ(∅) = 0, and ρ(N) = 1, for which the outermeasure ρ∗ constructed in Proposition 2.1.1 is identically equal to zero.

Exercise 2.1.5 Prove the uniqueness in Theorem 2.1.3

Exercise 2.1.6 ∗∗ Let G be a set of half-open intervals in R. Prove that⋃g∈G g is Lebesgue

measurable.

Exercise 2.1.7 Show that the Lebesgue measure can be obtained as the completion of anyBorel measure ω such that ω([a, b]) = b− a (∀a ≤ b).

Exercise 2.1.8 ∗ Prove that card(B(R)) < card(Σ(R)). Remark that in Exercise 2.2.6 wewill see that Σ(R) 6= P(R).

Exercise 2.1.9 Show that the distribution function F of the Borel measure µ given in(2.1.4) is increasing and right continuous.

Exercise 2.1.10 ∗ Prove Lemma 2.1.1.

Exercise 2.1.11 Let A ⊆ P(X) be an algebra, let µ0 be a σ-finite premeasure on A, andlet Ω be the σ-algebra generated by A. Show that there exists a unique extension of µ0 to ameasure µ on Ω.

Exercise 2.1.12 ∗∗ Prove Theorem 2.1.7.

Exercise 2.1.13 ∗∗ Show that the function µ defined in the subsection 2.1.5 is a premeasureon the algebra A.

Exercise 2.1.14 Show that if ∆ is infinite and R is taking with the Lebesgue measure thenthe measure µ constructed in 2.1.5 on

∏α∈∆

R does not satisfy 1.1.3.(d).

44 CHAPTER 2.

Exercise 2.1.15 Prove the formulas in (2.1.5), (2.1.6) and (2.1.7).

Exercise 2.1.16 Prove Lemma 2.1.4.

Exercise 2.1.17 ∗∗ Prove the Steinhaus theorem: for any Lebesgue measurable set A ⊆ Rsuch that µ(A) > 0 there exist nonempty open intervals I ⊆ A−A, J ⊆ A+A. Show that theconclusion of the theorem is true for the standard Cantor set in spite of it has measure zero.

Exercise 2.1.18 ∗ Let f : R → R be an additive function. Show that f is linear if f isbounded on a Lebesgue measurable set A ⊆ R, µ(A) > 0. Show that f is linear if f is boundedon a standard Cantor set. Show that f is linear if f is measurable on a nontrivial interval.Show that f is not Lebesgue measurable on any nontrivial interval if f is discontinuos at leastat one point.

Exercise 2.1.19 ∗∗ Let A, B be Lebesgue measurable subsets of Rn such that µ(A) > 0 andµ(B) > 0. Prove that

µ(x ∈ Rn : µ(A ∩ (B − x)) > 0) > 0.

Exercise 2.1.20 ∗∗ Let A be a subalgebra of P([0, 1]2) generated by all rectangles A × Bwhere A and B are Lebesgue measurable subsets of [0, 1]. Let ζ : A → R be the unique additiveextension of the function defined on rectangles by the formula:

ζ(A×B) = µ(A ∩B) ,

where µ is the Lebesgue measure. Show that ζ is a premeasure on A.

Exercise 2.1.21 Show that the statement of Lemma 2.1.6 may fail without the positivity ofthe function f .

2.2 Lebesgue Measure and Integral in Rn

In this section, we study Rn and functions from Rn to R from the point of viewof the Lebesgue measure and Lebesgue integration. All results presented belowpossess obvious C-valued analogs. Then we define and study Cantor sets whichare very interesting from the point of view of the set topology and the measuretheory. Cantor sets are closed Borel nowhere dense subsets of the interval [0, 1]or, more generally, of a Hausdorff space. The main advantage of Cantor sets is

2.2. LEBESGUE MEASURE AND INTEGRAL IN RN 45

that they are used in constructions of many examples in analysis.

2.2.1. The Lebesgue measure on Rn. The Lebesgue measure µn onRn is the completion of the product of the Lebesgue measure on R according toDefinition 2.1.3.

The domain Σn of µn (of course, B(Rn) ⊆ Σn) is the class of Lebesgue mea-surable sets in Rn. We write µ for µn and∫

f(x) dx :=

∫f dµ.

We extend some of the results of previous section to the n-dimensional case. IfE =

∏n1 Ej is a block in Rn, we call sets Ej ⊆ R the sides of the block E.

Theorem 2.2.1 Let E ∈ Σn. Then

(a) µ(E) = infµ(U) : E ⊆ U, U open = supµ(K) : K ⊆ E, K compact;

(b) E = A1 ∪ N1 = A2 \ N2, where A1 is an Fσ set, A2 is a Gδ set, andµ(N1) = µ(N2) = 0;

(c) If µ(E) <∞ then, for any ε > 0, there is a finite family RjNj=1 of disjoint

blocks, whose sides are intervals such that µ(E4⋃Nj=1 Rj) < ε.

Proof: By the definition of product measures, if E ∈ Σn and ε > 0, there is acountable family Tj∞j=1 of blocks such that E ⊆

⋃∞j=1 Tj and

∞∑j=1

µ(Tj) ≤ µ(E) + ε.

For each j, by applying Theorem 2.1.6 to the sides of Rj, we can find blocksUj ⊇ Fj whose sides are open sets such that µ(Uj) < µ(Tj)+ε2

−j. If U =⋃∞j=1 Uj

then U is open and

µ(U) ≤∞∑j=1

µ(Uj) ≤ µ(E) + 2ε.

This proves the first equation in part (a). The second equation and part (b)follow as in the proofs of Theorems 2.1.6 and 2.1.7.

46 CHAPTER 2.

Next, if µ(E) <∞ then µ(Uj) <∞ for all j. Since the sides of Uj are countableunions of open intervals, by taking suitable finite subunions, we obtain blocksVj ⊆ Uj whose sides are finite unions of intervals such that µ(Vj) ≥ µ(Uj)−ε2−j.If N is sufficiently large, we have

µ

(E \

N⋃j=1

Vj

)≤ µ

(N⋃j=1

Uj \ Vj

)+ µ

(∞⋃

j=N+1

Uj

)< 2ε

and

µ

(N⋃j=1

Vj \ E

)≤ µ

(∞⋃j=1

Uj \ E

)< ε,

so µ(E4⋃Nj=1 Vj) < 3ε. Since

⋃Nj=1 Vj can be expressed as a finite disjoint union

of rectangles whose sides are intervals, we have proved (c). 2

2.2.2. Lebesgue integrable functions on Rn. Let µ be the Lebesguemeasure in Rn. The set M(Rn, µ) of all real µ-measurable functions on Rn

is a real vector space (addition and scalar multiplication are pointwise). ByL1(Rn, µ) we denote its subspace of all Lebesgue integrable functions (with finiteintegral). Now write f ≈ g for f and g inM(Rn, µ), whenever f and g differ onlyon a µ-null set (a set of measure zero). It is easily seen that ≈ is an equivalencerelation. Let L0 = L0(Rn, µ) be the set of equivalence classes of functions inM(Rn, µ). We denote the equivalence classes of f, g, . . . by [f ], [g], . . .. The setL0 becomes a real vector space by defining [f ]+ [g] = [f +g] and α[f ] = [αf ] fora real α. Observe that these definitions do not depend on the choice of f and gin their equivalence classes. The same is true for the partial order in L0, if wedefine [f ] ≤ [g] to mean f(x) ≤ g(x) for all x ∈ Rn except a null set. In practice,the elements of L0 = L0(Rn, µ) are usually denoted by f, g, . . . and treated as ifthey were functions instead of equivalence classes of functions.

Theorem 2.2.2 If f ∈ L1(µ) and ε > 0, there is a simple function φ =∑Nj=1 αjχRj

, where each Rj is a product of intervals such that∫|f − φ| dµ < ε,

and there is a continuous function g vanishing outside of a bounded set such that∫|f − g| dµ < ε.

Proof: By the definition of Lebesgue integrable functions, we can approximatef by simple functions in L1-norm. Then use Theorem 2.2.1 to approximate asimple function by a function φ of the desired form. Finally, use the UrysohnLemma to approximate such φ by a continuous function. 2


Theorem 2.2.3 The Lebesgue measure on Rn is translation-invariant. Namely,let a ∈ Rn. Define the shift τa : Rn → Rn by τa(x) = x+ a.

(a) If E ∈ Ln then τa(E) ∈ Ln and µ(τa(E)) = µ(E);

(b) If f : Rn → R is Lebesgue measurable then so is f τa. Moreover, if eitherf ≥ 0 or f ∈ L1(µ) then ∫

(f τa) dµ =

∫f dµ.

Proof: Since τa and its inverse τ−a are continuous, they preserve the class ofBorel sets. The formula µ(τa(E)) = µ(E) follows easily from the trivial one-dimensional variant of this result if E is a block. For a general Borel set E, theformula µ(τa(E)) = µ(E) follows from the previous step, since µ is determinedby its action on blocks. Assertion (a) now follows immediately.

If f is Lebesgue measurable and B is a Borel set in R, we have f−1(B) = E∪N ,where E is Borel and µ(N) = 0. But τ−1

a (E) is Borel and µ(τ−1a (N)) = 0, so

(f τa)−1(B) ∈ Σn

and f is Lebesgue measurable. The equality∫(f τa) dµ =

∫f dµ

reduces to the equality µ(τ−a(E)) = µ(E) when f = χE. It is true for simplefunctions by linearity, and hence for nonnegative measurable functions by thedefinition of integral. Taking positive and negative parts of real and imaginaryparts, we obtain the result for f ∈ L1(µ). 2

2.2.3. The Lusin theorem. This theorem says that any Lebesgue mea-surable function on Rn can be approximated by a (partially defined) continuousfunction. More precisely.

Theorem 2.2.4 (Lusin’s theorem) If f is a Lebesgue measurable function onRn and ε > 0 then there exist a measurable set A ⊆ Rn such that µ(Rn \A) ≤ εand the restriction of f onto A is continuous.

48 CHAPTER 2.

Proof: First, assume that f vanishes outside of some Ik = [−k, k]n. Thereexists a sequence (ψm)m of simple functions vanishing outside Ik such that ψm →f pointwise. By Egoroff’s theorem, there exist G ⊆ Ik such that µ(G) > (2k)n−ε/2 and ψk → f uniformly on G. By Theorem 2.2.1, we can find for each m acompact Cm ⊆ G such that µ(Cm \ G) < 2−m−1ε and ψk is continuous on Ck.Take C = ∩∞m=1Cm, then

µ(Ik \ C) ≤ µ(Ik \G) + µ(G \ C) < ε.

Each ψm is continuous on C, and the sequence (ψm)m converges uniformly onC. Then ψm|C → ψ, which is continuous on C, and since ψm → f pointwisethen f |C is continuous.

In the general case when f is a Lebesgue measurable function on Rn we denotethe set (−k, k)n by Jk. As it was shown, for any k, there exists Rk ∈ Σn suchthat f |Rk

is continuous, Rk ⊆ Ik, and µ(Ik \Rk) < 2−kε. Denote

Ak := (Jk \ Ik−1) ∩Rk,

where I0 = ∅. Then the restriction of f onto A := ∪∞k=1Ak is continuous andµ(Rn \ A) < ε. 2

Corollary 2.2.1 If f ∈ L1(Rn) and ε > 0 then there exists a continuous func-tion g : Rn → R such that ∫

Rn

|f − g| dµ ≤ ε. 2

2.2.4. Cantor sets. Each x ∈ [0, 1] has a base-3 expansion

x =∞∑j=1

xj3−j, where xj ∈ 0, 1, 2.

This expansion is unique unless x is of the form m · 3−k for some integers m, k;in which case x has two expansions: one with xj = 0 for j > k and one withxj = 2 for j > k. Assuming m is not divisible by 3, one of these expansions willhave xk = 1 and the other will have xk = 0 or 2. If we agree always to use thelatter expansion, we see that

x1 = 1 ⇔ x ∈ (1/3, 2/3),


x1 6= 1 and x2 = 1 ⇔ x ∈ (1/9, 2/9) ∪ (7/9, 8/9),

and so forth. It will also be useful to observe that if

x =∞∑j=1

xj3−j & y =

∞∑j=1

yj3−j,

then x < y iff there exists n such that xn < yn and xj = yj for 1 ≤ j < n.

Definition 2.2.1 The standard Cantor set C is the set of all x ∈ [0, 1] havinga base-3 expansion x =

∑∞j=1 xj3

−j with xj 6= 1 for all j.

Thus, C =⋂∞n=1 ∆n, where

∆1 := [0, 1] \ (1/3, 2/3) , ∆2 := ∆1 \ ((1/9, 2/9) ∪ (7/9, 8/9)), etc.

Obviously, C is compact, nowhere dense, and totally disconnected (i.e. theonly connected subsets of C are single points). Moreover, C has no isolatedpoints. These simple properties of C we leave to the reader as an exercise.

2.2.5. Basic properties of the standard Cantor set. Other basicproperties of C are summarized as follows:

Proposition 2.2.1 Let C be the standard Cantor set, then C is a Borel setsuch that

µ(C) = 0 & card(C) = c.

Proof: C is obtained from [0, 1] by removing one interval of length 13, two

intervals of length 19, and so forth. Thus,

µ(C) = 1−∞∑j=0

2j

3j+1= 1− 1

3· 1

1− (2/3)= 0.

Suppose x ∈ C, so that x =∑∞

j=1 xj3−j, where xj = 0 or 2 for all j. Let

f(x) =∞∑j=1

yj2−j, (2.2.1)

50 CHAPTER 2.

where yj = xj/2. The series defining f(x) is the base-2 expansion of a numberin [0, 1], and any number in [0, 1] can be obtained in this way. Hence f maps Conto [0, 1] and card(C) is continuum. 2

2.2.6. Generalized Cantor sets. The construction of the standard Cantorset which is starting with [0, 1] and successively removing open middle thirds ofintervals has an obvious generalization. If I is a bounded interval and α ∈ (0, 1),let us call the open interval with the same midpoint as I, and length equal to αtimes the length of I the “open middle αth” of I. If (αj)

∞j=1 is any sequence of

numbers in (0, 1) then we define a decreasing sequence (∆j)∞j=1 of closed sets as

follows: ∆0 = [0, 1], and ∆j is obtained by removing the open middle αjth fromeach of the intervals that make up Kj−1. The resulting limiting set

K =∞⋂j=1

∆j

is called a generalized Cantor set. Generalized Cantor sets possess all prop-erties of the standard Cantor set C, but they do not have always the Lebesguemeasure zero. As for their Lebesgue measure, clearly

µ(∆j) = (1− αj)µ(∆j−1),

so

µ(K) =∞∏j=1

(1− αj).

If the αj are all equal to a fixed α ∈ (0, 1), we have µ(K) = 0. However, ifαj → 0 rapidly enough, µ(K) will be strictly positive and, for any β ∈ [0, 1),one can choose αj so that µ(K) = β.

2.2.7. Cantor functions. The map f : C → [0, 1] given by (3.1.1) has thefollowing property.

If x, y ∈ C, x < y, then f(x) < f(y) unless x and y are endpoints of one of theintervals removed from [0, 1] in obtaining C (in this case, f(x) = m·2−k for somenonnegative integers m, k; and f(x) and f(y) are the two base-2 expansions ofthis number).


Therefore we can extend f to a map f : [0, 1]→ [0, 1] by setting it to be constanton each interval removed from C. The map f , which is obviously increasing andcontinuous, is called the standard Cantor function. This notion can benaturally extended to a generalized Cantor function by using of the notionof the generalized Cantor set.

2.2.8. Applications of Cantor sets and functions in constructing ofcounterexamples. Generalized Cantor sets are examples of nowhere densesubsets of [0, 1] with Lebesgue measure 1− ε, where ε > 0 is arbitrarily. Let Ckbe a generalized Cantor set such that µ(Ck) > 1 − 1

k, then A =

⋃∞k=1Ak is a

set of the first category (i.e. the countable union of nowhere dense sets) andµ(A) = 1. By the Baire theorem, [0, 1] is a set of the second category (i.e.the set which is not of the first category) as a complete metric space. Therefore,[0, 1] \ A is an example of a dense subset of [0, 1] of the second category ofLebesgue measure zero.

By addition and multiplication, we define a Cantor like set K ⊆ [a, b] in aninterval [a, b] as:

K := a+ (b− a) ·G,where G is a some generalized Cantor set. The similar idea is used in thedefinition of a Cantor like function on an interval [a, b]. Now it is obvioushow (with the help of Cantor like sets) to construct an example of a dense subsetD ⊆ R of the second category of Lebesgue measure zero.

Cantor like functions give us the following quite simple construction of a con-tinuous function f : R → [0, 1] which is nowhere differentiable. Let ann be acountable dense subset of R, and suppose that an 6= an+1 for all n. For each n,take a Cantor like function fn on the interval [an, an+1], then the function

f :=∞∑j=1

2−nfn

possesses desired properties.


Exercise 2.2.1 ∗ Show that the Lebesgue measure in Rn is separable.

Exercise 2.2.2 ∗ Prove Corollary 2.2.1.

52 CHAPTER 2.

Exercise 2.2.3 ∗ Show that the Lebesgue measure in Rn is rotation invariant.

Exercise 2.2.4 Show that the family of all maps from Rn to Rn preserving the Lebesguemeasure in Rn is a group.

Exercise 2.2.5 ∗∗ Give an example of a transformation S : Rn → Rn, such that S is notequal a.e. to any affine transformation of R and S preserves the Lebesgue measure in Rn.Can such a transformation be continuous?

Exercise 2.2.6 ∗ Let Γ be a unit circle λ ∈ C : |λ| = 1 endowed with the Lebesgue measure.Then any λ ∈ Γ is of the form ei 2π α, where α ∈ [0, 1).

(a) Show that ≈ defined below is an equivalence relation on Γ:

ei 2π α1 ≈ ei 2π α1 if α1 − α2 ∈ Q .

(b) Let Γ =⋃κ Γκ be a representation of Γ as the union of equivalence classes with respect

to ≈ . By using of the axiom of choice, define a set Ω ⊆ Γ such that |Ω ∩ Γκ| = 1 for any κ.Let for any α ∈ R

Ωα := ei 2π αω : ω ∈ Ω.

Show that, for any two rationals r1 6= r2 from [0, 1), Ωr1 ∩ Ωr2 = ∅, and

Γ =⋃

r∈Q∩[0,1)

Ωr.

(c) By using (b), show that Ω is not Lebesgue measurable subset of Γ.(d) Construct non-measurable subsets of [0, 1], R, and Rn.

Exercise 2.2.7 Show that any generalized Cantor set is compact, nowhere dense, totally dis-connected, and has no isolated points.

Exercise 2.2.8 ∗ Show that for any β ∈ [0, 1) there is a generalized Cantor set K ⊆ [0, 1]such that µ(K) = β.

Exercise 2.2.9 ∗ Give an example of a subset in [0, 1] which is of the second category andhas Lebesgue measure one.

Exercise 2.2.10 ∗ Give an algorithm for constructing of a generalized Cantor function andshow that it is continuous.

Exercise 2.2.11 ∗∗ Construct a function f : [0, 1] → [0, 1] such that f ′ = 0 a.e. and f isconstant on no nonempty subinterval of [0, 1].


Exercise 2.2.12 ∗ Give an example of a set D ⊆ Rn of Lebesgue measure zero such thatD ∩ U is of the second category for any nonempty open U ⊆ Rn.

Exercise 2.2.13 ∗ Construct an example of a dense subset D ⊆ Rn of the second categoryand of Lebesgue measure zero.

Exercise 2.2.14 ∗∗∗ Construct a Borel subset A ⊆ R such that for each interval [a, b], wherea < b:

µ(A ∩ [a, b]) > 0 & µ((R \A) ∩ [a, b]) > 0.

Exercise 2.2.15 ∗∗∗ Construct a sequence (Ak)∞k=1 of pairwise disjoint Borel subsets Ak ⊆ Rsuch that for each interval [a, b], a < b, and k ∈ N µ(Ak ∩ [a, b]) > 0.

Exercise 2.2.16 ∗∗∗ Construct a sequence (Ak)k∈N of pairwise disjoint Borel subsets Ak ⊆Rn such that, for each rectangle

H = [a1, b1]× · · · × [an, bn] (∀j ∈ 1, . . . , n aj < bj)

and for each k ∈ N, we have that µ(Ak ∩H) > 0.

Exercise 2.2.17 ∗∗∗ Construct a nonnegative Lebesgue measurable function f on R, whichis finite everywhere and

∫ baf dµ =∞ for all a < b.

Exercise 2.2.18 ∗∗∗ Construct a sequence (fn)∞n=1 of pairwise disjoint nonnegative Lebesguemeasurable functions on R such that each fn is finite everywhere and∫ b

a

fn dµ =∞

for all a < b and n ∈ N.

Exercise 2.2.19 ∗∗ Let f ∈ L1(R, µ) such that for every open U ⊆ R∫U

f dµ =∫U

f dµ.

Show that f = 0 a.e.

54 CHAPTER 2.

Chapter 3

3.1 The Radon – Nikodym Theorem

In this section, we prove the Radon – Nikodym theorem and present several itsapplications, in particular, the Hahn decomposition theorem.

3.1.1. The Radon – Nikodym theorem. Let (X,B, µ) be a measurespace and let f : X → R be a nonnegative integrable function. Then µf definedby

µf (A) =

∫A

f(x) dµ (∀A ∈ A)

is a finite measure accordingly to 1.3.2 L4. There is a converse of the lastproperty. Namely, the following theorem is of great importance. We give itsdirect proof accordingly to A. Schep [10].

Theorem 3.1.1 (The Radon – Nikodym theorem) Let ν and µ be finitemeasures on (X,B). Then there exist D ∈ B with µ(D) = 0 and a nonnegativeµ-integrable function f0 such that

ν(E) = ν(E ∩D) +

∫E

f0 dµ (∀E ∈ B) .

First, we need the following lemma.

Lemma 3.1.1 Let ν and µ be finite measures on (X,B) satisfying 0 ≤ ν ≤ µon B. Then there exists a measurable function f0 , 0 ≤ f0 ≤ 1, such that

ν(E) =

∫E

f0 dµ (∀E ∈ B) .

55

56 CHAPTER 3.

Proof: Let

H = f : f is measurable , ν(E) ≥∫E

f dµ for all E ∈ B.

Note that H 6= ∅, since 0 belongs to H. Moreover, when f1, f2 ∈ H, thenmaxf1, f2 ∈ H. Indeed, if A = x : f1(x) ≥ f2(x) then∫E

maxf1, f2 dµ =

∫E∩A

f1 dµ+

∫E∩Ac

f2 dµ ≤ ν(E ∩ A) + ν(E ∩ Ac) = ν(E).

Let

M := sup∫

X

f dµ : f ∈ H.

Then 0 ≤ M < ∞, so there exist sequence (fn)∞n=1 ⊆ H with fn ↑ such that∫Xfn dµ > M − n−1. Let f0 = lim fn then f0 is a measurable function taking

values in [0, 1]. By the monotone convergence theorem, f0 ∈ H and∫Xf0 dµ ≥

M . Hence∫Xf0 dµ = M .

To complete the proof, we show that ν(E) =∫Ef0 dµ for all E ∈ B.

If not, there exists E ∈ B satisfying

ν(E) >

∫E

f0 dµ .

Denote E1 = x ∈ E : f0(x) = 1 and E0 = E \ E1. It follows from

ν(E0) + ν(E1) >

∫E

f0 dµ =

∫E0

f0 dµ+ µ(E1) ≥∫E0

f0 dµ+ ν(E1)

that ν(E0) >∫E0f0 dµ. Thus we may assume that f0(x) < 1 for all x ∈ E

(otherwise, we replace E with E0). Let Fn := x ∈ E : f0(x) < 1− n−1. ThenFn ↑ E, and ν(Fn)→ ν(E). Therefore, there exists n0 such that

ν(Fn0) >

∫Fn0

(f0 + n−10 χFn0

) dµ .

We assert that f0 + n−10 χF ∈ H for some measurable F ⊆ Fn0 , µ(F ) > 0.

If not, then every measurable subset F of Fn0 with positive µ-measure containsa measurable subset G with

∫G

(f0 + n−10 χF ) dµ =

∫G

(f0 + n−10 χG) dµ > ν(G).

Denote this property by P , namely,

A ∈ P ⇔ A ∈ B , A ⊆ Fn0 &

∫A

(f0 + n−10 χA) dµ > ν(A) .

3.1. THE RADON – NIKODYM THEOREM 57

The property P satisfies conditions of Theorem 1.1.2. Namely, every measurablesubset of Fn0 of non-zero measure contains a subset of non-zero measure withthe property P , and P is closed under at most countable unions of pairwisedisjoint families. Therefore by Theorem 1.1.2, Fn0 ∈ P , and

ν(Fn0) >

∫Fn0

(f0 + n−10 χFn0

) dµ > ν(Fn0),

which is a contradiction. Thus f0 + n−10 χF ∈ H for some measurable subset F

of Fn0 with a positive µ-measure. However,∫(f0 + n−1

0 χF ) dµ = M + n−10 µ(F ) > M .

This contradiction yields ν(E) =∫Ef0 dµ for all E ∈ B. 2

Proof of Theorem 3.1.1: Let λ = µ+ν. Then 0 ≤ ν ≤ λ so, by Lemma 3.1.1,there exists a Lebesgue integrable g : X → [0, 1], such that ν(E) =

∫Eg dλ for

all E in B. Then µ(E) =∫E

(1 − g) dλ for all E ∈ B. Let D = x : g(x) = 1,then µ(D) =

∫D

0 dλ = 0. Now, the equality

ν(E) =

∫E

g dν +

∫E

g dµ

implies ∫E

(1− g) dν =

∫E

g dµ

for all E ∈ B. Hence∫Xφ(1 − g) dν =

∫Xφg dµ for all nonnegative simple φ

and, thus,∫Xf(1− g) dν =

∫Xfg dµ for all nonnegative measurable f . Taking

f = (1 + g + · · ·+ gn)χE, we have∫E

(1− gn+1) dν =

∫E

(1 + g + · · ·+ gn)g dµ

for all E in B and all n ≥ 1. Since 0 ≤ g(x) < 1 on Dc, it follows from themonotone convergence theorem that

ν(E ∩Dc) = limn→∞

∫E∩Dc

(1− gn+1) dν = limn→∞

∫E∩Dc

(1 + g + · · ·+ gn)g dµ

=

∫E∩Dc

g(1− g)−1 dµ =

∫E

f0 dµ,

58 CHAPTER 3.

where f0 = g(1− g)−1χDc . The proof is completed. 2

In some textbooks the Radon – Nikodym theorem is given in a slightly differentform. We need the following definition.

Definition 3.1.1 Let µ and ν be signed measures on a σ-algebra A. Then ν iscalled µ-continuous if ν(A) = 0 for every A ∈ A such that µ(A) = 0. In thiscase, we write ν << µ.

Theorem 3.1.2 Let ν and µ be σ-finite signed measures on (X,B), let µ bepositive and ν << µ. Then there exists a locally µ-integrable function f (i.e.∫Af dµ <∞ for every A ∈ B, µ(A) <∞) such that

ν(E) =

∫E

f dµ (∀E ∈ B) . 2

The function f above is called the Radon – Nikodym derivative of ν withrespect to µ and denoted by ∂ν

∂µ.

3.1.2. Applications of the Radon – Nikodym theorem. We give twoimportant applications of the Radon – Nikodym theorem. First one is the proofof the Hahn decomposition theorem in the case of a finite signed measure. Weneed the following definition.

Definition 3.1.2 Let µ be a signed measure on a σ-algebra A. Then P ∈ A iscalled a positive set for µ if, for any B ∈ A such that B ⊆ P , µ(B) ≥ 0. Aset N ∈ A is called a negative set for µ if, for any B ∈ A such that B ⊆ N ,µ(B) ≤ 0.

Theorem 3.1.3 (The Hahn decomposition theorem) If ν is a signed mea-sure on (X,A), there exist a positive set P and a negative set N for ν such thatP ∪N = X and P ∩N = ∅.

Here we give a simple proof of this theorem based on Theorem 3.1.2 in the caseof finite measure ν.

3.1. THE RADON – NIKODYM THEOREM 59

Proof: Let ν = ν+−ν− be the Jordan decomposition of ν. Then obviously ν+

and ν− are |ν|-continuous measures, where |ν| = ν+ + ν−. Applying Theorem3.1.2 to them gives us two functions f1, f2 ∈ L1(|ν|) such that

ν+(E) =

∫E

f1 d|ν| , ν−(E) =

∫E

f2 d|ν| (∀E ∈ A).

Then P := x ∈ X : f1(x) > 0 and N := X \ P are required sets. 2

The next one is an application in the operator theory. It is related to thedefinition of one of the most important classes of Markov operators.

A transformation τ : Ω→ Ω is called measurable if τ−1(A) ∈ Σ for all A ∈ Σ.A measurable transformation τ : Ω→ Ω is called nonsingular if µ(τ−1(A)) = 0for all A ∈ Σ such that µ(A) = 0.

Let τ be a nonsingular transformation and let f ∈ L1(µ). Then the functionν : Σ→ R,

ν(E) :=

∫τ−1(E)

f dµ,

is a finite signed µ-continuous measure on Σ. Then, by the Radon – Nikodymtheorem, there exists Pf ∈ L1(µ) such that∫

A

Pf dµ =

∫τ−1(A)

f dµ (∀A ∈ Σ).

The map f → Pf is a linear operator from L1(µ) := L1(µ)/f : f = 0 a.e. toitself. This operator is called the Frobenius – Perron operator correspondingto τ . The elementary properties of Frobenius – Perron operators are collectedin Exercises 3.1.10 and 3.1.13.


Exercise 3.1.1 Let ν be a signed measure. Show that

|ν|(A) = sup|∫A

f d|ν| | : |f | ≤ 1

for any measurable A.

60 CHAPTER 3.

Exercise 3.1.2 ∗∗∗ Let ν be a σ-finite signed measure on (X,Σ), which is µ-continuous foranother σ-finite measure µ on (X,Σ). Show that

|ν|(A) :=∫A

∣∣∣∂ν∂µ

∣∣∣ dµ (∀A ∈ Σ).

Exercise 3.1.3 Show that a signed measure ν is µ-continuous iff, for any decreasing sequenceof measurable sets (An)∞n=1, µ(An)→ 0 implies ν(An)→ 0.

Exercise 3.1.4 ∗ Let ν be a finite signed measure and let µ be a positive measure on (X,Σ).Show that ν is µ-continuous iff, for any ε > 0, there exists δ > 0 such that |ν(A)| ≤ ε wheneverµ(A) ≤ δ. Give an example which shows that this assertion may fail if ν is not finite.

Exercise 3.1.5 Let ν and µ be σ-finite positive measures and let ν << µ. Show that if µ isseparable then ν is also separable.

Exercise 3.1.6 Let (γλ)λ∈Λ be a countable family of signed measures of bounded variationon (X,Σ). Show that there exists a measure γ on (X,Σ) such that every γλ is γ-continuous,and γ ∧ |γλ| > 0 for any λ ∈ Λ.

Exercise 3.1.7 Obtain the Hahn decomposition theorem for σ-finite signed measure as acorollary of this theorem for finite signed measure.

Exercise 3.1.8 ∗ Prove the Hahn decomposition theorem for arbitrary signed measure.

Exercise 3.1.9 Obtain the Jordan decomposition theorem as a corollary of the Hahn decom-position theorem.

Exercise 3.1.10 ∗ Let S be a nonsingular transformation of (X,Σ, µ) and let PS be thecorresponding Frobenius – Perron operator PS : L1(µ)→ L1(µ). Show that:

(a) PS is well defined;(b) PS is a linear operator;(c) PS preserves the L1-norm of any nonnegative f ∈ L1(µ).

Exercise 3.1.11 Let S be a nonsingular transformation of (X,Σ, µ). Show that Sn is non-singular transformation for any nonnegative integer n and that PSn = PnS for n ≥ 0.

Exercise 3.1.12 ∗∗ Let S be a transformation of [0, 1] with the Lebesgue measure. Find theFrobenius – Perron operator if:

(a) S(x) = 4x(1− x);(b) S(x) = 4x2(1− x2);(c) S(x) = sinπx;(d) S(x) = tan(x+ 1).

3.2. LP -SPACES 61

Exercise 3.1.13 ∗ Let S be a nonsingular transformation of (X,Σ, µ) and let PS be the corre-sponding Frobenius – Perron operator PS : L1(µ)→ L1(µ). Prove that, for every nonnegativef, g ∈ L1(µ), the condition supp(f) ⊆ supp(g) implies supp(PSf) ⊆ supp(PSg) a.e., wheresupp(ψ) := x : ψ(x) 6= 0 for ψ ∈ L1(µ), and supp(f) is defined for f ∈ L1 as follows:

supp(f) := A such that ψ(x) 6= 0 a.e. on A , ψ(x) = 0 a.e. on X \A

for any ψ ∈ [f ].

Exercise 3.1.14 ∗∗ In notations of Exercise 2.1.20, show that the unique extension ζ of thepremeasure ζ (which exists due to Caratheodory theorem) to the product σ-algebra Σ([0, 1])×Σ([0, 1])) is not µ×µ-continuous. Although the premeasure ζ on the subalgebra A ⊂ Σ([0, 1])×Σ([0, 1])) generated by all rectangles with measurable sides is µ×µ-continuous in the followingsense:

µ× µ(W ) = 0 ⇒ ζ(W ) = 0 (∀ W ∈ A) .

3.2 Lp-Spaces

Here we present basic results about Lp-spaces. We assume in this section thatall measurable functions are C-valued and all linear spaces are complex.

3.2.1. Lp-spaces for 0 ≤ p ≤ ∞. Let (X,A, µ) be a measure space and letp be a fixed real number 0 ≤ p <∞. We denote by Lp = Lp(X,A, µ) the set ofall measurable functions f : X → C such that∫

|f |p dµ <∞.

For any f ∈ Lp, we define a function ‖f‖p : Lp → R+ as

‖f‖p :=(∫|f |p dµ

) 1p.

This function introduces the natural topology on Lp(X,A, µ) by the base

U(f, ε) : f ∈ Lp, ε > 0,

whereU(f, ε) := g ∈ Lp : ‖f − g‖p < ε.

62 CHAPTER 3.

Here we don’t give the proof of this result. Below, in the next subsection, westate the important case by showing that if 1 ≤ p <∞ then Lp is a linear spaceand the function ‖ · ‖p is a seminorm.

The most popular examples of Lp-spaces are: Lp[0, 1] and Lp(R), where weconsider the Lebesgue measure on [0, 1] and on R.

Another type of examples is following. Let X be a nonempty set, let A = P(X)be the power set of X (i.e. the σ-algebra of all subsets of X), and let µ be thecounting measure on X. Every function f : X → C is measurable. A commonlyused notation of Lp(X,A, µ) in this case is `p(X). It is easy to verify that afunction f : X → C is in `p(X) iff

∑x∈X |f(x)|p < ∞, in the sense that the

finite subsumes have a finite upper bound.

The choices X = N and X = Z are of a special interest; these yield the spaceof absolutely p-summable sequences (resp., bilateral sequences). If X = N thenwe denote Lp(N) by `p.

In the case p = ∞, we cannot directly generalize the definition of our function‖ · ‖p, and we treat this case as follow. Set L∞ = L∞(X,A, µ), the set of allmeasurable functions f : X → C such that there exists a constant M ∈ R,

|f |(x) ≤M a.e.

It is obvious that L∞(X,A, µ) is a topological linear space for the topology givenby the seminorm ‖ · ‖∞ defined as

‖f‖∞ = esssup(|f |) = infM ∈ R : |f |(x) ≤M a.e..

The main examples of L∞-spaces are L∞[0, 1], L∞(R), and `∞, which are de-fined similarly to Lp[0, 1], Lp(R), and `p.

3.2.2. Inequalities. We begin with elementary analytic lemma.

Lemma 3.2.1 If a ≥ 0, b ≥ 0, and 0 < λ < 1 then

aλb1−λ ≤ λa+ (1− λ)b,

with equality iff a = b.

3.2. LP -SPACES 63

Proof: The result is obvious if b = 0; otherwise, dividing both sides by b andsetting t = a/b, we are reduced to showing that tλ ≤ λt+ (1− λ) with equalityiff t = 1. The derivative f ′(t) = λtλ−1−λ of the function f(t) = tλ−λt satisfies:f ′(t) > 0 if t < 1; f ′(1) = 0; f ′(t) < 0 if t > 1. Then f is strictly increasingfor t < 1 and strictly decreasing for t > 1, so its maximum value, namely 1− λ,occurs at t = 1. 2

If p > 1, we define q = pp−1

; thus q > 0 and satisfies p−1 +q−1 = 1, i.e. p+q = pq.

The following inequality which called the Elementary Young Inequality

αβ ≤ αp

p+βq

q

is true for all α, β ≥ 0 and p > 1. For the proof, it is enough to substitute λ = 1p,

α = a1p , and β = b

1q in Lemma 3.2.1.

Theorem 3.2.1 (Holder’s inequality) Let p > 1, f ∈ Lp, and g ∈ Lq. Thenf · g ∈ L1 and

‖f · g‖1 ≤ ‖f‖p‖g‖q. (3.2.1)

Proof: The result is trivial if ‖f‖p = 0 or ‖g‖q = 0, since f = 0 or g = 0a.e. Moreover, we observe that if (3.2.1) holds for a some f and g then it alsoholds for all scalar multiples of f and g; for it, f and g are replaced by a · f andb ·g, both sides of (3.2.1) change by a factor of |ab|. It suffices therefore to prove(3.2.1) when ‖f‖p = ‖g‖q = 1 with equality iff |f |p = |g|q a.e. To this end, weapply the Young inequality with α = |f(x)| and β = |g(x)| to obtain

|f(x)g(x)| ≤ p−1|f(x)|p + q−1|g(x)|q. (3.2.2)

Integration of the both sides yields

‖fg‖1 ≤ p−1

∫|f |p dµ+ q−1

∫|g|q dµ = p−1 + q−1 = 1 = ‖f‖p‖g‖q.

Equality holds here iff it holds a.e. in (3.2.1) and, by Lemma 3.2.1, this happensexactly iff |f |p = |g|q a.e. 2

One of the most important application of the Holder inequality is the followinginequality.

64 CHAPTER 3.

Theorem 3.2.2 (Minkowski’s inequality) Let p ≥ 1, f, g ∈ Lp. Thenf + g ∈ Lp and

‖f + g‖p ≤ ‖f‖p + ‖g‖p.

Proof: The result is obvious if p = 1 or f + g = 0 a.e.. Otherwise, we have

|f + g|p ≤ (|f |+ |g|)|f + g|p−1 = |f ||f + g|p−1 + |g||f + g|p−1 .

Taking the integral,

‖(f + g)p‖1 ≤ ‖f(f + g)p−1‖1 + ‖g(f + g)p−1‖1. (3.2.3)

Integration of the inequality

|f + g)|p ≤ 2p max|f |, |g|p ≤ 2p max|f |p, |g|p

gives us f + g ∈ Lp.

Since (p−1)q = p when q is the conjugate to p and applying Holder’s inequalityto the functions

φ = f ∈ Lp, ψ = (f + g)p−1 = (f + g)pq ∈ Lq,

we have

‖f(f + g)p−1‖1 ≤ ‖φ · ψ‖1 ≤ ‖φ‖p‖ψ‖q = ‖f‖p‖(f + g)p−1‖q. (3.2.4)

Similarly‖g(f + g)p−1‖1 ≤ ‖g‖p‖(f + g)p−1‖q. (3.2.5)

Combining (3.2.3), (3.2.4), and (3.2.5) gives us

‖f + g‖p =(∫|f + g|p dµ

) 1p

=

∫|f + g|p dµ

(∫|f + g|p dµ)

1q

=

∫|f + g|p dµ

(∫|f + g|(p−1)q dµ)

1q

=‖(f + g)p‖1

‖(f + g)p−1‖q≤ ‖g‖p + ‖f‖p . 2

Corollary 3.2.1 Lp is a vector space with respect to pointwise linear operations.The function f → ‖f‖p is a seminorm on Lp. 2

3.2. LP -SPACES 65

3.2.3. Completeness of Lp, where 1 ≤ p <∞ . A basic property of Lp isgiven in the following theorem.

Theorem 3.2.3 Let (fn)∞n=1 be a sequence in Lp = Lp(X,Σ, µ) satisfying‖fm−fn‖p → 0 as n,m→∞. Then there exists f ∈ Lp with lim

n→∞‖fn−f‖p = 0.

Proof: We prove this theorem for 1 ≤ p < ∞. The case p = ∞ is moreelementary and it is left for the reader as an exercise.

We may suppose, by passing to a subsequence, that

‖fn+1 − fn‖p ≤ 2−n (∀n ∈ N).

Thus we have∑∞

n=1 ‖fn+1 − fn‖p ≤ 1. With the convention f0 = 0, define

gn := |f1|+ |f2 − f1|+ · · ·+ |fn − fn−1| =n∑k=1

|fk − fk−1|.

By Theorem 3.2.2, gn ∈ Lp and

‖gn‖p ≤n∑k=1

‖fk − fk−1‖p = ‖f1‖p +n∑k=2

‖fk − fk−1‖p ≤ ‖f1‖p + 1 .

Thus gpn ∈ L1 for all n and∫X

|gn|p dµ = ‖gn‖pp ≤ (‖f1‖p + 1)p . (3.2.6)

Since (gn)∞n=1 is an increasing sequence and p ≥ 1, the sequence ((gn)p)∞n=1 isalso increasing. It follows from (3.2.6) and the monotone convergence theoremthat there exists h ∈ L1, h ≥ 0, such that (gn)p ↑ h a.e. Denoting g := h1/p, wehave g ∈ Lp and gn ↑ g a.e.

Let E ∈ Σ be such that µ(E) = 0 and gn(x) ↑ g(x) for all x ∈ X \ E. Then

g(x) = limn→∞

gn(x) = limn→∞

n∑k=1

|fk(x)− fk−1(x)| (∀x ∈ X \ E) .

66 CHAPTER 3.

Thus the series∑∞

k=1 |fk(x) − fk−1(x)| is convergent to g(x) for all x ∈ X \ E.Therefore the series

∑∞k=1(fk(x)−fk−1(x)) is also convergent for x ∈ X \E, and

satisfies:

limn→∞

fn(x) = limn→∞

n∑k=1

(fk(x)− fk−1(x)) ≤ g(x) (∀x ∈ X \ E) .

Define

f(x) := limn→∞

fn(x) for x ∈ X \ E , and f(x) := 0 for x ∈ E.

Then f is measurable, fn → f a.e., and |f | ≤ g. Therefore |f |p ≤ gp = h ∈ L1,and |f |p ∈ L1, i.e., f ∈ Lp.It remains to show that ‖fn − f‖p → 0. Given an ε > 0, choose Nε such that‖fm − fn‖p ≤ ε for all m,n ≥ Nε. Fix m ≥ Nε, then∫

X

|fm − fn|p dµ ≤ εp for all n ≥ Nε .

Since |fm − fn|p ∈ L1 for all n ≥ Nε, it follows from the Fatou lemma that∫X

(lim supn→∞

|fm − fn|p) dµ ≤ lim supn→∞

∫X

|fm − fn|p dµ ≤ εp . (3.2.7)

In particular, lim supn→∞

|fm − fn|p ∈ L1. But |fm − fn|p → |fm − f |p a.e. as

n→∞, thuslim sup

n→∞|fm − fn|p = |fm − f |p a.e.

So∫X|fm − f |p dµ ≤ εp by (3.2.7), i.e., ‖fm − f‖p ≤ εp for all m ≥ Nε. Since

ε > 0 has been chosen arbitrary, the proof is completed. 2

3.2.4. Lp-spaces for 1 ≤ p < ∞. Let now 1 ≤ p < ∞ and (X,A, µ) be ameasure space. Take the following relation ”≈” on Lp(X,A, µ):

f ≈ g ⇔ f(x) = g(x) a.e. (3.2.8)

Obviously ”≈” is an equivalence relation and f : f ≈ 0 = f : ‖f‖p = 0 is alinear subspace of Lp. Thus we may consider a quotient space

Lp(X,A, µ) := Lp(X,A, µ)/f : ‖f‖p = 0.

3.2. LP -SPACES 67

We will denote elements of Lp(X,A, µ) in the same way as functions of Lp(X,A, µ),so we will identify f ∈ Lp(X,A, µ) with [f ] ∈ Lp(X,A, µ). The seminorm ‖ · ‖pon Lp becomes the norm on Lp. We use the same symbol ‖·‖p for its denotation.

Accordingly to Corollary 3.2.1, all Lp-spaces are normed linear spaces. As adirect consequence of Theorem 3.2.3, we have the following theorem.

Theorem 3.2.4 Every Lp-space is a Banach space for 1 ≤ p < ∞, i.e. everyCauchy sequence in Lp has a norm limit. 2

As an application of the last theorem, we have the following important property

of Lp-spaces. Let 1 ≤ p < ∞. If fn ∈ Lp and∞∑i=1

‖fi‖p < ∞ then there exists

f ∈ Lp such that

limn→∞

∥∥∥∥∥g −n∑i=1

fi

∥∥∥∥∥p

= 0 .

In this case, we write g =∞∑i=1

fi.

In Lp-spaces, the norm convergence and the a.e. convergence are related to eachother as follows.

Theorem 3.2.5 Let 1 ≤ p <∞, fn∞n=1 ⊆ Lp, and let g ∈ Lp satisfy

fn(x)→ 0 a.e. & |fn(x)| ≤ g(x) a.e. (∀n).

Then ‖fn‖p → 0. 2

The proof of this theorem is a direct application of the dominated convergencetheorem.

Theorem 3.2.6 Let 1 ≤ p <∞, fn∞n=1 ⊆ Lp, and let M ∈ R satisfy

fn(x) ↑ a.e. & ‖fn‖p ≤M (∀n ∈ N).

Then there exists f ∈ Lp such that ‖fn − f‖ → 0. 2

68 CHAPTER 3.

The proof of this theorem is a direct application of the monotone convergencetheorem.

3.2.5.! Separable Lp-spaces. A Banach space X is called separable if thereexists a countable family xnn ⊆ X such that its norm closure is equal to X.It is easy to see that L∞(µ) is separable iff its dimension is finite.

Theorem 3.2.7 If the measure µ is σ-finite and separable and 1 ≤ p <∞ thenLp(µ) = Lp(X,Σ, µ) is separable.

Proof: We will prove that the countable collection Y of all simple functions,which are finite sums with rational complex coefficients of characteristic func-tions of measurable sets from a fixed countable family R (where R is a collectionof sets as in Definition 1.1.8), is dense in Lp(µ). Let f ∈ Lp(µ) and ε > 0 begiven. Since there exists an increasing sequence Xn ↑ X (each Xn of finite mea-sure), we have |f − f · χXn| ↓ 0 pointwise. Hence, there exists a set X ′ of finitemeasure (we may assume that X ′ = X1) such that if we define f1 := f · χX1

then ‖f − f1‖p ≤ ε/3.

The function f1 can be written as f = (g1− g2) + i(h1− h2), where g1, g2, h1, h2

are non-negative members of Lp vanishing outside of X1. For g1, there existsa sequence (g(n))∞n=1 of rational simple functions, non-negative and vanishingoutside of X1, such that g(n) ↑ g1 pointwise. Then |g1 − g(n)|p ↓ 0, so

‖g1 − g(n)‖p ≤ ε/12

for n sufficiently large. Similarly for g2, h1, and h2. By linear combination,we obtain a rational simple function f2 vanishing outside X1 and such that‖f1 − f2‖p ≤ ε/3.

The function f2 can be written as f2 =∑k

j=1 αjχAj, where all αj are rational

complex, all Aj are mutually disjoint, and ∪kj=1Aj = X1. For each Aj, thereexists a set Bj ∈ R such that

µ(Aj4Bj) ≤ εp/(3k|αj|)p ,

so‖χAj

− χBj‖p = µ(Aj4Bj)

1/p ≤ ε/(3k|αj|) .

3.2. LP -SPACES 69

Hence, writing f3 =∑k

j=1 αjχBj, we have

‖f2 − f3‖p ≤k∑j=1

|αj| · ‖χAj− χBj

‖p ≤ ε/3 .

It follows that ‖f−f3‖p ≤ ε. Since f3 ∈ Y , this shows that Lp(µ) is separable. 2

3.2.6. Hilbert spaces and L∞-spaces. Despite of the historically differentdefinitions of Hilbert spaces and L2-spaces, they are the same object and it is aquite reasonable idea to define Hilbert space as an L2-space. The main tool inthe investigation of Hilbert spaces is a notion of the scalar product:

(f, g) :=

∫f · g dµ (f, g ∈ L2).

The scalar product generates the norm ‖ · ‖ of L2 as ‖f‖ = |(f, f)| 12 .

The scalar product leads to the notion of orthogonality of functions. Two func-tions f, g ∈ L2 are called orthogonal if (f, g) = 0. The family fαα∈A is calledan orthonormal basis of L2 if ‖fα‖ = 1 for any α and (fα, fβ) = 0 for anyα 6= β. The following theorem is a generalization of the well known Pythagoriantheorem.

Theorem 3.2.8 Any L2-space has at least one orthonormal basis. If fαα∈Ais an orthonormal basis of L2 then for any f ∈ L2

f =∑α∈A

(f, fα)fα . 2

In Hilbert spaces, the Holder inequality is known as the Cauchy – Schwarzinequality:

|(f, g)| ≤ (f, f)12 · (g, g)

12 (f, g ∈ L2).

L∞-space is the quotient space of L∞ with respect to the equivalence relation”≈” defined in (3.2.8). The correspondent norm we will denote by ‖ · ‖∞. Thepoof of the next theorem is direct and routine and we refer for them to anytextbook on functional analysis.

70 CHAPTER 3.

Theorem 3.2.9 Every L∞-space is a Banach space. 2

It is an important remark that Theorem 3.2.5 and Theorem 3.2.6 fail for infinitedimensional L∞-spaces.

All Lp-spaces for 1 ≤ p ≤ ∞ are vector lattices and, since

|f | ≤ |g| ⇒ ‖f‖ ≤ ‖g‖ (∀f, g ∈ Lp),

they are Banach lattices.

3.2.7.! The duality and the weak convergence. Let X be a complexBanach space. A function ψ : X → C is called a linear functional if

ψ(αx+ βy) = αψ(x) + βψ(y) ∀x, y ∈ X, α, β ∈ C.

A linear functional ψ is called continuous if

‖xn − x‖ → 0 ⇒ ψ(xn − x)→ 0.

The set of all continuous linear functionals on the Banach space X is called thedual space (or adjoint) to X and denoted by X∗.

Theorem 3.2.10 The dual space X∗ of a Banach space X is a Banach spacewith respect to the norm

‖ψ‖ = sup|ψ(x)| : x ∈ X, ‖x‖ ≤ 1. 2

There is a standard way for constructing continuous linear functionals on Lp-spaces. Namely, for 1 ≤ p ≤ ∞ let q be such that

1/p+ 1/q = 1 .

Let g ∈ Lq, denote

g(f) = 〈f, g〉 :=

∫f · g dµ ∀f ∈ Lp.

By the Holder inequality, if p < ∞ (and, by the direct arguments, if p = ∞)g is a continuous linear functional on Lp. Moreover, it can be evaluated that‖g‖ = ‖g‖q. If p <∞ then this is the most general form of a continuous linearfunctional on Lp-space.

3.2. LP -SPACES 71

Theorem 3.2.11 Let 1 ≤ p <∞. Then, for every continuous linear functionalψ on Lp, there exists a unique g ∈ Lq such that ψ = g. Moreover, the dual spaceL∗p is isometrically isomorphic to Lq. 2

It is convenient to write L∗p = Lq. The structure of the dual of L∞ is morecomplicated. We say only that L∗∞ is much more bigger then L1 if dim(L∞) =∞.

For measurable functions and, in particular, for functions from Lp-spaces, wehave studied several modes of convergence: the convergence in measure, con-vergence in norm, uniform convergence, etc. Now we give one more type ofconvergence in Lp-spaces for 1 ≤ p <∞.

Definition 3.2.1 A sequence (fn)∞n=1 ⊆ Lp is weakly convergent to f ∈ Lpif

limn→∞

∫fn · g dµ =

∫f · g dµ (∀g ∈ Lq),

where, as usual, p−1 + q−1 = 1.

From the Holder inequality, we have

|〈fn − f, g〉| ≤ ‖fn − f‖p · ‖g‖q

and, thus, if ‖fn − f‖p converges to zero, so does 〈fn − f, g〉. Hence the conver-gence in norm implies the weak convergence.

In general, the weak convergence is strictly weaker then the norm convergence.There is only one interesting exception, namely.

Theorem 3.2.12 Let (ρn)∞n=1 be a sequence of elements of `1 which convergesweakly to ρ ∈ `1. Then ‖ρn − ρ‖ → 0. 2

The weak convergence generates the weak topology on Lp. This topology is,by the definition, the weakest topology on Lp in which all functionals g (whereg ∈ Lq, p−1 +q−1 = 1) are continuous. The following theorem is very important.Although the proof uses only ideas from the measure theory, it is quite involved,and we omit it in this book.

Theorem 3.2.13 Let g ∈ L+1 , then the set [0, g] ⊆ L1 is weakly compact. 2

72 CHAPTER 3.


Exercise 3.2.1 Prove the Chebyshev inequality:

µ(x : |f(x)| > ε) ≤‖f‖ppεp

,

where f ∈ Lp for 0 < p <∞, and ε > 0.

Exercise 3.2.2 ∗∗∗ Let 1 ≤ a < b < c ≤ ∞. Show that

La ∩ Lc ⊆ Lb & ‖f‖b ≤ ‖f‖αa · ‖f‖1−αc ,

where α is defined byb−1 = αa−1 + (1− α)c−1.

Exercise 3.2.3 ∗∗∗ Let µ be a finite measure and 1 ≤ a ≤ b ≤ ∞. Show that

Lb ⊆ La & ‖f‖a ≤ ‖f‖b · µ(X)(a−1−b−1).

This implies, in particular, that the strong convergence in Lb implies the strong convergence inLa with the same limit. Suppose that µ is infinite, construct an example of a sequence (fn)∞n=1

in La ∩ Lb and f ∈ La ∩ Lb such that ‖fn − f‖b → 0, but ‖fn − f‖a 6→ 0.

Exercise 3.2.4 ∗ Let (X,Σ, µ) be a finite measure space and let f ∈ L∞(X,Σ, µ). Show thatthe function

N (p) := ‖f‖p , (1 ≤ p <∞)

is continuous and limn→∞

N (n) = ‖f‖∞.

Exercise 3.2.5 Show that the linear subspace of all simple functions is dense in Lp-space forany 1 ≤ p ≤ ∞.

Exercise 3.2.6 Show that `p and Lp(R) are separable Banach spaces for any1 ≤ p <∞.

Exercise 3.2.7 Let fn = sinnx ∈ Lp[0, 1], where 1 ≤ p < ∞. Show that fn → 0 weakly, butfn 6→ 0 in measure.

Exercise 3.2.8 Let gn = nχ[0, 1n ] ∈ Lp[0, 1], where 1 ≤ p <∞. Show that gn → 0 in measure,but gn 6→ 0 weakly.

3.3. THE LIAPOUNOFF CONVEXITY THEOREM 73

Exercise 3.2.9 (Minkowski’s Inequality for Integrals. I.) ∗∗∗

Let 1 ≤ p < ∞, and let (X,M, µ), (Y,N , ν) be σ-finite measure spaces, and let f ≥ 0 be aM⊗N -measurable function on X × Y . Show that[ ∫ (

f(x, y) dν(y))pdµ(x)

]1/p≤(∫

f(x, y)p dµ(x))1/p

dν(y).

Exercise 3.2.10 (Minkowski’s Inequality for Integrals. II.) ∗∗∗

Let 1 ≤ p ≤ ∞, f(·, y) ∈ Lp(µ) for a.e. y, and let the function y → ‖f(·, y)‖p be in L1(ν).

Show that f(x, ·) ∈ L1(ν) for a.e. x; the function x→∫f(x, y) dν(y) is in Lp(µ); and∥∥∥∫ f(·, y) dν(y)

∥∥∥p≤∫‖f(·, y)‖p dν(y).

Exercise 3.2.11 ∗∗∗ Prove Theorem 3.2.12.


Exercise 3.2.13 ∗∗ Let An is a measurable subset of [0, 1] for each n, χAn∈ L1, and χAn

→ fweakly in L1. Show that f is not necessarily a characteristic function of some measurable set.

Exercise 3.2.14 Let (X,Σ, µ) be a measure space, and P be a Frobenius – Perron operatoron L1(X,Σ, µ). Show that the dual operator P ∗ on L∞(X,Σ, µ) satisfies the condition thatP ∗(χA) is a characteristic function of some measurable set for every A ∈ Σ.

3.3 The Liapounoff Convexity Theorem

In Exercise 1.2.8, we have seen that if µ is a non-atomic signed measure ofbounded variation on Σ then the range of µ is the closed interval

[infµ(A) : A ∈ Σ, supµ(A) : A ∈ Σ].

One of the central and nicest result about vector-valued measures is the Lia-pounoff convexity theorem, which says that the range of a non-atomic Rn-valuedmeasure of bounded variation is compact and convex. In this section, we give adirect proof of this theorem, which is due to D.A. Ross [8].

74 CHAPTER 3.

3.3.1. One-dimensional case. For the convenience of the reader, who willnot study the proof of main theorem of this section, we give here the proof ofthe result of Exercise 1.2.8, which is very partial, but important, case of theLiapounoff theorem.

First by the Jordan decomposition theorem, we may suppose that µ has nonnega-tive values only. Also we may suppose, that supµ(A) : A ∈ Σ = 1. So we haveto show that the range of µ coincides with the interval [0, 1]. It takes obviouslythe values 0 and 1. Let 0 < α < 1. Consider the family F of all pairs (A,B) ofmeasurable subsets of X satisfying µ(A) ≤ α, µ(B) ≤ 1 − α. The family F isordered by inclusion, and it is trivial to show that it satisfies conditions of theZorn lemma. Therefore, there exists a maximal element, say (A0, B0) in F . Ifµ(A0) < α, then µ(X\(A0∪B0)) > 0, and there exists a subset Y of X\(A0∪B0)(due to non-atomic condition on µ) with 0 < µ(Y ) < α − µ(A0). Thus we getthat a pair (A1, B0) is strictly greater then (A0, B0), where A1 := A0 ∪ Y . Theobtained contradiction with the maximality of (A0, B0) shows that µ(A0) = α,what is required.

3.3.2. Examples. First, let us show that the Liapounoff convexity theoremmay fail for E-valued measures, if E is an infinite dimensional Banach space(see, [3]).

Example 3.3.1 (J.J. Uhl, Jr.) A non-atomic vector-valued measure ofbounded variation, whose range is closed and is neither convex norcompact. Let B([0, 1]) be the Borel algebra on [0, 1] and µ be the Lebesguemeasure. Define

G : B([0, 1])→ L1(µ)

by G(E) := χE. If π ⊆ B([0, 1]) is a partition of [0, 1], it is evident that∑E∈π

‖G(E)‖1 =∑E∈π

µ(E) = 1.

By an easy application of the dominated convergence theorem, G(Σ) is closed inL1(µ). Show that G(Σ) is not a convex set. Indeed, 1

2χ[0,1] ∈ coG(Σ), but

‖G(E)− 1

2χ[0,1]‖1 = [µ([0, 1] \ E) + µ(E)]/2 =

1

2(∀E ∈ Σ).


To see that G(Σ) is not compact, let En = t ∈ [0, 1] : sin(2nπt) > 0 for eachpositive integer n. An easy computation shows that

‖G(Em)−G(En)‖ =1

4(∀m 6= n).

Hence G(Σ) is not compact.

Example 3.3.2 (A. Liapounoff) An `2-valued non-atomic measure ofbounded variation whose range is not convex. Let Σ = B[0, 2π] with theLebesgue measure µ. Accordingly to Exercise 3.3.4, select a complete orthogonalsystem (wn)∞n=0 in L2[0, 2π] such that each wn assumes only the values ±1 and

w0 = χ[0,2π] while∫ 2π

0wn(t) dt = 0 for n ≥ 1. For each n, define λn on Σ by

λn(E) = 2−n∫E

[(1 + wn(t))/2] dt,

E ∈ Σ. Define G : Σ→ `2 by

G(E) = (λ0(E), λ1(E), . . . , λn(E), . . .).

By Exercise 3.3.6, G is a non-atomic `1-valued measure. G has bounded varia-tion, since ‖G(E)‖`2 ≤ 2µ(E) for each E ∈ Σ.

Now considerG([0, 2π]) = (2π, π/2, π/4, . . .)

and suppose there is E ∈ Σ such that

G(E) = G([0, 2π])/2.

Then we have

π = λ0(E) =

∫E

dt = µ(E)

and for n ≥ 1

2−n−1π = λn(E) = 2−n∫E

[(1 + wn(t))/2] dt = 2−nµ(E ∩ Un),

where Un = s ∈ [0, 2π] : wn(s) = 1. From this and the identities

µ(Un) = µ(E) = π,

76 CHAPTER 3.

it follows

µ(E ∩ Un) = µ(E \ Un) = µ(Un \ E) = µ([0, 2π] \ (E ∪ Un)) = π/2

for n > 0. Define f = χE − χ[0,2π]\E. Then we have∫ 2π

0

f(t) · w0(t) dt = π − π = 0

and, for n > 0, we have∫ 2π

0

f(t)·wn(t) dt = µ(E∩Un)+µ([0, 2π]\(E∪Un)) = µ(E\Un)−µ(Un\E) = 0.

Since f ∈ L2([0, 2π]) and f 6= 0, this contradicts the completeness of (wn)∞n=0

and shows that G(Σ) is not convex.

3.3.3.! Liapounoff’s convexity theorem. Suppose that Ψ is an Rn-valuedmeasure on (X,Σ). The following theorem was stated by A. Liapounoff in [6].

Theorem 3.3.1 (Liapounoff convexity theorem) If Ψ is non-atomic andhas the bounded variation then

R(Ψ) := Ψ(A) : A ∈ Σ

is a closed convex subset of Rn.

We deduce this theorem from its special case, when R(Ψ) ⊆ Rn+. Suppose

that µ1, µ2, . . . , µn are finite non-atomic measures on (X,Σ). Denote by Φ =(µ1, . . . , µn) the corresponding Rn-valued measure.

Lemma 3.3.1 The set Φ(E) : E ∈ Σ ⊆ Rn+ is closed and convex.

The proof of this lemma is postponed until the next subsection.

Proof of Theorem 3.3.1: Let Ψ be non-atomic and have bounded variation.Then

Ψ = (ν1+ − ν1−, . . . , νn+ − νn−)


and, by the Hahn decomposition theorem (Theorem 3.1.3), there exists a familyPknk=1 ⊆ Σ such that Pk is a positive set for νk and X \Pk is a negative set forνk. Denote P 0

k := Pk and P 1k := X \ Pk for all k ∈ 1, n. It follows from Lemma

3.3.1 that, for any function f : 1, n→ 0, 1, the set

Rf (Ψ) := Ψ(A) : A ∈ Σ, A ⊆n⋂k=1

Pf(k)k

is closed and convex in Rn. Then our range R(Ψ) is a closed and convex subsetof Rn, as a finite sum of closed convex subsets of Rn:

R(Ψ) =∑

f is a function from 1,n to 0,1

Rf (Ψ). 2

3.3.4.! Proof of Lemma 3.3.1. This lemma is an application of the following,more elaborate, lemma the proof of which is postponed until the next subsection.

Lemma 3.3.2 Let µ1, µ2, . . . , µn be finite non-atomic measures on (X,Σ). De-note by Φ = (µ1, . . . , µn) the corresponding Rn-valued measure.

(LT1) For each E in Σ and r ∈ [0, 1], there is a subset A of E with A in Σand Φ(A) = r · Φ(E).(LT2) For each E in Σ, there is an r ∈ (0, 1) and a subset A of E with A inΣ and Φ(A) = r · Φ(E).(LT3) For each E in Σ, there is a subfamily Arr∈[0,1] of Σ such that Ar ⊆As ⊆ E whenever 0 ≤ r ≤ s ≤ 1 and Φ(Ar) = r · Φ(E) for each r in [0, 1].

Moreover, the conditions (LT1), (LT2), and (LT3) are equivalent

To see that Lemma 3.3.1 follows from Lemma 3.3.2, let E and F belong to Σ,and let 0 ≤ r ≤ 1. By (LT1), there are subsets E1 of E \ F and F1 of F \ Ewith Φ(E1) = r · Φ(E \ F ) and Φ(F1) = (1− r) · Φ(F \ E). Then

Φ(E1 ∪ [E ∩ F ] ∪ F1) = r · Φ(E) + (1− r) · Φ(F ).

This shows that R(Φ) is convex.

Now we show that R(Φ) is closed. It is known that any bounded closed convexsubset of Rn is the convex hull of the set of all its extreme points. Thus R(Φ) =

78 CHAPTER 3.

co(Ex(R(Φ))), and to show that R(Φ) is closed it is enough to show that z ∈R(Φ) for any z ∈ Ex(R(Φ)). Let z ∈ Ex(R(Φ)). There is (and unique up toscalar multiplication) yz ∈ Rn such that

(z, yz) = max(x, yz) : x ∈ R(Φ) = sup(x, yz) : x ∈ R(Φ). (3.3.1)

Consider a signed measure µz on (X,Σ) defined by

µz(A) := (Φ(A), yz) (A ∈ Σ).

Then, by (3.3.1),(z, yz) = supµz(A) : A ∈ Σ. (3.3.2)

It follows from (3.3.2) that there is a set E ∈ Σ such that

(z, yz) = µz(E) = (Φ(E), yz).

Since z is an extreme point of R(Φ) and yz satisfies (3.3.1), z = Φ(E) ∈ R(Φ),what is required. 2

3.3.5.! Proof of Lemma 3.3.2. The proof is carried out by an inductionon the number n of non-atomic measures. It is convenient to assume (LT3), thestrongest of the three statements, as the induction hypothesis for n measures,then prove the weakest of the statements, (LT2), for n+1 measures. We thereforeneed to show first that (LT1), (LT2), and (LT3) are equivalent, in the sense thata vector-valued measure Φ satisfying one of the statements must satisfy all three.The implications

(LT3)⇒ (LT1)⇒ (LT2)

are clear. Assume then that (LT2) holds, and fix a set E in Σ. If A is ameasurable subset of E, r is in [0, 1], and Φ(A) = r · Φ(E), then write rA = r.Let E be the family of all such A. Order E by A < B if A ⊆ B and rA < rB. LetC be a maximal chain in E with ∅ and E belonging to C. Put I = rA : A ∈ C.It suffices to show that I = [0, 1] (since we can take Arr∈[0,1] = C). Suppose(in search of a contradiction) that a ∈ (0, 1) \ I. Put

a∞ = sup(I ∩ [0, a)) ≤ a & b∞ = inf(I ∩ (a, 1]) ≥ a.

There exist An and Bn in E for n = 1, 2, 3, . . . such that rAn increases to a∞and rBn decreases to b∞. Put A∞ =

⋃nAn and B∞ =

⋂nBn. It is easy to see


that A∞ and B∞ are members of E with a∞ = rA∞ and b∞ = rB∞. Since Cis totally ordered, A∞ =

⋃A ∈ C : rA < a and B∞ =

⋂A ∈ C : rA > a.

Because C is maximal, A∞ and B∞ belong to C and (a∞, b∞)∩I = ∅. By (LT2),there is a subset C of B∞ \ A∞ with Φ(C) = r · Φ(B∞ \ A∞). Then A∞ ∪ C isin E and a∞ < r(A∞∪C) < b∞, so C ∪ A∞ ∪ C is a chain in E that properlyextends C, a contradiction. Thus

LT3⇔ LT1⇔ LT2 .

Given the measures µ1, . . . , µn, put

Φ′ = (µ1, µ1 + µ2, µ1 + µ2 + µ3, . . . , µ1 + . . .+ µn).

If A belongs to Σ and Φ′(A) = r · Φ′(E), then one can easy verify that Φ(A) =r · Φ(E). So we may always assume µ1 µ2 . . . µn.

When n = 1, (LT2) is trivial since µ1 is non-atomic (see Exercise 1.1.9), soLemma 3.3.2 is immediate. We proceed by induction on n. Suppose that Lemma3.3.2 holds for up to n measures, and that E ∈ Σ and measures µ1, µ2, . . . , µn+1

are given. Put ν = (µ2, . . . , µn+1). By two applications of (LT1), there aredisjoint sets E1, E2, and E3 in Σ with E = E1∪E2∪E3 and ν(Ei) = (1/3)·ν(E)for each i. By (LT3), for each i, there is a subfamily Airr∈[0,1] of Σ such thatAir ⊆ Ais ⊆ Ei whenever 0 ≤ r ≤ s ≤ 1 and

ν(Air) = r · ν(Ei) = (r/3) · ν(E).

We may assume that Ai1 = Ei. There are now three cases.

Case 1: µ1(Ei) = (r/3) · µ1(E) for some i. Since also ν(Ei) = (1/3) · ν(E),(LT2) holds with r = 1/3.

Case 2: for some choice of i1, i2, and i3,

µ1(Ei1) ≥ µ1(Ei3) > (1/3) · µ1(E) > µ1(Ei2).

For definiteness, take i1 = 1, i2 = 2, and i3 = 3. Define a subfamily Arr∈[0,1]

of Σ by

Ar =

A1

3r if 0 ≤ r ≤ 1/3;A1

1 ∪ A23r−1 if 1/3 < r ≤ 2/3;

A11 ∪ A2

1 ∪ A33r−2 if 2/3 < r ≤ 1.

One readily verifies that ν(Ar) = r·ν(E) for r in [0, 1]. If µ1(E) = 0 then Case 1applies. Otherwise the function φ given on [0, 1] by φ(r) = µ1(Ar)/µ1(E) is well

80 CHAPTER 3.

defined. Note that φ is increasing, and, the assumption µ1 µ2 ensures that φis continuous. Moreover,

φ(1/3) =µ1(A1

1)

µ1(E)=µ1(E1)

µ1(E)> 1/3,

and

φ(2/3) =µ1(A1

1 ∪ A21)

µ1(E)=µ1(E1 ∪ E2)

µ1(E)=µ1(E)− µ1(E3)

µ1(E)< 1− 1/3 = 2/3.

By the intermediate value theorem, φ(r) = r for some r ∈ (1/3, 2/3). In otherwords, µ1(Ar) = r · µ1(E). Since already ν(Ar) = r · ν(E), A = Ar ensures that(LT2) holds.

Case 3: for some choice of i1, i2, and i3,

µ1(Ei1) ≤ µ1(Ei3) < (1/3) · µ1(E) < µ1(Ei2).

In this case, the argument from Case 2 applies without change, except nowφ(1/3) < 1/3 and φ(2/3) > 2/3.

This exhausts the cases (since µ1(E) = µ1(E1) + µ1(E2) + µ1(E3)) and provesthe theorem. 2

3.3.6.! Exercises to Section 3.3.

Exercise 3.3.1 ∗ Let C := x = (x1, x2) ∈ R2 : |x1| ≤ 1 & |x2| ≤ 1. Construct a R2-valuedmeasure ν on B[0, 1] such that R(ν) = C.

Exercise 3.3.2 ∗∗∗ Let D := x = (x1, x2) ∈ R2 : x21 + x2

2 ≤ 1 be the closed unit disk in R2.Construct an R2-valued measure ν on the Borel algebra B[0, 1] whose range R(ν) is equal toD.

Exercise 3.3.3 ∗ Construct two R2-valued measures µ and ν of bounded variation on B[0, 1]such that

R(ν) ∪R(µ) ⊆ x = (x1, x2) ∈ R2 : x1 ≥ 0 & x2 ≥ 0,

and R(ν) +R(µ) 6= R(ν + µ).

Exercise 3.3.4 ∗∗ Show that in L2([0, 2π], µ) there exists a complete orthogonal system (wn)∞n=0

of −1, 1-valued functions such that w0 = χ[0,2π] while∫ 2π

0wn dµ = 0 for n ≥ 1.


Exercise 3.3.5 ∗∗∗ Show that the result of Exercise 3.3.4 can be extended to the case of anon-atomic measure space with finite measure. Namely, show that if (X,Σ, µ) is a measurespace with a non-atomic finite measure µ, then in L2(µ) there exists a complete orthogonalsystem (wn)∞n=0 of −1, 1-valued functions such that w0 ≡ 1 and

∫Xwn dµ = 0 for n ≥ 1.

Exercise 3.3.6 Show that the function G : B[0, 2π] → `2 constructed in Example 3.3.2 is anon-atomic `2-valued measure.

Exercise 3.3.7 Let µ be a finite signed measure on (X,Σ). Show that there exists A ∈ Σsuch that

µ(A) = supµ(B) : b ∈ Σ .

Prove the following facts about convex subsets of Rn which were used in the proof of Lemma3.3.1.

Exercise 3.3.8 Let A be a bounded closed convex subset of Rn. Show that A is the convexhull of all extreme points of A.

Exercise 3.3.9 Let A be a bounded convex subset of Rn and let a ∈ A be an extreme point ofthe closure A of A. Show that there is (and unique up to scalar multiplication) y ∈ Rn suchthat

(a, y) = max(x, y) : x ∈ A = sup(x, y) : x ∈ A.

Let A, a, and y be as before, and let b ∈ A. Show that

(b, y) = max(x, y) : x ∈ A ⇒ b = a .

82 CHAPTER 3.

Chapter 4

4.1 Function Spaces on Rn

In this section we describe some elementary notions which play an importantrole in the Fourier analysis and its applications to PDEs. Our aim is to giveonly a few basic constructions from measure theory concerning function spaceson Rn. We do not give any introduction to Fourier transform and to the theoryof distributions; for this propose, we send the reader to many special text-books(see, for example, [4] and [9]).

4.1.1. Multi-indexes. Let U be an open set in Rn and m ≥ 0, we de-note by Cm(U) the space of all real or complex valued functions on U whosepartial derivatives of order ≤ m exist and are continuous. We set C∞(U) =⋂∞m=1C

m(U). Furthermore, for any E ⊆ Rn, we denote by C∞00(E) the spaceof all C∞-functions on Rn whose support is compact and contained in E. IfE = Rn or U = Rn, we usually write without any confusion Lp = Lp(Rn),C00 = C00(Rn), and C∞ = C∞(Rn). The proof of the next simple lemma is leftto the reader as an exercise (Exercise 4.1.2).

Lemma 4.1.1 If f ∈ C00(Rn) then f is uniformly continuous. 2

If x = (xk)nk=1,y = (yk)

nk=1 ∈ Rn, we set

(x , y) :=n∑k=1

xk yk , ‖x‖ :=√

(x , x).

83

84 CHAPTER 4.

It will be convenient to have a compact notation for partial derivatives. Wewrite ∂j = ∂

∂xj, and we use multi-index notation for higher-order derivatives. A

multi-index is an ordered n-tuple of nonnegative integers. If α = (α1, . . . , αn)is a multi-index, we set

|α| =n∑j=1

αj, α! =n∏j=1

αj!, ∂α =

(∂

∂x1

)α1

· · · (

∂

∂xn

)αn

;

and if x ∈ Rn, then xα =n∏k=1

xαkk ∈ R.

Two subspaces of C∞(Rn) are of the great importance. The first one is the spaceC∞00 of C∞-functions with compact support. The existence of nonzero functionsin C∞00 is not quite obvious; the standard construction is based on the fact thatthe function η(t) = e−1/t · χ(0,∞)(t) is C∞-function even at the origin. If we set

ψ(x) = η(1− ‖x‖2) =

exp[(‖x‖2 − 1)−1)] if ‖x‖ < 1,0 if ‖x‖ ≥ 1,

(4.1.1)

We have ψ ∈ C∞ and supp(ψ) is the closed unit ball in Rn. The second impor-tant subspace of C∞ is the Schwartz space S consisting of those C∞-functionswhich, together with all their derivatives, vanish at infinity faster than any powerof ‖x‖. More precisely, for any nonnegative integer N and any multi-index α,we define

‖f‖(N,α) = supx∈Rn

(1 + ‖x‖)N |∂αf(x)|;

thenS = f ∈ C∞ : ‖f‖(N,α) <∞ for all N,α.

Examples of functions in S are fα(x) = xαe−‖x‖2, where α is any multi-index.

Also, clearly, C∞00 ⊆ S. Another important observation is that if f ∈ S then∂αf ∈ Lp for all α and all p ∈ [1,∞]. Indeed, |∂αf(x)| ≤ CN(1 + ‖x‖)−N for allN , and (1 + ‖x‖)−N ∈ Lp for all N > n/p.

Proposition 4.1.1 S is a Frechet space (= complete linear metric space)with the topology defined by the seminorms ‖ · ‖(N,α)N,α.

Proof: It is enough to prove the completeness of S. If (fk)∞k=1 is a Cauchy

sequence in S then ‖fj − fk‖(N,α) → 0 for all N,α. In particular, for each multi-index α, the sequence (∂αfk)

∞k=1 converges uniformly to a continuous function

4.1. FUNCTION SPACES ON RN 85

gα. Denoting by ej the vector (0. . . . , 1, . . . , 0) with the 1 in the j-th position,we have

fk(x + tej)− fk(x) =

∫ t

0

∂jfk(x + sej) ds.

Letting k →∞, we obtain

g0(x + tej)− g0(x) =

∫ t

0

gej(x + sej) ds.

Here we consider the vector ej as a multi-index. Then gej= ∂jg0, and an

induction on |α| yields gα = ∂αg0 for all α. Now it is easy to check that‖fk − g0‖(N,α) → 0 for all α. 2

4.1.2. Shift operation and convolution. Let f be a function on Rn andy ∈ Rn. We define the shift of f with respect to the vector y by

(τyf)(x) := f(x− y) .

Observe that ‖τyf‖p = ‖f‖p for 1 ≤ p ≤ ∞. It is almost obvious that acontinuous function f is uniformly continuous iff ‖τyf − f‖∞ → 0 as ‖y‖ → 0.

Proposition 4.1.2 If 1 ≤ p <∞, the translation is continuous in the Lp norm;that is, if f ∈ Lp and z ∈ Rn, then limy→0 ‖τy+zf − τzf‖p = 0.

Proof: Since τy+z = τyτz, replacing f by τzf it suffices to assume that z = 0.First, if g ∈ C00, for ‖y‖ ≤ 1 the functions τyg are all supported in a commoncompact set K. So, by Lemma 4.1.1,∫

Rn

|τyg − g|p dµ ≤ ‖τyg − g‖p∞ µ(K)→ 0 as ‖y‖ → 0 . (4.1.2)

Now suppose f ∈ Lp. If ε > 0 then, by the Urysohn lemma, there exists g ∈ C00

with ‖g − f‖p < ε/3, so

‖τyf − f‖p ≤ ‖τy(f − g)‖p + ‖τyg − g‖p + ‖g − f‖p <2

3ε+ ‖τyg − g‖p ,

and ‖τyg − g‖p < ε/3 if y is small enough accordingly to (4.1.2). 2

86 CHAPTER 4.

Let f and g be measurable functions on Rn. The convolution of f and g is thefunction f ∗ g defined by

(f ∗ g)(x) :=

∫Rn

f(x− y)g(y) dy

for all x ∈ Rn such that the integral exists. Various conditions can be imposedon f and g to guarantee that f ∗g is defined at least almost everywhere. For ex-ample, if f ∈ C00, g can be any locally integrable function i.e.

∫A|g| dµ <∞

for any compact A.

4.1.3. Basic properties of convolutions. The elementary properties ofconvolutions are summarized in the following proposition.

Proposition 4.1.3 Assuming that all integrals below exist, we have:(a) f ∗ g = g ∗ f ;(b) (f ∗ g) ∗ h = f ∗ (g ∗ h);(c) For z ∈ Rn, τz(f ∗ g) = (τzf) ∗ g = f ∗ (τzg);(d) If A is the closure of x + y : x ∈ supp(f),y ∈ supp(g)then supp(f ∗ g) ⊆ A.

Proof: (a): is proved by the substitution z = x− y:

f ∗ g(x) =

∫f(x− y)g(y) dy =

∫f(z)g(x− z) dz = g ∗ f(x).

(b): follows from (a) and Fubini’s theorem.(c):

τz(f ∗ g)(x) =

∫f(x− z− y)g(y) dy =

∫τzf(x− y)g(y) dy = ((τzf) ∗ g)(x)

and, by (a),τz(f ∗ g) = τz(g ∗ f) = (τzg) ∗ f = f ∗ (τzg).

(d): Observe that if x 6∈ A then, for any y ∈ supp(g), we have x− y 6∈ supp(f);hence f(x− y)g(y) = 0 for all y, so (f ∗ g)(x) = 0. 2

Theorem 4.1.1 (Young’s inequality) If f ∈ L1 and g ∈ Lp (1 ≤ p ≤ ∞)then (f ∗ g)(x) exists almost everywhere, f ∗ g ∈ Lp, and ‖f ∗ g‖p ≤ ‖f‖1‖g‖p.


Proof: By Minkowski’s inequality for integrals (see Exercise 3.2.10),

‖f ∗ g‖p =∥∥∥∫

Rn

f(y) · g(·,−y) dy∥∥∥p≤∫

Rn

|f(y)| · ‖τyg‖p dy = ‖f‖1 · ‖g‖p . 2

Proposition 4.1.4 If p−1 + q−1 = 1, f ∈ Lp, and g ∈ Lq then f ∗ g(x) existsfor every x ∈ Rn, f ∗ g is bounded and uniformly continuous, and ‖f ∗ g‖∞ ≤‖f‖p ‖g‖q. If 1 < p <∞ then f ∗ g ∈ C0(Rn).

Proof: The existence of f ∗g and the estimate for ‖f ∗g‖∞ follow immediatelyfrom Holder’s inequality. In view of Propositions 4.1.2 and 4.1.3, so does theuniform continuity of f ∗ g: If 1 ≤ p <∞,

‖τy(f ∗ g)− f ∗ g‖∞ = ‖(τyf − f) ∗ g)‖∞ ≤ ‖τyf − f‖p‖g‖q → 0 as y→ 0.

(If p = ∞, interchange the roles of f and g.) Finally, if 1 < p, q < ∞, choosesequences (fn)∞n=1 and (gn)∞n=1 of continuous functions with compact supportssuch that

‖fn − f‖p → 0 & ‖gn − f‖q → 0.

By Proposition 4.1.3 (d), fn ∗ gn ∈ C00. But

‖fn ∗ gn − f ∗ g‖∞ ≤ ‖fn − f‖p‖gn‖q + ‖f‖p‖gn − g‖q → 0,

so f ∗ g ∈ C0 by Exercise 4.1.1. 2

4.1.4. Approximate identity. The following theorem underlies manyimportant applications of convolutions on Rn. If φ is any function on Rn andt > 0, we set

φt(x) = t−nφ(t−1x). (4.1.3)

Remark that if φ ∈ L1 then∫

Rn φt(x) dx is independent on t, thus,∫Rn

φt(x) dx =

∫Rn

t−nφ(t−1x) dx =

∫Rn

φ(y) dy .

The proof of the following theorem is left to the reader as an exercise.

Theorem 4.1.2 Suppose φ ∈ L1 and∫

Rn φ(x) dx = 1.(a) If f ∈ Lp, where 1 ≤ p <∞ then lim

t→0‖f ∗ φt − f‖p = 0 .

(b) If f is bounded and uniformly continuous then limt→0‖f ∗ φt − f‖∞ = 0.

(c) If f ∈ L∞ and f is continuous on an open set U then f ∗ φt → f uniformlyon compact subsets of U as t→ 0. 2

88 CHAPTER 4.

The family φtt≥0 as in Theorem 4.1.2 is called an approximate identity.In some sense, an approximate identity plays the role of an identity operatorwhenever we are working with convolution as a multiplication operation.

Proposition 4.1.5 C∞00 (and hence also S) is dense in Lp (for 1 ≤ p < ∞)and in C0.

Proof: Given f ∈ Lp and ε > 0, there exists g ∈ C00 with ‖f − g‖p < ε/2 byExercise 4.2.5. Let φ be a function in C∞00 such that

∫Rn φ dµ = 1 (for example,

take φ = (∫

Rn ψ dµ)−1ψ as in (4.1.1)). Then g ∗φt ∈ C∞00 by Proposition 4.1.3(d)and Exercise 4.1.9, and

‖g ∗ φt − g‖p < ε/2

for sufficiently small t by Theorem 4.1.2. The same argument works if Lp isreplaced by C0, and ‖ · ‖p, by ‖ · ‖∞. 2

Theorem 4.1.3 (The C∞-Urysohn lemma) If K ⊆ Rn is compact and U isan open set containing K, there exists f ∈ C∞00 such that 0 ≤ f ≤ 1, f = 1 onK, and supp(f) ⊆ U .

Proof: Let δ be the distance from K to U c (δ is positive since K is compact)and let

V = x ∈ Rn : ρ(x, K) < δ/3.Choose a nonnegative φ ∈ C∞00 such that

∫Rn φ dx = 1 and φ(x) = 0 for ‖x‖ >

δ/3 (for example, (∫

Rn ψ dx)−1 · ψδ/3 with ψ as in (4.1.1)), and set f = χV ∗ φ.Then f ∈ C∞00 by Proposition 4.1.3 (d) and Exercise 4.1.9. It is easy to see that0 ≤ f ≤ 1, f = 1 on K, and

supp(f) ⊆ x : ρ(x, K) ≤ 2δ/3 ⊆ U. 2

Theorem 4.1.2 does not say anything about the pointwise convergence f ∗ φt tof . Indeed, the situation with the pointwise convergence is much more delicate.To illustrate problems which arise here, we give without a proof the followingtheorem.

Theorem 4.1.4 Suppose |φ(x)| ≤ C(1 + ‖x‖)−n−ε for some C, ε > 0. Thenφ ∈ L1 and

∫Rn φ(x) dx = 1. If f ∈ Lp (1 ≤ p < ∞) then f ∗ φt(x) → f(x) as

t → 0 for every x in the Lebesgue set of f . In particular, for almost every xand for every x at which f is continuous. 2


By the Lebesgue set Lf of a locally integrable function f : Rn → R (for shortf ∈ Lloc(Rn)) we understood here

Lf :=

x ∈ Rn : limt→∞

1

µ(B(x, t−1))

∫B(x,t−1)

|f(y)− f(x)|dy = 0.


Exercise 4.1.1 Show that the space C0(Rn) of functions vanishing at infinity is uniformclosure of C∞00 .

Exercise 4.1.2 Show that any f ∈ C0(Rn) is uniformly continuous.

Exercise 4.1.3 ∗ Show that for x, y ∈ R:

(x+ y)k =∑|α|=k

k!α!xα1yα2 α = (α1, α2).

Exercise 4.1.4 ∗ Show that for x ∈ Rn:

(x1 + x2 + ...+ xn)k =∑|α|=k

k!α!xα.

Exercise 4.1.5 ∗ Show that for x,y ∈ Rn:

(x + y)α =∑

β+γ=α

α!β! · γ!

xβ · yγ .

Exercise 4.1.6 ∗ Prove the product rule for partial derivatives:

∂α(xβf) = xβ∂αf +∑

cγδxδ∂γf, xβ∂αf = ∂α(xβf) +∑

c′γδ∂(xδf),

where cγδ = c′γδ = 0 unless |γ| < |α| and |δ| < |β|.

Exercise 4.1.7 ∗∗ Let f ∈ L1(R) and f be nonnegative. Use the result of Exercise 2.1.19 toshow that:(a) if f ∗ f = 0 then f(t) = 0 a.e. on R;(b) if f ∗ f = f then f(t) = 0 for all t ∈ R;(c) the equality (f ∗ f)(t) = 1 a.e. is impossible.

Exercise 4.1.8 ∗∗ Prove Theorem 4.1.2.

Exercise 4.1.9 ∗∗ Let f ∈ L1, g ∈ Ck and let ∂αg be bounded for |α| ≤ k. Show thatf ∗ g ∈ Ck and ∂α(f ∗ g) = f ∗ (∂αg) for |α| ≤ k.


90 CHAPTER 4.

4.2 The Riesz Representation Theorem

In this section, we study the structure of the dual space to the spaces of continu-ous functions on a locally compact Hausdorff space. We treat only the R-valuedcase. The complex case is essentially the same.

4.2.1. The space C00(X) and positive functionals. Let X be a locallycompact Hausdorff space (LCH-space, for short). This means that, for everyx 6= y in X, there exist open neighborhoods U of x and V of y such that U andV are compact and U∩V = ∅. From the point-set topology, it is known that anyLCH-space is normal, so we may use the Urysohn lemma in its investigation.

We denote by C00(X) the space of continuous functions on X with compactsupport, i.e. f ∈ C00(X) if f is continuous and supp(f) = clx : f(x) 6= 0is compact. A linear functional Ψ on C00(X) is called positive if Ψ(f) ≥ 0whenever f ≥ 0. It is an important fact that every positive linear functionalon C00(X) is continuous with respect to the uniform convergence on compactsubsets of X. There are many ways for proving this, one of them is given below.

Lemma 4.2.1 If Ψ is a positive linear functional on C00(X). Then, for eachcompact K ⊆ X, there is CK < ∞ such that |Ψ(f)| ≤ CK‖f‖∞ for all f ∈C00(X) with supp(f) ⊆ K.

Proof: Given a compact K, choose φ ∈ C00(X) with values in [0, 1] such thatφ|K ≡ 1 by Urysohn’s lemma. If supp(f) ⊆ K, we have

‖f‖∞φ− f ≥ 0 & ‖f‖∞φ+ f ≥ 0.

Thus

‖f‖∞Ψ(φ)−Ψ(f) ≥ 0 & ‖f‖∞Ψ(φ) + Ψ(f) ≥ 0,

so |Ψ(f)| ≤ CK‖f‖∞ with CK = Ψ(φ). 2

If µ is a Borel measure on X such that µ(K) < ∞ for every compact K ⊆ X,then C00(X) ⊆ L1(µ), so

f →∫X

f dµ

4.2. THE RIESZ REPRESENTATION THEOREM 91

is a positive linear functional on C00(X). The main result of this section saysthat every positive linear functional on C00(X) arises in this way. Moreover, wegive some regularity conditions on µ, under which µ is unique.

Let µ be a Borel measure on X and let E be a Borel subset of X. The measureµ is called outer regular on E if

µ(E) = infµ(U) : E ⊆ U, U open;

µ is called inner regular on E if

µ(E) = supµ(K) : K ⊆ E, K compact.

If µ is outer and inner regular on all Borel sets, µ is called regular (comparewith 2.1.5). The regularity of a Borel measure sometimes is too strong condition.

Definition 4.2.1 A Borel measure that is finite on compact sets, outer regularon Borel sets, and inner regular on open sets is called a Radon measure.

4.2.2. The Riesz representation theorem for C00(X). This theoremhistorically is one of the first results in functional analysis. Originally it wasstated by F. Riesz in 1909 for C[0, 1]. Then it was generalized by several authorsfor C(K), where K is a compact Hausdorff space, and for C00(X), where X isan LCH-space. Here we begin with the general case and derive from it the Rieszrepresentation theorem for C(K). We use the following notation. If U is openin X and f ∈ C00(X), 0 ≤ f ≤ 1, and supp(f) ⊆ U , then we write f ≺ U .

Theorem 4.2.1 If Ψ is a positive linear functional on C00(X), where X is anLCH-space. Then there exists a unique Radon measure µ on X such that

Ψ(f) =

∫X

f dµ (∀f ∈ C00(X)) . (4.2.1)

Moreover,

µ(U) = supΨ(f) : f ∈ C00(X), f ≺ U for all open U ⊆ X (4.2.2)

and

µ(K) = infΨ(f) : f ∈ C00(X), f ≥ χK for all compact K ⊆ X. (4.2.3)

92 CHAPTER 4.

Sketch of the proof: First, we prove uniqueness. Let µ be a Radon measuresatisfying (4.2.1). Let U ⊆ X be an open subset of X. Then Ψ(f) ≤ µ(U)whenever f ≺ U . If K is a compact subset of U , by Urysohn’s lemma, there isf ∈ C00(X) such that f ≺ U and f |K ≡ 1. Thus

µ(K) ≤∫X

f dµ = Ψ(f) ≤ µ(U) .

This proves (4.2.2) since µ is inner regular on U . Thus µ is uniquely determinedby Ψ on open sets, and hence on all Borel sets because of outer regularity. Thisproves the uniqueness of µ and suggests how to prove existence.

We defineν(U) := supΨ(f) : f ∈ C00(X), f ≺ U

for any open U , and we define ν∗(E) for an arbitrary E ⊆ X by

ν∗(E) := infν(U) : U ⊇ E, U open.

Clearly, ν(U) ≤ ν(V ) if U ⊆ V , and hence ν∗(U) = ν(U) if U is open. Theoutline of the proof is now as follows. We only fix several important steps andleave the proofs of them to the reader.

(1) Show that ν∗ is an outer measure on P(X).

(2) Show that every open set is ν∗-measurable.

Now it follows from Caratheodory’s theorem that every Borel set is ν∗-measurableand µ := ν∗|BX is a Borel measure (note that µ(U) = ν∗(U) = ν(U)) for anyopen U). Moreover, µ is outer regular and satisfies (4.2.2) by the definition.

(3) Show that µ satisfies (4.2.3).

Since X is an LCH-space, µ is finite on compact sets. Prove the inner regularityon open sets. Let U be an open set and µ(U) > α. Fix ε > 0, µ(U) > α + ε.Choose f ∈ C00(X) such that f ≺ U and Ψ(f) > α + ε, and let

K := supp(f) ⊆ U .

If g ∈ C00(X) and g ≥ χK then g ≥ f and hence Ψ(g) ≥ Ψ(f) > α + ε. Butthen µ(K) ≥ α + ε > α by (4.2.3), so µ is inner regular on U .

(4) Show that µ satisfies (4.2.1).


With this last step, the proof of the theorem is completed. 2

4.2.3. The Riesz representation theorem for C(K) . Let K be a compactHausdorff space and let C(K) be a vector space of all continuous R-valuedfunctions on K. It is well known that C(K) is a Banach space with respect tothe sup-norm. The following theorem, which is historically also called the Rieszrepresentation theorem describes the structure of its dual space space C(K)∗.We prove this theorem by using of Theorem 4.2.1.

Theorem 4.2.2 Let K be a compact Hausdorff space and let ψ ∈ C(K)∗. Thenthere are two unique Radon measures µ1 and µ2 on X such that

ψ(f) =

∫K

f dµ1 −∫K

f dµ2 (∀f ∈ C(K)) .

Moreover, µ1 = µ+ and µ2 = µ− for some signed measure µ of bounded variationon the Borel algebra B(K).

Proof: Let ψ ∈ C(K)∗. Then the function ϑ : C+(K)→ R+, which is definedby the formula

ϑ(f) := supψ(g) : 0 ≤ g ≤ f (f ∈ C+(K)) , (4.2.4)

is additive on C+(K) (see Exercise 4.2.14) . Every such an additive functionϑ : C+(K)→ R+ possesses a unique extension ψ+ on C(K), which is a positivelinear functional (see Exercise 4.2.15) . This extension is given by the formula:

ψ+(f) := ϑ(f+)− ϑ(f−) (f ∈ C(K)).

Let ψ− be the positive functional correspondent to −ψ, i.e., ψ− := (−ψ)+.

Now we apply Theorem 4.2.1 to ψ+ ∈ C(K)∗ and ψ− ∈ C(K)∗. Then we have

ψ+(f) =

∫K

f dµ1 (∀f ∈ C(K))

and

ψ−(f) =

∫K

f dµ2 (∀f ∈ C(K)),

where µ1 and µ2 are unique Radon measures on X. It is routine to show that

µ1 = µ+ & µ2 = µ− (4.2.5)

for a signed Borel measure µ := µ1 − µ2. 2

94 CHAPTER 4.


Exercise 4.2.1 ∗∗ Let X be a locally compact Hausdorff space. Show that, for any compactK ⊆ X such that K = ∩∞n=1Un and Un is open for any n ∈ N, there exists f ∈ C00(X) suchthat K = f−1(1).

Exercise 4.2.2 Let fn(t) = t2n be the sequence of R-valued functions on R. These functionsdefine Borel measures µn on R as follows

µn(A) :=∫A

t2ndt (∀A ∈ B(R)).

(a) Show that µn is a Radon measure for each n.

(b) Show that µk is µj-continuous for every k, j ∈ N.

(c) Find the Radon – Nikodym derivative of µk with respect to µj.

(d) Let A ∈ B(R) with the Lebesgue measure µ(A) 6= 0. Show that there is no constantM ∈ R such that µn(A) ≤M for all n.

Exercise 4.2.3 Show that any Radon measure µ is inner regular on each σ-finite subset of X,where A is called σ-finite if there exists a sequence An of measurable sets such that µ(An) <∞for any n and A = ∪∞n=1An.

Exercise 4.2.4 Let Y be the one-point compactification of a set X with the discrete topologyand let µ be a Radon measure on Y . Show that supp(µ) is countable, where

supp(µ) = Y \ [∪U ⊆ Y : µ(U) = 0] .

Exercise 4.2.5 ∗ Show that C00(X) is dense in any Lp(µ) for any 1 ≤ p <∞ and any Radonmeasure µ on X.

Exercise 4.2.6 Prove the properties (1)− (4) in the proof of Theorem 4.2.1.

Exercise 4.2.7 ∗∗ Show that a sequence (fn)∞n=1 in C0(X) converges weakly to 0 iffsup ‖fn‖∞ <∞ and fn → 0 pointwise.

Exercise 4.2.8 Let µ be a Radon measure on X such that µ(x) = 0 for all x ∈ X and aBorel set A ∈ B(X) satisfies µ(A) < ∞. Show that, for any α, 0 ≤ α ≤ µ(A), there existsB ∈ B(X), B ⊆ A, such that µ(B) = α.


Exercise 4.2.9 Let ν be a non-atomic Radon measure on Rn and let µ be another measureon Rn such that µ is ν-continuous. Show that µ is a Radon measure.

Exercise 4.2.10 ∗ Let ν be a Radon measure on Rn and let µ be another measure on Rnsuch that µ is ν-continuous. Show that µ is a Radon measure.

Exercise 4.2.11 ∗ Let ν be a σ-finite Radon measure on a LCH-space and let µ be anothermeasure such that µ is ν-continuous. By using of the Radon – Nikodym theorem, show that µis a Radon measure.

Exercise 4.2.12 ∗∗∗ Let ν be a Radon measure on a LCH-space and let µ be another measuresuch that µ is ν-continuous. Show that µ is a Radon measure.

Exercise 4.2.13 Let X = N with the discrete topology. Show that C0(X)∗ ∼= `1.

Exercise 4.2.14 Show that the function ψ+ in (4.2.4) is well defined and additive on C+(K).

Exercise 4.2.15 Show that every additive on C+(K) R-valued function possesses a uniqueextension on C(K) which is a positive linear functional.

Exercise 4.2.16 Prove the formula (4.2.5).

96 CHAPTER 4.

Bibliography

[1] C.D. Aliprantis; O. Burkinshaw, Principles of real analysis. Third edi-tion, Academic Press, Inc., San Diego, CA, (1998), xii+415 pp.

[2] Chae, Soo Bong Lebesgue integration. Second edition, Universitext.Springer-Verlag, New York, (1995), xiv+264 pp.

[3] J. Diestel; J.J.,Jr. Uhl, Vector measures, Math. Surv. No. 15. AmericanMath. Soc., Providence, R.I., (1977), xiii+322 pp.

[4] G.B. Folland, Real analysis. Modern techniques and their applications.Second edition, Pure and Applied Mathematics (New York). A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, (1999),xvi+386 pp.

[5] P.R. Halmos, Measure Theory, D. Van Nostrand Company, Inc., NewYork, N. Y., (1950), xi+304 pp.

[6] A. Liapounoff, Sur les fonctions-vecteurs completement additives, Bull.Acad. Sci. URSS. Ser. Math. 4, (1940), 465–478.

[7] W.A.J. Luxemburg and A.C. Zaanen, Riesz spaces. Vol.I., North-Holland Mathematical Library. North-Holland Publishing Co.,Amsterdam-London; American Elsevier Publishing Co., New York,(1971), xi+514 pp.

[8] D.A. Ross, An elementary proof of Lyapunov’s theorem, Amer. Math.Monthly 112 (2005), no. 7, 651–653.

[9] W. Rudin, Functional analysis. Second edition, International Series inPure and Applied Mathematics. McGraw-Hill, Inc., New York, (1991),xviii+424 pp.

97

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[10] A.R. Schep, And still one more proof of the Radon-Nikodym theorem,Amer. Math. Monthly 110 (2003), no. 6, 536–538.

[11] A.C. Zaanen, Riesz spaces. Vol.II., North-Holland Mathematical Li-brary, 30. North-Holland Publishing Co., Amsterdam, (1983), xi+720pp.

[12] A.C. Zaanen, Introduction to operator theory in Riesz spaces, Springer-Verlag, Berlin, (1997), xii+312 pp.

Index

(Xn)n-bounded sequence, 10C∞-Urysohn lemma, 88E-valued measure, 13L1-convergence, 24µ-continuous measure, 58µ-integrable function, 58σ-additivity of measure, 7σ-algebra, 5σ-algebra generated by G, 6σ-finite measure, 14σ-finite measure space, 8σ-finite premeasure, 43σ-finite subset, 94ξ-measurable set, 32

a.e., 21additive function, 19adjoint space, 70algebra of subsets, 34almost everywhere, 21almost everywhere convergence, 9approximate identity, 88

Baire theorem, 51Banach lattice, 19, 70block, 39Borel algebra of X, 6Borel measure, 8

Cantor like function, 51Cantor like set, 51

Caratheodory extension theorem, 31Caratheodory theorem, 33Cauchy – Schwarz inequality, 69Cauchy in measure sequence, 27Chebyshev inequality, 72compact support, 90complete measure, 8complex measure, 14continuous linear functional, 70convergence in measure, 27convex hull, 30convolution, 86counting measure, 8

Dedekind completeness, 19difference of sets, 5distribution function, 43dominated convergence theorem, 26dual space, 70

Egoroff theorem, 9elementary Young inequality, 63exhaustion argument, 10exhaustion theorem, 10

Fatou lemma, 25finite measure, 14finite measure space, 8Frobenius – Perron operator, 59Fubini theorem, 31, 40

generalized Cantor function, 51

99

100 INDEX

generalized Cantor set, 50

h-interval, 36Holder inequality, 63Hahn decomposition theorem, 58

inner regular measure, 91integrable function, 22

Jordan decomposition theorem, 18

LCH-space, 90Lebesgue σ-algebra, 36Lebesgue – Stieltjes Measure, 31Lebesgue integral, 21Lebesgue measurable set in Rn, 45Lebesgue measure, 9, 31Lebesgue measure on R, 36Lebesgue measure on Rn, 45Lebesgue set, 89Liapounoff, 73, 75Liapounoff convexity theorem, 76linear functional, 70locally compact Hausdorff space, 90locally integrable function, 86Lusin theorem, 47

measurable function, 6measurable set, 7measurable transformation, 59measure, 7measure of bounded variation, 16measure space, 7Minkowski inequality, 64, 73monotone class, 41monotone convergence theorem, 24monotone function, 19monotone function on A, 35multi-index, 84

negative set, 58non-atomic measure, 11, 12, 20

orthogonal functions, 69orthonormal basis, 69outer measure, 31outer regular measure, 91

partial ordering, 17partially ordered vector space, 17pointwise convergence, 9positive linear functional, 90positive set, 58premeasure, 34probabilistic measure space, 8probability, 8product measure, 39product of measure spaces, 39purely atomic measure, 12

Radon – Nikodym derivative, 58Radon – Nikodym theorem, 55Radon measure, 91regular measure, 91Riesz representation theorem, 91, 93Ross, 73

scalar product, 69Schep, 55Schwartz space, 84separable Banach space, 68separable measure, 8, 14set of the first category, 51set of the second category, 51shift, 47, 85side of block, 45signed measure, 14simple function, 21standard Cantor function, 51

INDEX 101

standard Cantor set, 49submeasure, 31symmetric difference of sets, 5

Tonelli theorem, 42total variation of measure, 16totally disconnected set, 49

Uhl, Jr., 74uniform convergence, 9Urysohn lemma, 46, 85, 90, 92

variation of measure, 15vector lattice, 17vector-valued measure, 13

weak topology on Lp, 71weakly convergent sequence, 71

Young inequality, 86

introduction to measure theory and lebesgue integration

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