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INTRODUCTION

In these notes we present the fundamentals of the Lebesguetheory of integration and an introduction to some of itsapplications.

Let us recall the definition of the Riemann integral of a real-valued function f defined on the interval [a, b]. Let a = x0 <

x1 < · · · < xn = b, and set Ik = [xk−1, xk], with |Ik|, the lengthof Ik, equal to xk − xk−1. Given these Ik, write

S =n∑

k=1

{supIk

f (x)} · |Ik|

and

S =n∑

k=1

{infIk

f (x)} · |Ik|,

and define

(1)∫ b

af (x)dx = inf S,

∫ b

a

f (x)dx = sup S;

here the suprema and infima are taken over all partitions of[a, b]. A function f is said to be Riemann-integrable if

∫fdx =∫

fdx, and in that case the common value is denoted by∫

fdx,and called the Riemann integral of f over [a, b].

The major deficiency of this concept of integral is that notenough functions have integrals. For example, we show thatthe pointwise limit of a uniformly bounded, monotone decreas-ing sequence of integrable functions may not be integrable.

v

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vi LECTURES ON MEASURE AND INTEGRATION

Let {rn} be an enumeration of the rationals on [a, b] anddefine fn by

fn(x) ={

0 if x = r1, r2, . . . , rn

1 otherwise.

Each fn is Riemann-integrable since it has only a finite numberof discontinuities, but the limit function

f (x) ={

0 if x rational1 if x irrational

is not; in fact∫

fdx = 0 and∫

fdx = b − a.We shall take a seemingly devious route in our quest to

define an integral on a wider class of functions which will beclosed at least under monotone limits on [a, b]. Rather thandeal with functions themselves, we spend Chapter I discussinghow to give a definition of length (measure) which can bedefined on a large number of sets. Of course, the situationwould be best if any subset E of [a, b] could be assigned ameasure |E|. If we then allowed I1, . . . , In to be any pairwisedisjoint collection of subsets whose union is [a, b] and defined∫

and∫

as in (1), we would have the following pleasant result.

THEOREM. Under the above assumptions, let f be boundedon [a, b]. Then f has an integral; i.e.,

∫f (x)dx =

∫f (x)dx.

Proof. We assume m ≤ f (x) < M for all x. For each n,define

Ei = {x: m + i − 1

n(M − m) ≤ f (x) < m + i

n(M − m)}

(i = 1, 2, . . . , n).

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INTRODUCTION vii

On each Ei,

supEi

f − infEi

f ≤ M − m

n

and so

S − S ≤ M − m

n

∑|Ei|

= M − m

n· (b − a).

This shows that∫ ≤ ∫

since we may let n → ∞. The oppositeinequality is proved, as in the case of the Riemann integral,by considering refinements of partitions. We needn’t give thedetails.

Actually, it is not necessary that |E| be defined for all setsE. Of course, the proof of the above pseudo-theorem showsthat the larger the number of sets that can be assigned ameasure, the more functions will be integrable. But, as weshall see, the theory is quite satisfactory if we assume onlythat |E| is defined for a class of sets closed under the countableset-theoretic operations.

We construct measures on such classes of sets in Chapter I;in subsequent chapters we discover that the functions inte-grable in the sense above are indeed closed under the mostcommon operations, and that the integral itself enjoys usefulproperties. In the last chapter, integration theory is applied tothe study of Fourier series.

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TABLE OF CONTENTS

Chapter I: Measures

Definition of measure 3; regular measure 4; outer mea-sure 6; measurable set 8; Hahn extension theorem 10; mono-tone families of sets 12; completion of measures 14; Borel setsand construction of Lebesgue measure in the real line 15; theCantor set 16; a non-measurable set 16.

Chapter II: Integration

Measure spaces and measurable functions 19; definition ofthe integral for non-negative functions 22; the integral as ameasure 23; linearity of the integral 25; monotone convergencetheorem 28; Fatou’s lemma 30; the integral for real-valuedfunctions 32; the integral for complex-valued functions 33;dominated convergence theorem 34; bounded convergencetheorem 35; Egoroff’s theorem 36; convergence in measure 37;Lp-convergence 39; the Lebesgue and Riemann integrals 41.

Chapter III: The Theorems of Fubini

Definition of X ×Y 45; simple functions 49; interchangeof integration for non-negative functions 51; interchange ofintegration for L1 functions 51; completion of X × Y 52;a theorem on change of variables in integration 54.

Chapter IV: Representations of Measures

Definition of |μ| 58; Jordan decomposition theorem 60;Hahn decomposition theorem 61; the integral for complexmeasures 63; functions of bounded variation 64; the

ix

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x CONTENTS

Cantor function 66; absolutely continuous measures 67;Radon-Nikodyn theorem 68; the Radon-Nikodynderivative 72; mutually singular measures 73; Lebesguedecomposition theorem 73.Chapter V: The Lebesgue Spaces

Holder’s inequality 77; completeness of Lp 79; functionson regular measure spaces 82; continuity of translation in Lp

norm 83; continuous linear functionals 84; weak and strongconvergence of functionals 86; the dual spaces of Lp 89; thedual space of C0(S) 93.Chapter VI: Differentiation

Derivative of a measure 105; Vitali covering theorem 106;upper and lower derivates 108; regularity of finite Borelmeasures 109; existence of Dμ(x) a.e. 112; convergence ofmeasures 114; points of density and dispersion 115; differenti-ation of integrals 116; differentiation of singular measures 117;the Lebesgue set 118; integration by parts 120.Chapter VII: Fourier Series

Orthogonal systems of functions 121; definition of Fourierseries and Fourier coefficients 122; Bessel’s inequality 124;Riesz-Fischer theorem 125; complete systems 126; complete-ness of exponential and trigonometric systems 127; Poissonkernel 130; positive kernels 134; Riemann-Lebesgue theo-rem 137; the kernel Dn(u) 139; Dirichlet-Jordan theorem 140;integration of Fourier series 145; Abel summability of Fourierseries 147; (C, 1) summability of Fourier series 147; (C, 1)summability of Fourier series on L1 149; a continuous functionwhose Fourier series diverges at 0 151; uniform boundednesstheorem 154; Lp convergence of Fourier series 155; M. Rieszinterpolation theorem 161; Hausdorff-Young theorem 164.

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CHAPTER I

MEASURES

Definition. Let S be a set and � a collection of subsets of S. �

is said to be a field if:

(i) ∅ ε � (∅ denotes the empty set);(ii) A ε � implies A′ ε � (A′ denotes the complement of A);

(iii) A ε �, B ε � implies A ∪ B ε �.

Thus � contains the empty set, the complement of each ofits members, and finite unions of its members. A field � issometimes also referred to as an algebra of sets.

Definition. An additive set function μ on � is a function withdomain � such that

μ(A ∪ B) = μ(A) + μ(B) if A ∩ B = ∅.

The range can be any set which has a binary operation(group, field, vector space); however we shall consider onlythe real numbers, complex numbers, and the non-negative realswith ∞ adjoined (the extended non-negative reals, henceforthdenoted by R∗) as ranges of set functions.

Some trivial properties of an additive set functionμ: � → R∗ are

(i) μ(∅) = 0 unless μ is identically ∞;(ii) A ⊂ B implies μ(A) ≤ μ(B);

(iii) if A ⊃ B and μ(B) < ∞, then

1

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2 LECTURES ON MEASURE AND INTEGRATION

μ(A − B) = μ(A) − μ(B);

(iv) Ai ε �, i = 1, . . . , n implies

μ

(n⋃

i= 1

Ai

)≤

n∑

i= 1

μ(Ai).

Proof of (iv). We haven⋃

i= 1

Ai = A1 ∪ (A2 − A1) ∪ · · · ∪[

An −n−1⋃

i= 1

Ai

].

Thus, since the right side is now a disjoint union,

μ

(n⋃

i= 1

Ai

)= μ(A1)+μ(A2 −A1)+· · ·+μ

[An −

n−1⋃

i= 1

Ai

],

and so by (ii)

μ

(n⋃

i= 1

Ai

)≤ μ(A1) + μ(A2) + · · · + μ(An).

Examples: (i) The finite unions of half open intervals

[a, b) ⊂ [A, B)

form a field � of subsets of [A, B) (finite unions of closed oropen subsets of an interval do not form a field). Let φ be anon-decreasing function on [A, B) and define

μφ([a, b)) = φ(b) − φ(a).

μφ gives rise, in the obvious manner, to an additive set functionon �. Of particular interest is the case φ(x) = x, which willlater lead to the familiar “dx” in integration.

(ii) Let S be any set and � be the collection of all subsetsof S. Define μ: � → Z+ ∪ {∞} by letting μ(A) be the numberof elements in A (here Z+ denotes the non-negative integers).This μ is called the counting measure.

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MEASURES 3

Definitions. A field � is called a σ -field if Ai ε �, i = 1, 2, . . .implies ∪∞

i= 1 Ai ε �. An additive set function μ defined ona field �, and whose range is also a topological space, iscalled σ -additive or countably additive on �, if when the Ai

are pairwise disjoint and ∪∞i= 1 Ai is in �, then �∞

i= 1μ(Ai) isdefined and

μ

( ∞⋃

i= 1

Ai

)=

∞∑

i= 1

μ(Ai).

In any space S there are at least two σ -fields, namely thoseconsisting of all subsets of S and of S and φ only.

Definition. An R∗-valued countably additive set function on afield � is called a measure on �.

Consider the non-negative additive set function μφ definedpreviously. We have

[a, b − δ

2)⋃

{ ∞⋃

k=1

[b − δ

2k, b − δ

2k+1)

}= [a, b)

where δ is so chosen that b− δ/2 > a. If μφ is to be countablyadditive, then we must have

φ(b) − φ(a) = μφ([a, b))

= μφ

[[a, b − δ

2)⋃

{ ∞⋃

k=1

[b − δ

2k, b − δ

2k+1)

}]

= limk→∞

φ

(b − δ

2k

)− φ(a) = φ(b−) − φ(a),

implying that φ(b) = φ(b−) and so φ must be left continuous.This condition is also sufficient for the countable additivityof μφ . However we may remove the restriction that φ be left

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4 LECTURES ON MEASURE AND INTEGRATION

continuous by changing our definition of μφ to read

μφ([a, b)) = φ(b−) − φ(a−);

then μφ is a measure for all non-decreasing functions φ.

Definition. Let μ be an additive R∗-valued function on a field� of subsets of a topological space S. μ is said to be regular ifgiven ε > 0 and A ε �, we can find E, F ε � with E compact,E ⊂ A ⊂ F0, μ(A − E) < ε, and μ(F − A) < ε. (Here F0

denotes the interior of F.)

Example. The μφ are regular:

From the definition of μφ it suffices to consider only setsof the field � of the form A = [a, b). Let F = [a − δ, b), E =[a, b − δ), where the appropriate δ > 0 is to be chosen.

μφ(A − E) = μφ([b − δ, b)) = φ(b−) − φ(b − δ−)

μφ(F − A) = μφ([a − δ, a)) = φ(a−) − φ(a − δ−).

Since limδ→0

φ(a − δ−) = φ(a−) and limδ→0

φ(b − δ−) = φ(b−)

we may pick δ > 0 sufficiently small so that μφ(A − E) < ε

and μφ(F − A) < ε.

THEOREM 1. If μ is a regular additive set function on afield �, then μ is a measure (i.e., μ is countably additive).

Proof. Let Ai ε � be pairwise disjoint with ∪∞i= 1 Ai ε �.

For each n,

μ

( ∞⋃

i= 1

Ai

)≥ μ

(n⋃

i= 1

Ai

)=

n∑

i= 1

μ(Ai),

whence

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MEASURES 5

μ

( ∞⋃

i= 1

Ai

)≥

∞∑

i= 1

μ(Ai).

Now we must get this inequality going in the other direction.Let E ε �, E compact, with E ⊂ ∪∞

i= 1 Ai and μ(∪∞i= 1

Ai − E) < ε. Choose Fi ε �, such that F0i ⊃ Ai, μ(Fi − Ai) <

ε/2i. The F0i cover E so that for some M, we have ∪M

i= 1Fi ⊃ E.Then

μ

( ∞⋃

i= 1

Ai

)= μ(E) + μ

( ∞⋃

i= 1

Ai − E

)≤ μ

(M⋃

i= 1

Fi

)+ ε

≤M∑

i= 1

{μ(Ai) + μ(Fi − Ai)} + ε

≤∞∑

i= 1

μ(Ai) + 2ε.

Since ε > 0 was arbitrary we have

μ

( ∞⋃

i= 1

Ai

)≤

∞∑

i= 1

μ(Ai)

and the theorem is proved.

THEOREM 2. Let � be a collection of subsets of a set S.There exists a unique smallest σ -field of subsets of S contain-ing �. (We call this the σ -field generated by � and write thisas σ(�)).

Proof. Let � be the intersection of all the σ -fields contain-ing �. (We know there is at least one.) Since the intersectionof σ -fields is a σ -field, � is the required σ -field.

Given a measure μ on a field � of subsets of a set S, thequestion of extending μ to a measure on the σ -field σ(�)

generated by � naturally arises. The answer is that we canin fact extend μ to a measure on a σ -field containing σ(�).

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6 LECTURES ON MEASURE AND INTEGRATION

Our technique of constructing this extension will be to givea postulational description of what is called an outer measurefunction. In general this will not be a measure function, butwill have for its domain the class of all subsets of S. We shallprove that a suitable restriction of an outer measure function toa smaller domain, which is however a σ -field, always yields ameasure function. Our goal thus becomes to construct an outermeasure on all subsets of S if we are given a measure on somefield of subsets of S.

Definition. Let λ: 2S → R∗, where 2S denotes the collectionof all subsets of S. We say that λ is an outer measure if thefollowing are satisfied:

(i) λ(∅) = 0;(ii) if A ⊂ B, then λ(A) ≤ λ(B);

(iii) λ(⋃∞

i= 1 Ai)≤∑∞

i= 1 λ(Ai) for any sets A1, A2, . . . ε S.

It is a simple exercise to show that an outer measure whichis finitely additive is also countably additive (use (iii) and aninequality derived at the beginning of the proof of Theorem 1).

THEOREM 3. Let μ be a measure which is defined on a field� of S. Define μ∗: 2S → R∗ by

μ∗(E) = inf∞∑

i= 1

μ(Ai),

where the inf is taken over all sequences {Ai} ε � satisfying⋃∞i= 1 Ai ⊃ E. Then μ∗ is an outer measure and μ∗ = μ on �.

Proof. We first show that μ∗ = μ on �. For A ε �, clearly

μ∗(A) ≤ μ(A).(1)

By letting B1 = A1, and

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MEASURES 7

Bn = An −n−1⋃

i= 1

Ai

we have⋃∞

i= 1 Ai =⋃∞

i= 1 Bi with the Bi’s disjoint, whence

μ(A) =∞∑

i= 1

μ(A ∩ Bi) ≤∞∑

i= 1

μ(Ai).

Therefore

μ(A) ≤ inf{Ai}

∞∑

i= 1

μ(Ai) = μ∗(A),

and by using (1) we have μ(A) = μ∗(A).For the first statement of the theorem we prove that proper-

ties (i)–(iii) of outer measures belong to μ∗.(i) μ∗(∅) = μ(∅) = 0 since ∅ ε �;(ii) Let A ⊂ B, A, B ε 2S and let {Bj} ⊂ � be such that⋃∞

j=1Bj ⊃ B and∞∑

j=1

μ(Bj) ≤ μ∗(B) + ε.

But⋃∞

j=1 Bj ⊃ A, so that

μ∗(A) ≤∞∑

j=1

μ(Bj) ≤ μ∗(B) + ε.

Since this holds for all ε > 0 we have

μ∗(A) ≤ μ∗(B).

(iii) Let Ej be a countable collection of subsets of S. Foreach j choose a countable covering {Aij} of Ej with Aij ε � and

μ∗(Ej) + ε

2j≥

∑

i

μ(Aij

).

Since ⋃

j

⋃

i

Aij ⊃⋃

j

Ej,

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8 LECTURES ON MEASURE AND INTEGRATION

we have

μ∗⎛

⎝⋃

j

Ej

⎞

⎠ ≤∑

j

∑

i

μ(Aij

) ≤∑

j

μ∗(Ej) + ε;

again, since ε > 0 was arbitrary we have the desired result.

Definition. Let λ be an outer measure on S. A set A ismeasurable (with respect to λ, or λ-measurable) if

λ(E) = λ(A ∩ E) + λ(A′ ∩ E)

for all E ⊂ S.It follows that if A is measurable and if E is taken to be S

then

λ(S) = λ(A) + λ(A′).The measurability of a set depends on the outer measurefunction under consideration. There may well be two outermeasures for the same space such that a given set A ismeasurable with respect to one, but not to the other. Thefollowing is the important fact we shall need concerning outermeasures.

THEOREM 4. The measurable subsets of S form a σ -field�0, on which λ is a measure.

Proof. ∅ ε �0 since λ(E) = λ(∅) + λ(E). If A ε �0, thenby the symmetry of the relation

λ(E) = λ(A ∩ E) + λ(A′ ∩ E)(2)

it trivially follows that A′ ε �0. Now suppose that A and B arein �0. Then

λ(E) = λ(A ∩ E) + λ(A′ ∩ E)(3)

= λ(A ∩ B ∩ E) + λ(A ∩ B′ ∩ E)

+ λ(A′ ∩ B ∩ E) + λ(A′ ∩ B′ ∩ E).

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MEASURES 9

Alsoλ([A ∩ B]′ ∩ E) = λ(A ∩ [A ∩ B]′ ∩ E) + λ(A′ ∩ [A ∩ B]′ ∩ E)

= λ(A ∩ B′ ∩ E) + λ(A′ ∩ E),

and by the measurability of B this is

λ(A ∩ B′ ∩ E) + λ(A′ ∩ B ∩ E) + λ(A′ ∩ B′ ∩ E).

This identity and (3) give

λ([A ∩ B]′ ∩ E) + λ(A ∩ B ∩ E) = λ(E)

implying that A ∩ B ε �0, and that �0 is a field.Suppose A, B ε �0, A ∩ B = ∅. If we replace E by

(A ∪ B) ∩ E in (2) we obtain

λ([A ∪ B] ∩ E) = λ(A ∩ E) + λ(B ∩ E).

By induction, if A1, . . . , An ε �0 and are pairwise disjoint, thenfor any E ⊂ S

λ

([n⋃

i= 1

Ai

]∩ E

)=

n∑

i= 1

λ(Ai ∩ E).(4)

Now let {Ai} (i = 1, 2, . . .) be a countable collection ofmembers of �0 which are pairwise disjoint. For any n

λ(E) = λ

([n⋃

i= 1

Ai

]∩ E

)+ λ

([n⋃

i= 1

Ai

]′∩ E

)

≥ λ

([n⋃

i= 1

Ai

]∩ E

)+ λ

([ ∞⋃

i= 1

Ai

]′∩ E

)

=n∑

i= 1

λ(Ai ∩ E) + λ

([ ∞⋃

i= 1

Ai

]′∩ E

)

by (4). Since this is true for any n we deduce

λ(E) ≥∞∑

i= 1

λ(Ai ∩ E) + λ

([ ∞⋃

i= 1

Ai

]′∩ E

),

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10 LECTURES ON MEASURE AND INTEGRATION

and by (iii) this is

≥ λ

([ ∞⋃

i= 1

Ai

]∩ E

)+ λ

([ ∞⋃

i= 1

Ai

]′∩ E

)

≥ λ(E).

We therefore have equality in all of the above statements,implying that ∪∞

i= 1 Ai ε �0 and that �0 is a σ -field. Nowsubstituting

∞⋃

i= 1

Ai

for E we get

λ

( ∞⋃

i= 1

Ai

)=

∞∑

i= 1

λ(Ai) + λ(∅) =∞∑

i= 1

λ(Ai)

so that λ is indeed a measure on �0.

Definition. A measure μ on a field � of subsets of S is saidto be σ -finite if there is a sequence of sets {Ai} ⊂ � with∪∞

i= 1 Ai = S and each μ(Ai) < ∞.

THEOREM 5 (HAHN EXTENSION THEOREM). A measureμ on a field � can be extended to a measure on σ(�), theσ -field generated by �. If μ is σ -finite the extension is unique.

Proof. Extend μ to an outer measure μ∗ on S as inTheorem 4, with �0 the associated σ -field of measurable sets.We shall show that �0 ⊃ σ(�), so that μ∗ will be the requiredextension of μ to �0. It clearly suffices to show that �0 ⊃ �.Let A ε � and E ⊂ S. There exists a collection {Ai}, Ai ε �

with ∪i Ai ⊃ E and �iμ(Ai) ≤ μ∗(E) + ε. The A ∩ Ai coverA ∩ E and the A′ ∩ Ai cover A′ ∩ E, whence

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MEASURES 11

μ∗(E) ≤ μ∗(A ∩ E) + μ∗(A′ ∩ E) ≤ �i

μ∗(A ∩ Ai) + �i

μ∗(A′ ∩ Ai)

= �i

μ(A ∩ Ai) + �i

μ(A′ ∩ Ai) = �i

μ(Ai)

≤ μ∗(E) + ε.

Therefore μ∗(E) = μ∗(A ∩ E) + μ∗(A′ ∩ E) and A ε �0.

In order to demonstrate the uniqueness of this extension wefirst need two lemmas, important in themselves.

LEMMA 1. Let μ be a measure on a σ -field �. Then

(i) if Ai ↑ (i.e., Ai ⊂ Aj for i ≤ j), then

μ

( ∞⋃

i= 1

Ai

)= lim

i→∞μ(Ai).

(ii) if Ai ↓ (i.e., Ai ⊃ Aj for i ≤ j) and μ(A1) < ∞, then

μ

( ∞⋂

i= 1

Ai

)= lim

i→∞μ(Ai).

Proof. (i) We may assume each μ(Ai) < ∞ for otherwiseboth sides equal ∞. Supposing this we write

∞⋃

i= 1

Ai = A1 ∪ (A2 − A1) ∪ (A2 − A3) ∪ · · · .

Since the sets on the right are pairwise disjoint,

μ

( ∞⋃

i= 1

Ai

)= μ(A1) + μ(A2 − A1) + μ(A3 − A2) + · · · .

Using the fact that Ai ⊂ Ai+1, this gives

μ

( ∞⋃

i= 1

Ai

)= lim

i→∞μ(Ai).

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12 LECTURES ON MEASURE AND INTEGRATION

(ii) We have

A1 = (A1 − A2)⋃

(A2 − A3) ∪ . . . ∪( ∞⋂

i=1

Ai

),

implying

μ(A1) = μ(A1) − limi→∞

μ(Ai) + μ

( ∞⋂

i=1

Ai

),

whence

limi→∞

μ(Ai) = μ

( ∞⋂

i=1

Ai

).

Remarks. (1) In (ii) it was only necessary that μ(Ak) < ∞ forsome k.

(2) If μ1 and μ2 are finite measures on a σ -field, then itfollows from the lemma that the set {A: μ1(A) = μ2(A)} isclosed under monotone limits.

Definitions. A collection of sets is called monotone if itis closed under monotone limits of sequences. If � is afamily of subsets of S, m(�) is the monotone family itgenerates. (Thus m(�) consists of � itself together with allsets obtainable from � by monotone operations.) The chiefadvantage of introducing m(�) is that its elements are so easilycharacterized; its usefulness is a consequence of

LEMMA 2. If � is a field, m(�) is a σ -field and thereforeis equal to σ(�).

Proof. If {Ai} is a countable collection of sets belongingto m(�), then by setting Bk = ⋃k

i=1 Ai we have⋃∞

i=1 Ai =⋃∞k=1 Bk, with Bk ↑. To show that m(�) is a σ -field it therefore

suffices to show that m(�) is a field. Define

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MEASURES 13

= {A: A ε m(�), A′ ε m(�)}.We shall show that is monotone. Let Ai be in , with Ai ↑.We have

⋃Ai ε m(�) since m(�) is monotone. But A′

i are alsoin m(�) and A′

i ↓. Thus ∩i A′i ε m(�), that is (∪i Ai)

′ ε m(�).Thus ∪i Ai ε . Similarly if Ai ε and Ai ↓, then ∩Ai ε .Consequently is monotone. Since ⊃ � we deduce = m(�). This is exactly the statement that A ε m(�) impliesA′ ε m(�). Thus m(�) is closed under complementation andwe now proceed to show it is closed under unions. For A ε � letA = {B ε m(�): A∪B ε m(�)}. A is a monotone family. Forlet Bi ε A and Bi ↑ B. Each A∪Bi ε m(�) and A ∪ Bi ↑ A ∪ B,whence A ∪ B ε m(�) and thus B ε A. A ⊃ � since � isa field and therefore A = m(�). If A ε m(�) we can repeatthis argument, using what we have just proved to show in thepresent situation that A ⊃ �, and thus conclude again thatA = m(�). This is just the statement that A ε m(�) andB ε m(�) imply A ∪ B ε m(�). Thus m(�) is a field.

We now complete the proof of Theorem 5. If μ is afinite measure, it follows immediately from Lemma 1 (seeRemark 2) and Lemma 2 that the extension is unique. In caseμ is σ -finite

S =∞⋃

i=1

Ai, with Ai ε �, μ(Ai) < ∞.

We can assume without loss of generality that the Ai aredisjoint. Let μ1 and μ2 be two extensions of μ to σ(�). ForE ε σ (�) set

μij(E) = μj

(Ai

⋂E)

From the finite case it follows that μi1 = μi

2 for each i. Butμ1 = �iμ

i1, μ2 = �iμ

i2, whence the theorem is proved.

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14 LECTURES ON MEASURE AND INTEGRATION

The following simple example shows that the σ -finitenessof μ was essential to ensure the uniqueness of the extension.Again consider the space [A, B) and the field � of finite unionsof half open intervals. Define μ: � → R∗ by μ(∅) = 0and μ([a, b)) = ∞. μ is clearly a measure on �. We maydefine two extensions μ1 and μ2 by letting μ1 be the countingmeasure on σ(�) and μ2 the measure which is defined by

μ2(E) ={∞ if E �= ∅0 if E = ∅.

Suppose μ is a σ -finite measure on a σ -field � of subsetsof S. Then if we apply the procedure of the Hahn extensiontheorem, we get a σ -field �0 of measurable subsets. How are� and �0 related? We know � ⊂ �0, but in fact the inclusionmay be proper, and we shall see what new sets are obtained bythe process.

Define �0 = {E F: E ε �, F is a subset of some G ε �

with μ(G) = 0}. Here E F (read symmetric difference of Eand F) is (E − F) ∪ (F − E). �0 is easily seen to be a σ -field.Now set

μ0(E F) = μ(E).

(μ0, �0) is said to be the completion of (μ, �). Since everysubset of a set of μ-measure zero is μ∗-measurable, we havethat �0 ⊂ �0. The following theorem establishes the equalityof the two σ -fields.

THEOREM 6. Let μ be a σ -finite measure on a σ -field �,with �0 the σ -field of μ∗-measurable subsets of S and �0 thecompletion of �. Then �0 = �0.

Proof. Let E ε �0, and Ai,n a covering of E with μ∗(E) ≥�iμ(Ai,n) − 1

n and Ai,n ε �. Then

μ

( ⋃

i

Ai,n

)≤ μ∗(E) + 1

n

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MEASURES 15

and if A = ⋂n

⋃i Ai,n, then A ε � and

μ∗(E) ≤ μ(A) ≤ μ∗(E),

so that

μ∗(A − E) = 0.

We have E = A − (A − E). By replacing E with A − E andrepeating this construction we find a B ε �, B ⊃ A − E andμ(B) = 0, whence E ε �0 and �0 = �0.

Definition. The smallest σ -field containing all the closed setsof a given topological space S is called the Borel field of S, andthe sets in this field are called the Borel sets.

Consider the collection of subsets of R of the form [a, b),(−∞, a), and [a,∞) with a, b real. Finite unions of theseclearly form a field � of subsets of R. Define |[a, b)| =|b − a|, and |(−∞, a)| = |[a,∞)| = ∞. Then by aslight modification of Theorem 1, | |: � → R∗ is ameasure on �. Using the Hahn extension theorem | | maybe extended to the σ -field �0 of measurable sets. We call�0 the Lebesgue measurable sets of R and | | Lebesguemeasure. In a similar fashion we may define μφ on the samefield � and then extend to the collection of μφ-measurablesets. The measure thus obtained is called Lebegue-Stieltjesmeasure.

Remark. It follows from Theorem 6 that the Lebesgue measur-able sets may be defined as the sets in the completion of (| |,Borel sets).

It is not a difficult exercise to prove that every countableset is measurable with Lebesgue measure zero. Are there alsouncountable subsets of the real line which have measure zero?The answer is yes and the following is an example of sucha set.

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16 LECTURES ON MEASURE AND INTEGRATION

Consider the unit interval [0, 1]. From it remove the openmiddle third (1/3, 2/3). From [0, 1/3] ∪ [2/3, 1] remove themiddle thirds (1/9, 2/9), (7/9, 8/9). Continue in this fashionindefinitely. What remains of [0, 1] is known as the Cantor set,and is denoted by C. It is uncountable, since it consists of thosereals which have a ternary expansion consisting only of zerosand twos. Moreover μ(C) = 0, for

μ(C′) = 1/3 + 2/9 + 4/27 + · · · = 1.

C is closed since it is the complement of an open set. It cannothave any interior as otherwise it would contain points whoseternary expansions have ones in them. Thus C is also nowheredense.

We are now in a position to determine the cardinality ofthe class M of Lebesgue measurable sets of the real line. Thecardinality is at most 22ℵ0 since M is contained in the classof all subsets of R. But C is a set with measure zero, whenceevery subset of it is measurable. It follows that the cardinalityof M is 22ℵ0, since C has cardinality 2ℵ0 .

Remark. The class of Borel subsets of R, in contrast with theLebesgue sets, have cardinality 2ℵ0 .

THEOREM 7. Not every set is Lebesgue measurable.

Proof. Define an equivalence relation “∼” on [0, 1) byp ∼ q if and only if p − q ε Q, where Q denotes the rationalnumbers. This divides [0, 1) into disjoint equivalence classesAα, with [0, 1) = ∪α Aα. By the axiom of choice, there existsa set B which consists of one element from each Aα. B is not ameasurable set.

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MEASURES 17

To show this we assume on the contrary that B is measur-able with μ(B) = β. If we write Q = ∪{rn} then

[0, 1) =∞⋃

n=1

(B + rn) (mod 1)(4)

where

(B + rn)(mod 1)

= {x ε [0, 1): x ≡ y + rn(mod 1), y ε B}.The right hand side of (4) is a disjoint union. We leave it to thereader to verify that Lebesgue measure is translation invariant;i.e., μ(E + a) = μ(E) for E measurable and real a. Using thisfact it is clear that

1 = μ([0, 1)) =∞∑

n=1

μ [(B + rn) (mod 1)] =∞∑

n=1

μ(B).

If β �= 0 then 1 = ∞, and if β = 0 then 1 = 0, a contradictionin either case.

The complexity of the above construction indicates thatnon-measurable sets are indeed difficult to find, and in factnone have been constructed without the axiom of choice.

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ISBN-13:ISBN-10:

978-0-486-81028-70-486-81028-3

9 780486 810287

5 1 4 9 5

$14.95 USA

WWW.DOVERPUBLICATIONS.COM

MATHEMATICS

These well-known and concise lecture notes present the fundamentals of the Lebesgue theory of integration and an introduction to some

of the theory’s applications. Suitable for advanced undergraduates and graduate students of mathematics, the treatment also covers topics of interest to practicing analysts.

Author Harold Widom emphasizes the construction and properties of measures in general and Lebesgue measure in particular as well as the definition of the integral and its main properties. The notes contain chapters on the Lebesgue spaces and their duals, differentiation of measures in Euclidean space, and the application of integration theory to Fourier series.

Dover republication of the edition originally published by the Van Nostrand Reinhold Co., New York, 1969.

Cover design by John M

. Alves

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