AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUE … · AN INTRODUCTION TO MEASURE THEORY AND THE...

AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUEINTEGRAL

MARCO M. PELOSO

Contents

1. First elements of measure theory 11.1. Measure spaces 21.2. Measures 42. Abstract integration theory 72.1. Measurable functions 72.2. Integration of non-negative functions 122.3. Integration of complex-valued functions 172.4. The space L1(µ) 193. The Lebesgue measure on R 213.1. Outer measures 213.2. The Lebesgue measure on R 273.3. Examples 333.4. The Cantor set and its generalizations 343.5. Integrals depending on a parameter 363.6. More on L1(m) 384. Product measure spaces and the Lebesgue integral in Rn 414.1. Product measure spaces 414.2. Integration on product measure spaces 444.3. The Lebesgue integral in Rn 504.4. Polar coordinates in Rn 535. Hausdor↵ measaures 575.1. A quick review of submanifolds in Rn 575.2. Hausdor↵ measures 585.3. Hausdor↵ measures in Rn 616. The fundamental theorem of calculus and the theorems of Gauss and Green 656.1. The fundamental theorem of calculus and Gauss theorem 656.2. Extesion to domains with lower regularities 716.3. Green’s Theorem and identities 72References 74

Notes for the course Analisi Matematica 4 per i corsi di Laurea presso Dipartimento di Matematicadell’Universita di Milano, a.a. 2017/18.

A † denotes the proofs and the parts that are not stricly required for the exam for the a.a. 2017/18.– March 6, 2018.

ii M. M. PELOSO

MEASURE THEORY AND LEBESGUE INTEGRAL 1

In these notes we present a concise introduction to abstract measure theory and to theLebesgue integral in euclidean spaces. These notes should be considered only as a supportfor the preparation for the exam, and not as detailed introduction to the subject.

1. First elements of measure theory

We would like to indroduce a notion of measure as a function µ that assigns to every subsetE of Rn a value µ(E) 2 [0,+1] in such a way the following conditions are satisfied:

(i) if E1, . . . , Ek

. . . are disjoint subsets, then µ([k

Ek

) =P

k

µ(Ek

);

(ii) if F is congruent to E, that is, obtained from E by a translation, rotation or reflection,then µ(F ) = µ(E);

(iii) µ(Q) = 1, where Q = [0, 1)n denotes the unit cube.

Unfortunately, these three conditions are mutually incompatible, as the next example shows.

Example 1.1. Assume that n = 1. Let us introduce an equivalent relation ⇠ in [0, 1) by settingx ⇠ y if x � y is rational. Let N be a subset of [0, 1) containing exactly one element for eachequivalent class.1 Let R = Q \ [0, 1) and for r 2 R define

Nr

=�x+ r : x 2 N \ [0, 1� r)

[ �x+ r � 1 : x 2 N \ [1� r, 1)

.

Then we have:

(a)S

r2R Nr

= [0, 1);

(b) if r 6= s are element of R, then Nr

\Ns

= ;.In order to show (a), we notice that clearly each N

r

✓ [0, 1). Next, let x 2 [0, 1) and we showthat it belongs to at least one N

r

. Indeed, let y 2 N be the element that is in the same equivalentclass as x. If x � y, then x 2 N

r

, where x � y = r 2 R, while if x < y, then x 2 Nr

, wherex � y + 1 = r 2 R. Thus, x 2 N

r

for some r 2 R. This proves (a). Finally, if x 2 Nr

\ Ns

,then we would have x� r (or x� r + 1) and x� s (or x� s+ 1) as distinct elements in N butbelonging to the same equivalent class, against the definition of N .

Suppose then that µ : P(R) ! [0,+1] satisfy (i)-(iii). By (i) and (ii) we have that

µ⇣�

x+ r : x 2 N \ [0, 1� r) ⌘

+ µ⇣�

x+ r : �1x 2 N \ [1� r, 1) ⌘

= µ(Nr

)

for every r 2 R. Moreover, by (a), i.e. since [0, 1) is the disjoint union of the Nr

and these setsare countably many, by (ii) and (iii) we have

1 = µ�[0, 1)

�= µ

� [r2R N

r

�=Xr2R

µ(Nr

) =Xr2R

µ(N) .

But this is impossible, since the right-hand-side equals +1 if µ(N) > 0, or it equals 0 ifµ(N) = 0.

This example easily generilizes to the case n > 1.

1To obtain this, we need to assume the validity of the axiom of choice.

2 M. M. PELOSO

1.1. Measure spaces. Led by the previous example, we try to define such function µ on adomain which is strictly contained in P(Rn) but still satisfying (i)-(iii). To this end we introducethe following definitions.

Definition 1.2. Given a set X we call algebra a non-empty collection A of subsets of X thatis closed under finite unions and complements, that is, if the following conditions are satisfied:

(i) if E1, . . . , Em

2 A, then [m

k=1Ek

2 A;

(ii) if E 2 A, then cE 2 A.

Notice that, if E 2 A, X = E [ cE 2 A, so that also ; = cX 2 A.A non-empty collection A of subsets of X is called a �-algebra if it is an algebra and it is

closed under countable unions. In other words, if the following conditions are satisfied:

(i’) if E1, E2, . . . ,2 A, then [+1k=1Ek

2 A;

(ii’) if E 2 A, then cE 2 A.

Notice that a �-algebra A is closed under countable intersections, since�\+1

j=1 Ej

�= c

�[+1j=1

cEj

� 2 A.We observe that an algebra A that is closed under countable unions of disjoint subsets of X ,

is a �-algebra. Indeed, given a sequence {Ek

} of elements in A, define

F1 = E1, Fk

= Ek

\⇣ k�1[

j=1

Ej

⌘= E

k

\c

⇣ k�1[j=1

Ej

⌘. (1)

Then the Fk

are disjoint and [+1k=1Ek

= [+1k=1Fk

, which belongs to A by assumption.

Example 1.3. Let X be any set.

(1) P(X ) is a �-algebra;

(2) if {;,X} is a �-algebra;

(3) if X is uncountable and we set A =�E : E is countable, or cE is countable2

, then A

is a �-algebra.

It is easy to see that the intersection of �-algebras is again a �-algebra. Therefore, thefollowing definition makes sense.

Definition 1.4. Given any subset E of P(X ), we call M(E) the �-algebra generated by E asthe smallest �-algebra containing E , that is, the intersection of all �-algebras containing E .

Observe that P(X ) is a �-algebra containing E , so the above intersection is not empty. Thefollowing lemma is elementary, but it deserves its own statement.

Lemma 1.5. If A is a �-algebra, and E ✓ M(F) ✓ A, then M(E) ✓ M(F).

Proof. Since M(F) is a �-algebra containing E , it contains the smallest �-algebra containing E ,i.e. M(E). ⇤

The above definition leads us to the following fundamental notion.

Definition 1.6. If X is a topological space, we call the Borel �-algebra in X , and we denote itby BX , the �-algebra generated by the collections of open sets in X , and a set E 2 BX a Borelset in X .

2In this case we say that E is co-countable.


Observe that if F is a closed set, then F 2 BX , so are countable unions of closed sets, theircomplements, countable unions of such sets, etc.

We call a countable intersection of open sets a G�

-set, and countable union of closed sets anF�

-set.If X = R is endowed with its natural topology, we denote by BR the �-algebra of Borel sets in

R (with respect to this topology). It is going to play a fundamental role in remeinder of thesenotes.

The following are elementary consequences of the definition.

Proposition 1.7. The �-algebra BR is generated by each of the following sets:

(1) the collection of open intervals E1 =�(a, b) : a < b

;

(2) the collection of closed intervals E2 =�[a, b] : a < b

;

(3) the collection of half-open intervals E3 =�(a, b] : a < b

, or E 0

3 =�[a, b) : a < b

;

(4) the collection of open rays E4 =�(�1, b) : b 2 R

, or E 0

4 =�(a,+1) : a 2 R

;

(5) the collection of closed rays E5 =�(�1, b] : b 2 R

, or E 0

5 =�[a,+1) : a 2 R

.

Proof. The elements of the Ej

and E 0j

are all open, or closed, or intersection of open and closedsets, so that each of the �-algebras in (1)-(5) is contained in BR. Conversely, it is clear thatM(E1) ◆ BR since each open set is countable union of elements of E1. In all remaining cases,it su�ces to show that open sets are contained in M(E

j

) or M(E 0j

). By symmetry, we consider

only the cases of M(Ej

), j = 1, . . . , 5. Clearly, (a, b) = [+1k=1

⇥a� 1

k

, b� 1k

⇤, so (2) follows. Next,

(a, b) = [+1k=N

(a, b� 1k

], where N is chosen so that a < b� 1N

. This shows (3). To prove (4) weconsider [b,+1) = c(�1, b) and (�1, a) \ [b,+1) = [b, a), if b < a. Thus, (4) follows fromthe case (3) (for E 0

3). Case (5) is analogous and left to the reader. ⇤A similar description can be given for the �-algebra of Borel sets in Rn. If X1, . . . ,Xn

is acollection of non-empty sets, X =

Qn

j=1Xj

, we denote by ⇡j

: X ! Xj

the coordinate maps. IfM

j

is a �-algebra on Xj

, j = 1, . . . , n, we setnO

j=1

Mj

= M⇣�⇡�1j

(Ej

) : Ej

2 Mj

, j = 1, . . . , n ⌘

that is, the �-algebra generated by ⇡�1j

(Ej

) where Ej

2 Mj

, j = 1, . . . , n.

Proposition 1.8. We have that

BRn =nO

j=1

BR .

Proof. (†) Let Ej

2 BR, j = 1, . . . , n. Then ⇡�11 (E1) = E1 ⇥ R ⇥ · · ·R, and similarly for

⇡�1j

(Ej

). ThenQ

n

j=1Ej

= \n

j=1⇡�1j

(Ej

). Hence, ⌦n

j=1BR contains the �-algebra generated

by�Q

n

j=1Ej

: Ej

open in R

=: F , i.e.⌦n

j=1BR ◆ M(F). On the other hand, for each

j = 1, . . . , n, the set�E

j

✓ R : ⇡�1j

(Ej

) 2 M(cF ) is a �-algebra, that contains BR. Hence,

nOj=1

BR = M⇣� nY

j=1

Ej

: Ej

open in R ⌘

.

This implies that ⌦n

j=1BR ✓ BRn .

4 M. M. PELOSO

To see the reverse inclusion, let E be any open set in Rn and consider its points with rationalcoordinates, which is a dense subset in E. Take a ngbh, contained in E, of any such point thatis a cartesian product of open sets in R in each coordinate. Then E is union of open sets thatare of the form

Qn

j=1Ej

with Ej

open in R. Thus,

BRn ✓ M⇣� nY

j=1

Ej

: Ej

open in R ⌘

=nO

j=1

BR .

This proves the proposition. ⇤

1.2. Measures. Let X be a set and M a �-algebra in P(X ). The pair (X ,M) is called ameasurable space.

Definition 1.9. Given a measurable space (X ,M), a function µ : M ! [0,+1] is called ameasure on X if

(i) µ(;) = 0;

(ii) if E1, E2, · · · 2 M, are disjoint, then µ⇣+1[

j=1

Ej

⌘=

+1Xj=1

µ(Ej

).

The triple (X ,M, µ) is called a measure space.

Property (ii) is called countable additivity, in contrast with an analogous property

(ii’) if E1, E2, . . . , Em

2 M, are disjoint, then µ⇣[m

j=1 Ej

⌘=P

m

j=1 µ(Ej

),

which will be called finite additivity. Notice that a measure µ satisfies also (ii’) since we maytake E

k

= ; for all k � 1.Given a measure space (X ,M, µ), µ is said to be finite if µ(X ) < +1 and is said to be �-finite

if there exists a collection of sets {Ej

}, with Ej

2 M, [+1j=1Ej

= X and such that µ(Ej

) < +1.In other words, µ is said to be �-finite if X is countable union of sets of finite measure. Ourtreatment will essentially concern only with �-finite measures.

Here are some simple examples of measure spaces.

Example 1.10. (1) Let X be any set, and let M = P(X ). Define the counting measure onX by setting

µ(E) =

8><>:+1 if E is infinite

m if E contains exactly m elements

0 if E = ; .Then µ is a measure. Notice that µ({x}) = 1 for all x 2 X , and that it is �-finite if Xis countable, and it is finite if X is finite.

(2) A particular case of (1) is when X is countable, or more in particular, equals N or Z.(3) Instead, a generalization of (1) is the following. Let f : X ! [0,+1] be given and define

µ(E) =Xx2E

f(x) .

Since f(x) is non-negative, there is no ambiguity in the above definition even if E isuncountable. For, if there exist uncountably many x 2 E such that f(x) > 0, then thereexists n 2 N for which E

n

=�x 2 E : f(x) > 1/n

is infinite. Then,

Px2E f(x) �

MEASURE THEORY AND LEBESGUE INTEGRAL 5Px2E

n

f(x) = +1. If there exist at most countably many x 2 E such that f(x) > 0,then the sum becomes a series, and the notion of convergence is the standard one. It iseasy to chech again that such µ is a measure.

A particular case is when f is the function that equals 1 at a given x0 2 X and it is0 anywhere else. Such measure, is called the Dirac delta at x0.

(4) if X is uncountable, let A be the �-algebra of countable or co-countable sets. Define

µ(E) =

(0 if E is countable

1 if E is co-countable .

Again, it is easy to check that such µ is a measure on A.

The first elementary properties of measures are given in the next result.

Proposition 1.11. Let (X ,M, µ) be a measure space. Then, the following properties hold true.

(i) Monotonicity: If E,F 2 M with E ✓ F , then µ(E) µ(F ).

(ii) Subadditivity: If {Ej

} ⇢ M, then

µ⇣+1[

j=1

Ej

⌘

+1Xj=1

µ(Ej

) .

(iii) Continuity from below: If {Ej

} ⇢ M and E1 ✓ E2 ✓ · · · , then

limj!+1

µ(Ej

) = µ⇣+1[

j=1

Ej

⌘.

(iv) Continuity from above: If {Ej

} ⇢ M and E1 ◆ E2 ◆ · · · and µ(E1) < +1 then

limj!+1

µ(Ej

) = µ⇣+1\

j=1

Ej

⌘.

Proof. (i) Since F ◆ E, we have that

µ(F ) = µ�(F \ E) [ (F \ E)

�= µ

�(F \ E) [ E

�= µ(F \ E) + µ(E) � µ(E) .

(ii) Given {Ej

}, we define {Fj

} as in (1), that is, we set

F1 = E1, Fk

= Ek

\⇣ k�1[

j=1

Ej

⌘for k � 2 .

Since the {Fj

} are disjoint and [+1j=1Fj

= [+1j=1Ej

, by (i) we have that

µ⇣+1[

j=1

Ej

⌘= µ

⇣+1[j=1

Fj

⌘=

+1Xj=1

µ(Fj

) +1Xj=1

µ(Ej

) .

(iii) Let now {Ej

} ⇢ M and E1 ✓ E2 ✓ · · · . Setting E0 := ;, we have that the sets Ej

\Ej�1

are disjoint and [+1j=1Ej

\ Ej�1 = [+1

j=1Ej

. Then,

µ⇣+1[

j=1

Ej

⌘=

+1Xj=0

µ⇣E

j

\ Ej�1

⌘= lim

n!+1

nXj=0

µ⇣E

j

\ Ej�1

⌘= lim

n!+1µ(E

n

) .

6 M. M. PELOSO

(iv) Set Fj

= E1 \Ej

. Then F1 ✓ F2 ✓ · · · , and µ(E1) = µ(Fj

) + µ(Ej

), since Ej

and Fj

aredisjoint, and [+1

j=1Fj

= E1 \ \+1j=1Ej

. Finally, using (iii) we have

µ(E1) = µ⇣+1\

j=1

Ej

⌘+ lim

j!+1µ(F

j

) = µ⇣+1\

j=1

Ej

⌘+ lim

j!+1µ(E1)� µ(E

j

) .

Since µ(E1) < +1, we can substract it from both side of the equation to obtain (iv). ⇤

We remark that the assumption µ(E1) < +1 could be replaced by µ(En

) < +1 for somen, but the finiteness of some µ(E

n

) is necessary. For, consider the case of (N,P(N), µ), whereµ is the counting measure, and the sets E

j

= {n : n � j}. Then, µ(Ej

) = +1 for all j butµ� \+1

j=1 Ej

�= µ(;) = 0.

Let (X ,M, µ) be a measure space. A very important class of sets, is the class of sets of measurezero, also called null sets, that is, the sets E such that µ(E) = 0. By countable subadditivity,a countable union of null sets is again a null set. By monotonicity, if E 2 M, F ✓ E, F 2 Mand µ(E) = 0, then also µ(F ) = 0. But, in general, it is not true that F 2 M.

Definition 1.12. A measure space (X ,M, µ) is said to be complete if M contains all subsetsof null sets, that is, for every E 2 M with µ(E) = 0 and F ✓ E, we have F 2 M.

The next result shows that any measure space can be extended to a complete measure space.

Theorem 1.13. Let (X ,M, µ) be a measure space and let

M = M [N =�E : E = E [N with E 2 M, N 2 N

,

where N =�E 2 P(X ) : there exists F 2 M with µ(F ) = 0

. If E [ N 2 M, with E 2 M

and N 2 N , define

µ�E [N

�= µ(E) .

Then (X ,M, µ) is a complete measure space, called the completion of (X ,M, µ).

Observe that the completion of (X ,M, µ) is uniquely determined, in the sense that once Mis constructed, there exists a unique measure µ on M that is complete and restricted to Mcoincides with µ.

Proof. We begin by showing that M is a �-algebra. If {Ej

} ✓ M, then Ej

= Ej

[ Nj

, whereE

j

2 M and Nj

2 N , for all j. Since Nj

2 N , there exists Fj

2 M such that Nj

✓ Fj

andµ(F

j

) = 0. Therefore, [+1j=1Nj

✓ [+1j=1Fj

=: F , where µ(F ) P+1j=1 µ(Fj

) = 0. Therefore,

[+1j=1Nj

2 N . Hence,

+1[j=1

Ej

=+1[j=1

�E

j

[Nj

�=⇣+1[

j=1

Ej

⌘[⇣+1[

j=1

Nj

⌘2 M [N = M .

Next, let E 2 M and N 2 N . We need to show that c(E [ N) 2 M. Then there existsF 2 M such that N ✓ F and µ(F ) = 0. By possibly replacing N by N \ E, we may assumethat E \ N = ;, and then, replacing F by F \ cE, also that E \ F = ;. In this case,


E [N =�E [ F

� \ �cF [N

�and3

c

�E [N

�= c

�E [ F

� [ �F \N�

.

Notice that c

�E [ F

� 2 M, while F \N ✓ F , with µ(F ) = 0, so that F \N 2 N . This showsthat M is a �-algebra.

We now show that µ is well defined and is a complete measure. If E1 [ N1 = E2 [ N2

with Ej

2 M and Nj

2 N , with Nj

✓ Fj

and µ(Fj

) = 0, j = 1, 2, then E1 ✓ E2 [ F2 andµ(E1) µ(E2)+µ(F2) = µ(E2). Thus, µ(E1) µ(E2). Arguing in the same way we obtain thereverse inequality so that µ(E1) = µ(E2). Therefore, µ is well defined.

Next, let {Ej

[Nj

} be a sequence of disjoint sets in M. If Fj

2 M, Nj

✓ Fj

and µ(Fj

) = 0,then

µ⇣+1[

j=1

(Ej

[Nj

)⌘= µ

⇣�+1[j=1

Ej

� [ �+1[j=1

Nj

�⌘= µ

⇣+1[j=1

Ej

⌘=

+1Xj=1

µ(Ej

) =+1Xj=1

µ(Ej

[Nj

) .

It follows that µ is a measure, and it is clear that it is complete. ⇤

2. Abstract integration theory

In order to begin our approach to the theory of integration, we need to discuss the notion ofmeasurable functions.

2.1. Measurable functions. In analogy with the definition of continuous functions, as mor-phisms between topological spaces, we have the following definition.

Definition 2.1. Let (X ,M), (Y,N ) be measurable spaces, and f : X ! Y be given. We saythat f is (M,N )-measurable if f�1(E) 2 M for all E 2 N .

It is clear that if f : X ! Y is (M,N )-measurable and g : Y ! Z is (N ,O)-measurable,then g � f : X ! Z is (M,O)-measurable. Also, observe that if N is a �-algebra in P(Y),then

�f�1(E) : E 2 N

is a �-algebra in P(X ). In fact, [j

f�1(Ej

) = f�1� [

j

Ej

�and

c

�f�1(E)

�= f�1( cE), and the conclusion follows.

Proposition 2.2. Let f : X ! Y be (M,N )-measurable and suppose that N is generated by E.Then f is (M,N )-measurable if and only if f�1(E) 2 M for all E 2 E.

Hence, if X ,Y are topological spaces, and f : X ! Y continuous, then f is (BX ,BY)-measurable.

Proof. First of all, we observe that the implication “only if” is trivial. Conversely, suppose thatf�1(E) 2 M for all E 2 E . It is easy to check that

�E ✓ Y : f�1(E) 2 M

is a �-algebra inP(Y) that contains E . Hence, it contains the �-algebra generated by E , that is, it contains Nand f is (M,N )-measurable. The second conclusion now is obvious. ⇤

3Indeed, recall that A \B = A \ cB, then, since E \ F = ;,

E [N = (E [ F ) \ (F \N) = (E [ F ) \ c(F \ cN) = (E [ F ) \ ( cF [N).

8 M. M. PELOSO

Definition 2.3. If (X ,M) is a measurable space, then f : X ! R is said to be measurable if itis (M,BR)-measurable.

It is however convenient to consider functions that take value in the extended reals, that is,in R = R [ {±1} = [�1,+1]. In this case, the �-algebra of Borel sets BR is defined as�E ✓ R : E \R 2 BR

. Then we say that f : X ! R is measurable if it is (M,BR)-measurable.

Proposition 2.4. Let (X ,M) be a measurable space, and let f : X ! R be given. Then thefollowing conditions are equivalent.

(i) f is measurable, that is, it is (M,BR)-measurable.

(ii) f�1�[�1, b)

� 2 M for every b 2 R.(iii) f�1

�[�1, b]

� 2 M for every b 2 R.(iv) f�1

�(a,+1]

� 2 M for every a 2 R.(v) f�1

�[a,+1]

� 2 M for every a 2 R.

Proof. If f : X ! R, the conclusions follow at once from Prop.’s 1.7 and 2.2. If f takes valuesin [�1,+1], the conclusions are a simple consequence of Def. 2.3. ⇤

The following simple result will be needed later on.

Lemma 2.5. Let (X ,M) be a measurable space, and let f : X ! C be given. Then f ismeasurable if and only if Re f and Im f are measurable.

Proof. (†) We identify f with (f1, f2) : X ! R2. Observe that the coordinate maps ⇡j

:R2 ! R, j = 1, 2 are measurable. Since composition of measurable functions is measurable andf1 = Re f = ⇡1 � f , f2 = Im f = ⇡2 � f , f measurable implies Re f, Im f are measurable.

Conversely, suppose Re f and Im f are measurable. By Prop. 1.8 we know that BC = BR2 =BR ⌦ BR. Then, if E 2 BC, E = E1 ⇥ E2, with E1, E2 2 BR and f�1(E) = f�1

1 (E1) \ f�12 (E2)

that is in M by assumption. ⇤We remark that when we require that f takes values in R we include the case of f having

finite values.

Theorem 2.6. Let (X ,M) be a measurable space.

(i) If f, g : X ! R are measurable, then f + g, fg are measurable.

(ii) If fj

: X ! R are measurable, j = 1, 2, . . . , then

g1 = supj

fj

, g2 = infj

fj

, g3 = lim supj!+1

fj

, g4 = lim infj

fj

are measurable.

(iii) f, g : X ! R are measurable, then

max(f, g), min(f, g)

are measurable.

(iv) If fj

: X ! C are measurable, j = 1, 2, . . . , and g(x) = limj!+1 f

j

(x) exists, then g ismeasurable.

Proof. (i) Observe that f + g = S � F , where F (x) = (f(x), g(x)) and S : C ⇥ C ! C is thesum-funciton, i.e. S(z + w) = z + w. Arguing as in Lemma 2.5 we see that F is measurable.


Since S is continuous is measurable, and then f + g is measurable. Replacing S by P , whereP (z, w) = zw, we obtain the measurability of fg.

(ii) Notice that

g�11

�(a,+1]

�=�x 2 X : sup

j

fj

(x) > a =

+1[j=1

f�1j

�(a,+1]

�,

while

g�12

�[�1, b)

�=�x 2 X : inf

j

fj

(x) < b =

+1[j=1

f�1j

�[�1, b)

�.

The measurability of g1 and g2 follows from Prop. 2.4.Next, we observe that4

g3(x) = lim supj!+1

fj

(x) = infk

�supj�k

fj

(x)�

Then hk

(x) = supj<k

fj

(x) are measurable, so is infk

hk

(x) = g3(x). The argument for g4 isanalogous, since

g4(x) = lim infj!+1

fj

(x) = supk

�infj�k

fj

(x)�.

Finally, (iii) and (iv) are trivial consequences of (ii) ⇤

Corollary 2.7. If f : X ! R is measurable, then f+ = max(f, 0) and f� = �min(f, 0) aremeasurable. If g : X ! C is measurable, then |g| and sgn g := g/|g| are measurable.

We point out that f = f+ � f�, |f | = f+ + f�, and g = sgn g · |g|.Proof. We only need to prove the statements for |g| and sgn g. But these are elementary andwe leave the details to the reader. ⇤

Definition 2.8. Let (X ,M) be a measurable space, and let E 2 M. We define the characteristicfunction of E as the function

�E

(x) =

(1 if x 2 E

0 otherwise .

We call a simple function a finite linear combination with complex coe�cients of characteristicfuncitons of measurable sets E

j

f(x) =nX

j=1

cj

�E

j

(x) . (2)

Observe that simple functions can characterized as the measurable functions whose range isfinite (that is, they attain at most finitely many values). We remark that the representation (2)of a simple function f is not unique, but it becomes unique if we write

f(x) =mXj=1

dj

�F

j

(x) .

4We recall that, given a sequence {an

}, lim supn

an

= s⇤, where s⇤ is sup of the limit points of {an

}. Then,s⇤ = inf

k

(supn�k

an

) = limk!1(sup

n�k

an

). Analogously, if we set s⇤ = lim infn

an

, then s⇤ = supk

(infn�k

an

) =

limk!1(inf

n�k

an

).

10 M. M. PELOSO

where Fj

= f�1({dj

}), dj

6= 0, and�d1, . . . , dm

= f

�X � \ {0} is the range of f taken away thevalue 0.

Notice also that finite sums and products of simple functions are again simple functions.

The main result of this section is that measurable can be suitable approximated with simplefunctions, as the next result shows.

Theorem 2.9. Let (X ,M) be a measurable space.(1) Let f : X ! [0,+1] be a measurable funtion. Then there exists an increasing sequence of

non-negative measurable functions 0 s1 s2 · · · f such that sn

(x) ! f(x) as n ! +1and the convergence is uniform on all sets where f is bounded.

(2) Let f : X ! C be a measurable funtion. Then there exists a sequence of complex-valuedsimple measurable functions {'

n

} such that 0 |'1| |'2| · · · |f | such that 'n

(x) ! f(x)as n ! +1 and the convergence is uniform on all sets where f is bounded.

Proof. (1) For n = 0, 1, 2, . . . and k integer, 0 k 22n � 1 we define the sets

Fn

= f�1�(2n,+1]

�E

n,k

= f�1�(k2�n, (k + 1)2�n]

�,

and set

sn

(x) = 2n�F

n

(x) +22n�1Xk=0

k2�n�E

n,k

(x) .

Clearly {sn

} are measurable and non-negative. In order to check that {sn

} is increasing, noticethat�

k2�n, (k + 1)2�n

⇤=�2k2�(n+1), (2k + 2)2�(n+1)

⇤=�2k2�(n+1), (2k + 1)2�(n+1)

⇤ [ �(2k + 1)2�(n+1), (2k + 2)2�(n+1)

⇤.

This implies that

En,k

= En+1,2k [ E

n+1,2k+1 . (3)

Moreover,

�2n, 2n+1

⇤=

22(n+1)�1[j=22n+1

�j2�(n+1), (j + 1)2�(n+1)

⇤,

so that

Fn

\ Fn+1 =

22(n+1)�1[j=22n+1

En+1,j . (4)


Therefore, using (3) and (4),

sn

(x) = 2n�F

n

(x) +22n�1Xk=0

k2�n

⇣�E

n+1,2k(x) + �E

n+1,2k+1(x)⌘

= 2n��F

n+1(x) + �F

n

\Fn+1

(x)�+

22n�1Xk=0

2k2�(n+1)⇣�E

n+1,2k(x) + �E

n+1,2k+1(x)⌘

2n��F

n+1(x) + �F

n

\Fn+1

(x)�

+22n�1Xk=0

2k2�(n+1)�E

n+1,2k(x) +22n�1Xk=0

(2k + 1)2�(n+1)�E

n+1,2k+1(x)

= 2n�F

n+1(x) + 2n22(n+1)�1Xj=22n+1

�E

n+1,j (x) +22n+1�1X

`=0

`2�(n+1)�E

n+1,`(x)

2n+1�F

n+1(x) +22(n+1)�1Xj=22n+1

j2�(n+1)�E

n+1,j (x) +22n+1�1X

`=0

`2�(n+1)�E

n+1,`(x)

= sn+1(x) .

Hence, {sn

} is monotone increasing. It is clear that sn

(x) f(x) for all x. Finally, notice that forx /2 F

n

, 0 f(x)� sn

(x) 2�n. Then, sm

! f uniformly on each set cFn

=�x : f(x) 2n

.

On F = \n

Fn

, f(x) = +1 and sn

(x) ! +1 for x 2 F . This proves (1).(2) now follows easily from (1). If f : X ! C, we write f = u+ iv, u = u+�u�, v = v+� v�.

Then, there exists increasing sequences {s±n

} and {�±n

} of non-negative simple functions suchthat

s±n

! u±, and �±n

! v± ,

as in (1). Then, setting

'n

= s+n

� s�n

+ i��+n

� ��n

�we have that '

n

! u+ � u� + i(v+ � v�) = f , pointwise, and uniformly on all sets where f isbounded. Finally, it also holds that

|'n

| = ⇥(s+

n

� s�n

)2 + (�+n

� ��n

)2⇤1/2

=⇥(s+

n

)2 + (s�n

)2 + (�+n

)2 + (��n

)2⇤1/2

⇥(s+

n+1)2 + (s�

n+1)2 + (�+

n+1)2 + (��

n+1)2⇤1/2

|'n+1|

for all n, where we have used the obvious fact that g+g� = 0 for all g. ⇤

Definition 2.10. Given a measure space (X ,M, µ), we say that a property (P) is valid µ-a.e.(or, simply, a.e. if the measure µ is understood by the context), if there exists a set F 2 M,with µ(F ) = 0 such that the property (P) holds on cF .

One convenience of working with complete measures is the following result.

Proposition 2.11. Let (X ,M, µ) be a measure space The following implications are true if andonly if the measure µ is complete.

12 M. M. PELOSO

(1) If f is measurable and g = f µ-a.e., then g is measurable.

(2) If fn

are measurable, n = 1, 2, . . . and fn

! f pointwise µ-a.e., then f is measurable.

Proof. (†) Exercise. ⇤On the other hand, we have the following result.

Proposition 2.12. Let (X ,M, µ) be a measure space and let (X ,M, µ) be its completion. If fis M-measurable, then there exists g M-measurable function such that f = g µ-a.e.

Proof. The proof is indicative of a method that will be often used. Suppose first f is a charac-teristic function of an M-measurable set E = E [ N , with E 2 M, N ✓ F , µ(F ) = 0. Thenf = �

E

= �E

µ-a.e. and the conclusion holds true in this case. In the same way we see that theconclusion follows in the case of a simple function f =

Pn

j=1 cj�E

j

.

In the general case, given f an M-measurable function, by Thm. 2.9 we can find a sequenceof M-measurable functions {'

n

} pointwise converging to f . For each n, let n

be an M-measurable function,

n

= 'n

except on a set Nn

2 M with µ(Nn

) = 0. Then, there existFn

2 M, Fn

◆ Nn

and µ(Fn

) = 0. Set F = [n

Fn

and define g = limn!+1 � c

F

n

. Then g isM-measurable by the previous Prop. 2.11 (2), and g = f on cF , that is, µ-a.e. ⇤

2.2. Integration of non-negative functions. We begin the construction of the integral withthe simple functions.

Definition 2.13. Let (X ,M, µ) be a measure space. We define

L+ =�f : X ! [0,+1], M�measurable

. (5)

If s 2 L+ is a simple function s =P

n

j=1 cj�E

j

we define the integral of s with respect to µ,Zs dµ =

nXj=1

cj

µ(Ej

) ,

(with the standard convention that 0 ·1 = 0). More generally, if E 2 M we also defineZE

s dµ =

Z�E

s dµ =nX

j=1

cj

µ(E \ Ej

) .

We will use the following notation to denote the integral of s w.r.t. µ:Zs dµ =

Zs =

ZXs dµ =

ZXs(x) dµ(x)

and similarly in the case ofRE

s dµ.

The following properties are easy to check.

Proposition 2.14. Let (X ,M, µ) be a measure space and s,� be non-negative simple functionson X . Then, the following properties hold true.

(i) If c � 0, then

Zcs dµ = c

Zs dµ ;

(ii)

Z(s+ �) dµ =

Zs dµ+

Z� dµ;


(iii) if s �, then

Zs dµ

Z� dµ;

(iv) the map M 3 E 7!ZE

s dµ is a measure.

We are now in the position to define the integral of non-negative measurable functions.

Definition 2.15. Let (X ,M, µ) be a measure space. We define

L+ =�f : X ! [0,+1], M�measurable

. (6)

If f 2 L+ we set Zf dµ = sup

nZs dµ : 0 s f, s simple

o. (7)

For any E 2 M we also define ZE

f dµ =

Zf�

E

dµ .

It follows at once that, if c � 0 and f 2 L+, then cf 2 L+ andRcfdµ = c

Rfdµ. It also

follows at once that if f, g 2 L+ and f g, thenZf dµ

Zg dµ . (8)

The next result is a fundamental building block of this theory.

Theorem 2.16. (Monotone Convergence Theorem.) Let (X ,M, µ) be a measure space{f

n

} be given such that fn

: X ! [0,+1], 0 f1 f2 · · · . Setting f(x) = limn!+1 f

n

(x),we have Z

f dµ = limn!+1

Zfn

dµ .

Notice that the conclusion can we written in the suggestive formZ ⇣lim

n!+1fn

⌘dµ = lim

n!+1

Zfn

dµ .

Proof. Since 0 f1(x) f2(x) · · · is a monotone sequence the limit function f exists, and itis measurable by Thm. 2.6 (iv). Moreover, by the mononicity of the integral, property (8),Z

f1 dµ Z

f2 dµ · · · ,

that is, the sequence� R

fn

dµ is increasing, so that it has a limit and since

Rfn

dµ Rfdµ for

all n4, by the comparison test

limn!+1

Zfn

dµ Z

f dµ .

In order to prove the reverse inequality, let s be a simple function such that 0 s f andlet 0 < ↵ < 1. Since lim

n!+1 fn

(x) = f(x), setting

En

=�x : f

n

(x) � ↵s(x) ,

we have that E1 ✓ E2 ✓ · · · , and [+1n=1En

= X . Therefore, by the monotonicity of the integral,

↵

ZE

n

s dµ ZE

n

fn

dµ =

Z�E

n

fn

dµ Z

fn

dµ . (9)

14 M. M. PELOSO

By Prop. 2.14 (iv), since {En

} is an increasing sequence of measurable sets whose union is X ,lim

n!+1RE

n

s dµ =Rs dµ, therefore, passing to the limit in (9) we obtain

↵

Zs dµ lim

n!+1

Zfn

dµ .

This holds for all 0 < ↵ < 1 so thatZs dµ lim

n!+1

Zfn

dµ .

Passing to the supremum on the left hand side for all simple functions s with 0 s f weobtain Z

f dµ limn!+1

Zfn

dµ ,

and we are done. ⇤

Corollary 2.17. (1) Let f, g 2 L+. ThenZ(f + g) dµ =

Zf dµ+

Zg dµ .

(2) If fn

2 L+, n = 1, 2, . . . , thenZ ⇣ +1Xn=1

fn

⌘dµ =

+1Xn=1

Zfn

dµ

Notice that part (1) is just the statement of the additivity of the integral. In the presenttheory, in order to obtain such property we had to recourse to the Monotone ConvergenceTheorem (that we will abbreviate as MCT in what follows).

Proof. Let {sn

}, {�n

} be sequences of non-negative simple functions, monotonically convergingto f and g, respectively. Then, {s

n

+ �n

} is a sequence of non-negative simple functions,monotonically converging to f + g. By the MCT,Z

(f + g) dµ = limn!+1

Z(s

n

+ �n

) dµ = limn!+1

Zsn

dµ+ limn!+1

Z�n

dµ =

Zf dµ+

Zg dµ .

This proves (1). In order to prove (2), notive that by (1) and induction we have thatZ ⇣ NXn=1

fn

⌘dµ =

NXn=1

Zfn

dµ ,

for every N . Observe that�P

N

n=1 fn is a sequence of functions in L+ that converges mono-

tonically aP+1

n=1 fn. Hence, applying the MCT again we obtainZ ⇣ +1Xn=1

fn

⌘dµ = lim

n!+1

Z ⇣ NXn=1

fn

⌘dµ = lim

n!+1

NXn=1

Zfn

dµ =+1Xn=1

Zfn

dµ ,

as we wished to show. ⇤

Corollary 2.18. Let f 2 L+. ThenRfdµ = 0 if and only if f = 0 µ-a.e. Therefore, if

f, g 2 L+ and f = g a.e., then Zf dµ =

Zg dµ .


Proof. Assume first the f is simple, f =P

j

cj

�E

j

. Then f = 0 clearly impliesRfdµ = 0, while

if 0 =Rfdµ =

Pj

cj

µ(Ej

), it must be either cj

= 0 or µ(Ej

) = 0, for each j. In any case, f = 0µ-a.e.

Let now f 2 L+ and assume first f = 0 µ-a.e. If s is simple, 0 s f , then s = 0 µ-a.e.,Rsdµ = 0, and passing to the supremum of such functions we obtain that

Rfdµ = 0. Finally,

supposeRfdµ = 0. Then�

x : f(x) > 0 =

+1[n=1

�x : f(x) > 1/n

=:

+1[n=1

En

,

Since E1 ✓ E2 ✓ · · · ,µ⇣�

x : f(x) > 0 ⌘

= limn!+1

µ(En

) .

If f is not equal to 0 a.e., then it must be µ(En

) > 0 for some n, but thenZf dµ �

ZE

n

f dµ �ZE

n

1

ndµ =

1

nµ(E

n

) > 0 ,

a contradiction. Then,Rfdµ = 0 implies f = 0 µ-a.e., and we are done. ⇤

The conclusion of the MCT holds true also if we just assume that {fn

} is a sequence in L+

monotonically convergent a.e.

Corollary 2.19. Let {fn

} is a sequence in L+ monotonically convergent a.e. to f 2 L+, thenZf dµ = lim

n!+1

Zfn

dµ .

Proof. Let E 2 M be such that µ(E) = 0, and f1(x) f2(x) · · · and limn!+1 f

n

(x) = f(x)for x 2 cE. Setting g

n

= � c

E

fn

and g = � c

E

f , we observe that gn

= fn

a.e. and g = f a.e. sothat

Rgn

dµ =Rfn

dµ, for all n andRgdµ =

Rfdµ, by Cor. 2.18.

We may apply the MCT to {gn

} and obtain thatZf dµ =

Zg dµ = lim

n!+1

Zgn

dµ = limn!+1

Zfn

dµ

as we wish to show. ⇤In the MCT the key assumption, besides the non-negativity of the functions, was the mono-

tonicity of the sequence. Such monotonicity guaranteed the existence of the limits. In thegeneral case, we have the following.

Theorem 2.20. (Fatou’s Lemma.) Let {fn

} ✓ L+, thenZ ⇣lim infn!+1

fn

⌘dµ lim inf

n!+1

Zfn

dµ .

Proof. As in the proof of Thm. 2.6 (iv), if we set gk

(x) = infjk

fj

(x), for k = 1, 2, . . . , we havethat {g

k

} is an increasing sequence in L+ and limk

gk

= lim infn!+1 f

n

. Moreover, gk

fk

sothat

limk!+1

Zgk

dµ lim infn!+1

Zfn

dµ .

Therefore, by the MCT,Z ⇣lim infn!+1

fn

⌘dµ =

Z ⇣lim

k!+1gk

⌘dµ = lim

k!+1

Zgk

dµ lim infn!+1

Zfn

dµ ,

16 M. M. PELOSO

as we wished to show. ⇤The following result follows at once from Fatou’s Lemma and Cor. 2.19.

Corollary 2.21. Let {fn

} ✓ L+ and suppose fn

! f a.e. ThenZf dµ lim inf

n!+1

Zfn

dµ .

Proof. Let E 2 M be a set such that fn

! f in E and µ( cE) = 0. Set gn

= �E

fn

and g = �E

f .Then g

n

! g, fn

= gn

a.e. and f = g a.e. By Fatou’s Lemma,Zf dµ =

Zg dµ lim inf

n!+1

Zgn

dµ = lim infn!+1

Zfn

dµ . ⇤

Remark 2.22. We observe that the previous result cannot be improved to have equality evenif we assume that the limit of the sequence

� Rfn

dµ exists.

Indeed, consider the measure space (N,P(N), µ), where µ is the counting measure. Let sn

be the numerical sequence {sn,k

} that is equal to 1 if k = n and equals 0 otherwise. Then thesequence of functions on N {s

n

} converges to 0 pointwise, butZNsn

dµ =+1Xk=1

sn,k

= 1

for all n. Therefore,

0 =

ZNs dµ lim

n!+1

ZNsn

dµ = 1 .

Remark 2.23. We also observe that there exists a version of the MCT for decreasing sequences{f

n

}, but one needs to assume thatRf1dµ < +1. Precisely, if {f

n

} ✓ L+, f1 � f2 � · · · , andRf1dµ < +1, then Z

f dµ = limn!+1

Zfn

dµ .

Proof. We first observe that if 0 g f , then f � g � 0 andZ �f � g) dµ =

Zf dµ�

Zg dµ .

Next, since {fn

} is monotone and non-negative, the limit f exists it is non-negative, andfn

� f for each n. Set gn

= f1 � fn

. Then {gn

} is a sequence of non-negative functions suchthat g

n

gn+1 for all n and converging to f1 � f . Applying the MCT we obtainZ

f1 dµ�Z

f dµ =

Z ⇣f1 � f) dµ =

Zg dµ = lim

n!+1

Zgn

dµ = limn!+1

⇣Zf1 dµ�

Zfn

dµ�.

Therefore, sinceRf1dµ < +1 we can subtractive from both sides to obtainZ

f dµ = limn!+1

Zfn

dµ ,

as we wished to prove. ⇤


2.3. Integration of complex-valued functions. We now pass to consider generic real, orcomplex, valued functions. If (X ,M, µ) is a measure space, f : X ! C is measurable, andu = Re f , v = Im f , we write

f = u+ � u� + i(v+ � v�) .

Then, u±, v± : X ! [0,+1). If f is real-valued, we simply write f = f+ � f�.

Definition 2.24. Given a measure space (X ,M, µ), given f : X ! C is measurable, we saythat f is absolutely integrable, or simply integrable, ifZ

|f | dµ < +1;

equivalently, if

0 Z

u± dµ,

Zv± dµ < +1 .

For f integrable we setZf dµ =

Zu+ dµ�

Zu� dµ+ i

✓Zv+ dµ�

Zv� dµ

◆.

In particular, if f is real-valued, we haveZf dµ =

Zf+ dµ�

Zf� dµ .

Remark 2.25. Notice that, if f is integrable, f = u+ iv, then 0 u±, v± |f | so that

0 Z

u± dµ,

Zv± dµ

Z|f | dµ < +1 .

Conversely, if 0 Ru± dµ,

Rv± dµ < +1, then |f | |u|+ |v| = u+ + u� + v+ + v�, so thatZ

|f | dµ Z

u+ dµ+

Zu� dµ+

Zv+ dµ+

Zv� dµ < +1 .

Moreover, the set of integrable functions is a complex vector space, as it is easy to check, andfor integrable f, g and a, b 2 C we haveZ �

af + bg) dµ = a

Zf dµ+ b

Zg dµ .

Proposition 2.26. The following properties hold true.

(i) If f is integrable, then�� Z f dµ

�� Z|f |dµ .

(ii) If f, g are integrable, then

ZE

f dµ =

ZE

g dµ for every E 2 M if and only if |f � g| = 0

a.e. if and only if

Z|f � g| dµ = 0.

Proof. (i) IfRf = 0 the result is trivial. Next, if f is real�� Z f dµ

�� = �� Z f+ dµ�Z

f� dµ�� Z

f+ dµ+

Zf� dµ =

Z|f | dµ .

18 M. M. PELOSO

If f is complex-valued, andRf 6= 0, from Cor. 2.7 that sgn f, |f | are measurable and we set

↵ = | R fdµ|/(R fdµ). Then �� Z f dµ�� = ↵

Zf dµ =

Z↵f dµ

so that, in particularR↵f dµ is non-negative. Therefore,�� Z f dµ�� = Re

Z↵f dµ =

ZRe(↵f) dµ

Z|↵f | dµ =

Z|f | dµ ,

since |↵| = 1.(ii) If f, g 2 L1(µ), from Cor. 2.18 we know that

R |f � g| dµ = 0 if and only if |f � g| = 0a.e., that is, if and only if f = g a.e. Now, if

R |f � g| dµ = 0, then, for every E 2 M,�� ZE

f dµ�ZE

g dµ�� Z

�E

|f � g| dµ Z

|f � g| dµ = 0 ,

so that ZE

f dµ =

ZE

g dµ for every E 2 M .

Finally, suppose that the above condition holds, and assume for simplicity that f and g arereal-valued. Consider the function f � g = m and its decomposition m = m+ �m�. Then, byassumption we have that

RE

mdµ = 0 for all E 2 M. In particular if we take E+ =�x : m(x)

0 , then we have

0 =

ZE+

mdµ =

ZE+

m+ dµ =

Zm+ dµ ,

since m+ = 0 on cE+. Thus, m+ = 0 a.e. and, with a similar argument we obtain also m� = 0a.e. Hence, m = 0 a.e., i.e. f = g a.e., as we wished to show. ⇤

From now on we will identify functions that di↵er only on a set of measure 0. Indeed, wehave the following definition.

Definition 2.27. Given a measure space (X ,M, µ), we consider the equivalent relation ⇠ onthe space of integrable saying that f ⇠ g if f = g a.e. We define the space L1(µ) as the spaceof equivalent classes of integrable function modulo the relation ⇠ and we define the norm

kfkL

1(µ) =

Z|f | dµ .

Remark 2.28. We need to check that the above definition gives in fact a norm. It is clear thatkfk

L

1(µ) � 0 and that f = 0 in L1(µ) (that is, f = 0 a.e.) implies kfkL

1(µ) = 0. Conversely, if0 = kfk

L

1(µ) =R |f |dµ, then Cor. 2.18 implies that |f | = 0 a.e., ie. f = 0 a.e. and f = 0 in L1.

The symmetry and triangular inequality of the norm follow easily.In Thm. 2.31 we will show that L1 is a complete normed space, that is, a Banach space.

Therefore, we have a norm on L1(µ), hence a metric, given by the expression

d(f, g) = kf � gkL

1(µ) .

Then, given {fn

} ✓ L1(µ) we say that fn

! f in L1(µ) (or, simply in L1 if the measure µ isunderstood) if

kfn

� fkL

1 ! 0 as n ! +1 .


Theorem 2.29. (Dominated Convergence Theorem.) Let (X ,M, µ) be a measure spaceand {f

n

} ✓ L1(µ). Suppose that fn

! f pointwise a.e. and that there exists a non-negativeg 2 L1(µ) such that g � |f

n

| for all n. Then, fn

! f in L1(µ) andZf dµ = lim

n!+1

Zfn

dµ .

Proof. Since fn

! f pointwise a.e., using Prop.’s 2.11 and 2.12, it follows that f we can redefinef on a set of measure 0 in such a way that it is measurable. For, if E 2 M is such that f

n

! fon E and µ( cE) = 0, then (�

E

fn

) ! (�E

f) pointwise, �E

fn

are measurable, so that �E

f ismeasurable, and �

E

f = f a.e. Thus, possibly by replacing f by �E

f we may assume that thelimit function f is measurable. Moreover, since |f

n

| g for all n, by the comparison theoremwe have that |f | g. Therefore, f 2 L1(µ).

Observing that |fn

� f | 2g, we apply Fatou’s Lemma to the sequence {2g � |fn

� f |} andobtain Z

2g dµ lim infn!+1

Z2g � |f

n

� f | dµ =

Z2g dµ+ lim inf

n!+1

⇣�Z

|fn

� f | dµ⌘

=

Z2g dµ� lim sup

n!+1

Z|f

n

� f | dµ .

SinceR2g dµ < +1, we can subtract it from both sides of the inequality and obtain that

lim supn!+1

Z|f

n

� f | dµ 0 .

This implis that limn!+1

R |fn

� f | dµ = 0, that is, fn

! f in L1(µ). Finally,�� Z fn

dµ�Z

f dµ�� Z

|fn

� f | dµ ! 0

and the last conclusion follows. ⇤

2.4. The space L1(µ). Our next goal is to show that L1(µ) is a Banach space, that is it iscomplete in its norm. We recall that a normed space (X, k · k

X

) is complete if it is complete as ametric space w.r.t. the metric d(f, g) = kf �gk

X

. We also recall that a seriesP

n

fn

of elementsin X is said to be absolutely convergent in X if the numerical series

Pn

kfn

kX

is convergent.We have the following result.

Theorem 2.30. A normed space (X, k · kX

) is a Banach space if and only if every absolutelyconvergent series

Pn

fn

is convergent in X.

Proof. Suppose first X is a Banach space and thatP

n

fn

is an absolutely convergent series. Let{s

N

} be the sequence of partial sums, that is, sN

=P+1

n=1 fn. Then, given " > 0, there existsn"

such that for n"

< M < N ,

d(sN

, sM

) = ksN

� sM

kX

=�� NXn=M+1

fn

��X

NX

n=M+1

kfn

kX

< " .

Hence, {sN

} is a Cauchy sequence in X, that converges, since X is complete.Conversely, suppose every absolutely convergent series is convergent in X. Let {g

n

} be aCauchy sequence in X. Hence, for every k = 1, 2, . . . , there exists an integer N

k

such that ifm,n � N

k

, d(gn

, gm

) = kgn

� gm

kX

< 2�k. Notice that we may assume that Nk+1 > N

k

for all

20 M. M. PELOSO

k. Then, the subsequence {gN

k

} is such that kgN

k+1�gN

k

kX

< 2�k. Then,P+1

k=1 kgNk+1�gN

k

kX

converges, so that by assumption,P+1

k=1 gNk+1 � gN

k

converges to an element g 2 X. But, the

seriesP+1

k=1 gNk+1 � gN

k

is telescopic and the partial sum

MXk=1

gN

k+1 � gN

k

= gN

M+1 � gN1 ,

so that gN

k

� gN1 ! g as k ! +1. Hence, the original sequence must converge too, and the

conclusion follows. ⇤

Theorem 2.31. Let {fn

} be a sequence of functions in L1(µ) such thatP+1

n=1 kfkL1(µ) < +1.

Then, there exists f 2 L1(µ) suchP+1

n=1 fn converges to f in L1(µ) andZf dµ =

+1Xn=1

Zfn

dµ .

As a consequence, L1(µ) is a Banach space. Finally, the simple functions are dense in L1(µ).

Proof. By Cor. 2.17 (ii) we know thatZ Xn

|fn

|dµ =Xn

Z|f

n

|dµ =+1Xn=1

kfn

kL

1(µ) < +1 .

Then, setting g =P+1

n=1 |fn|, we have g 2 L1(µ), and hence it is finite a.e. This implies thatP+1n=1 fn converges but on a set of measure 0, call f such limit. We apply the DCT to the

sequence of partial sums sn

=P

n

k=1 fk. Clearly, sn ! f pointwise, and also

|sn

| =�� nXk=1

fk

�� nXk=1

|fk

| = g .

Hence, the DCT applies to give that sn

! f in L1(µ) and limn!+1

Rsn

dµ =Rfdµ, that isZ

f dµ = limn!+1

Z ⇣ nXk=1

fk

dµ⌘= lim

n!+1

nXk=1

Zfk

dµ =+1Xn=1

Zfn

dµ .

In particular, every absolutely convergent series also converges in L1(µ), hence L1(µ) is complete.Finally, let f 2 L1(µ). Then, |f | is finite a.e. and the sequence of simple functions {'

n

} inThm. 2.9 (2), is such that '

n

! f pointwise, and |'n

| |f |. Therefore, we can apply the DCTto obtain that '

n

! f in L1(µ). ⇤

We conclude this part by comparing L1(µ) with L1(µ). If f 2 L1(µ) and g is given by by Prop.2.12, then g = f µ-a.e. and g 2 L1(µ). Conversely, if g 2 L1(µ), then g is also M-measurableand

R |g|dµ =R |g|dµ. Then g 2 L1(µ). Therefore, we have shown that the spaces L1(µ) and

L1(µ) can be identified and we will do in what follows.

AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUE … · AN INTRODUCTION TO MEASURE THEORY AND THE...

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