measure theory master - Rhodes University€¦ · 2010 MEASURE THEORY ALP Introduction In...

MEASURE THEORY NOTES

(2010)

Mr. Andrew L Pinchuck

Department of Mathematics (Pure & Applied)

Rhodes University

Contents

Introduction 1

1 Riemann integration 2

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Basic definitions and theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Fundamental theorem of calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Measure spaces 14

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 Notation and preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.3 Measure on a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.4 Outer measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5 Lebesgue measure on R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3 The Lebesgue integral 31

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2 Measurable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2.1 Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2.2 Convergence of sequences of measurable functions . . . . . . . . . . . . . . . . . 36

3.3 Definition of the integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.3.1 Integral of a nonnegative simple function . . . . . . . . . . . . . . . . . . . . . . 41

3.3.2 Integral of a nonnegative measurable function . . . . . . . . . . . . . . . . . . . . 45

3.3.3 Integrable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.4 Lebesgue and Riemann integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.5 Lp spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.5.1 Holder and Minkowski inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 61

4 Decomposition of measures 66

4.1 Signed measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

1

2010 MEASURE THEORY ALP

Introduction

In mathematics, more specifically in measure theory, a measure on a set is a systematic way to assign

to each suitable subset a number, intuitively interpreted as the size of the subset. In this sense, a measure

is a generalization of the concepts of length, area, volume, etc. A particularly important example is the

Lebesgue measure on a Euclidean space, which assigns the conventional length, area and volume of Eu-

clidean geometry to suitable subsets of Rn. Most measures met in practice in analysis (and in many cases

also in probability theory) are Radon measures. Radon measures have an alternative definition in terms of

linear functionals on the locally convex space of continuous functions with compact support. In this course,

however, we will mainly concentrate on the Lebesgue measure.

Measure theory was developed in successive stages during the late 19th and early 20th century by

Emile Borel, Henri Lebesgue, Johann Radon and Maurice Frchet, among others. The main applications

of measures are in the foundations of the Lebesgue integral, in Andrey Kolmogorov’s axiomatisation of

probability theory and in ergodic theory. In integration theory, specifying a measure allows one to define

integrals on spaces more general than subsets of Euclidean space; moreover, the integral with respect to

the Lebesgue measure on Euclidean spaces is more general and has a richer theory than its predecessor,

the Riemann integral. Probability theory considers measures that assign to the whole set, the size 1, and

considers measurable subsets to be events whose probability is given by the measure. Ergodic theory

considers measures that are invariant under, or arise naturally from, a dynamical system. The importance

of integration and how measure theory puts integration and probability theory on an axiomatic foundation

is a principle motivation for the development of this theory. Measure theory is useful in functional analysis

in defining Lp function spaces, which cannot otherwise be defined properly.

These notes are intended to be an introduction to measure theory and integration. I make no claims of

originality with regards to this material, and I have used a number of different sources as references in the

compilation of these notes. Chapter 1 deals with the theory of Riemann integration and highlights some of

its shortcomings. In Chapter 2, we define the fundamental concepts of measure theory and present some

well-known, significant results. In Chapter 3, we define the Lebesgue integral and consider some results that

follow from this and lastly, in Chapter 4, we provide a brief introduction to the study of signed measures.

Throughout the course notes you will find exercises that need to be attempted for complete under-

standing of the text. Once again, I should mention that I have found wikipedia to be a useful resource

and recommend its use to students. All the course information and material can be found on the Rhodes

mathematics web site: www.ru.ac.za/mathematics by following the appropriate links.

1

Chapter 1

Riemann integration

1.1 Introduction

In the branch of mathematics known as real analysis, the Riemann integral, created by Bernhard Riemann

(1826-1866), was the first rigorous definition of the integral of a function on an interval. The Riemann

integral was clearly an improvement upon the way in which integration was initially conceived by Leibnitz

and Newton, i.e., simply as the anti-derivative. While the Riemann integral is one of the easiest integrals to

define, and has obvious uses in numerical analysis, it is unsuitable for many theoretical purposes. Some of

these technical deficiencies can be remedied by the RiemannStieltjes integral, and most of them disappear

in the Lebesgue integral. The idea behind the Riemann integral is to use very simple approximations for the

area of S. By taking better and better approximations, we can say that ‘in the limit’, we get exactly the area

of S under the curve.

1.2 Basic definitions and theorems

In this section we discuss the construction of the Riemann integral. This discussion reveals some of the

shortcomings of the Riemann integral and we present these as a motivation for the construction of the

Lebesgue integral.

1.2.1 Definition

Let Œa; b� be a closed interval in R. A partition of Œa; b� is a set P D fx0; x1; x2; : : : ; xng of points in R

such that

a D x0 < x1 < x2 < � � � < xn D b:

Let f be a real-valued function which is bounded on Œa; b� and let P D fx0; x1; x2; : : : ; xng be a partition

of Œa; b�. Denote by

M D supa�x�b

f .x/ and m D infa�x�b

f .x/:

Since f is bounded on Œa; b�, it is bounded on each subinterval Œxi�1; xi�, for each i D 1; 2; : : : ; n. Let

Mi D supff .x/ W xi�1 � x � xig and mi D infff .x/ W xi�1 � x � xig;

for each i D 1; 2; : : : ; n. Clearly,

m � mi � Mi � M; for each i D 1; 2; : : : ; n:

We can now form the sums

U.f;P / DnX

iD1

Mi.xi � xi�1/ and L.f;P / DnX

iD1

mi.xi � xi�1/:

2


1.2.2 Definition

The sums U.f;P / and L.f;P / are called, respectively, the upper and the lower sum of f relative to the

partition P .

1.2.3 Remark

[1] It is important to note that U.f;P / and L.f;P / depend on the partition P . If f is nonnegative on

Œa; b�, then the upper sum U.f;P / is the sum of the areas of the rectangles whose heights are Mi and

whose bases are Œxi�1; xi�. Similarly, L.f;P / is the sum of the areas of the rectangles whose heights

are mi , and whose bases are Œxi�1; xi�.

[2] It also follows that U.f;P / � L.f;P /.

1.2.4 Theorem

Let f be a real-valued function which is bounded on Œa; b� and let P D fx0; x1; x2; : : : ; xng be a partition

of Œa; b�. Then

m.b � a/ � L.f;P / � U.f;P / � M.b � a/:

PROOF.

Since Mi � M , for each i D 1; 2; : : : ; n, it follows that

U.f;P / DnX

iD1

Mi.xi � xi�1/ �nX

iD1

M.xi � xi�1/ D M

nX

iD1

.xi � xi�1/ D M.b � a/:

Similarly, since m � mi , for each i D 1; 2; : : : ; n, it follows that

L.f;P / DnX

iD1

mi.xi � xi�1/ �nX

iD1

m.xi � xi�1/ D m

nX

iD1

.xi � xi�1/ D m.b � a/:

2

This theorem says that the set A D fU.f;P / W P is a partition of Œa; b�g is bounded below by m.b � a/.

Hence, A has an infimum,†.f /, say. That is,

†.f / D infP

U.f;P /;

where the infimum is taken over all possible partitions P of Œa; b�. This theorem also shows that the set

B D fL.f;P / W P is a partition of Œa; b�g is bounded above by M.b � a/ and hence B has a supremum,

�.f /, say. That is,

�.f / D supP

L.f;P /;

where the supremum is taken over all possible partitions P of Œa; b�. It follows that

m.b � a/ � †.f / � M.b � a/; and

m.b � a/ � �.f / � M.b � a/:

1.2.5 Definition

Let f be a real-valued function which is bounded on Œa; b�. The upper integral of f on Œa; b� is defined by

bZ

a

f .x/dx D infP

U.f;P /;

3


and the lower integral of f on Œa; b� is defined by

bZ

a

f .x/dx D supP

L.f;P /;

where, once again, the infimum and supremum are taken over all possible partitions P of Œa; b�.

It is intuitively clear thatR b

af .x/dx �

R b

af .x/dx. Of course, this is not enough and we will therefore

prove this fact shortly.

1.2.6 Definition

Let P D fx0; x1; x2; : : : ; xng be a partition of Œa; b�. A partition P � of Œa; b� is called a refinement of

P , denoted by P � P �, if xi 2 P �, for each i D 0; 1; 2; : : : ; n. A partition P � is called a common

refinement of the partitions P1 and P2 of Œa; b� if P � is a refinement of both P1 and P2.

The next theorem states that refining a partition decreases the upper sum and increases the lower sum.

1.2.7 Lemma (Refinement Lemma)

Let f be a real-valued function which is bounded on Œa; b�. If P � is a refinement of a partition P Dfx0; x1; x2; : : : ; xng of Œa; b�, then

L.f;P / � L.f;P �/ � U.f;P �/ � U.f;P /:

PROOF.

Suppose that P � has one more point that P , say a point x� which lies in the subinterval Œxr�1; xr �. Let

L1 D supff .x/ W xr�1 � x � x�g;L2 D supff .x/ W x� � x � xr g and

l1 D infff .x/ W xr�1 � x � x�g; l2 D infff .x/ W x� � x � xrg:

Recalling that

Mr D supff .x/ W xr�1 � x � xr g; and mr D infff .x/ W xr�1 � x � xrg;

we observe that

mr � l1;mr � l2;L1 � Mr ; and L2 � Mr :

It follows that

mr .xr � xr�1/ D mr .xr � x�/C mr .x� � xr�1/ � l2.xr � x�/C l1.x

� � xr�1/:

Hence,

L.f;P �/ Dr�1X

jD1

mj .xj � xj�1/C l1.x� � xr�1/C l2.xr � x�/C

nX

jDrC1

mj .xj � xj�1/

�r�1X

jD1

mj .xj � xj�1/C mr .xr � xr�1/CnX

jDrC1

mj .xj � xj�1/

DnX

jD1

mj .xj � xj�1/ D L.f;P /:

Similarly,

Mr .xr � xr�1/ D Mr .xr � x�/C Mr .x� � xr�1/ � L2.xr � x�/C L1.x

� � xr�1/;

4


and so

U.f;P / Dr�1X

jD1

Mj .xj � xj�1/C L1.x� � xr�1/C L2.xr � x�/C

nX

jDrC1

Mj .xj � xj�1/

�r�1X

jD1

Mj .xj � xj�1/C Mr.xr � xr�1/CnX

jDrC1

Mj .xj � xj�1/

DnX

jD1

Mj .xj � xj�1/ D U.f;P /:

the case where P � contains k � 2 more points than P can be proved by simply repeating the above

argument k times.

2

1.2.8 Theorem

Let f be a real-valued function which is bounded on Œa; b�. Then,

bZ

a

f .x/dx �bZ

a

f .x/dx:

PROOF.

Let P1 and P2 be any two partitions of Œa; b� and let P � be their common refinement. Then by Lemma

1.2.7,

L.f;P1/ � L.f;P �/ � U.f;P �/ � U.f;P2/:

Since P1 is any partition of Œa; b�, it follows that

supP

L.f;P / � U.f;P2/;

and since P2 is any partition of Œa; b�, we have that

supP

L.f;P / � infP

U.f;P /;

where the infimum and the supremum are taken over all possible partitions P of Œa; b�. Thus,

bZ

a

f .x/dx �bZ

a

f .x/dx:

2

1.2.9 Remark

Implicit in the proof of Theorem 1.2.8, is the fact that no lower sum can exceed an upper sum. That is,

every lower sum is less than or equal to every upper sum.

1.2.10 Definition

Let f be a real-valued function on Œa; b�. We say that f is Riemann-integrable on Œa; b� if f is bounded

on Œa; b� and

bZ

a

f .x/dx DbZ

a

f .x/dx:

5


If f is Riemann-integrable on Œa; b�, we define the integral of f on Œa; b� to be the common value of the

upper and lower integrals, i.e,

bZ

a

f .x/dx DbZ

a

f .x/dx DbZ

a

f .x/dx:

We shall denote by RŒa; b� the collection of all functions that are Riemann-integrable on Œa; b�.

1.2.11 Remark

In the definition of the integral of f on Œa; b�, we have tacitly assumed that a < b. If a D b, thenR b

af .x/dx D

R a

af .x/dx D 0. Also if b < a, then define

aZ

b

f .x/dx D �bZ

a

f .x/dx:

1.2.12 Examples

[1] Show that if f is a constant function on Œa; b�, then f 2 RŒa; b� and find it’s integral.

Solution: Let f .x/ D k , for all x 2 Œa; b� and let P D fx0; x1; x2; : : : ; xng by any partition of Œa; b�.

Then

Mi D supff .x/ W xi�1 � x � xig D k and mi D infff .x/ W xi�1 � x � xig D k;

for each i D 1; 2; : : : ; n. Therefore,

U.f;P / DnX

iD1

Mi.xi � xi�1/ D k

nX

iD1

.xi � xi�1/ D k.b � a/ and

L.f;P / DnX

iD1

mi.xi � xi�1/ D k

nX

iD1

.xi � xi�1/ D k.b � a/

Since P is any partition of Œa; b�, it follws that U.f;P / D L.f;P / D k.b � a/, for all partitions P

of Œa; b�. Therefore,

bZ

a

f .x/dx D k.b � a/ DbZ

a

f .x/dx:

That is, f is integrable on Œa; b� and

bZ

a

f .x/dx D k.b � a/:

[2] Let f be a function defined by

[3]

f .x/ D�

1 if x 2 Q \ Œ0; 1�0 if x 62 Q \ Œ0; 1�:

Show that f is not Riemann-integrable on Œ0; 1�.

6


Solution: Firstly we observe that f is bounded on Œ0; 1�. Let P D fx0; x1; x2; : : : ; xng be any

partition of Œ0; 1�. Since for each i D 1; 2; : : : ; n, the subinterval Œxi�1; xi� contains both rational and

irrational numbers, we have that

Mi D supff .x/ W xi�1 � x � xig D supf0; 1g D 1; and

mi D infff .x/ W xi�1 � x � xig D inff0; 1g D 0;

for each i D 1; 2; : : : ; n. Thus,

U.f;P / DnX

iD1

Mi.xi � xi�1/ D .1/

nX

iD1

.xi � xi�1/ D 1 � 0 D 1, and

L.f;P / DnX

iD1

mi.xi � xi�1/ D .0/

nX

iD1

.xi � xi�1/ D 0:

Since P is any partition of Œ0; 1�, it follows that U.f;P / D 1 � 0 D 1 and L.f;P / D 0, for all

partitions of P of Œ0; 1�. Therefore,

1Z

0

f .x/dx D 0 and

1Z

0

f .x/dx D 1;

and hence f is not Riemann-integrable on Œ0; 1�.

Notice that in the last example, the given function f is discontinuous at every point.

1.2.13 Theorem (Darboux’s integrability condition)

Let f be a real-valued function which is bounded on Œa; b�. Then f is integrable on Œa; b� if and only if, for

any � > 0, there exists a partition P � of Œa; b� such that

U.f;P �/ � L.f;P �/ < �:

PROOF.

Assume that f is integrable on Œa; b� and let � > 0. Since

bZ

a

f .x/dx DbZ

a

f .x/dx D supP

L.f;P /;

there is a partition P1 of Œa; b� such that

bZ

a

f .x/dx � �

2< L.f;P1/:

Again, since

bZ

a

f .x/dx DbZ

a

f .x/dx D infP

U.f;P /;

there exists a partition P2 of Œa; b� such that

U.f;P2/ <

bZ

a

f .x/dx C �

2:

7


Let P � be a common refinement of P1 and P2. Then

bZ

a

f .x/dx � �

2< L.f;P1/ � L.f;P �/ � U.f;P �/ � U.f;P2/ <

bZ

a

f .x/dx C �

2:

It now follows that

U.f;P �/ � L.f;P �/ < �:

For the converse, assume that given � > 0, there is a partition P � of Œa; b� such that

U.f;P �/ � L.f;P �/ < �:

Then,bZ

a

f .x/dx D infP

U.f;P / � U.f;P �/; and

bZ

a

f .x/dx D supP

L.f;P / � L.f;P �/:

Thus,

0 �bZ

a

f .x/dx �bZ

a

f .x/dx � U.f;P �/� L.f;P �/ < �:

Since � > 0 is arbitrary, we have that

bZ

a

f .x/dx DbZ

a

f .x/dx:

That is, f 2 RŒa; b�.

2

We now highlight the following important fact which is contained in the first part of the proof of Dar-

boux’s integrability condition:

1.2.14 Theorem

If f is integrable on Œa; b�, then for each � > 0, there exists a partition P of Œa; b� such that

bZ

a

f .x/dx � � < L.f;P / � U.f;P / <

bZ

a

f .x/dx C �:

1.2.15 Theorem

If f is continuous on Œa; b�, then it is integrable there.

PROOF.

Since f is continuous on Œa; b�, we have that f is bounded on Œa; b�. Furthermore, f is uniformly

continuous on Œa; b�. Hence, given � > 0, there is a ı > 0 such that

jf .x/ � f .y/j < �

b � a, whenever x; y 2 Œa; b� and jx � yj < ı:

Let P D fx0; x1; x2; : : : ; xng be any partition of Œa; b� such that xi � xi�1 < ı, for each i D 1; 2; : : : ; n.

By the Extreme-Value Theorem (applied to f on Œxi�1; xi�, for each i D 1; 2; : : : ; n), there exist points tiand si in Œxi�1; xi�, for each i D 1; 2; : : : ; n such that

f .ti / D supff .x/ W xi�1 � x � xig D Mi and f .si/ D infff .x/ W xi�1 � x � xig D mi :

8


Since xi � xi�1 < ı, it follows that jti � si j < ı, and so

Mi � mi D f .ti /� f .si / D jf .ti / � f .si/j <�

b � a; for all i D 1; 2; : : : ; n:

Thus,

U.f;P / � L.f;P / DnX

iD1

Mi.xi � xi�1/�nX

iD1

mi.xi � xi�1/ DnX

iD1

.Mi � mi/.xi � xi�1/

�nX

iD1

jf .ti / � f .si/j.xi � xi�1/ <

nX

iD1

�

�

b � a

�

.xi � xi�1/

D�

b � a.b � a/ D �:

It now follows from Theorem 1.2.13, that f is integrable on Œa; b�.

2

1.2.16 Theorem

If f is monotone on Œa; b�, then f is integrable there.

PROOF.

Assume that f is monotone increasing on Œa; b� and f .a/ < f .b/. Since f .a/ � f .x/ � f .b/, for

all x 2 Œa; b�, f is clearly bounded on Œa; b�. We want to show that, given any � > 0, there is a partition

P of Œa; b� such that U.f;P / � L.f;P / < �. Let � > 0 be given an let P D fx0; x1; x2; : : : ; xng be any

partition of Œa; b� such that xi � xi�1 <�

f .b/�f .a/, for each i D 1; 2; : : : ; n. Since f is increasing on Œa; b�,

we have that

Mi D supff .x/ W xi�1 � x � xig D f .xi/ and mi D infff .x/ W xi�1 � x � xig D f .xi�1/;

for each i D 1; 2; : : : ; n. Hence,

U.f;P / � L.f;P / DnX

iD1

Mi.xi � xi�1/�nX

iD1

mi .xi � xi�1/ DnX

iD1

.Mi � mi/.xi � xi�1/

�nX

iD1

Œf .xi/� f .xi�1/�.xi � xi�1/ <�

f .b/ � f .a/

nX

iD1

Œf .xi /� f .xi�1/�

D �

f .b/ � f .a/ Œf .b/ � f .a/� D �:

It then follows from Theorem 1.2.13, that f is Riemann-integrable on Œa; b�. The case where f is monotone

decreasing can be proved in exactly the same way.

2

1.2.17 Theorem

If f is integrable on Œa; b� and a � c < d � b, then f is integrable on Œc; d �.

PROOF.

Since f is integrable on Œa; b�, it is bounded there. Hence f is bounded on Œc; d �. Furthermore, given

any � > 0, there is a partition P of Œa; b�, such that

U.f;P / � L.f;P / < �:

Let P � D P [ fc; dg. Then P � is a refinement of P , and hence

U.f;P �/ � L.f;P �/ � U.f;P / � L.f;P / < �:

9


Let Q1 D P � \ Œa; c�;Q2 D P � \ Œc; d �;Q3 D P � \ Œd; b�. Then P � D Q1 [ Q2 [ Q3, and so,

U.f;P �/ D U.f;Q1/C U.f;Q2/C U.f;Q3/; and

L.f;P �/ D L.f;Q1/C L.f;Q2/C L.f;Q3/:

Hence,

ŒU.f;Q1/� L.f;Q1/�C ŒU.f;Q2/� L.f;Q2/�C ŒU.f;Q3/� L.f;Q3/� D U.f;P �/� L.f;P �/ < �:

Note that all terms on the left are nonnegative. Therefore Q2 is a partition of Œc; d � with the property that

U.f;Q2/ � L.f;Q2/ < �:

This implies that f is integrable on Œc; d �.

2

1.2.18 Corollary

If f is integrable on Œa; b� and a < c < b, then f is integrable on both Œa; c� and Œc; b�.

The following theorem states that the converse of Corollary 1.2.18 also holds.

1.2.19 Theorem

If a < c < b and f is integrable on both Œa; c� and Œc; b�, then f is integrable on Œa; b� and

bZ

a

f .x/dx DcZ

a

f .x/dx CbZ

c

f .x/dx:

PROOF.

Let � > 0 be given. Since f is integrable on Œa; c� and on Œc; b�, there are partitions P1 and P2 of Œa; c�

and Œc; b� respectively, such that

U.f;P1/ � L.f;P1/ <�

2; and

U.f;P2/� L.f;P2/ <�

2:

Let P D P1 [ P2. Then P is a partition of Œa; b� and

U.f;P / � L.f;P / D U.f;P1/C U.f;P2/� L.f;P1/ � L.f;P2/

D U.f;P1/ � L.f;P1/C U.f;P2/ � L.f;P2/ <�

2C �

2D �:

Therefore f is integrable on Œa; b�. Furthermore,

bZ

a

f .x/dx � U.f;P / D U.f;P1/C U.f;P2/

<

�

L.f;P1/C �

2

�

C�

L.f;P2/C �

2

�

D L.f;P1/C L.f;P2/C �

�cZ

a

f .x/dx CbZ

c

f .x/dx C �:

10


Since � is arbitrary, it follows that

bZ

a

f .x/dx �cZ

a

f .x/dx CbZ

c

f .x/dx (1.1)

Also,bZ

a

f .x/dx � L.f;P / D L.f;P1/C L.f;P2/

>

�

U.f;P1/ � �

2

�

C�

U.f;P2/� �

2

�

D U.f;P1/C U.f;P2/ � �Since � is arbitrary, it follows that

bZ

a

f .x/dx �cZ

a

f .x/dx CbZ

c

f .x/dx: (1.2)

Combining (1.1) and (1.2), we have that

bZ

a

f .x/dx DcZ

a

f .x/dx CbZ

c

f .x/dx:

2

1.2.20 Corollary

Let f be defined on Œa; b� and suppose that a D c0 < c1 < � � � < cn�1 < cn D b. Then f is integrable on

Œa; b� if and only if f is integrable on Œck�1; ck �, for each k D 1; 2; : : : ; n. In this case,

bZ

a

f .x/dx DnX

kD1

ckZ

ck�1

f .x/dx:

1.2.21 Corollary

If f is continuous at all but a finite set of points in Œa; b�, then f is Riemann-integrable on Œa; b�.

1.3 Fundamental theorem of calculus

1.3.1 Theorem (Fundamental theorem of calculus)

Let f be a bounded real-valued integrable function on Œa; b�. Suppose that F W Œa; b�! R is continuous on

Œa; b� and differentiable on .a; b/ and that

F 0.x/ D f .x/; for each x 2 .a; b/:

Then,bZ

a

f .x/dx D F.b/ � F.a/:

11


PROOF.

Let P D fx0; x1; x2; : : : ; xng be a partition of Œa; b�.

Mi D supff .x/ W xi�1 � x � xig and mi D infff .x/ W xi�1 � x � xig;

for each i D 1; 2; : : : ; n. For each i D 1; 2; : : : ; n, F is continuous on Œxi�1; xi� and differentiable on

.xi�1; xi/. By the Mean Value Theorem, there is a �i 2 .xi�1; xi/ such that

F.xi / � F.xi�1/ D F 0.�i/.xi�1 � xi/ D f .�i /.xi�1 � xi/:

Since, for each i D 1; 2; : : : ; n mi � f .�i / � Mi , it follows that

mi.xi � xi�1/ � f .�i/.xi�1 � xi/ � Mi.xi � xi�1/ , mi.xi � xi�1/ � F.xi / � F.xi�1/ � Mi.xi � xi�1/

)nX

iD1

mi.xi � xi�1/ �nX

iD1

ŒF.xi/ � F.xi�1/� �nX

iD1

Mi .xi � xi�1/

, L.f;P / � F.b/ � F.a/ � U.f;P /:

Since P is an arbitrary partition of Œa; b�, it follows that F.b/�F.a/ is an upper bound for the set fL.f;P / WP is a partition of Œa; b�g and a lower bound for the set fU.f;P / W P is a partition of Œa; b�g. Therefore,

bZ

a

f .x/dx D supP

L.f;P / � F.b/ � F.a/ � infP

U.f;P / DbZ

a

f .x/dx:

Since f is integrable on Œa; b�, we have that

bZ

a

f .x/dx DbZ

a

f .x/dx DbZ

a

f .x/dx;

and consequently

bZ

a

f .x/dx D F.b/ � F.a/

2

1.3.2 Exercise

[1] Let f be the function on Œ0; 1� given by

f .x/ D�

x if x is rational

0 if x is irrational:

Show that f is not Riemann-integrable on Œ0; 1�.

[2] Let f be the function on Œ0; 1� given by

f .x/ D�

1 if 0 � x < 12

x � 12

if 12

� x < 1:

(a) Show, from first principles, that f is Riemann-integrable on Œ0; 1�.

12


(b) Quote a result that assures us that f is Riemann-integrable.

(c) Find

1Z

0

f .x/dx:

(d) Let fr1; r2; r3; : : : g be an enumeration of rationals in the interval Œ0; 1�. For each n 2 N, define

fn.x/ D�

1 if x 2 fr1; r2; r3; : : : ; rng0 otherwise:

Show that .fn/ is a nondecreasing sequence of functions that are Riemann-integrable on Œ0; 1�.

Show also that the sequence .fn/ converges pointwise to the function

f .x/ D�

1 if x 2 Q \ Œ0; 1�0 otherwise;

and that f is not Riemann-integrable.

13

Chapter 2

Measure spaces

2.1 Introduction

We now concern ourselves with the idea of a defining a measure on the real numbers, which puts the intuitive

idea of length on an axiomatic foundation. This turns out to be more difficult than you might expect and we

begin with the motivation for developing measure theory. If we are to properly define the definite integral

of a real-valued function, we need a meaningful notion of the length of a set A � R.

If we consider the set of real numbers, we would like to be able to define a function that measures the

‘length’ of a set. That is a function m W P.R/ ! Œ0;1� such that:

[1] �.Œa; b�/ D b � a, for a � b.

[2] � is translation invariant, i.e., �.A/ D m.x C A/, for A � R and x 2 R.

[3] �.;/ D 0.

[4] If .Ai / is a sequence of mutually disjoint sets in .R/ (i.e., Ai \ Aj D ;, for every i 6D j ), then

�

� 1[

iD1

Ai

�

D1X

iD1

Ai :

14


This turns out to be impossible. That is, you cannot define such a function whose domain consists of

the whole of P.R/. To show this we present the following example by Giuseppe Vitali (1905):

2.1.1 Example

A Vitali set V , is a subset of Œ0; 1�, which for each real number r , V contains exactly one number v such

that v� r is rational (This implies that V is uncountable, and also, that v� u is irrational, for any u; v 2 V ,

u 6D v). Such sets can be shown to exist provided we assume the axiom of choice.

To construct the Vitali set V , consider the additive quotient group R=Q. Each element of this group is

a ‘shifted copy’ of the rational numbers. That is, a set of the form Q C r , for some r 2 R. Therefore, the

elements of this group are subsets of R and partition R. There are uncountably many elements. Since each

element intersects Œ0; 1�, we can use the axiom of choice to choose a set V � Œ0; 1� containing exactly one

representative out of each element R=Q.

A Vitali set is non-measurable. To show this, we argue by contradiction and assume that V is measur-

able. Let q1; q2; : : : be an enumeration of the rational numbers in Œ�1; 1�. From the construction of V , note

that the translated sets Vk D V C qk D fv C qk W v 2 V g, k D 1; 2; : : : are pairwise disjoint, and further

note that Œ0; 1� �S

k Vk � Œ�1; 2� (to see the first inclusion, consider any real number r in Œ0; 1� and let v

be the representative in V for the equivalence class Œr �, the r �v D q for some rational number q in Œ�1; 1�).

Now assume that there exists a � that satisfies the four conditions listed above. Apply � to these

inclusions using condition [4]:

1 �1X

kD1

�.Vk / � 3:

Since � is translation invariant, �.Vk / D �.V / and

1 �1X

kD1

�.V / � 3:

This is impossible. Summing infinitely many copies of the constant �.V / yields either zero or infinity,

according to whether the constant is zero or positive. In neither case is the sum in Œ1; 3�. So, V cannot have

been measurable after all, i.e., � must not define any value for �.V /.

It is important to stress that this whole argument depends on the assumption of the axiom of choice.

2.2 Notation and preliminaries

Let X be a set.

� We will use the standard notation for the compliment of A:

Ac D fx 2 X W x 62 Ag

� We denote the complement of B with respect to A by AnB D A \ Bc .

� We denote by P.X /, the powerset of X , i.e., the set of all subsets of X .

� Let A and B be in P.X /. The symmetric difference of A and B, denoted by A4B, is given by

A4B D .AnB/ [ .BnA/:

� A subset C of P.X / is said to be:

15


[1] closed under finite intersections if, whenever fA1;A2; : : : ;Ang is a finite collection of elements

of C, then

n\

jD1

Aj 2 C.

[2] closed under finite unions if, whenever fA1;A2; : : : ;Ang is a finite collection of elements of C,

then

n[

jD1

Aj 2 C.

[3] closed under countable intersections if, whenever .An/ is a sequence of elements of C, then1\

jD1

Aj 2 C.

[4] closed under countable unions if, whenever .An/ is a sequence of elements of C, then

1[

jD1

Aj 2

C.

� Let .An/ be a sequence in P.X /. We denote by

lim supn!1

An D1\

nD1

1[

mDn

Am, and lim infn!1

An D1[

nD1

1\

mDn

Am:

2.2.1 Exercise

Let X be a nonempty set. Prove that if a collection C of P.X / is closed under countable intersections (resp.,

unions), then it is closed under finite intersections (resp., unions).

2.2.2 Definition

An algebra (in X ) is a collection† of subsets of X such that:

(a) ;;X 2 †.

(b) If A 2 †, then Ac 2 †.

(c) If A1; : : : ;An 2 †, then [niD1

Ai and \niD1

Ai are in†.

† is a �-algebra (or �-field) if in addition:

(d) If A1;A2; : : : are in †, then [1iD1

Ai and \1iD1

Ai are in †.

2.2.3 Examples

[1] Let X be any nonempty set. Then there are two special �-algebras on X : †1 D f;;X g and †2 DP.X / (the power set of X ).

[2] Let X D Œ0; 1�. Then † D f;;X; Œ0; 12�; .1

2; 1�g is a �-algebra on X .

[3] Let X D R and let

† D fA � R W A is countable or Ac is countableg:† is a �-algebra on X . Parts (a) and (b) of the definition are easy to show. Suppose A1;A2; : : : are

all in†. If each of the Ai’s are countable, then [iAi is countable,and so in†. If Aci0

is countable for

some i0, then

.[iAi/c D \iA

ci � Ac

i0

is countable, and again [iAi is in †. Since \iAi D .[iAci /

c , then the countable intersection of sets

in † is again in †.

2.2.4 Remark

Let X D R and let † be a �-algebra on X . If .An/ is a sequence of sets in †, then lim infn!1 An 2 †.

16


If S is any collection of subsets of a set X , then by taking the intersection of all �-algebras containingS,

we obtain the smallest �-algebra containing S. Such a �-algebra, denoted by �.S/ is said to be generated

by S. It is clear that if S is a �-algebra, then �.S/ D S.

2.2.5 Definition

The smallest �-algebra conataining all open subsets of R is called the Borel �-algebra of R and is denoted

by B.R/. Elements of this �-algebra are called Borel sets.

2.2.6 Proposition

Let I D fŒa; b/ W a; b 2 R; a � bg. Then �.I/ D B.R/.

PROOF.

If a; b 2 R and a � b, then since

Œa; b/ D1\

nD1

�

a � 1

n; b�

;

it follows that Œa; b/ 2 B.R/ and consequently, I � B.R/. It now follows that �.I/ � B.R/.

Conversely, let A be an open subset of R. Then A is a countable union of open intervals. Since any

open interval .a; b/ can be expressed in the form

.a; b/ D1[

nDn0

h

a C 1

n; b�

;

where 1n0< b � a, it follows that A 2 �.I/. Thus �.I/ D B.R/.

2

2.2.7 Exercise

Let X be a nonempty set and S the smallest �-algebra in X which contains all singletons fxg. Show that

S D fA 2 P.X / W either A or Ac is countableg:

2.3 Measure on a set

2.3.1 Definition

A measurable space is a pair .X; †/, where X is a set and † is a �-algebra on X . The elements of the

�-algebra † are called measurable sets.

2.3.2 Definition

Let .X; †/ be a measurable space. A measure on .X; †/ is a function � W † ! Œ0;1� such that:

(a) �.;/ D 0, and

(b) if .Ai / is a sequence of mutually disjoint sets in † (i.e., Ai 2 †, for each i D 1; 2; 3; : : : , and

Ai \ Aj D ; for every i 6D j ), then

�

� 1[

iD1

Ai

�

D1X

iD1

Ai :

Property (b) in the above definition is known as the countable additivity property of the measure �. If

(b) is replaced by the condition that �.Sn

iD1 Ai / DPn

iD1 Ai for every finite sequence A1;A2; : : : ;An of

pairwise disjoint sets in †, then we say that � is finitely additive. It follows immediately, that if a measure

17


is countably additive, then it must also be finitely additive. If you have a finite sequence A1;A2; : : : ;An in

†, then you can consider it to be an infinite sequence .Ai/ with AnC1 D AnC2 D : : : ;. Then

�

� n[

iD1

Ai

�

D �

� 1[

iD1

Ai

�

D1X

iD1

Ai DnX

iD1

Ai :

2.3.3 Definition

A measure space is a triple .X; †; �/, where X is a set, † a �-algebra in X , and � a measure on †.

2.3.4 Remark

Let .X; †; �/ be a measure space. If �.X / D 1, then .X; †; �/ is called a probability space and � is

called a probability measure.

2.3.5 Examples

[1] Let X be a nonempty set and † D P.X /. Define � W † ! Œ0;1� by

�.A/ D�

n if A has n elements

1 otherwise:

Then � is a measure on X called the counting measure.

[2] Let .X; †/ be a measurable space. Choose and fix x 2 X . Define �x W † ! Œ0;1� by

�x.A/ D�

1 if x 2 A

0 otherwise:

Then �x is a measure on X called the unit point mass at x.

2.3.6 Theorem

Let .X; †; �/ be a measure space. Then the following statements hold:

[1] If A;B 2 † and A � B, then �.A/ � �.B/.

[2] If .An/ is a sequence in †, then

�

� 1[

nD1

An

�

�1X

nD1

�.An/:

[3] If .An/ is a sequence in † such that An � AnC1, for each n 2 N, then

�

� 1[

nD1

An

�

D limn!1

�.An/:

[4] If .An/ is a sequence in † such that An � AnC1, for each n 2 N and �.A1/ < 1, then

�

� 1\

nD1

An

�

D limn!1

�.An/:

PROOF.

[1] Write B D A [ .BnA/, a disjoint union. Since � is finitely additive, we have that

�.B/ D �.A/C �.BnA/ � �.A/:

18


[2] Set B1 D A1, and for each natural number n > 1, let Bn D AnnSn�1

iD1 Ai . Then

Bn Dn�1\

iD1

.AnnAi / D An

n�1\

iD1

Aci :

Therefore, for each n D 1; 2; 3; : : : , we have that Bn 2 †. Clearly, Bn � An, for each n D1; 2; 3; : : : . Let l and k be integers such that l < k . Then Bl � Al , and so

Bl \ Bk � Al \ Bk D Al \ Ak

k�1\

iD1

Aci

D Al \ Ak \ � � � \ Acl \ � � �

D Al \ Acl \ Ak \ � � � \ � � � D ; \ Ak \ � � � \ � � �

D ;:

That is, Bi

T

i 6Dj Bj D ;:Claim:

S1nD1 An D

S1nD1 Bn:

Since Bn � An, for each n D 1; 2; 3; : : : ; we have thatS1

nD1 Bn �S1

nD1 An.

Let x 2S1

nD1 An: Then x 2 An for some n. Let k be the smallest index such that x 2 An. Then

x 2 Ak and x 62 Aj , for each j D 1; 2; : : : ; k � 1. Therefore, x 2 Acj , for each j D 1; 2; : : : ; k � 1.

That is, x 2 Bk D Ak

Tk�1jD1 Ac

j , whence x 2S1

nD1 Bn, which proves the claim.

Therefore,

�

� 1[

nD1

An

�

D �

� 1[

nD1

Bn

�

D1X

nD1

�.Bn/ �1X

nD1

�.An/;

by part [1].

[3] Let B1 D A1 and Bn D AnnAn�1 for all n > 1. Then Bi

T

i 6Dj Bj D ;;Ak DSk

nD1 Bn andS1

nD1 An DS1

nD1 Bn. Therefore,

�

� 1[

nD1

An

�

D �

� 1[

nD1

Bn

�

D1X

nD1

�.Bn/ D limk!1

kX

nD1

�.Bn/ D limk!1

�.Ak /:

[4] Let A DT1

nD1 An. Then

A1nA D A1n1\

nD1

An D1[

nD1

.A1nAn/;

and A1nAn � A1nAnC1 , for each natural number n. Therefore, by [3] above,

�.A1nA/ D �

� 1[

nD1

.A1nAn/

�

D limn!1

�.A1nAn/: (2.1)

Since A � A1 and �.A1/ < 1, we have that �.A/ < 1. Also, since An � A1, for each natural

number n, it follows that �.A/ < 1, for each n. Writing A1 as a disjoint union A1 D .A1nA/ [ A,

we have that �.A1/ D �.A1nA/C �.A/. It now follows that

�.A1nA/ D �.A1/ � �.A/: (2.2)

19


Similarly, by expressing A1 as A1 D .A1nAn/[ An, we get that

�.A1nAn/ D �.A1/ � �.An/: (2.3)

From equations 2.1, 2.2 and 2.3, we have that

�.A1/ � �.A/ D limn!1

Œ�.A1/ � �.An/� D �.A1/� limn!1

�.An/:

Since �.A1/ < 1, we have that

�.A/ D limn!1

�.An/; that is �

� 1\

nD1

An

�

D limn!1

�.An/:

2

2.3.7 Theorem (Borel-Cantelli Lemma)

Let .X; †; �/ be a measure space. If .An/ is a sequence of measurable sets such thatP1

nD1 �.An/ < 1,

then �

�

lim supn!1 An

�

D 0.

PROOF.

Since lim supn!1 An DT1

nD1

S1kDn Ak , it follows that for each n 2 N,

�

�

lim supn!1

An

�

� �

� 1[

nD1

Ak

�

�1X

nD1

�.Ak / ! 0

as n ! 1.

2

2.3.8 Definition

Let .X; †; �/ be a measure space. The measure � is said to be:

[1] finite if �.X / < 1, and

[2] �-finite if there is a sequence .Xn/ of sets in†with�.Xn/ < 1, for each n such that X DS1

nD1 Xn.

2.4 Outer measure

Constructing a measure is a nontrivial exercise as illustrated in the introduction to this chapter. The basic

idea that we will use is as follows: We will begin by defining an ‘outer measure’ on all subsets of a set.

Then we will restrict the outer measure to a reasonable class of subsets and in so doing we will define a

measure. This is indeed how Lebesgue solved the problems associated with measure and integration. Both

were published as part of his dissertation in 1902.

2.4.1 Definition

Let X be a set. An outer measure on X is a set function �� W P.X / ! Œ0:1� satisfying the following

properties:

(a) ��.;/ D 0:

(b) If A � B, then ��.A/ � ��.B/.

(c) For any sequence .An/ of subsets of X , ��

�

S1nD1 An

�

�P1

nD1 ��.An/:

Property (b) above is referred to as monotonicity, and property (c) is countable subadditivity

20


2.4.2 Examples

[1] Let X be a set. Define �� on P.X / by:

��.A/ D�

1 if A 6D ;0 if A D ;:

Then �� is an outer measure on X .

[2] Let X be an uncountable set. Define �� on P.X / by:

��.A/ D�

0 if A is countable

1 if A is uncountable:

Then �� is an outer measure on X .

We can now use the concept of outer measure to construct a measure.

2.4.3 Definition

Let X be a set and �� an outer measure on X . A subset E of X is said to be ��-measurable if, for every

subset A of X , we have

��.A/ D ��.A \ E/C ��.A \ Ec/:

2.4.4 Remark

[1] Since A D .A \ E/[ .A \ Ec/ and �� is subadditive, we always have that

��.A/ � ��.A \ E/C ��.A \ Ec/:

Therefore, a subset E of X is ��-measurable if, for every subset A of X ,

��.A/ � ��.A \ E/C ��.A \ Ec/:

[2] Replacing E by Ec in the definition of ��-measurability, we have that

��.A/ D ��.A \ Ec/C ��ŒA \ .Ec/c � D ��.A \ Ec/C ��.A \ E/:

Thus, E is ��-measurable if and only if Ec is ��-measurable.

2.4.5 Lemma

Let X be a set, �� an outer measure on X and E a subset of X . If ��.E/ D 0, then E is ��-measurable.

PROOF.

Let A be any subset of X . Then A \ E � E. Since �� is monotone, ��.A \ E/ � ��.E/ D 0. Also,

since A \ Ec � A, it follows that ��.A \ Ec/ � ��.A/. Therefore,

��.A/ � ��.A \ E/C ��.A \ Ec/;

and therefore, E is ��-measurable.

2

2.4.6 Corollary

The empty set ; is ��-measurable.

2.4.7 Lemma

Let X be a set, �� an outer measure on X , and E and F ��-measurable subsets of X . Then E [ F is

��-measurable.

21


PROOF.

Let A be any susbset of X . Then since E is ��-measurable,

��.A/ D ��.A \ E/C ��.A \ Ec/:

Also, since F is ��-measurable,

��.A \ Ec/ D ��Œ.A \ Ec/\ F �C ��Œ.A \ Ec/ \ F c �:

From the two equations above we get:

��.A/ D ��.A \ E/C ��Œ.A \ Ec/ \ F �C ��Œ.A \ Ec/ \ F c �:

Therefore,��.A/ D ��.A \ E/C ��Œ.A \ Ec/ \ F �C ��Œ.A \ Ec/ \ F c �

� ��Œ.A \ E/[ ..A \ Ec/\ F/�C ��Œ.A \ Ec/ \ F c �

D ��ŒA \ .E [ F/�C ��ŒA \ .E [ F/c �;

whence E [ F is �� measurable.

2

2.4.8 Theorem

Let X be a set and �� an outer measure on X . Denote by M, the collection of all ��-measurable subsets

of X . Then M is an algebra on X .

PROOF.

This follows from Remark 2.4.4 [2], Corollary 2.4.6 and Lemma 2.4.7

2

2.4.9 Proposition

Let X be a set and �� an outer measure on X . If E1;E2 2 M and E1 \ E2 D ;, then

��ŒA \ .E1 [ E2/� D ��.A \ E1/C ��.A \ E2/:

PROOF.

Since E1 is ��-measurable,

��.A/ D ��.A \ E1/C ��.A \ Ec1/;

for each subset A of X . Replacing A by A \ .E1 [ E2/, we have

��ŒA \ .E1 [ E2/� D ��Œ.A \ .E1 [ E2// \ E1�C ��Œ.A \ .E1 [ E2// \ Ec1�

D ��Œ..A \ E1/ [ .A \ E2// \ E1�C ��Œ..A \ E1/ [ .A \ E2// \ Ec1�

D ��Œ.A \ E1 \ E1/ [ .A \ E2 \ E1/�C ��Œ.A \ E1 \ Ec1/ [ .A \ E2 \ Ec

1/�

D ��.A \ E1/C��.A \ E2/:

2

2.4.10 Theorem

Let X be a set and �� an outer measure on X . Suppose that .En/ be a sequence of mutually disjoint sets in

M. Then for any n � 1 and any subset A of X ,

��

�

A \n[

jD1

Ej

�

DnX

jD1

��.A \ Ej /: (2.4)

22


PROOF.

We prove this theorem by induction. The case n D 1 is obviously true. The case n D 2 has been proved

in Proposition 2.4.9. Assume that

��

�

A \n�1[

jD1

Ej

�

Dn�1X

jD1

��.A \ Ej /: (2.5)

Firstly, we note that

A \� n[

jD1

Ej

�

\ En D A \ En, and

A \� n[

jD1

Ej

�

\ Ecn D A \

� n�1[

jD1

Ej

�

:

Since En is ��-measurable,

��

�

A \� n[

jD1

Ej

��

D ��

�

A \� n[

jD1

Ej

�

\ En

�

C ��

�

A \� n[

jD1

Ej

�

\ Ecn

�

D ��.A \ En/C ��

�

A \� n�1[

jD1

Ej

��

D ��.A \ En/Cn�1X

jD1

��.A \ Ej / (by 2.5)

DnX

jD1

��.A \ Ej /:

2

By taking A D X in (2.4), we obtain

��

� n[

jD1

Ej

�

DnX

jD1

��.Ej /:

i.e., �� is finitely additive on M.

The following theorem is the key to using an outer measure to construct a measure.

2.4.11 Theorem

Let X be a set and �� an outer measure on X . Denote by M, the collection of all ��-measurable subsets

of X . Then

[1] M is a �-algebra, and

[2] the restriction of ��jM of �� to M is a measure on .X;M/.

PROOF.

[1] We have already shown in Theorem 2.4.8 that M is an algebra. It remains to show that M is closed

under countable unions. We will show that M is in fact closed under pairwise disjoint countable

23


unions. Let .En/ be a pairwise disjoint sequence of sets in M, Sn DSn

jD1 Ej and S D \1jD1Ej .

Then, since Sn 2 M, we have that for any A � X ,

��.A/ D ��.A \ Sn/C ��.A \ Scn /

� ��.A \ Sn/C ��.A \ Sc/ (since Scn � Sc and �� is monotone)

DnX

jD1

��.A \ Ej /C ��.A \ Sc / (by Theorem 2.4.10).

Since this is true for every natural number n, it follows that

��.A/ �1X

jD1

��.A \ Ej /C ��.A \ Sc/

� ��.A \ S/C ��.A \ Sc/ (since �� is countably subadditive).

Therefore,S1

jD1 Ej 2 M.

[2] By definition of the outer measure, �.;/ D 0. We have to show that �� is countably additive. We

do this as follows, let .En/ be a pairwise disjoint sequence of sets in M. As shown in [1], for any

A � X ,

��.A/ �1X

jD1

��.A \ Ej /C ��

�

A \� 1[

jD1

Ej

�c�

:

If A is replaced by [1jD1

Ej , we have that

��

� 1[

jD1

Ej

�

�1X

jD1

��.Ej /: (2.6)

Since �� is, by definition, countably subadditive, we have that

��

� 1[

jD1

Ej

�

�1X

jD1

��.Ej /: (2.7)

From (2.6) and (2.7), we conclude that

��

� 1[

jD1

Ej

�

D1X

jD1

��.Ej /:

Thus, �� is a measure on M.

2

The measure � D ��jM on X is often referred to as the measure on X induced by the outer measure

��.

2.4.12 Exercise

[1] In proving Theorem 2.4.11 [1], we showed that M is closed under pairwise disjoint countable unions.

Deduce from this that M is closed under countable unions.

[2] Let �� be an outer measure on a set X such that ��.A/ D 0, for some subset A of X . Show that if

B is a subset of X , then ��.A [ B/ D ��.B/.

24


2.5 Lebesgue measure on R

We now use the approach discussed in the previous section to construct the Lebesgue measure on R. The

Lebesgue measure will lead to the definition of the Lebesgue integral in the next chapter. Let I be a

collection of open intervals in R. For I 2 I , let l.I / denote the length of I .

2.5.1 Theorem

For any subset A of R, define m� W P.R/ ! Œ0;1� by

m�.A/ D inf

�

X

n

l.In/ W In 2 I , for each n and A �[

n

In

�

:

Then m� is an outer measure on R.

PROOF.

(i) Since ; � .a; a/, for any real number a, it follows that 0 � m�.;/ � a � a D 0. Thus, m�.;/ D 0.

(ii) Let A and B be subsets of R such that A � B. If B �S

n In, where .In/ is a sequence of open

intervals in R, then A �S

n In. Therefore, m�.A/ �P1

nD1 l.In/. It follows that m�.A/ � m�.B/.

(iii) Let .An/ be a sequence of subsets of R. If m�.An/ D 1 for some n, then

m�

� 1[

nD1

An

�

�1X

nD1

m�.An/:

Suppose that m�.An/ < 1, for each n. Then given � > 0, there is a sequence .Ink /k of open

intervals in R such that An �S1

kD1 Ink , and

1X

kD1

l.Ink / < m�.An/C �

2n:

Now, sinceS1

nD1 An �S1

nD1

S1kD1 Ink , it follows that

m�

� 1[

nD1

An

�

�1X

nD1

1X

kD1

l.Ink / <

1X

nD1

�

m�.An/C �

2n

�

D1X

nD1

m�.An/C �:

Since � is arbitrary, we have that m�

�

S1nD1 An

�

�P1

nD1 m�.An/.

2

The outer measure defined in Theorem 2.5.1 is called the Lebesgue outer measure.

2.5.2 Definition

A set E � R is said to be Lebesgue-measurable (or simply measurable) if

m�.A/ D m�.A \ E/C m�.A \ Ec/;

for every set A � R.

Denote by M, the set of all Lebesgue-measurable subsets of R. We have already shown in Theorem

2.4.11, that M is a �-algebra and that m D m�jM is a measure on M. This measure is called the Lebesgue

measure on R.

The Lebesgue outer measure has other important properties that are listed in the following propositions.

25


2.5.3 Proposition

The outer measure of an interval I is its length. That is, m�.I / D l.I /:

PROOF.

Case 1: Let I be a closed and bounded interval, say I D Œa; b�. Then Œa; b� � .a � �; b C �/, for each

� > 0. Thus,

m�.Œa; b�/ � l.a � �; b C �/ D b � a C 2�:

Since � is arbitrary, m�.I / D m�.Œa; b�/ � b � a. It remains to show that m�.I / � b � a. Suppose that

I �S1

kD1 Ik , where Ik is open for each k 2 N. Since I is compact, there is a finite subcollection fIkgmkD1

of intervals such that I �Sm

kD1. Since a 2 I , there is an interval .a1; b1/ in the collection fIkgmkD1

such

that a1 < a < b1. If b � b1, then we are done. Otherwise, a < b1 < b and so b1 2 I . Therefore, there

is an interval .a2; b2/ in the collection fIkgmkD1

such that a2 < b1 < b2. If b < b2, then we are done.

Otherweise, a < b2 < b. That is, b2 2 I . So there is an interval .a3; b3/ in the collection fIkgmkD1

such that

a3 < b2 < b3. Continuing in this fashion, we obtain a sequence of intervals .a1; b1/; .a2; b2/; : : : ; .ar ; br/

from the collection fIkgmkD1

such that ai < bi�1 < bi for i D 2; 3; : : : ; r . Since fIkgmkD1

is a finite

collection, this process must terminate after s � m steps with b < bs.

Now,

1X

jD1

l.Ij / �mX

jD1

l.Ij / �sX

jD1

l.aj ; bj/

D .bs � as/C .bs�1 � as�1/C .bs�2 � as�2/C � � � C .b1 � a1/

D bs � .as � bs�1/ � .as�1 � bs�2/ � � � � � .a2 � b1/ � a1

> bs � a1 > b � a:

It follows from this that m�.I / � b � a D l.I /.

Case 2: Let I be an interval of the form .a; b/, .a; b� or Œa; b/. Then for each 0 < � < .b � a/ with

Œa C �

2; b � �

2� � I � Œa � �

2; b C �

2�:

Hence,

m�Œa C �

2; b � �

2� � m�.I / � m�Œa � �

2; b C �

2�

,l�

Œa C �

2; b � �

2��

� m�.I / ��

Œa � �

2; b C �

2��

,b � a � � � m�.I / � b � a C � ( by case 1).

Since � is arbitrary, it follows that m�.I / D b � a D l.I /:

Case 3: Suppose that I is an unbounded interval. Then for any real number k , there is a closed interval

J � I such that l.J / D k . Hence,

m�.I / � m�.J / D l.J / D k:

Thus, m�.I / D 1.

2

2.5.4 Proposition

For each x 2 R, m�.fxg/ D 0:

PROOF.

26


Given any � > 0, fxg � .x � �2; x C �

2/. Thus

0 � m�.fxg/ � m�.x � �

2; x C �

2/ D l.x � �

2; x C �

2/ D �:

Since � is arbitrary, we have that m�.fxg/ D 0.

2

2.5.5 Corollary

Every countable subset of R has outer measure zero.

PROOF.

Let A be a countable subset of R. Then A DS

nfang. Therefore,

0 � m�.A/ D m�

�

[

n

fang�

�X

n

m�.fang/ D 0:

Thus, m�.A/ D 0.

2

2.5.6 Corollary

The interval Œ0; 1� is uncountable.

2.5.7 Proposition

The Lebesgue outer measure is translation-invariant. That is, m�.A C x/ D m�.A/ for all A � R and

x 2 R.

PROOF.

Let .In/ be a sequence of open intervals such that A �S

n In. Then A C x �S

n.In C x/. Thus,

m�.A C x/ �X

n

l.In C x/ DX

n

l.In/:

That is, m�.A C x/ is a lower bound for the sumsP

n l.In/, with A �S

n In. Therefore,

m�.A C x/ � m�.A/: (2.8)

To prove the reverse inequality, we only need to notice that

m�.A/ D m�.A C x C .�x// � m�.A C x/;

where the inequality follows from (4.1). Thus, m�.A C x/ D m�.A/.

2

2.5.8 Proposition

For any set A � R and any � > 0, there is an open set V such that A � V and m�.V / < m�.A/ C �.

PROOF.

By definition of m�.A/, there is a sequence .In/ of open intervals such that A �S

n In and

X

n

l.In/ < m�.A/C �

2:

If In D .an; bn/, let Jn D .an � �

2nC1 ; bn/. Then A �S

n Jn. Let V DS

n Jn. Then V is an open set in R

and

m�.V / �X

n

l.Jn/ DX

n

l.In/C �

2< m�.A/ C �:

2

27


2.5.9 Proposition

For any a 2 R, the set .a;1/ is measurable.

PROOF.

Let A be any subset of R. We must show that

m�.A/ � m�.A \ .a;1// C m�.A \ .a;1/c / D m�.A \ .a;1// C m�.A \ .�1; a�/:

Let A1 D A \ .a;1/ and A2 D A \ .�1; a�. Then we need to show that

m�.A/ � m�.A1/C m�.A2/:

If m�.A/ D 1, then there is nothing to prove. Assume that m�.A/ < 1. From the definition of m�, given

� > 0, there is a sequence .In/ of open intervals such that A �S

n In and

m�.A/ C � >X

n

l.In/:

Let Jn D In \ .a;1/ and J 0n D In \ .�1; a�. Then Jn and J 0

n are disjoint (possibly empty) intervals and

In D Jn [ J 0n. Therefore,

l.In/ D l.Jn/C l.J 0n/ D m�.Jn/C m�.J 0

n/:

Since A1 �S

n Jn and A2 �S

n J 0n, it follows that

m�.A1/ � m�

�

[

n

Jn

�

�X

n

m�.Jn/ DX

n

l.Jn/; and

m�.A2/ � m�

�

[

n

J 0n

�

�X

n

m�.J 0n/ D

X

n

l.J 0n/:

Therefore,

m�.A1/C m�.A2/ �X

n

Œl.Jn/C l.J 0n/� D

X

n

l.In/ < m�.A/ C �:

Since � is arbitrary, m�.A1/C m�.A2/ � m�.A/.

2

2.5.10 Corollary

[1] Every interval in R is measurable.

[2] Every open set in R is measurable.

[3] Every closed set in R is measurable.

[4] Every set that is a countable intersection of open sets in R is measurable.

[5] Every set that is a countable union of closed sets in R is measurable.

PROOF.

[1] We have already shown that for each a 2 R, the interval .a;1/ is measurable. It follows from the

fact that M is an algebra that .�1; a� D .a;1/c is measurable for each a 2 R.

Since Œa;1/ D fag [ .a;1/ and m�.fag/ D 0, it follows from Lemma 2.4.5 that fag is measur-

able. Therefore, Œa;1/ is measurable. Since M is an algebra, we have that .�1; a/ D Œa;1/c is

measurable for each a 2 R.

If a and b are real numbers such that a < b, then .a; b/ D .�1; b/ \ .a;1/. Since .�1; b/ and

.a;1/ are both measurable, it follows thta .a; b/ is also measurable.

If a and b are real numbers such that a < b, then Œa; b/ D fag [ .a; b/, .a; b� D .a; b/ [ fbg, and

Œa; b� D fag [ .a; b/[ fbg. Therefore, the sets Œa; b/, .a; b�, and Œa; b� are all measurable.

28


[2] Every open set in R is a countable union of open intervals. Since each open interval is measurable

and M is a �-algebra, it follows that the union of open intervals is also measurable.

[3] A closed set F in R is a complement of an open set V in R. That is F D V c for some open set V in

R. Since V is measurable, V c D F is also measurable.

[4] Let U DT1

iD1 Vi , where Vi is open in R for each i 2 N. Then by [2], Vi is measurable, for each

i 2 N, and so is V ci for each i 2 N. Since M is a �-algebra, we have that

U D1\

iD1

Vi D� 1[

iD1

V ci

�c

is measurable.

[5] Let F DS1

iD1 Fi , where Fi is closed in R, for each i 2 N. Then F ci is open in R, for each

i 2 N. By [4], we have thatT1

iD1 F ci is measurable. Since M is a �-algebra, it follows that

�

T1iD1 F c

i

�c

DS1

iD1 Fi D F is measurable.

2

2.5.11 Definition

(a) A set that is a union of a countable collection of closed sets is called a F� -set.

(b) A set that is an intersection of a countable collection of open sets is called a Gı-set.

2.5.12 Remark

[1] The previous corollary says, amongst other things, that every F� -set and every Gı-set is measurable.

[2] The previous corollary also states that all ‘nice’ sets in R are measurable. There are however sets that

are not measurable. The point being that these sets are certainly not ‘natural’ but somewhat strange.

[3] The previous corollary shows that every open set in R is measurable. The smallest �-algebra con-

taining all the open sets is called the Borel �-algebra.

The following result states that any measurable subset of R can be approximated by the ‘nice’ ones.

2.5.13 Proposition

Let E � R and m� the Lebesgue outer measure on R. The following statements are equivalent:

[1] E is measurable.

[2] For each � > 0, there is an open set V such that E � V and m�.V nE/ < �.

[3] There is a Gı-set G such that E � G and m�.GnE/ D 0.

[4] For each � > 0, there is a closed set F such that F � E and m�.EnF/ < �.

[5] There is an F� -set H such that H � E and m�.EnH/ D 0.

PROOF.

[1] ) [2]: Suppose that E is measurable and let � > 0. If m.E/ < 1, then by Proposition 2.5.8, there is an

open set V such that E � V and

m�.V / < m�.E/C �:

Since V and E are measurable, we can rewrite the equation above as

m.V / < m.E/C �:

29


Hence, m.V nE/ D m.V / � m.E/ < �. Suppose that m.E/ D 1. Write R DS

n2N In, a union

of disjoint finite intervals. For each n 2 N, let En D E \ In. Then E DS

n2N En, and for each

n 2 N, m.En/ < 1, Therefore, by the first part, there is an open set Vn such that En � Vn and

m.VnnEn/ <�

2n . Write V DS

n2N Vn. Then V is an open set and

V nE D[

n2N

Vnn[

n2N

En D[

n2N

.VnnEn/:

Hence

m.V nE/ �X

n2N

m.VnnEn/ <X

n2N

�

2nD �:

[2] ) [3]: For each n 2 N, let Vn be an open set such that E � Vn and m�.VnnE/ < 1n

. Let G DT

n2N Vn.

Then G is a Gı-set, E � G, and for each n 2 N,

m�.GnE/ D m�

�

\

n2N

.VnnE/

�

� m�.VnnE/ � 1

n:

Thus, m�.GnE/ D 0:

[3] ) [1]: Since G is a Gı-set, it is measurable. Also GnE is measurable since m�.GnE/ D 0. Therefore, the

set E D Gn.GnE/ is also measurable.

[1] ) [4]: Suppose that E is measurable and let � > 0. Then Ec is also measurable. Since [1] implies [2],

we have, by [2], that there is an open set V such that Ec � V and m�.V nEc/ < �. Note that

V nEc D E \ V D EnV c . Take F D V c . Then F is a closed set, F � E, and m�.EnF/ < �.

[4] ) [5]: For each n 2 N, let Fn be a closed set such that Fn � E and m�.EnFn/ <1n

. Let H DS

n2N Fn.

Then H is an F� -set, H � E, and for each n 2 N,

m�.EnH/ D m�

�

En[

n2N

Fn

�

D\

n2N

.EnFn/ � m�.EnFn/ <1

n:

Thus, m�.EnH/ D 0.

[5] ) [1]: Since H is an F� -set, it is measurable. Also EnH is measurable since m�.EnH/ D 0. Therefore,

the set E D .EnH/ [ H is also measurable.

2

30

Chapter 3

The Lebesgue integral

3.1 Introduction

We now turn our attention to the construction of the Lebesgue integral of general functions, which, as

already discussed, is necessary to avoid the technical deficiencies associated with the Riemann integral.

3.2 Measurable functions

3.2.1 Basic notions

The extended real number system, R, is the set of real numbers together with two symbols �1 and C1.

That is, R D R [ f�1;C1g D Œ�1;1�. The algebraic operations for these two infinities are:

[1] For r 2 R, ˙1 C r D ˙1.

[2] For r 2 R, r .˙1/ D ˙1, if r > 0 and r .˙1/ D �1 if r < 0.

[3] C1 C .C1/ D C1 and �1 C .�1/ D �1.

[4] 1 C .�1/ is undefined.

[5] 0 � .˙1/ D 0.

3.2.1 Definition

Let .X; †/ be a measurable space and E 2 †. A function f W E ! R is said to be measurable if for each

˛ 2 R, the set fx 2 E W f .x/ > ˛g is measurable.

3.2.2 Proposition

Let .X; †/ be a measurable space and E 2 †, and f ! R. Then the following are equivalent:

[1] f is measurable.

[2] For each ˛ 2 R, the set fx 2 E W f .x/ � ˛g is measurable.

[3] For each ˛ 2 R, the set fx 2 E W f .x/ < ˛g is measurable.

[4] For each ˛ 2 R, the set fx 2 E W f .x/ � ˛g is measurable.

PROOF.

31


[1] ) [2]: If f is measurable, then the set

fx 2 E W f .x/ � ˛g D\

n2N

�

x 2 E W f .x/ > ˛ � 1

n

�

;

which is an intersection of measurable sets, is measurable.

[2] ) [3]: If the set fx 2 E W f .x/ � ˛g is measurable, then so is the set

Enfx 2 E W f .x/ � ˛g D fx 2 E W f .x/ < ˛g:

[3] ) [4]: If the set fx 2 E W f .x/ < ˛g is measurable, then so is the set

fx 2 E W f .x/ � ˛g D\

n2N

�

x 2 E W f .x/ < ˛ C 1

n

�

since it is an intersection of measurable sets.

[4] ) [1]: If the set fx 2 E W f .x/ � ˛g, is measurable, so is its complement in E. Hence,

fx 2 E W f .x/ > ˛g D Enfx 2 E W f .x/ � ˛g

is measurable. It follows from this that f is measurable.

2

3.2.3 Examples

[1] The constant function is measurable. That is, if .X; †/ is a measurable space, c 2 R, and E 2 †,

then the function f W E ! R given by f .x/ D c, for each x 2 E, is measurable.

Let ˛ 2 R. If ˛ � c, then the set fx 2 E W f .x/ > ˛g D ;, and is thus measurable.

If ˛ < c, then the set fx 2 E W f .x/ > ˛g D E, and is therefore measurable. Hence, f is

measurable.

[2] Let .X; †/ be a measurable space and let A 2 †. The characteristic function, �A

, is defined by

�A.x/ D

�

1 if x 2 A

0 if x 62 A:

The characteristic function �A

is measurable. This follows immediately from the observation that,

for each ˛ 2 R and E 2 †, the set fx 2 E W �A> ˛g is either E, A or ;.

[3] Let B be the Borel �-algebra in R and E 2 B. Then, any continuous function f W E ! R is

measurable. This is an immediate consequence of the fact that, if f W E ! R is continuous and

˛ 2 R, then the set fx 2 E W f .x/ > ˛g is open and hence belongs to B.

3.2.4 Proposition

[1] If f and g are measurable real-valued functions defined on a common domain E 2 † and c 2 R,

then the functions

(a) f C c,

(b) cf ,

(c) f ˙ g,

(d) f 2,

32


(e) f � g,

(f) jf j,(g) f _ g,

(h) f ^ g are also measurable.

[2] If .fn/ is a sequence of measurable functions defined on a common domain E 2 †, then the functions

(a) supn fn,

(b) infn fn,

(c) lim supn fn,

(d) lim infn fn are also measurable.

PROOF.

[1] (a) For any real number ˛,

fx 2 E W f .x/C c > ˛g � fx 2 E W f .x/ > ˛ � cg:

Since the set on the right-hand side is measurable, we have that f C c is measurable.

(b) If c D 0, then cf is obviously measurable. Assume that c < 0. Then, for each real number ˛,

fx 2 E W cf .x/ > ˛g D�

x 2 E W f .x/ < ˛

c

�

:

Since the set fx 2 E W f .x/ < ˛cg is measurable, it follows that cf is also measurable.

(c) Let ˛ be a real number. Since the rationals are dense in the reals, there is a rational number r

such that

f .x/ < r < ˛ � g.x/;

whenever f .x/C g.x/ < ˛. Therefore,

fx 2 E W f .x/C g.x/ < ˛g D[

r2Q

.fx 2 E W f .x/ < r g \ fx 2 E W g.x/ < ˛ � r g/:

Since the sets fx 2 E W f .x/ < r g and fx 2 E W g.x/ < ˛ � r g are measurable, so is the set

fx 2 E W f .x/ < r g \ fx W g.x/ < ˛ � r g, and consequently, the set fx 2 E W f .x/C g.x/ <

˛g, being a countable union of measurable sets, is also measurable.

If g is measurable, it follows from (b) that .�1/g is also measurable. Hence, so is f C.�1/g Df � g.

(d) Let ˛ 2 R and E 2 †. If ˛ < 0, then fx 2 E W f 2.x/ > ˛g D E, which is measurable.

If ˛ � 0, then

fx 2 E W f 2.x/ > ˛g D fx 2 E W f .x/ >p˛g [ fx 2 E W f .x/ < �

p˛g:

Since the two sets on the right hand side are measurable, it follows that the set fx 2 E Wf 2.x/ > ˛g is also measurable. Hence f 2 is measurable.

(e) Since f � g D 14Œ.f C g/2 � .f � g/2�, it follows from (b), (c), and (d) that f � g is measurable.

(f) Let ˛ 2 R and E 2 †. If ˛ < 0, then fx 2 E W jf .x/j > ˛g D E, which is measurable.

If ˛ � 0, then

fx 2 E W jf .x/j > ˛g D fx 2 E W f .x/ > ˛g [ fx 2 E W f .x/ < �˛g:

Since the two sets on the right hand side are measurable, it follows that the set fx 2 E Wf .x/2 > ˛g is also measurable. Thus, jf j is measurable.

33


(g) It is sufficient to observe that f _ g D 12ff C g C jf � gjg. It now follows from (b), (c), and

(d) that f _ g is measurable.

(h) It is sufficient to observe that f ^ g D 12ff C g � jf � gjg. It now follows from (b), (c), and

(d) that f ^ g is measurable.

[2] (a) Let ˛ 2 R. Then

fx 2 E W supnfn > ˛g D

1[

nD1

fx 2 E W fn.x/ > ˛g:

Since for each n 2 N, fn is measurable, it follows that the set fx 2 E W fn.x/ > ˛g is

measurable for each n 2 N. Therefore, the set fx 2 E W supn fn.x/ > ˛g is measurable as it is

a countable union of measurable sets.

(b) It is sufficient to note that infn fn D � supn.�fn/.

(c) Notice that lim supn fn D infn�1

�

supk�n fk

�

and use (a) and (b) above.

(d) Notice that lim infn fn D supn�1

�

infk�n fk

�

and use (a) and (b) above.

2

3.2.5 Corollary

Let .X; †/ be a measurable space. If f is a pointwise limit of a sequence .fn/ of measurable functions

defined on a common domain E 2 †, then f is measurable.

PROOF.

If the sequence .fn/ converges pointwise to f , then, for each x 2 E,

f .x/ D limn!1

fn.x/ D lim supn

fn.x/:

Now, by Proposition 3.2.4 [2], f is measurable.

2

3.2.6 Definition

Let f be a real-valued function defined on a set X . The positive part of f , denoted by f C, is the function

f C D maxff; 0g D f _0 and the negative part of f , denoted by f �, is the function f � D maxf�f; 0g D.�f / _ 0.

Immediately we have that if f is a real-valued function defined on X , then

f D f C � f � and jf j D f C C f �:

Note that

f C D 1

2Œjf j C f � and f � D 1

2Œjf j � f �:

Let .X; †/ be a measurable space and f W X ! R. It is trivial to deduce from Proposition 3.2.4 that f is

measurable if and only if f C and f � are measurable.

3.2.7 Definition

Let .X; †/ be a measurable space. A simple function on X is a function of the form � DPn

jD1 cj�Ej,

where, for each j D 1; 2; : : :n, cj is an extended real number and Ej 2 †.

Since �Ej

is measurable for each j D 1; 2; : : : ; n, it follows from Proposition 3.2.4 that � is also

measurable.

34


The representation of� in Definition 3.2.7 is not unique. If� assumes distinct nonzero values a1; a2; : : : ; an,

then

� DnX

jD1

aj�Aj;

where Aj D fx 2 X W �.x/ D aj ; j D 1; 2; : : : ; ng. This is called the cannonical representation of �.

Note that the sets Aj are pairwise disjoint and X DSn

iD1 Aj .

What Corollary 3.2.5 is saying is that the pointwise limit of a sequence of simple functions is a mea-

surable function. The converse also holds. That is, every measurable function is a pointwise limit of a

sequence of simple functions which we state as a theorem.

3.2.8 Theorem

Let .X; †/ be a measurable space and f a nonnegative measurable function. Then there is a monotonic

increasing sequence .�n/ of nonnegative simple functions which converge pointwise to f .

PROOF.

For each n 2 N and for k D 1; 2; : : : ; n2n, let

Enk D�

x 2 X W k � 1

2n� f .x/ <

k

2n

�

; and

Fn D fx 2 X W f .x/ � ng:

Then fEnk W k D 1; 2; : : :n2ng [ fFn W n 2 Ng is a collection of pairwise disjoint measurable sets whose

union is X .

For each n 2 N, define �n by

�n Dn2nX

kD1

k � 1

2n�

EnkC n�

Fn:

Then the �n are nonnegative simple functions, �n � f , and �n � �nC1, for each n 2 N. If x 2 X is such

that f .x/ < 1, then, if f .x/ < n,

0 � f .x/ � �n.x/ <k

2n� .k � 1/

2nD 1

2n:

Since limn!11

2n D 0, we have that limn!1 �n.x/ D n, and so limn!1 �n.x/ D f .x/:

2

3.2.9 Corollary

Let .X; †/ be a measurable space and f a measurable function. Then there is a sequence .�n/ of simple

functions which converge pointwise to f .

PROOF.

The functions f C and f � are both nonnegative and measurable. By Theorem 3.2.8, there exist two

sequences .�n/ and . n/ of nonnegative simple functions which converge pointwise to f C and f � respec-

tively. The sequence .�n/, where, for each n 2 N, �n D �n � n, is a sequence of simple functions which

converges pointwise to f C � f � D f .

2

3.2.10 Definition

Let .X; †; �/ be a measure space and E 2 †. A property P .x/, where x 2 E, is said to hold almost

everywhere (abbreviated, a.e.) on E if the set N D fx 2 E W P .x/ failsg belongs to † and �.N / D 0.

3.2.11 Examples

[1] f D g a.e. if �.fx 2 X W f .x/ 6D g.x/g/ D 0.

35


[2] f � g a.e. if �.fx 2 X W f .x/ > g.x/g/ D 0.

3.2.12 Definition

Let .X; †; �/ be a measure space. A function f W X ! R is said to be almost everywhere real-valued,

denoted by a.e. real-valued, if

�.fx 2 X W jf .x/j D 1g/ D 0:

We call a set of measure zero a null set.

3.2.13 Definition

A measure space .X; †; �/ is said to be complete if† contains all subsets of sets of measure zero. That is,

if E 2 †, �.E/ D 0 and A � E, then A 2 †.

It follows from Theorem 2.3.6 [1] that if a measure space .X; †; �/ is complete, E 2 †, �.E/ D 0

and A � E, then �.A/ D 0.

3.2.14 Proposition

Let .X; †; �/ be a complete measure space and f D g a.e. If f is measurable on E 2 †, then so is g.

PROOF.

Let ˛ 2 R and N D fx 2 E W g.x/ 6D f .x/g. Then N 2 † and �.N / D 0. Now,

fx 2 E W g.x/ > ˛g D fx 2 EnN W g.x/ > ˛g [ fx 2 N W g.x/ > ˛gD fx 2 EnN W f .x/ > ˛g [ fx 2 N W g.x/ > ˛g:

The first set on the right hand side is measurable since f is measurable. The second set is measurable since

it is a subset of the null set N and the measure is complete.

2

3.2.2 Convergence of sequences of measurable functions

We now consider various notions of convergence of a sequence of measurable functions examine the rela-

tionships that exist between them.

3.2.15 Definition

Let .X†;�/ be a measure space. A sequence .fn/ of a.e. real-valued measurable functions on X is said to

[1] converge almost everywhere to an a.e. real-valued measurable function f . denoted by fn !a.e. f ,

if for each � > 0 and each x 2 X , there is a set E 2 † and a natural number N D N.�/ such that

�.E/ < � and

jfn.x/ � f .x/j < �, for each x 2 X nE and each n � N:

[2] converge almost uniformly to an a.e. real-valued measurable function f , denoted by fn !a.u. f , if

for each � > 0, there is a set E 2 † and a natural number N D N.�/ such that �.E/ < � and

kfn � f k1 D supx2X nE

jfn.x/� f .x/j < �, for each n � N:

[3] converge in measure to an a.e. real-valued measurable function f , denoted by fn !� f , if for

every � > 0,

limn!1

�.fx 2 X W jfn.x/ � f .x/j � �g/ D 0:

(if � is a probability measure, then this mode of convergence is called convergence in probability).

3.2.16 Proposition

Let .X; †; �/ be a complete measure space and .fn/ a sequence of measurable functions on E 2 † which

converges to f a.e. Then f is measurable on E.

36


PROOF.

Let ˛ 2 R and N D fx 2 E W limn!1 fn.x/ 6D f .x/g. Then N 2 †;�.N / D 0 and

limn!1 fn.x/ D f .x/, for each x 2 EnN . Since E and N are measurable, so is EnN . For each

n 2 N, define gn and g by

gn.x/ D�

0 if x 2 N

fn.x/ if x 2 EnN :

and

g.x/ D�

0 if x 2 N

f .x/ if x 2 EnN :

Then gn D fn a.e. and f D g a.e. It follows from Proposition 3.2.4, that for each n 2 N, gn is measurable.

If x 2 N , thenlim

n!1gn.x/ D lim

n!10 D 0 D g.x/; (3.1)

and if x 2 EnN , then

limn!1

gn.x/ D limn!1

fn D 0 D f .x/ D g.x/ (3.2)

It follows from (3.1) and (3.2) that the sequence .gn/ converges pointwise to g everywhere on E. By Corol-

lary 3.2.5, that g is measurable. Since g D f a.e., we have, by Proposition 3.2.14, that f is measurable.

2

3.2.17 Theorem

Let .X; †; �/ be a measure space and .fn/ be a sequence of real-valued measurable functions on X . If

the sequence .fn/ converges almost uniformly to f , then it converges in the measure to f . That is, almost

uniform convergence implies convergence in the measure.

PROOF.

Let � > 0 be given. Since the sequence .fn/ converges almost uniformly to f , there is a measurable set

E and a natural number N such that �.E/ < � and

jfn.x/� f .x/j < �, for all x 2 X nE and all n � N:

It now follows that, for all n � N; fx 2 X W jfn.x/� f .x/j � �g � E. Therefore, for all n 2 N,

�.fx 2 X W jfn.x/ � f .x/j � �g/ < �:

2

3.2.18 Theorem

Let .X; †; �/ be a measure space and .fn/ be a sequence of a.e. real-valued measurable functions on X .

If the sequence .fn/ converges almost uniformly to f , then it converges almost everywhere to f . That is,

almost uniform convergence implies convergence almost everywhere.

PROOF.

Suppose that the sequence .fn/ converges almost uniformly to f . Then, for each n 2 N, there is a

measurable set En, with �.En/ <1n

such that fn ! f uniformly on X nEn. Let A DS1

nD1.X nEn/.

Then

�.X nA/ D �

� 1\

nD1

En

�

� �.En/ D 1

n! 0:

37


That is, �.X nA/ D 0. Furthermore, for each x 2 A, fn.x/ ! f .x/ as n ! 1. Hence, fn !a.e. f .

2

In general, the converses of the previous two theorems do not hold. The following is an example of a

sequence that converges almost everywhere and in measure, but does not converge almost uniformly.

3.2.19 Example

Let X D Œ0;1/ with the Lebesgue measure �. For each n 2 N, let fn D �Œn;nC 1

n �, i.e.,

fn.x/ D�

1 if x 2 Œn; n C 1n�

0 otherwise:

Then, for each x 2 X , fn.x/ !n!1 0. That is, the sequence .fn/ converges pointwise on X to the zero

function. Since fx 2 X W fn.x/ 6! 0g D ;, it follows that �.fx 2 X W fn.x/ 6! 0g/ D 0, Hence,

fn !a.e. 0.

The sequence .fn/ converges in measure to the zero function. Indeed, for each � > 0,

�.fx 2 X W jfn.x/� 0j � �g/ D1

n!n!1 0:

We show that the sequence .fn/ does not converge almost uniformly to the zero function. Assume, on the

contrary that .fn/ converges almost uniformly to the zero function. Then, there is a Lebesgue measurable

set E with �.E/ D 0 such that fn ! 0 such that fn ! 0 uniformly on Ec . Therefore, with � D 1, there

is an N 2 N such that for all x 2 Ec and all n � N ,

kfn � 0k1 D kfnk1 < 1 , jfn.x/j D fn.x/ < 1:

For each n 2 N, let An D Œn; n C 1n�. Then, Ec \

S

n�N An D ;, and soS

n�N An � E. Thus,

0 D �.E/ � �

�

[

n�N

An

�

D1X

nDN

�.An/ D1X

nDN

1

nD 1;

which is absurd. Therefore .fn/ does not converge almost uniformly to the zero function.

3.2.20 Theorem (Egoroff’s Theorem)

Let .X; †; �/ be a finite measure space and .fn/ be a sequence of a.e. real-valued measurable functions on

X . If the sequence .fn/ converges almost everywhere to f , then it converges almost uniformly to f .

PROOF.

Without loss of generality, we assume that .fn/ converges to f everywhere on X . For m; n 2 N , let

Emn D

1[

kDn

fx 2 X W jfk.x/ � f .x/j � 1

mg:

Then, for m; n 2 N, Emn is a measurable set and Em

nC1� Em

n . That is, .Emn /n is a decreasing sequence of

measurable subsets of X . Since fn.x/ !n!1 f .x/ for each x 2 X , we have that

1\

nD1

Emn D ;. Also,

since �.X / < 1, we have that �.Em1/ < 1. By Theorem 2.3.6 [4], it follows that

0 D �.;/ D �

� 1\

nD1

Emn

�

D limn!1

�.Emn /:

38


For ı > 0, let km 2 N be such that �.Emkm/ < ı

2m . Let Eı D1[

mD1

Emkm

. Then, Eı is a measurable set and

�.Eı/ �1X

mD1

�.Emkm/ < ı:

Notice that

X nEı D X n1[

mD1

Emkm

D1\

mD1

.X nEmkm/

D1\

mD1

�

X n1[

kDkm

�

x 2 X W jfk .x/ � f .x/j � 1

m

��

D1\

mD1

1\

kDkm

�

x 2 X W jfk.x/ � f .x/j < 1

m

�

:

Let � > 0. Choose m 2 N such that 1m< �. Then for any x 2 X nEı and any k � km,

jfk.x/ � f .x/j < 1

m< �:

That is, .fn/ converges uniformly to f on X nEı .

2

3.2.21 Corollary

Let .X; †; �/ be a finite measure space and .fn/ be a sequence of a.e. real-valued measurable functions on

X . If the sequence .fn/ converges almost everywhere to f , then it converges in measure to f .

PROOF.

This follows from Theorem 3.2.17 and Theorem 3.2.20.

2

The hypothesis that �.X / < 1 in Egoroff’s Theorem is essential. Let X D R with the Lebesgue

measure �. For each n 2 N, let fn D �Œn;1/

, i.e.,

fn.x/ D�

1 if x 2 Œn;1/

0 otherwise:

Then, for each x 2 R, fn.x/ !n!1 0. That is, the sequence .fn/ converges pointwise on R to the zero

function. Since fx 2 R W fn.x/ 6! 0g D ;, it follows that �.fx 2 R W fn.x/ 6! 0g/ D 0. Thus, fn !a.e. 0.

We show that the sequence .fn/ does not converge almost uniformly to the zero function. Assume, on the

contrary, that .fn/ converges almost uniformly to the zero function. Then, there is a Lebesgue measurable

set E with �.E/ D 0 such that fn ! 0 uniformly on Ec. Therefore, with � D 1, there is an N 2 N such

that for all x 2 Ec and all n � N ,

kfn � 0k1 D kfnk1 < 1 , jfn.x/j D fn.x/ < 1:

For each n 2 N, let An D Œn;1/. Then, Ec \[

n�N

An D Ec \ AN D ;, and so AN � E.

Hence, 0 D �.E/ � �.AN / D 1, which is a contradiction. Therefore .fn/ does not converge almost

uniformly to the zero function.

39


3.2.22 Theorem (F. Riesz)

Let .X; †; �/ be a measurable space and .fn/ be a sequence of a.e. real-valued measurable functions on X .

If the sequence .fn/ converges in measure to f , then there is a subsequence .fnk/ of .fn/ which converges

almost everywhere to f .

PROOF.

Let n1 2 N such that for all n � n1,

�.fx 2 X W jfn.x/ � f .x/j � 1g/ < 1

2:

Now, choose n2 2 N such that n2 > n1 and for all n � n2,

�

�

fx 2 X W jfn.x/ � f .x/j �1

2g�

<1

22:

Next, choose n3 2 N such that n3 > n2 and for all n � n3,

�

�

fx 2 X W jfn.x/ � f .x/j � 1

3g�

<1

23:

Continue this process obtaining an increasing sequence .nk / of natural numbers with

�

�

fx 2 X W jfnk.x/ � f .x/j � 1

kg�

<1

2k

For each k 2 N, let

Ak D fx 2 X W jfnk.x/� f .x/j � 1

kg:

Since, for each k 2 N, �.Ak / <1

2k , we have that the series

1X

kD1

�.Ak / converges. By the Borel-Cantelli

Lemma (Theorem 2.3.7), �.lim sup Ak / D 0. Let

A D lim sup Ak D1\

kD1

1[

jDk

Aj :

Choose x 2 X nA. Then there is an index jx 2 N such that x 2 X nAjx , Note that

X nAjx D X n1[

kDjx

fx 2 X W jfnk.x/ � f .x/j � 1

kg

D1\

kDjx

�

X nfx 2 X W jfnk.x/ � f .x/j � 1

kg�

D1\

kDjx

fx 2 X W jfnk.x/� f .x/j � 1

kg:

Hence, for all k � jx; jfnk.x/� f .x/j < 1

k. Let � > 0 be given. Choose k0 � jx such that 1

k0< �. Then,

for all k � k0,

jfnk.x/ � f .x/j < 1

k< �:

Therefore, for each x 2 X nA, fnk.x/ ! f .x/, and hence, fnk

!a.e. f .

2

40


3.3 Definition of the integral

In this section, we define a Lebesgue integral. This is done in three stages: firstly, we define the integral

of a nonnegative simple function; then, using the integral of a nonnegative simple function, we define the

integral or any measurable function. We also prove three key results: Monotone Convergence Theorem,

Dominated Convergence Theorem and Fatou’s Lemma.

Unless otherwise specified, we will work in the measure space .X; †; �/.

3.3.1 Integral of a nonnegative simple function

3.3.1 Definition

Let � be a nonnegative simple function with the canonical representation � DnX

iD1

ai�Ei. The Lebesgue

integral of � with respect to �, denoted byR

X�d�, is the extended real number

Z

X

�d� DnX

iD1

ai�.Ei/:

If A 2 †, we defineZ

A

�d� DZ

X

�A�d�:

The function � is integrable ifR

X�d� is finite.

We also writeR

�d� forR

X�d�.

It is straightforward to show that if A is a measurable set and � is as above, then

Z

A

�d� DnX

iD1

ai�.A \ Ei/:

Furthermore,Z

A

d� DZ

A

�A

d� DZ

X

�A

d� D �.A/:

It is necessary to show that the above definition of the Lebesgue integral is unambiguous, i.e, the integral is

independent of the representation of �. Assume that

� DnX

iD1

ai�EiD

mX

jD1

bj�Fi;

where Ei \ Ek D ;, for all 1 � i 6D k � n, X Dn[

iD1

Ei; Fk \ Fl D ;, for all 1 � j 6D l � m, and

m[

jD1

Fj . Then for each i D 1; 2; : : : ; n and each j D 1; 2; : : : ;m,

Ei D Ei \ X D Ei \� m[

jD1

Fj

�

Dm[

jD1

.Ei \ Fj /, a disjoint union, and

Fj D Fj \ X D� n[

iD1

Ei

�

Dn[

iD1

.Ei \ Fj /, a disjoint union.

41


If Ei \ Fj 6D ;, then, for x 2 Ei \ Fj , ai D �.x/ D bj . That is, ai D bj in this case. If Ei \ Fj D ;,

then �.Ei \ Fj / D 0. Thus, ai�.Ei \ Fj / D bj�.Ei \ Fj /, for all 1 � i � n and for all 1 � j � m.

ThereforenX

iD1

ai�.Ei/ DnX

iD1

ai�

� m[

jD1

.Ei \ Fj /

�

DnX

iD1

ai

mX

jD1

�.Ei \ Fj /

DnX

iD1

mX

jD1

ai�.Ei \ Fj / DnX

iD1

mX

jD1

bj�.Ei \ Fj /

DmX

jD1

bj

nX

iD1

�.Ei \ Fj / DmX

jD1

bj�

� n[

iD1

.Ei \ Fj /

�

mX

jD1

bj�.Fj /:

Hence the definition of the integral of � is unambiguous.

3.3.2 Proposition

Let .X; †; �/ be a measure space, � and nonnegative simple functions, and c a nonnegative real number.

Then

[1]R

X.� C /d� D

R

X�d�C

R

X d�.

[2]R

Xc�d� D c

R

X�d�:

[3] If � � , thenR

X�d� �

R

X d�.

[4] If A and B are disjoint measurable sets, then

Z

A[B

�d� DZ

A

�d�CZ

B

�d�:

[5] The set function � W † ! Œ0;1� defined by

�.A/ DZ

A

�d�;

for A 2 †, is a measure on X .

[6] If A 2 † and �.A/ D 0, thenR

A�d� D 0.

PROOF.

Let � DPn

iD1 ai�Ei, and D

PmjD1 bj�Fj

be canonical representations of � and respectively.

Then Ei \ Ek D ;, for all i � i 6D k � n, Fj \ Fl D ;, for all 1 � j 6D l � m, X D [niD1Ei, and

X D [mjD1Fj .

[1] For 1 � i � n and 1 � j � m, let Gij D Ei \ Fj . Then, for each x 2 Gij , �.x/C .x/ D ai C bj ,

i.e., the function � C takes the values ai C bj on Ei \ Fj . For all 1 � i; k � n and 1 � j ; l � m,

Gij \ Gkl D .Ei \ Fj /\ .Ek \ Fl /

D .Ei \ Ek/ \ .Fj \ Fl /

D ; \ ; D ;:

42


That is, fGij W 1 � i � n and 1 � j � mg is a collection of nm pairwise disjoint sets. Furthermore,

Ei D Ei \ X D Ei \� m[

jD1

Fj

�

Dm[

jD1

.Ei \ Fj / Dm[

jD1

Gij ;

Fj D Fj \ X D Fj \� n[

iD1

Ei

�

Dn[

iD1

.Ei \ Fj / Dn[

iD1

Gij , and

X D X \ X D� n[

iD1

Ei

�

\� m[

jD1

Fj

�

Dn[

iD1

m[

jD1

.Ei \ Fj / Dn[

iD1

m[

jD1

Gij

Therefore,

� C DnX

iD1

mX

jD1

.ai C bj /�Gij:

With this representation of �C , the numbers ai C bj are not necessarily distinct. Let c1; c2; : : : ; ck

be the distinct values assumed by � C . Then

Z

X

.� C /d� DkX

rD1

cr�.fx W �.x/C .x/ D crg/

DkX

rD1

cr�

[

ai Cbj Dcr

Ei \ Fj

!

DkX

rD1

cr

X

ai Cbj Dcr

�.Ei \ Fj /

DnX

iD1

mX

jD1

.ai C bj /�.Gij /

DnX

iD1

mX

jD1

ai�.Gij /CnX

iD1

mX

jD1

bj�.Gij /

DnX

iD1

ai

mX

jD1

�.Gij /CnX

iD1

bj

mX

jD1

�.Gij /

DnX

iD1

ai�.Ei/CmX

jD1

bj�.Fj /

DZ

X

�d�CZ

X

d�:

[2] If c D 0, then c� vanishes identically and hence,R

Xc�d� D c

R

X�d�. Assume that c > 0.

If � DnX

iD1

ai�Eiis the canonical representation of �, then c� D

PniD1 cai�Ei

is the canonical

representation of c�. Therefore,

Z

X

c�d� DnX

iD1

cai�.Ei/ D c

nX

iD1

ai�.Ei/ D c

Z

X

�d�:

43


[3] As shown above,

Z

X

� DnX

iD1

ai�.Ei/ DnX

iD1

ai�

� m[

jD1

Ei \ Fj

�

DnX

iD1

mX

jD1

ai�.Ei \ Fj /, and

Z

X

DmX

jD1

bj�.Fj / DmX

jD1

bj�

� n[

iD1

Fj \ Ei

�

DmX

jD1

nX

iD1

bj�.Ei \ Fj /:

Now suppose that � � . Then, for each x 2 Ei \ Fj , ai D �.x/ � .x/ D bj . i.e., ai � bj , for

all i; j such that Ei \ Fj 6D ;. It follows thatR

X�d� �

R

X d�.

[4] Let � be as above. Then

Z

A[B

�d� DnX

iD1

ai�..A [ B/\ Ei/

DnX

iD1

ai�Œ.A \ Ei/ [ .B \ Ei/�

DnX

iD1

ai Œ�.A \ Ei/C �.B \ Ei/�

DnX

iD1

ai�.A \ Ei/CnX

iD1

�.B \ Ei/

DZ

A

�d�CZ

B

�d�:

[5] We have that, for A 2 †,

�.A/ DZ

A

�d� DnX

iD1

ai�.A \ Ei/:

If A D ;, then A \ Ei D ;, for each i D 1; 2; : : : ; n. Hence, �.A \ Ei/ D 0, for i D 1; 2; : : : ; n

and so �.;/ D 0.

Let .Ak / be a sequence of pairwise disjoint measurable sets and A D1[

kD1

Ak : Then, for each i D

1; 2; : : : ; n,

Ei \ A D Ei \� 1[

kD1

Ak

�

D1[

kD1

.Ei \ Ak /, a disjoint union.

44


Thus,

�.A/ DnX

iD1

nX

iD1

ai�.A \ Ei/

DnX

iD1

ai�

� 1[

kD1

Ei \ Ak

�

DnX

iD1

ai

1X

kD1

�.Ei \ Ak /

D1X

kD1

nX

iD1

ai�.Ei \ Ak /

D1X

kD1

�.Ak /:

That is, �

� 1[

kD1

Ak

�

D1X

kD1

�.Ak /:

[6] We have by [5], that

0 �Z

A

�d� DnX

iD1

ai�.A \ Ei/ �nX

iD1

ai�.A/ D 0:

Thus,R

A�d� D 0.

2

The next result asserts that changing a simple function on a null set does not change the integral.

3.3.3 Corollary

Let .X; †; �/ be a measure space and � a nonnegative simple function. If A and B are measurable sets

such that A � B and �.BnA/ D 0, then

Z

A

�d� DZ

B

�d�:

PROOF.

Noting that B D A [ .BnA/, a disjoint union, we have by Proposition 3.3.2 [4] and [6], that

Z

B

�d� DZ

A[.BnA/

�d� DZ

A

�d�CZ

BnA

�d� DZ

A

�d�C 0 DZ

A

�d�:

2

3.3.2 Integral of a nonnegative measurable function

We showed in Theorem 3.2.8 that every nonnegative measurable function is a pointwise limit of an increas-

ing sequence of of nonnegative simple functions. Using this fact, we are able to define the integral of a

nonnegative measurable function using the integrals of nonnegative simple functions.

45


3.3.4 Definition

Let f be a nonnegative measurable function. The Lebesgue integral of f with respect to �, denoted byR

Xfd�, is defined as

Z

X

fd� D sup

� Z

X

�d� W 0 � � � f; � is a simple function

�

:

If E 2 †, then we define the Lebesgue integral of f over E with respect to � as

Z

E

fd� DZ

f �E

d�:

Notice that if f is a nonnegative simple function, then Definitions 3.3.1 and 3.3.4 coincide.

3.3.5 Proposition

Let f and g be nonnegative measurable functions and c a nonnegative real number.

[1]R

Xcfd� D c

R

Xfd�:

[2] If f � g, thenR

Xfd� �

R

Xgd�:

[3] If A and B are measurable sets such that A � B, then

Z

A

fd� �Z

B

fd�:

PROOF.

[1] If c D 0, then the equality holds trivially. Assume that c > 0. Then

Z

X

cfd� D sup

�Z

X

�d� W 0 � � � cf; � a simple function

�

D sup

�

c

Z

X

�

cd� W 0 � �

c� f; � a simple function

�

D c sup

�Z

X

�

cd� W 0 � �

c� f; � a simple function

�

D c

Z

X

f ı�:

[2] Since f � g, it follows that f� W 0 � � � f; � a simple functiong � f� W 0 � � � g; � a simple functiong.

Thus,Z

X

fd� D sup0��f

Z

X

�d� � sup0��g

Z

X

�d� DZ

X

gd�:

[3] If A � B, then �A

� �B

. Therefore, for any nonnegative measurable function f , we have that

f �A

� f �B

. By [2], it follows that

Z

A

fd� DZ

X

f �A

d� �Z

X

f �B

d� D2B fd�:

46


2

3.3.6 Proposition (Chebyshev’s inequality)

Let .X; †; �/ be a measure space, f a nonnegative measurable function and c a positive real number. Then

�.fx 2 X W f .x/ � cg/ � 1

c

Z

X

fd�:

PROOF.

Let A D fx 2 X W f .x/ � cg. Then, since c � f on A and A � X ,

�A

� �Af � �

Xf D f:

By Proposition 3.3.5, we have thatZ

X

�A

cd� �Z

X

fd�

) c

Z

X

�A

d� �Z

X

fd�

) c

Z

A

d� �Z

X

fd�

) c�.A/ �Z

X

fd�

) �.A/ � 1

c

Z

X

fd�

i.e., �.fx 2 X W f .x/ � cg/ � 1c

R

Xfd�:

2

3.3.7 Proposition

Let .X; †; �/ be a measure space and f a nonnegative measurable function on X . Then,R

Xfd� D 0 if

and only if f D 0 a.e. on X .

PROOF.

Assume thatR

Xfd� D 0. For each n 2 N, let En D

�

x 2 X W f .x/ � 1n

�

: Then by Chebyshev’s

inequality (Proposition 3.3.6),

0 � �.En/ � n

Z

X

fd� D 0:

Hence, �.En/ D 0 for each n 2 N. Since En � EnC1, for each n 2 N and

E D fx 2 X W f .x/ > 0g D1[

nD1

En;

it follows, by Theorem 2.3.6, that

�.E/ D �

� 1[

nD1

En

�

D limn!1

�.En/ D 0:

That is, f D 0 a.e. on X .

47


Conversely, assume that f D 0 a.e. on X . Let � DnX

iD1

ai�Eibe a nonnegative simple function,

written in canonical form, such that � � f . Then � D 0 a.e. on X . Since Ei D fx 2 X W �.x/ D ai; i D

1; 2; : : : ; ng; we have that �

� n[

iD1

Ei

�

D 0. But since

�

� n[

iD1

Ei

�

DnX

iD1

�.Ei/;

we have that �.Ei/ D 0, for each i D 1; 2; : : : ; n. Therefore,

Z

X

�d� DnX

iD1

ai�.Ei/ D 0:

It then follows thatZ

X

fd� D sup0��f

Z

X

�d� D 0:

2

3.3.8 Proposition

Let .X; †; �/ be a measure space and f a nonnegative measurable function. If A 2 † and �.A/ D 0, then

Z

A

fd� D 0:

PROOF.

Let � be a nonnegative simple function such that � � f . Then, by Proposition 3.3.2 [6], we have thatR

A�d� D 0. Therefore

Z

A

fd� D sup0��f

Z

A

�d� D sup0��f

f0g D 0:

2

3.3.9 Theorem (Monotone convergence theorem)

Let .X; †; �/ be a measure space, .fn/ a sequence of measurable functions on X such that 0 � fn � fnC1,

for every n 2 N and limn!1 fn.x/ D f .x/, for each x 2 X . Then f is measurable and

Z

X

fd� DZ

X

�

limn!1

fn

�

D limn!1

Z

X

fnd�:

PROOF.

We have already shown in Corollary 3.2.5 that f is measurable.

Since 0 � fn � fnC1 � f , for each n 2 N, it follows, by Proposition 3.3.5, that

Z

X

fnd� �Z

X

fnC1d� �Z

X

fd�;

for each n 2 N. Hence,

limn!1

Z

X

fnd� �Z

X

fd� (3.3)

48


We now prove the reverse inequality. Let � be a simple function such that 0 � � � f . Choose and fix

� such that 0 < � < 1. For each n 2 N, let

An D fx 2 X W fn.x/ � .1 � �/�.x/g:

Now, for each n 2 N, An is measurable, and since .fn/ is an increasing sequence, we have that An � AnC1.

Furthermore X D1[

nD1

An. Since for each n 2 N, An � X , it follows that

1[

nD1

An � X . For the reverse

containment, let x 2 X . If f .x/ D 0, then �.x/ D 0, and therefore x 2 A1. If on the other hand,

f .x/ > 0, then .1 � �/�.x/ < f .x/. Since fn " f and 0 < � < 1, there is a natural number N such that

.1 � �/�.x/ � fn.x/, for all n � N . Thus, x 2 An, for all n � N , and so, X �1[

nD1

An. Now, for each

n 2 N,Z

X

fnd� �Z

An

fnd� � .1 � �/Z

An

�d�:(3.4)

Define � W † ! Œ0;1� by

�.E/ DZ

E

�d�; E 2 †:

We have already shown in Proposition 3.3.2 [5], that � is a measure on †. Since .An/ is an increasing

sequence of measurable sets such that X D1[

nD1

An, we have, by Theorem 2.3.6 [3], that

�.X / D �

� 1[

nD1

An

�

D limn!1

�.An/:

That is,R

X�d� D limn!1

R

An�d�. Letting n ! 1 in (3.4), we have that

limn!1

Z

X

fnd� � .1 � �/

Z

X

�d�:

Since � is arbitrary, it follows that

limn!1

Z

X

fnd� �Z

X

�d�;

for any nonnegative simple function � such that � � f . Then,

limn!1

Z

X

fnd� � sup

�Z

X

�d� W � is simple and 0 � � � f

�

DZ

X

fd�: (3.5)

From (3.3) and (3.5), we have that

limn!1

Z

X

fnd� DZ

X

fd�:

2

3.3.10 Corollary

Let f and g be nonnegative measurable functions. Then

Z

X

.f C g/d� DZ

X

fd�CZ

X

gd�:

49


PROOF.

By Theorem 3.2.8, there are monotonic increasing sequences .�n/ and . n/ of nonnegative simple

functions such that �n ! f and n ! g, as n ! 1. Then .�n C n/ is a monotonic increasing sequence

which converges to f C g. By the Monotone Convergence Theorem (Theorem 3.3.9), we have that

Z

X

.f C g/d� D limn!1

Z

X

.�n C n/d�

D limn!1

�Z

X

�nd�CZ

X

nd�

�

D limn!1

Z

X

�nd�C limn!1

Z

X

nd�

DZ

X

fd�CZ

X

gd� (by the Monotone Convergence Theorem)

2

3.3.11 Corollary

Let .X; †; �/ be a measure space and .fn/ a sequence of nonnegative measurable functions on X . Then,

for each n 2 N,Z

X

nX

kD1

fkd� DnX

kD1

Z

X

fkd�:

PROOF.

(Induction on n). If n D 1, then the result is obviously true. If n D 2, then, by Corollary 3.3.10, the

result holds. Assume thatZ

X

n�1X

kD1

fkd� Dn�1X

kD1

Z

X

fkd�:

thenZ

X

nX

kD1

fkd� DZ

X

� n�1X

kD1

fk C fn

�

d�

DZ

X

n�1X

kD1

fk CZ

X

fnd� (by Corollary 3.3.10)

Dn�1X

kD1

Z

X

fkd�CZ

X

fnd� (by the induction hypothesis)

DnX

kD1

Z

X

fkd�:

2

3.3.12 Corollary (Beppo Levi)

Let .X; †; �/ be a measure space .fn/ a sequence of nonnegative measurable functions on X and f D

50


P1nD1 fn. Then f is measurable and

Z

X

fd� DZ

X

� 1X

nD1

fn

�

d� D1X

nD1

Z

X

fnd�:

PROOF.

For each m 2 N, let gm DPm

nD1 fn: Then, for each m 2 N, gm is measurable, 0 � gm � gmC1 and

gm ! f as m ! 1. By the Monotone Convergence Theorem (Theorem 3.3.9), we have thatZ

X

fd� D limm!1

Z

X

gmd�

D limm!1

Z

X

mX

nD1

fnd�

D limm!1

mX

nD1

Z

X

fnd� (by Corollary 3.3.11)

D1X

nD1

Z

X

fnd�

2

3.3.13 Corollary

Let .X; †; �/ be a measure space and f a nonnegative measurable function on X . If N is a null set in †,

thenZ

X

fd� DZ

N c

fd�:

PROOF.

By definition,

f �N.x/ D

�

f .x/ if x 2 N

0 if x 2 N c :

That is, f�ND 0 a.e. By Proposition 3.3.7 and Corollary 3.3.10, we have that

Z

X

fd� DZ

X

.f �N

C f �N c / D

Z

X

f �N

d�CZ

X

f �N c d�:

2

In the hypothesis if the Monotone Convergence Theorem (Theorem 3.3.9), we required that the mono-

tone sequence of nonnegative measurable functions converge pointwise to the function f . We can weaken

the ‘pointwise convergence’ to a.e. convergence and require that the limit be a measurable function.

3.3.14 Theorem (Monotone Convergence Theorem)

Let .X; †; �/ be a measure space, .fn/ a sequence of measurable functions on X such that 0 � fn � fnC1,

for every n 2 N and fn !a.e. f , where f is measurable. Then

limn!1

Z

X

fnd� DZ

X

fd�:

51


PROOF.

Since fn !a.e., there is a set N 2 † such that �.N / D 0 and fn ! f pointwise on N c . By Theorem

3.3.9,

limn!1

Z

N c

fnd� DZ

N c

fd�:

From Corollary 3.3.13, it follows thatZ

X

fd� DZ

N c

fd� D limn!1

Z

N c

fnd� D limn!1

Z

X

fnd�:

2

3.3.15 Proposition

Let .X; †; �/ be a measure space and f a nonnegative measurable function on X . Then the set function

� W † ! Œ0;1� defined by

�.A/ DZ

A

fd�; A 2 †

is a measure on X . Furthermore, for any nonnegative measurable function g,Z

X

gd� DZ

X

gfd�:

PROOF.

We note that

�.A/ DZ

A

fd� DZ

X

�Afd�:

Since f � 0, it follows that �.A/ � 0, for each A 2 †. If A D ;, then �A D 0 and so �Af D 0. Therefore

�.;/ D �.A/ DZ

X

�Afd� D

Z

X

0d� D 0:

Let .An/ be a sequence of pairwise disjoint measurable sets and A DS1

nD1 An. Then

�A

D �S1nD1

AnD

1X

nD1

�An:

Therefore, �Af D

P1nD1 �An

f , and so

�.A/ DZ

X

�Afd� D

Z

X

1X

nD1

�Anfd�

D1X

nD1

Z

X

�Anfd� (by Corollary 3.3.12)

D1X

nD1

Z

An

fd�

D1X

nD1

�.An/:

52


That is, �

�

S1nD1 An

�

DP1

nD1 �.An/.

If g D �A

, for some A 2 †, i.e., g is a characteristic function, thenZ

X

gd� DZ

X

�A

d� DZ

A

d� D �.A/

andZ

X

gfd� DZ

X

�Afd� D

Z

A

fd� D �.A/:

Thus,R

Xgd� D

R

Xgfd�. Assume that g D �, a nonnegative simple function and let � D

PniD1 ai�Ei

be the canonical representation of �. Then,

Z

X

gd� DZ

X

�d� DZ

X

� nX

iD1

ai�Ei

�

d�

DnX

iD1

ai

Z

X

�Ei

d� (by Corollary 3.3.11 and Proposition 3.3.5)

DnX

iD1

ai

Z

Ei

d�

DnX

iD1

ai�.Ei/;

andZ

X

gfd� DZ

X

�fd� DZ

X

� nX

iD1

ai�Ei

�

fd�

DnX

iD1

ai

Z

X

�Eifd� (by Corollary 3.3.11 and Proposition 3.3.5)

DnX

iD1

ai

Z

Ei

fd�

DnX

iD1

ai�.Ei/:

Therefore, if g is a nonnegative simple function, thenR

Xgd� D

R

Xgfd�.

Let g be a nonnegative measurable function. Then, by Theorem 3.2.8, there is an increasing sequence

.�n/ of nonnegative simple functions such that �n !n!1 g. Then �nf !n!1 gf . Clearly the �nf are

nonnegative and measurable functions for each n 2 N (see Proposition 3.2.4). By the Monotone Conver-

gence Theorem (Theorem 3.3.9), we have thatZ

X

gd� D limn!1

Z

X

�nd�

D limn!1

Z

X

�nfd�

DZ

X

gfd�:

53


2

3.3.16 Theorem (Fatou’s Lemma)

Let .X; †; �/ be a measure space and .fn/ a sequence of nonnegative measurable functions on X . Then

Z

X

lim infn!1

fnd� � lim infn!1

Z

X

fnd�:

PROOF.

For each k 2 N, let gk D infn�k fn. Then each gk is a nonnegative measurable function. Furthermore,

for each k 2 N, gk � gkC1 and limk!1 gk D lim infn!1 fn. By the Monotone Convergence Theorem

(Theorem 3.3.9), we have thatZ

X

lim infn!1

fnd� D limk!1

Z

X

gkd�: (3.6)

Since gk � fn for all n � k , it follows, by Proposition 3.3.5, that

Z

X

gkd� �Z

X

fnd�; (for all n � k/:

That is, the numberR

Xgkd� is a lower bound for the sequence of numbers

�

R

Xfnd�

�

n�k

. Hence,

Z

X

gkd� � infn�k

Z

X

fnd�;

and consequently

limk!1

Z

X

gkd� � lim infn!1

Z

X

fnd�: (3.7)

From (3.6) and (3.7), we have thatZ

X

lim infn!1

fnd� � lim infn!1

Z

X

fnd�:

2

3.3.3 Integrable functions

Let f be a measurable function on X . Recall that the positive part of f , denoted by f C, is the nonnegative

measurable function

f C D maxff; 0g;and the negative part of f , denoted f �, is the nonnegative measurable function

f � D maxf�f; 0g:

We have already seen that

f D f C � f � and jf j D f C C f �:

Suppose now that f D g � h, where g and h are nonnegative measurable functions. Then f C � g and

f � � h. Indeed, if f � 0, then f C D f . Hence, for any x 2 X ,

f C.x/ D f .x/ D g.x/ � h.x/ � g.x/:

54


If f < 0, then f C D 0 and so, for any x 2 X ,

f C.x/ D 0 � g.x/:

A similar argument shows that f � � h.

3.3.17 Definition

Let .X; †; �/ be a measure space and f a measurable function on X . IfR

Xf Cd� < 1 or

R

Xf �d� < 1,

then we define the integral of f , denoted byR

Xfd�, as the extended real number

Z

X

fd� DZ

X

f Cd� �Z

X

f �d�:

IfR

Xf Cd� < 1 and

R

Xf �d� < 1, then f is said to be (Lebesgue) integrable on X . The set of all

integrable functions is denoted by L1.X; �/.

It is clear that f is integrable if and only ifR

Xjf jd� < 1. In this case,

Z

X

jf jd� DZ

X

f Cd�CZ

X

f �d�:

3.3.18 Remark

Let .X; †; �/ be a measure space and f a measurable function on X . If f is integrable on X , then jf j < 1a.e. on X . Indeed, if A 2 † with �.A/ > 0 and jf j D 1 on A, then, for each n 2 N, jf j > n�

A. Hence,

Z

X

jf jd� �Z

X

n�A

d� D n

Z

A

d� D n�.A/:

Letting n tend to infinity, we have thatR

Xjf jd� D 1.

3.3.19 Theorem

Let .X; †; �/ be a measure space, f; g 2 L1.X; �/ and let c 2 R. Then

[1] cf 2 L1.X; �/ andR

X.cf /d� D c

R

Xfd�.

[2] f C g 2 L1.X; �/ andR

X.f C g/d� D

R

Xfd�C

R

Xgd�.

PROOF.

[1] Assume that c � 0. Then,

.cf /C D maxfcf; 0g D c maxff; 0g D c.f C/ and

.cf /� D maxf�.cf /; 0g D c maxf�f; 0g D c.f �/

Therefore,Z

X

.cf /Cd� DZ

X

cf Cd� D c

Z

X

f Cd� < 1 and

Z

X

.cf /�d� DZ

X

cf �d� D c

Z

X

f �d� < 1:

55


Hence, cf is integrable and

Z

X

.cf /d� DZ

X

.cf /Cd��Z

X

.cf /�d�

D c

Z

X

f Cd�� c

Z

X

f �d�

D c

�Z

X

f Cd��Z

X

f �d�

�

D c

Z

X

fd�:

If c < 0, then c D �k for some k > 0. Therefore

.cf /C D maxfcf; 0g D maxf�kf; 0g D k maxf�f; 0g D kf � D �cf � and

.cf /� D maxf�.cf /; 0g D maxf�cf; 0g D maxfkf; 0g D k maxff; 0g D kf C D �cf C:

From this it follows thatZ

X

.cf /Cd� DZ

X

.�cf �/d� D �c

Z

X

f �d� < 1 and

Z

X

.cf /�d� DZ

X

.�cf C/d� D �c

Z

X

f Cd� < 1:

Therefore, cf is integrable and

Z

X

.cf /d� DZ

X

.cf /Cd� �Z

X

.cf /�d�

D �c

Z

X

f �d�� .�c/

Z

X

f Cd�

D �c

Z

X

f �d�C c

Z

X

f Cd�

D c

�Z

X

f Cd��Z

X

f �d�

�

D c

Z

X

fd�:

[2] Sincef C g D .f C g/C � .f C g/� and

f C g D .f C � f �/C .gC � g�/ D .f CgC/� .f � C g�/;

we have that

.f C g/C � f CgC and .f C g/� � f � C g�:

56


By Proposition 3.3.5 [2], it follows that

Z

X

.f C g/Cd� �Z

X

.f C C gC/d� �Z

X

f Cd�CZ

X

gCd� < 1 and

Z

X

.f C g/�d� �Z

X

.f � C g�/d� �Z

X

f �d�CZ

X

g�d� < 1:

Thus, f C g is integrable. Since .f C g/C � .f C g/� D f C g D f C � f � C gC � g�, we have

that

.f C g/C C f � C g� D .f C g/� C f C C gC:

By Corollary 3.3.10, it follows that

Z

X

Œ.f C g/C C f � C f ��d� DZ

X

Œ.f C g/� C f C C gC�d�

,Z

X

.f C g/Cd�CZ

X

f �d�CZ

X

g�d� DZ

X

.f C g/�d�CZ

X

f Cd�CZ

X

gCd�

,Z

X

.f C g/Cd� �Z

X

.f C g/�d� DZ

X

f Cd�CZ

X

f �d�CZ

X

gCd� �Z

X

g�d�

,Z

X

.f C g/d� DZ

X

fd�CZ

X

gd�:

2

3.3.20 Corollary

Let .X; †; �/ be a measure space. Then L1.X; �/ is a vector space with respect to the usual operations of

addition and scalar multiplication.

3.3.21 Proposition

Let .X; †; �/ be a measure space, f; g 2 L1.X; �/. If f � g a.e. on X , then

Z

X

fd� �Z

X

gd�:

PROOF.

Since g � f � 0 a.e. on X , we have that

Z

X

gd� �Z

X

fd� DZ

X

.g � f /d� � 0:

It now follows thatR

Xfd� �

R

Xgd�.

2

3.3.22 Lemma

Let .X; †; �/ be a measure space. If g is a Lebesgue integrable function on X and f is a measurable

function such that jf j � g a.e., then f is Lebesgue integrable on X .

PROOF.

57


Since jf j D f C C f �, it follows that f C � g and f � � g. Hence,

0 �Z

X

f Cd� �Z

X

gd� and 0 �Z

X

f �d� �Z

X

gd�:

Therefore f C and f � are both integrable.

2

The following theorem, known as Lebesgue’s Dominated Convergence Theorem, provides a useful

criterion for the interchange of limits and integrals and is of fundamental importance in measure theory. As

a motivation for the measure theoretic approach to integration, think about how careful one must be when

interchanging the limits and integrals in Riemann integration.

3.3.23 Theorem (Lebesgue’s Dominated Convergence Theorem)

Let .X; †; �/ be a measure space, .fn/ a sequence of measurable functions on X such that fn !n!1 f

a.e., for f a measurable function. If there is a Lebesgue integrable function g such that for each n 2 N,

jfnj � g a.e., then f is Lebesgue integrable and

limn!1

Z

X

fnd� DZ

X

fd�:

PROOF.

Since jfnj � g a.e., for each n 2 N, it follows that jf j � g a.e. By Lemma 3.3.22, we have that f is

Lebesgue integrable.

Since g ˙ fn � 0, for each n 2 N, we have by Fatou’s Lemma (Theorem 3.3.16), that

Z

X

gd�CZ

X

fd� DZ

X

.g C f /d� � lim infn

Z

X

.g C fn/d�, and (3.8)

Z

X

gd� �Z

X

fd� DZ

X

.g � f /d� � lim infn

Z

X

.g � fn/d�: (3.9)

Now,

lim infn

Z

X

.g C fn/d� DZ

X

gd�C lim infn

Z

X

fnd�, and (3.10)

lim infn

Z

X

.g � fn/d� DZ

X

gd� � lim supn

Z

X

fnd�: (3.11)

Since g is Lebesgue integrable, we have, from (3.8) and (3.10), that

Z

X

fd� � lim infn

Z

X

fnd�: (3.12)

Similarly, from (3.9) and (3.11), we have that

lim supn

Z

X

fnd� �Z

X

fd� (3.13)

From (3.12) and (3.13), it follows that

lim supn

Z

X

fnd� �Z

X

fd� � lim infn

Z

X

fnd�;

58


whence, limn!1

Z

X

fnd� DZ

X

fd�.

2

Note that in the previous theorem, if we replace the almost everywhere convergence with pointwise

convergence, then we can remove the assumption that f is measurable since f would be guaranteed to

be measurable in that case. Furthermore, if we have a complete measure space .X; †; �/, then the almost

everywhere limit of .fn/ is definitely measurable.

3.3.24 Remark

The existence of a Lebesgue-integrable dominating function g in Lebesgue’s Dominated Convergence The-

orem (Theorem 3.3.23) is essential. Let � be the Lebesgue measure on Œ0; 1� and, for each n 2 N, let

fn D n�Œ0; 1

n �. Then fn !a.e. 0 and fn !� 0, but

R 1

0fnd� D 1 6D 0.

For another example, let � again be the Lebesgue measure on R. For each n 2 N, let fn D 1n�

Œ0;n�.

Then fn ! 0 uniformly on R, butR

Rfnd� D 1 6D 0.

3.4 Lebesgue and Riemann integrals

In this section, we show that the Lebesgue integral is a generalization of the Riemann integral. the phrase

‘f is Lebesgue-integrable on Œa; b�’ in this section means that f is integrable with respect to the Lebesgue

measure on Œa; b�.

3.4.1 Theorem

Let f be a bounded real-valued function on Œa; b�. If f is Riemann-integrable on Œa; b� then f is Lebesgue-

integrable on Œa; b� and the two integrals coincide.

PROOF.

Let Œa; b� be a closed and bounded interval of real numbers, f a bounded real-valued function on Œa; b�

and P D fx0; x1; : : : ; xng a partition of Œa; b�. Let

Mi D supff .x/ W xi�1 � x � xig, and mi D infff .x/ W xi�1 � x � xig;

for each i D 1; 2; : : : ; n. Recall that the upper sum of f relative to the partition P is

U.f;P / DnX

iD1

Mi.xi � xi�1/

and the lower sum of f relative to the partition P is

u.f;P / DnX

iD1

mi.xi � xi�1/:

Define u.f;P / W Œa; b� ! R and l.f;p/ W Œa; b� ! R by

u.f;P / D Mi if xi�1 � x � xi ; i D 1; 2; : : : ; n

l.f;P / D mi if xi�1 � x � xi ; i D 1; 2; : : : ; n:

Note that u.f;P / and l.f;P / are simple functions since they can be expressed in the form:

u.f;P / DnX

iD1

Mi�Œxi�1 ;xi �and l.f;P / D

nX

iD1

mi�Œxi�1 ;xi �:

59


Let m be the Lebesgue measure on Œa; b�. It is clear that the upper sum of f relative to the partition P ,

U.f;P /, is just the Lebesgue integral of the simple function u.f;P / and the lower sum of f relative to the

partition P , L.f;P /, is the Lebesgue integral of the function simple l.f;P /.

Let .Pn/ be a sequence of partitions of Œa; b� such that, for each n 2 N, PnC1 is a refinement of Pn and

maxi 4xj D maxj .xi � xi�1/ tends to zero as n tends to infinity. For each n 2 N, let �n D u.f;Pn/ and

n D l.F;Pn/. Then,

1 � 2 � � � � � f � � � � � �2 � �1:

That is, the sequence .�n/ is decreasing and bounded below, while the sequence . n/ is increasing and

bounded above. Hence, in the limit, � �, where � D limn!1 �n D infn �n and limn!1 n Dsupn n. Note that � and are measurable functions. If jf j is bounded by M , then so are the functions

j�nj and j nj, for each n 2 N. Since a constant function M is Lebesgue-integrable on Œa; b�, it follows, by

Lebesgue’s Dominated Convergence Theorem (Theorem 3.3.23), that

Z

Œa;b�

�d� D limn!1

Z

Œa;b�

�nd� D limn!1

U.f;Pn/, and

Z

Œa;b�

d� D limn!1

Z

Œa;b�

nd� D limn!1

L.f;Pn/

Since f is Riemann-integrable on Œa; b�, limn!1 U.f;Pn/ DR b

af .x/dx D limn!1 L.f;Pn/. This

implies thatZ

Œa;b�

d� DZ

Œa;b�

�d� DZ

Œa;b�

�d� ,Z

Œa;b�

.� � /d� D 0:

By Proposition 3.3.7, we have that � D a.e. with respect to the Lebesgue measure � on Œa; b�. Since

� f � �, it follows that D f D � a.e. [�]. It follows that f is Lebesgue integrable and

Z

Œa;b�

d� DZ

Œa;b�

fd� DbZ

a

f .x/dx:

2

3.5 Lp spaces

Let .X; †; �/ be a measure space and 1 � p < 1. Denote by Lp.X; �/ the set of all measurable functions

f W X ! R such that jf jp is integrable; i.e.,

Lp.X; �/ D ff W X ! R W f is measurable and

Z

X

jf jpd� < 1g:

If ˛ 2 R and f 2 Lp.X; �/, then f is measurable and

Z

X

j f jpd� D j˛jpZ

X

jf jpd� < 1;

i.e., f 2 Lp.X; �/.

If f; g 2 Lp.X; �/, then f C g is measurable and, since

jf C gjp � 2p�1.jf jpjgjp/;

60


it follows thatZ

X

jf C gjpd� �Z

X

2p�1.jf jp C jgjp/d� D 2p�1

�Z

X

jf jpd�CZ

X

jgjpd�

�

< 1;

and so f C g 2 Lp.X; �/.

3.5.1 Definition

Let .X; †; �/ be a measure space. A measurable function f W X ! R is said to be essentially bounded

(with respect to �) if there is a constant M > 0 such that

jf .x/j � M a.e. on X (3.14)

The smallest constant M such that (3.14) holds is called the essential supremum of jf j and is denoted by

kf k1; i.e.,

kf k1 D ess supX jf j WD inffM 2 R W jf j � M a.e. on X g:

3.5.2 Remark

Note that f is essentially bounded if there is a constant M > 0 such that

�.fx 2 X W jf .x/j > M g/ D 0I

i.e., the sets of points in X where f fails to be bounded is a null set.

Denote by L1.X; �/ the set of all essentially bounded functions on a measure space .X:†; �/.

3.5.3 Proposition

Let .X; †; �/ be a measure space and 1 � p � 1. Then Lp.X; �/ is a linear space.

PROOF.

Exercise.

2

3.5.1 Holder and Minkowski inequalities

3.5.4 Definition

Let p and q be positive real numbers. If 1 < p < 1 and1

pC 1

qD 1, or if p D 1 and q D 1, or if p D 1

and q D 1, then we say that p and q are conjugate exponents.

3.5.5 Lemma (Young’s inequality)

Let p and q be conjugate exponents, with 1 < p; q < 1, ˛ � 0 and ˇ � 0. Then

˛ˇ � ˛p

pC ˇq

q:

�If p D 2 D q, then the inequality follows from the fact that .˛ � ˇ/2 � 0.

Consider the function f W Œ0;1/ ! R given by

f .˛/ D ˛p

pC ˇq

q� ˛ˇ; for fixed ˇ > 0:

Then f .0/ D ˇq

q> 0; f .ˇq=p/ D 0; and lim

˛!C1f .˛/ D C1. Hence f has a global minimum on

Œ0;1/, i.e., there is an ˛0 2 Œ0;1/ such that f .˛0/ � f .˛/ for all ˛ 2 Œ0;1/: Now, since f 0.˛0/ D 0, it

follows that

0 D f 0.˛0/ D ˛p�10

� ˇ ” ˇ D ˛p�10

” ˛0 D ˇ1

p�1 D ˇq=p :

61


Hence, for all ˛ 2 Œ0;1/,

0 D f .ˇq=p/ � f .˛/ D ˛p

pC ˇq

q� ˛ˇ ” ˛ˇ � ˛p

pC ˇq

q: �

3.5.6 Theorem (Holder’s inequality)

Let 1 � p; q � 1 be conjugate exponents, .X; †; �/ a measure space, f 2 Lp.X; �/ and g 2 Lq.X; �/.

Then fg 2 L1.X; �/ andZ

X

jfgjd� � kf kpkgkq:

PROOF.

Firstly, we consider the case that 1 < p < 1. If kf kp D 0 or kgkq D 0, then the result follows

trivially. We therefore assume that kf kp > 0 and kgkq > 0. Let F D fkf kp

and G D gkgkq

. Then,

F 2 Lp.X; �/ and G 2 Lq.X; �/ and kFkp D 1kGkq . By Young’s inequality (Lemma 3.5.5), we have

that

jFGj � 1

pjF jp C 1

qjGjq :

It follows thatZ

x

jFGjd� � 1

p

Z

X

jF jpd�C 1

q

Z

X

jGjqd� D 1

pC 1

qD 1;

and so,Z

X

jfgjd� � kf kpkgkq:

Now assume that p D 1 and q D 1. Let M D kgk1. Then jfgj � M jf j a.e. on X and soZ

X

jfgjd� �Z

X

M jf jd� � M

Z

X

jf jd� D kgk1kf k1:

2

3.5.7 Theorem

Let .X; †; �/ be a finite measure space and 1 � r < p � 1. Then Lp.X; �/ � Lr .X; �/.

PROOF.

Let t D pr

and s D pp�r

. Then 1t

C 1s

D 1; i.e., s and t are conjugate exponents. Let f 2 Lp.X; �/.

Then, by Holder’s inequality (Theorem 3.5.6),

kf krr D

Z

X

jf jrd� ��Z

X

.jf jr/t d��

1t�Z

X

1sd�

�1s

D�Z

X

.jf jr /pr d�

�rp�Z

X

1sd�

�1s

D kf krpŒ�.X /�

1s :

Hence,

kf kr � kf kp Œ�.X /�1

srD kf kpŒ�.X /�

1r � 1

p < 1;

and so f 2 Lr .X; �/.

2

62


3.5.8 Theorem (Minkowski’s inequality)

Let 1 � 0 � 1. If f; g 2 Lp.X; �/, then f C g 2 Lp.X; �/ and

kf C gkp � kf kp C kgkp:

PROOF.

We have shown that f C g 2 Lp.X; �/. Let q D pp�1

. IfR

Xjf C gjpd� D 0, then the result follows

trivially. Assume then that 2X jf C gjpd� 6D 0. Then

Z

X

jf C gjpd� DZ

X

jf C gjjf C gjp�1d�

�Z

X

jf jjf C gjp�1s� CZ

X

jgjjf C gjp�1d�

��Z

X

jf C gj.p�1/q

�1q��Z

X

jf jpd�

�1p

C�Z

X

jgjpd�

�1p�

D�Z

X

jf C gjpd�

�1q

Œkf kp C kgkp�:

Dividing both sides by

�

R

Xjf C gjpd�

�1q

, gives the desired result, that is,

�Z

X

jf C gjpd�

�1p

D�Z

X

jf C gjpd�

�1� 1q

� kf kp C kgkq , kf C gkp � kf kp C kgkp :

If p D 1, then for each x 2 X ,

jf .x/C g.x/j � jf .x/j C jg.x/j � kf k1 C kgk1:

Thus, f C g 2 L1.X; �/ and

kf C gk1 � kf k1 C kgk1:

2

Let .X; †; �/ be a measure space and 1 � p < 1. Define a real-valued function k � kp on Lp.X; �/ by

kf kp WD�Z

X

jf jpd�

�

;

for f 2 Lp.X; �/.

Note that if f 2 Lp.X; �/, then

kf kp D 0 ,Z

X

jf jpd� D 0 , jf j D 0 a.e. on X , f D 0 a.e. on X:

This shows that k � kp does not define a norm on Lp.X; �/.

The way that we deal with this, is we define a relation � on Lp.X; �/ as follows: For f; g 2 Lp.X; �/,

f � g , f D g a.e. on X:

63


It is easy to show that � defines an equivalence relation on Lp.X; �/. The relation � identifies elements of

Lp.X; �/ that only differ on a null set. We denote by Lp.X; �/ the quotient space Lp.X; �/= �. That is

Lp.X; �/ D Lp.X; �/=N ;

where N D ff 2 Lp.X; �/ W f D 0 a.e. on X g. Note that Lp.X; �/ is a set of equivalence classes

of functions. For f 2 Lp.X; �/, denote by Œf � the equivalence class of f . That is, Œf � is the set of all

functions in Lp.X; �/ that differ from f on a null set.

Similarly, denote L1.X; �/, the quotient space obtained by identifying functions in L1.X; �/ that

coincide almost everywhere.

3.5.9 Remark

We can consider the elements of Lp.X; �/ as functions. This is technically incorrect since Lp.X; �/ is

really the set of equivalence classes of functions. Never-the-less, we will continue to abuse the language

by not distinguishing between elements Œf � of Lp.X; �/ and the function f representing the equivalence

class Œf �.

3.5.10 Theorem

Let .X; †; �/ be a measure space and 1 � p � 1. Define addition and scalar multiplication in Lp.X; �/

as follows:

Œf �C Œg� D Œf C g� and ˛Œf � D Œ f �; where f; g 2 Lp.X; �/ and ˛ 2 R:

Then Lp.X; �/ is a real vector space.

PROOF.

Exercise.

2

Now that we know that Lp.X; �/ is a linear space, we want to define a norm on it. If 1 � p < 1 and

f 2 Lp.X; �/, define k � kp W Lp.X; �/ ! Œ0;1/ by

kŒf �kp D kf kp D�Z

X

jf jpd�

�1p

;

and for p D 1, define

kŒf �kp D kf k1 D esssupX jf j:

Note that it is important that the definition of k � kp (or k � k1) is independent of the choice of a

representative of the equivalence class Œf �, for f 2 Lp.X; �/ (or f 2 Lp.X; �/). Indeed, let f; g 2Lp.X; �/ such that g 2 Œf �. Then Œf � D Œg� and f D g a.e. on X . Therefore,

Z

X

fd� DZ

X

gd�:

3.5.11 Theorem

Let .X; †; �/ be a measure space. Then .Lp.X; �/; k � kp/ and .L1.X; �/; k � k1/ are normed linear

spaces.

3.5.12 Example

Let X D N; † D P.N/, and � be the counting measure on .N;P.N//. Then the functions on N are se-

quences, integration is summation, and integrable functions correspond to absolutely summable sequences.

64


For 1 � p < 1, then Lp.N; �/ D lp.N/, the spaces of real sequences .xn/ such thatP

n2N jxnjp < 1.

The norm on lp.N/ is

k.xn/kp D�

X

n2N

jxnjp�

1p

:

If p D 1, then L1.N�/ D l1.N/, then the spaces of bounded real sequences .xn/ and

k.xn/k1 D supn2N

jxnj:

65

Chapter 4

Decomposition of measures

4.1 Signed measures

In this chapter, we discuss the Hahn decomposition, Jordan decomposition, Radon-Nikodym Theorem and

the Lebesgue decomposition Theorem, which are amongst the most important results in measure theory.

4.1.1 Definition

Let .X; †/ be a measurable space. A signed measure on † is an extended real-valued set function � such

that

(a) � assumes at most one of the values 1 and �1,

(b) �.;/ D 0,

(c) for any sequence .An/ of disjoint measurable sets,

�

�

[

n2N

An

�

DX

n2N

�.An/:

It follows from the above definition that every measure space is a signed measure, but not conversely.

4.1.2 Example

Let .X; †; �/ be a measure space and f a measurable function on X such that f C or f � is integrable.

Define � on † by

�.A/ DZ

A

fd�;

for A 2 †. Then � is a signed measure on †.

4.1.3 Definition

Let � be a signed measure on a measurable space .X; †/. A set A 2 † is said to be a

[1] positive set with respect to � if �.A \ E/ � 0, for any E 2 †.

[2] negative set with respect to � if �.A \ E/ � 0, for any E 2 †.

4.1.4 Remarks

[1] It can be shown tht a set A 2 † is positive (resp., negative) with respect to � if and only if �.E/ � 0

(resp., �.E/ � 0) for every measurable subset E of A.

[2] One can also show that A 2 † is a null set if and only if �.E/ D 0, for all E 2 † such that E � A.

66


[3] Any measurable subset of a positive (resp., negative) set is positive (resp., negative). Also, a countable

union of positive (resp., negative) sets is positive (resp., negative).

We show the latter for positive sets: Let .An/ be a sequence of positive sets in †. then there is a

sequence .Bn/ in† such that Bn � An, for each n 2 N, Bi \Bj D ; if i 6D j and[

n2N

An D[

n2N

Bn.

Let E 2 † such that E �[

n2N

An. Then, since

E D E \�

[

n2N

An

�

D E \�

[

n2N

Bn

�

D[

n2N

.E \ Bn/ (a disjoint set);

it follows that

�.E/ D1X

nD1

�.E \ Bn/:

Since An is positive for each n 2 N and E \ Bn � An, it follows that �.E \ Bn/ � 0, for each

n 2 N. Hence, �.E/ � 0. That is,[

n2N

An is a positive set.

[4] If A is a positive for each set with respect to �, then �.A/ D 0. The converse is not true in general.

4.1.5 Theorem

Let .X; †/ be a measurable space and � a signed measure on †. If E 2 † and 0 < �.E/ < 1, then E

contains a positive set A with �.A/ > 0.

PROOF.

If E is positive with respect to �, then take A D E. Assume that E is not positive. Then there is a set

B 2 † with B � E and �.B/ < 0. Let k1 be the smallest positive integer such that there is an E1 2 †

with E1 � E and �.E1/ < � 1k1

. Since

0 < �.E/C 1

k1

< �.E/ � �.E1/ D �.EnE1/ < 1;

we can re[eat the above argument used for E in the case of EnE1. Let k1 be the smallest positive integer

such that there is an E2 2 † with E2 � EnE1 and �.E2/ < � 1k2

. Then

0 < �.EnE1/C 1

k2

< �.EnE1/ � �.E2/ D �.En.E1 [ E2// < 1:

Continuing in this fashion, we obtain a sequence .kn/ or positive integers such that for each n 2 N, kn is

the smallest positive integer for which there is a measurable set En such that

En � Enn�1[

jD1

Ej

and �.En/ � EnSn�1

jD1 Ej and �.En/ < � 1kn

. If the process stops at ns, then let A D EnSs

jD1 Ej . Then

A is a positive set and �.A/ > 0 (if �.A/ D 0, then �.E/ DsX

jD1

�.Ej / < 0, contradicting the fact that

�.E/ > 0). Otherwise, let A D En1[

jD1

Ej . Then E D A [S1

jD1 �.Ej /, as disjoint union. Hence,

�.E/ D �.A/ C1X

jD1

�.Ej /:

67


If �.A/ D 0, then �.E/ DP1

jD1 �.Ej / < 0, contradicting the hypothesis that �.E/ > 0. Similarly, if

�.A/ < 0, then �.E/ < 0. Thus �.A/ > 0. Since 0 < �.E/ < 1, the series

1X

jD1

�.Ej / converges.

Therefore �1 <

1X

jD1

�.Ej /, whence

1X

jD1

1

kj

< 1. In particular, we have that kj ! 1 as j ! 1. We

show that A is a positive set. Since limn!1 kj D 1, there is a j0 2 N such that kj > 1, for all j � j0.

Let B be a measurable subset of A and j > j0. Then B � Enj[

iD1

Ei and so

�.B/ � � 1

kj � 1:

This is true for all j � j0. Letting j ! 1, we get that �.B/ � 0. Hence A is a positive set.

2

4.1.6 Theorem (Hahn decomposition)

Let � be a signed measure on a measurable space .X; †/. Then there is a pair .P;N / of measurable subsets

of X such that

[1] P is a positive set and N a negative set.

[2] P \ N D ;.

[3] X D P [ N .

PROOF.

Since � assumes at most one of the values of �1 or 1, we may assume that 1 is the value that is

never attained. That is, �.E/ < 1, for all E 2 †. Let } be the collection of all sets in † that are positive

with respect to �. Then } is nonempty since ; 2 }. Let

˛ D supf�.A/ W A 2 }g:

Then ˛ � 0 since ; 2 }. Let .An/ be a sequence in} such that ˛ D limn!1 �.An/ and let P DS

n2N An.

Since a countable union of positive sets is positive, we have that P is positive with respect to �. By the

definition of ˛, we have that ˛ � �.P /. Since PnAn � P , for each n 2 N and P is positive with respect

to �, it follows that �.PnAn/ � 0, for each n 2 N. Therefore,

�.P / D �.An/C �.PnAn/ � �.An/;

for each n 2 N. Thus, ˛ � �.P /. Hence, �.P / D ˛.

Let N D X nP . We show that N is negative with respect to �. Let E be a measurable subset of N . If

�.E/ > 0, then, by Theorem 4.1.5, E contains a subset B such that �.B/ > 0. But then P [ B (a disjoint

union) would be a positive set with �.P [ B/ D �.P / C �.B/ > ˛, contradicting the definition of ˛.

It is clear from the above that P \ N D ; and X D P [ N .

2

4.1.7 Definition

The pair of sets .P;N / as discussed in Theorem 4.1.6, is called a Hahn decomposition of X with respect

to the signed measure �.

The Hahn decomposition .P;N / is not unique. Indeed, if M is a null set with respect to �, then the

pairs .P [ M;N nM / and .PnM;N [ M / are also Hahn decompositions of X with respect to �. Despite

this lack of uniqueness, we have the following proposition.

68


4.1.8 Proposition

Let .X; †/ be a measurable space and � a signed measure on X . If .P1;N1/ and .P2;N2/ are Hahn

decompositions of X with respect to �, then, for all E 2 †,

�.E \ P1/ D �.E \ P2/ and �.E \ N1/ D �.E \ N2/:

PROOF.

We firstly note that since

P1nP2 � P1 and P1nP2 D P1 \ P c2 D P1 \ N2;

we have that, for any E 2 †,

E \ .P1nP2/ � E \ P1 \ P1 and E \ .P1nP2/ D E \ .P1 \ N2/ � N2:

Therefore,

�.E \ .P1nP2// � 0 and �.E \ .P1nP2// � 0:

It now follows that �.E \ .P1nP2// D 0. A similar argument shows that �.E \ .P2nP1// D 0.

Now,E \ P1 D E \ Œ.P1nP2/[ .P1 \ P2/�

D .E \ .P1nP2// [ .E \ .P1 \ P2// (a disjoint union):

Thus,

�.E \ P1/ D �.E \ .P1nP2// C �.E \ .P1 \ P2// D �.E \ .P1 \ P2//:

Similarly,

E \ P2 D E \ Œ.P2nP1/[ .P1 \ P2/�

D .E \ .P2nP1// [ .E \ .P1 \ P2// (a disjoint union):

Therefore

�.E \ P2/ D �.E \ .P2nP1// C �.E \ .P1 \ P2// D �.E \ .P1 \ P2//:

It follows from this that

�.E \ P1/ D �.E \ P2/:

The proof that �.E \ N1/ D �.E \ N2/ is left as an exercise to the reader.

2

4.1.9 Definition

Let � be a signed measure on a measurable space .X; †/ and let fP;N g be a Hahn decomposition of X

with respect to �. The positive variation �C, negative variation ��, and total variation j�j of � are finite

measures defined on † by

�C.E/ D �.E \ P /

��.E/ D ��.E \ N /

j�j.E/ D �.E \ P /� �.E \ N / D �C.E/C ��.E/:

It is clear from Proposition 4.1.8, that the definitions of �C , ��, and j�j are independent of any Hahn

decomposition of X with respect to the signed measure �.

Let .P;N / be a Hahn decomposition of X with respect to a signed measure �. Then X D P [ N and

P \ N D ;. Let E 2 †, then

E D E \ X D E \ .P [ N / D .E \ P /[ .E \ N /, a disjoint union.

69


Therefore,

�.E/ D �.E \ P /C �.E \ N / D �C.E/ � ��.E/:

Hence,

� D �C � ��: (This is called the Jordan decompostion of �.)

Since � assumes at moest one of the values �1 and 1, either �C or �� must be finite. If both �C and ��

are finite, then � is called a finite signed measure.

4.1.10 Example

Let .X; †; �/ be a measure space and f 2 L1.�/. Let � be a set function defined on † by

�.E/ DZ

E

fd�;E 2 †:

Show that � is a signed measure on X and

�C.E/ DZ

E

f Cd�; ��.E/ DZ

E

f �d�; and j�j.E/ DZ

E

jf jd�:

Solution: That � is a signed measure on X follows from Proposition 3.3.15.

Let P D fx 2 X W f .x/ > 0g and P D fx 2 X W f .x/ � 0g. Then X D P [ N and P \ N D ;. Let

E 2 †. Then

�.E \ P / DZ

E\P

fd� DZ

fx2EWf .x/>0g

fd�

DZ

E

f Cd� � 0;

and

�.E \ N / DZ

E\N

fd� DZ

fx2X Wf .x/�0g

fd�

D �Z

E

f �d� � 0:

Thus, P is a positive set with respect to � and N is a negative set with respect to �. that is, .P;N / is a Hahn

decomposition of X with respect to �. By definition of �C; ��, and j�j we have that, for each E 2 †,

�C.E/ D �.E \ P / DZ

E

f Cd�;

��.E/ D ��.E \ N / DZ

E

f �d�;

j�j.E/ D �C.E/C ��.E/ DZ

E

f Cd�CZ

E

f �d� DZ

E

.f C C f �/d� DZ

E

jf jd�:

4.1.11 Definition

Two measures � and � on a measurable space .X; †/ are said to be mutually singular, denoted by � ? �,

if there is an A 2 † such that �.A/ D 0 D �.Ac/.

If � and � are signed measures, then � ? � if j�j ? j�j.

70


4.1.12 Example

Let � be a signed measure on a measurable space .X; †/. Then the measures �C and �� are mutually

singular. Indeed, let .P;N / be a Hahn decomposition of X with respect to �. Then X D P [ N and

P \ N D ;. Therefore,

�C.N / D �.N \ P / D �.;/ D 0 and

��.P / D ��.N \ P / D ��.;/ D 0:

Thus �C ? ��.

4.1.13 Definition

Let � and � be measures on a measurable space .X; �/. We say that � is absolutely continuous with

respect to �, denoted by � � �, if �.E/ D 0, for every E 2 † for which �.E/ D 0.

If � and � are signed measures, then � � � if j�j � j�j.

4.1.14 Example

Let .X; †; �/ be a measure space and f a nonnegative measurable function on X . We have shown in

Proposition 3.3.15 that the set function

�.E/ DZ

E

fd�; E 2 † (4.1)

defines a measure on X . The measure � will be finite if and only if f is integrable. Since the integral over

a set E 2 †, with �.E/ D 0 is zero, it follows that � is absolutely continuous with respect to �.

4.1.15 Proposition

Let �; �, and � be measures on a measurable space .X; †/ and c1; c2 numbers. Then

[1] If � � � and � � �, then .c1�C c2�/ � �.

[2] If � ? � and � ? �, then .c1�C c2�/ ? �.

[3] If � ? � and � ? �, then � D 0.

PROOF.

We prove [3] and leave [1] and [2] as exercises.

Since � ? �, there is a set E 2 † such that �.E/ D 0 D �.Ec/. Using the fact that � � �, it follows

that �.Ac/ D 0. Therefore,

�.X / D �.A [ Ac / D �.A/C �.Ac/ D 0:

Hence, � D 0.

2

The following theorem, known as the Radon-Nikodym Theorem shows that if .X; †; �/ is a �-finite

measure space, then every measure on X that is absolutely continuous with respect to� arises in the manner

indicated by (4.1).

4.1.16 Theorem (Radon-Nikodym Theorem)

Let � and � be �-finite measures on a measurable space .X; †/. Suppose that � is absolutely continuous

with respect to �. Then there is a nonnegative measurable function f on X such that, for each E 2 †,

�.E/ DZ

E

fd�:

Moreover, the function f is uniquely determined almost everywhere Œ��.

71


The function f as described in the Radon-Nikodym Theorem is also called the Radon-Nikodym

derivative of � with respect to �. We write f D d�d�

or d� D fd�.

4.1.17 Theorem (Lebesgue Decomposition Theorem)

Let .X; †; �/ be a �-finite measure space and � a �-finite measure on †. Then there is a unique pair of

measures �a and �s , with � D �a C �s such that �a � � and �s ? �.

PROOF.

Let � D �C�. Then � is a �-finite measure with� � � and � � �. By the Radon-Nikodym Theorem,

there are nonnegative measurable functions f and g on X such that

�.E/ DZ

E

fd� and �.E/ DZ

E

gd�; E 2 †:

Let A D fx 2 X W f .x/ > 0g and B D fx 2 X W f .x/ D 0g. Then X D A [ B, A \ B D ;, and

�.B/ DR

Bfd� D 0. Define �a and �s by

�a.E/ D �.E \ A/ and �s.E/ D �.E \ B/:

Let E 2 †. Then

E D E \ X D E \ .A [ B/ D .E \ A/ [ .E \ B/, a disjoint union.

Therefore,

�.E/ D �.E \ A/C �.E \ B/ D �a.E/C �s.E/ D .�a C �s/.E/:

Hence, � D �a C �s .

Since, �s.A/ D �.A \ B/ D �.;/ D 0, we have that �s.A/ D 0 D �.B/. That is, �s ? �. We now

show that �a � �. Let E 2 † such that �.E/ D 0. Then

0 D �.E/ DZ

E

fd�:

Therefore, f D 0 a.e. Œ��. Now,

0 �Z

A\E

fd� �Z

E

fd� D 0:

Hence, by Proposition 3.3.7,R

A\Efd� D 0. Since f > 0 on A \ E, we must have that �.A \ E/ D 0.

Since � � �, it follows that �.A \ E/ D 0 and therefore �a.E/ D 0. That is, �a � �.

For uniqueness, ;et !a and !s be another pair of measures with � D !a C !s such that !a � � and

!s ? �. Then

�a C �s D !a C !s , �a � !a D !s � �s:

Hence, .�a � !a/ � � and .!s � �s/ ? �. But since �a � !a D !s � �s , it follows that .�a � !a/ ? �.

That is,.�a � !a/ � �; .�a � !a/ ? � and

.!s � �s/ � � .!s � �s/ ? �:

By Proposition 4.1.15, we have that �a � !s D 0 and !s � �s D 0. Hence, �a D !a and �s D !s .

2

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