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Chapter 3. Lebesgue integral and the monotone convergence

theorem

3.1 Starting point

Remark3.1.1. What we need to define integrals ofR-valued functions, apart from a considerable

amount of the terminology we have come across so far, is that

we have R and the Borel -algebra on it; this will come in for the values of the functionswe consider;

on the domain we need a measure, and a measure needs a -algebra. We will use (R,L, ),where L is the -algebra of Lebesgue measurable sets and : L [0, ] is the measuregiven by (F) = m

(F) for F L

.

In fact, we could equally well have a more general domain Xand we would need a -algebra of subsets of X together with a measure : [0, ]. That is we could treat

Xf d =

Xf(x) d(x), integrals of (certain) functions f: X R with general domains X, provided we

have (X, , ), but we will mostly stick to f: R R where (X, , ) = (R,L, = m|L).One advantage of sticking to domain X = R is that we can picture our functions as graphs

y = f(x) in R2. We could still have that advantage if we allowed X R, for instance to define[a,b]

f d where f: [a, b] R. Actually this case will be included rather easily, but anotherperspective is to generalise to multiple integrals, that is to integrals of functions f: Rn R. Weare not ready for that because we only did m as length outer measure on R. For R2, we would

need to do something similar to what we did with length to define area measure. The startingpoint would be areas of rectangles, and we would need to replace what we did with the interval

algebra I by stuff about a rectangle algebra. Some of the details at the start would be a bit

different but many of the same proofs can be used in R2, and for volume outer measure in R3

and more generally for n-dimensional measure in Rn.However, what we do in this chapter can be done in considerable generality and we set out

the general definitions concisely.

Definition 3.1.2. If X is a set then a collection P(X) is called a -algebra (of subsets ofX) if it satisfies:

(i) (ii) E Ec

(iii) E1, E2, . . .

n=1 En .A pair (X, ) of a set Xand a -algebra of subsets ofXis sometimes called a measurable space.

Definition 3.1.3. If is a -algebra (of subsets of set X) and : [0, ] is a function, thenwe call a measure on if it satisfies

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2 201112 Mathematics MA2224

(a) () = 0(b) is countably additive, that is whenever E1, E2, . . .

are disjoint, then (n=1 En) =

n=1 (En).

The combination (X, , ) is called a measure space and the subsets E Xwith E arecalled measurable subsets (with respect to the given ). If in addition (X) = 1, then is calleda probability measure and the combination (X, , ) is called a probability space. In the contextof probability, measurable subsets E are called events and (E) [0, 1] the probability ofthe event E.

Examples 3.1.4. 1. Our primary focus will be (X, , ) = (R,L, ) with = m|L beingLebesgue length measure on the -algebra Lof Lebesgue measurable subsets ofR.

2. IfX

L then we can define a -algebra on X and a measure :

R by

= {F X : F L}and

(F X) = (F X) = m(F X).It may be helpful to have a notation LX for this (though that is not a standard notation).

Proof. There is nothing difficult in this except perhaps that there is a possibility of confu-

sion between two notions of complement. Perhaps this can be solved if we write R \ Eforthe complement ofE R and X\ Efor the complement ofE X as a subset ofX. Thenotation we usually use ofEc for the complement ofE is not very adaptable to more than

one meaning of complement.

Notice that, since X L, we could also say = {E L : E X}

(because E Lwith E X implies E = E X, and on the other hand F L impliesthat E = F X L and has E X). So , E X\ E = X (R \ E) .Moreover E1, E2, . . . implies

n=1 En .

The properties required for to be a measure are just true because we know is a measure.

3. If X = [0, 1], then the previous example turns into an example of a probability space([0, 1], , ).

4. If (X, , 1) and (X, , 2) are two measure spaces with the same underlying set and-algebra, and if c1, c2 0, then we can get a new measure = c11 + c22 on bydefining

(E) = c11(E) + c22(E) (E )(This is easy to check and we can leave it as an exercise.)

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Lebesgue integral 3

3.2 Measurable functions

Definition 3.2.1. A function f: R

R is called Lebesgue measurable (or measurable with

respect to the measurable space (R,L)) if for each a R{x R : f(x) a} = f1((, a]) L.

Examples 3.2.2. (i) Continuous functions f: R R are always measurable, because (, a]is a closed subset and so f1((, a]) is closed, hence in L.If you prefer to think about open sets, (a, ) is open, so f1((a, )) is open, hence in L,hence its (closed) complement (f1((a, )))c = f1((a, )c) = f1((, a]) is in L.

(ii) The characteristic function Q of the rationals is measurable (but not continuous anywhere).

The reason is that the set QL (it is a countable union of single point sets, hence a

Borel set, hence in L; another way is to use m(Q) = 0 to get Q L) and so then is itscomplement R \Q L. For f = Q, we can see that

f1((, a]) = {x R : f(x) a} =

ifa < 0R \Q if0 a < 1R ifa 1

and in all cases we get a set in L.

(iii) For a subset E R, its characteristic function E is a measurable function if and only ifE L (which means E is a measurable set). As in the previous case, for f = E, we cansee that

f1((, a]) = {x R : f(x) a} =

ifa < 0R \ E if0 a < 1R ifa 1

Certainly ,R L and we have R \ E L E L.Remark 3.2.3. Our aim will be to define integrals for all measurable functions, but we will not

quite succeed in that.

Proposition 3.2.4. Iff: R

R is Lebesgue measurable, then

f1(B) L

for each Borel setB.

Proof. Let

f = {E R : f1(E) L}.We claim that f is a -algebra.

That is quite easy to verify.

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(i) f1() = L f(ii) E

f

f1(E)

L

f1(Ec) = (f1(E))c

L

Ec

f

(iii) E1, E2, . . . f f1(En) L for n = 1, 2, . . .. From this we have

n=1 f1(En)

L and so

f1

n=1

En

=

n=1

f1(En) Ln=1

En f

As we know f is Lebesgue measurable, we have

(, a] f (a R)and that means that f contains the -algebra generated by {(, a] : a R} which weknow to be the Borel -algebra by Corollary 2.3.12.Proposition 3.2.5. If f, g : R R are Lebesgue measurable functions and c R, then the

following are also Lebesgue measurable functions

cf,f2, f + g,fg, |f|, max(f, g)The idea here is to combine functions by manipulating their values at a point. So f g : R R

is the function with value at x R given by (f g)(x) = f(x)g(x), and similarly for the otherfunctions.

Proof. cf: First ifc = 0, cf is the zero function, which is measurable (very easy to check that

directly). Let h = cf (so that h(x) = cf(x)). For c > 0, then

h1((, a]) = {x R : cf(x) a} = {x R : f(x) a/c} Land so h = cf is Lebesgue measurable. For c < 0,

h1((, a]) = {x R : f(x) a/c} = f1([a/c, )) Lusing Proposition 3.2.4 and the fact that [a/c, ) is a Borel set.

f2: Here we say

{x

R : f(x)2

a

}= ifa < 0

f1([a, a]) ifa 0f + g: (This is perhaps the most tricky part.) Fix a R and we aim to show that {x R :

f(x) + g(x) a} L. Taking the complement, this would follow from {x R :f(x) + g(x) > a} L.For any x where f(x) + g(x) > a then f(x) > a g(x) and there is a rational q Q sothat f(x) > q > a g(x). Then g(x) > a q. So

x Sq = f1((q, )) g1((a q, ))

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Lebesgue integral 5

for some q Q. On the other hand iff(x) > qand g(x) > a q, then f(x) + g(x) > a,from which we conclude that

{x R : f(x) + g(x) > a} = qQ

Sq.

By Proposition 3.2.4 Sq L for each q and hence {x R : f(x) + g(x) > a} L(because it is countable union of sets Sq L).

f g: This follows from the previous parts because f g = ((f + g)2 (f g)2)/4.|f|: This is easy as {x R : |f(x)| a} = ifa < 0 but is f1([a, a]) ifa 0.max(f, g): {x R : max(f(x), g(x)) a} = f1((, a]) g1((, a]).

Remark 3.2.6. It is convenient to allow for functions that sometimes have the value andsometimes . For that we need to introduce the extended real line [, ], which is Rwith two extra elements and + added. This is similar to our earlier use of[0, ].

We order [, ] so that < x < for each x R. We introduce arithmetic on[, ] in a similar way to what we did before for [0, ] but we do not allow subtraction offrom itself or from itself. We extend the usual arithmetic operations on R with the rules

x + = + x = x + () = () + x = () (for x R)

x

=

x =

x() = ()x = () (for 0 = x R)0 = 0 = 0

0() = ()0 = 0From the point of view of the ordering of [, ], there is a strictly monotone increasing

bijection : R (1, 1) given by(x) =

x

1 + |x|and both and 1 are continuous (so that is a homeomorphism). We can extend to get anorder-preserving bijection : [, ] [1, 1] by defining () = 1, (x) = (x) forx R and () = 1. Since we know that nonempty subsets of [1, 1] have both a supremumand an infimum, it follows by applying 1 that all nonempty subsets of [, ] also have asupremum (least upper bound) and infimum (greatest lower bound).

We define convergence on [, ] such that becomes a homeomorphism onto [1, 1](which means for example that limn xn = has the standard meaning if all the terms xn andthe limit are finite, but limn xn = means limn (xn) = () = 1 or that given anyM < , there is N so that M < xn holds for all n > N). It is also true that monotoneincreasing sequences (xn)

n=1 in [, ] always converge (to supn xn in fact) and so also do

monotone decreasing sequences always converge (to their infimum).

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Definition 3.2.7. A function f: R [, ] is called Lebesgue measurable (or measurablewith respect to the measurable space (R,L)) if for each a R

{x R : f(x) a} = f1([, a]) L.Note that this definition agrees with the earlier Definition 3.2.1 iff(x) R always (because

then f1([, a]) = f1((, a])).Proposition 3.2.8. Letf: R [, ] be a function and letA = {x R : f(x) = } =f1({}), B = {x R : f(x) = } = f1({}). Then the following are equivalent(a) f is Lebesgue measurable

(b) A, B Land the function g : R R given by

g(x) =

f(x) if < x < 0 ifx A B

is Lebesgue measurable.

Proof. (a) (b): Note thatA =

n=1

f1([, n]) L

(because -algebras are closed under the operation of taking countable intersections) and

R \ B =n=1

f1([, n]) L B L

(countable unions and complements stay in L).

To show that g is measurable, note that

g1((, a]) =

f1([, a]) B ifa 0f1([, a]) \ A ifa < 0

So g1((, a]) L for each a.(b)

(a): Note that

f1([, a]) =

g1((, a]) \ B ifa 0g1((, a]) A ifa < 0

and so f1([, a]) L for each a (using (b)).Remark 3.2.9. We can also define measurability of functions defined on Lebesgue measurable

subsets X R, as follows. This would include for example the situation where X is an intervalof any kind, or any Borel set (which includes open subsets X ofR, in particular).

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Lebesgue integral 7

Definition 3.2.10. IfX L, then a function f: X [, ] is called Lebesgue measurableon X if for each a R

{x X : f(x) a} = f1([, a]) L.

Proposition 3.2.11. LetX Landf: R [, ]. Then the following are equivalent

(a) f is Lebesgue measurable on X

(b) the function g : R [, ] given by

g(x) =

f(x) ifx X0 ifx

R

\X

is Lebesgue measurable on R.

Proof. For a R we have

g1([, a]) = {x R : g(x) a}

=

(R \ X) {x X : f(x) a} ifa 0{x X : f(x) a} ifa < 0

So (a)

(b). For the converse use

f1([, a]) =

{x R : g(x) a} \ (R \ X) ifa 0{x R : g(x) a} ifa < 0

We can extend Proposition 3.2.5 to functions with values in [, ] except for one diffi-culty.

Proposition 3.2.12. If X L, c R and f, g : R [, ] are Lebesgue measurablefunctions on X, then the following are also Lebesgue measurable functions (on X)

cf,f2, f g , |f|, max(f, g)

If{x : f(x) = }{x : g(x) = } = = {x : f(x) = }{x : g(x) = }, then f+ gis measurable (on X).

The proof (which we leave as an exercise) is to use Propositions 3.2.11, 3.2.8, and 3.2.5 for

cf (c = 0), f2, f + g, f g and |f|. For max(f, g) a similar proof can be used.

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3.3 Limits of measurable functions

The next result shows that taking (pointwise) limits of sequence of measurable functions still

give a measurable limit.

Definition 3.3.1. If fn : R [, ] (n = 1, 2, . . .) is an infinite sequence of functions, thenwe say that f: R [, ] is the pointwise limitof the sequence (fn)n if we have

f(x) = limn

fn(x)

for each x R.For any sequence fn : R [, ] we can define limsupn fn as the function with value

at x given by

limsupn

fn(x) = limn

supkn

fk(x)

(something that always makes sense because supkn fk(x) decreases as n increases or at leastdoes not get any bigger as n increases).

Proposition 3.3.2. Let fn : R [, ] (n = 1, 2, . . .) be Lebesgue measurable functions.Then

(a) the function x supn1 fn(x) is Lebesgue measurable;

(b) the function x infn1 fn(x) is Lebesgue measurable;

(c) limsupn fn is Lebesgue measurable; and

(d) if the pointwise limit function f: R [, ] exists, then f is Lebesgue measurable.

Proof. (a) Ifg(x) = supn1 fn(x), then g1([, a]) = n=1 f1n ([, a]) L (since the

-algebra L is closed under taking countable intersections).

(b) Ifh(x) = infn1 fn(x), then g(x) = supn1(fn(x)) is measurable by (a) and so then isg(x) measurable.

(c) We have limsupn fn(x) = infn1 gn(x) where gn(x) = supkn fk(x) is a decreasingsequence. But each gn is measurable by (b) and then limsupn fn(x) is measurable by(a).

(d) If the pointwise limit f(x) exists, then f(x) = limsupn fn(x), which is measurable by(c).

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Lebesgue integral 9

3.4 Simple functions

Definition 3.4.1. A function f: R

R is called a simple function if the range f(R) is a finite

set.

Note that we require finite values.

Proposition 3.4.2. Each simple function f: R R has a representation

f =ni=1

aiEi

as a linear combination of finitely many characteristic functions of disjoint sets E1, E2, . . . , E nwith coefficients a1, a2, . . . , an R.

Moreover we can assume a1, a2, . . . , an are distinct values, that the Ej are all nonempty andthatnj=1 Ej = R. With these assumptions the representation is called the standard representa-

tion and is unique apart from the order of the sum.

A simple function is Lebesgue measurable if and only if the sets Ej in the standard represen-tation are all in L.

Proof. Since f(R) is a finite set (and cant be empty) we can list the elements f(R) = {y1, y2, . . . , yn}with n 1. Take Ej = f1({yj}) = {x R : f(x) = yj} and then we get the standard repre-sentation f =

ni=1 yiEi .

If f is measurable then Ej L for each j (because one point sets {aj} are Borel sets andwe can use Proposition 3.2.4).

Conversely, if each Ej is inL

, then each Ej measurable, and then so is f measurable byProposition 3.2.5.

Notation 3.4.3. We will develop integration first for nonnegative measurable functions f: R [0, ].

To reduce the case of general (Lebesgue measurable) f: R [, ] to the positive casewe will use the two associated functions f+ : R [0, ] and f : R [0, ] given by

f+ = max(f, 0) =1

2(|f| + f), f = max(f, 0) = 1

2(|f| f),

both of which we know to be measurable. Notice that f = f+

f and

|f

|= f+ + f.

We call f+ the positive partoff and f the negative part(but note that f is positive also).

Definition 3.4.4. Iff: R [0, ) is a nonnegative measurable simple function, with standardrepresentation f =

nj=1 ajEj , then we define

R

f d =n

j=1

aj(Ej)

(where (Ej) means the Lebesgue measure ofEj). (Note that the integral makes sense in [0, ].)

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For E L, we define

Ef d = R Ef d(noting that Ef is also a nonnegative measurable simple function).

Examples 3.4.5. (i) For f = [0,1]+2[4,) the possible values are 0, 1 and 2, and the standardrepresentation is

f = [0,1] + 2[4,) + 0(,0)(1,4)

so thatR

f d = 1([0, 1]) + 2([4, )) + 0((, 0) (1, 4)) = 1 + + 0 =

(ii) For f = Q, the standard representation is f = Q + 0R\Q andR

f d = 1(Q) + 0(R \Q) = 0 + 0 = 0

(iii) (Exercise) Show that ifE L, then R

Ed = (E).

Proposition 3.4.6. If f, g : R [0, ) are measurable simple functions and [0, ), thenf, f + g are measurable simple functions and

R

f d =

R

f d,

R

(f + g) d =

R

f d +

R

g d.

Proof. We know from Proposition 3.2.5 that f, f + g are measurable.If = 0, then 0f has standard representation 0 = 0R and integralR

f d =R

0R d =0(R) = 0 = 0

R

f d = R

f d. If > 0 and f(R) = {a1, a2, . . . , an} let Fj = f1({aj})so that f has standard representation f =

nj=1 ajFj , while f has standard representation

f =n

j=1 ajFj . SoR

f d =n

j=1

(aj)(Fj) = n

j=1

aj(Fj) =

R

f d.

To cope with h = f+g, let f(R) = {a1, a2, . . . , an}, Fj = f1({aj}), g(R) = {b1, b2, . . . , bm},Gk = g

1(

{bk

}). The range ofh = f + g is certainly finite because it is contained in

{aj + bk :

1 j n, 1 k m}. So h is a simple function. Write h(R) = {c1, c2, . . . , cp} and then foreach 1 p we have

H = h1(c) =

{(j,k):aj+bk=c}

Fj Gk.

The sets making up H are a disjoint union and so note that finite additivity of gives us

(H) =

{(j,k):aj+bk=c}

(Fj Gk)

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Lebesgue integral 11

We now have standard representations f =n

j=1 ajFj , g =m

k=1 bkGk , h = f + g =

p

=1 cH andR

h d =

p=1

c(H)

=

p=1

c

{(j,k):aj+bk=c}

(Fj Gk)

=

{(j,k):FjGk=}

(aj + bk)(Fj Gk)

=

1jn,1km

(aj + bk)(Fj Gk)

(since () = 0)=

nj=1

mk=1

aj(Fj Gk) +mk=1

nj=1

bk(Fj Gk)

=n

j=1

aj

mk=1

Fj Gk

+mk=1

bk

n

j=1

Fj Gk

(using finite additivity of)

=n

j=1

aj(Fj) +mk=1

bk(Gk)

=R

f d +R

g d

Remark 3.4.7. The tricky part in the above proof is to stick to the definition of the integral of a

simple function in terms of the standard representation. We do this so that the definition is not

ambiguous, but now that we have proved this result we can say that if we have any representation

of a simple function f =n

i=1 yjEj with yj 0 and Ej measurable, thenR

f d =ni=1

yj

R

Ej d =ni=1

yj(Ej)

Corollary 3.4.8. If f, g : R [0, ) are measurable simple functions with f(x) g(x) foreach x, then

R

f d R

g d

Proof. h = g f is measurable (by Proposition 3.2.5), simple (can only have finitely manyvalues) and also h(x) 0. So

Rh d is defined and by Proposition 3.4.6,

R

g d =

R

(f + h) d =

R

f d +

R

h d R

f d.

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Proposition 3.4.9. If f: R [0, ) is a measurable simple function, then we can define ameasure f : L [0, ] by

f(E) =E

f d

Proof. We know we can write f in the form f =n

j=1 ajEj with aj 0 and Ej L(1 j n). So

Ef =n

j=1

ajEEj =n

j=1

ajEEj

and

f(E) =

E

f d =

R

Ef d =n

j=1

aj(E Ej)

We know that Ej (E) = (E Ej) defines a measure on L (Tutorial 6, Q2) and it follows fromExamples 3.1.4-4 that f =

nj=1 ajEj is a measure.

Proposition 3.4.10. If : L [0, ] is a measure, then it satisfies(a) If E1, E2, . . . L with E1 E2 (that is a monotone increasing sequence of

measurable sets) then

n=1

En

= lim

n(En) = sup

n

(En)

(b) If E1, E2, . . . L with E1 E2 (that is a monotone decreasing sequence ofmeasurable sets) and if(E1) < then

n=1

En

= lim

n(En) = inf

n(En)

Proof. (a) (See similar proof on tutorial sheet 5 (for the special case = = m. Also look at aVenn diagram to see what is happening.) If(En)

n=1 is a monotone increasing sequence in L

then we can take F1 = E1, F2 = E2\E1, F3 = E3\E2, in general Fn = En\En1 for n 2,so that Fn L for each n, (Fn)n=1 a disjoint sequence (because E1 E2 E3 meansthat Fn = En \ n1i=1 Ei and so Fn Ei = for i < n, and that gives Fn Fi = asFi Ei).Now

ni=1

Fi = En

for each n andi=1

Fi =n=1

En.

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By countable additivity of (and disjointness of the Fi),

i=1Ei

=

i=1Fi

=

i=1

(Fi) = supn

ni=1

(Fi) = supn

n

i=1Fi

= supn

(En)

(b) Assume now that (En)n=1 is a monotone decreasing sequence in L. Let Gn = E1 \ En sothat E1 = Gn En, a disjoint union of sets on L, and

(E1) = (Gn) + (En).

Using (E1) < , this implies

(En) = (E1)

(Gn).

Since G1 G2 G3 we have

n=1

Gn

= lim

n(Gn) = sup

n

(Gn)

by (a). Now,

E1 \

n=1En = E1

n=1En

c=

n=1(E1 Ecn) =

n=1(E1 \ En) =

n=1Gn

and, as above,

(E1) =

E1 \

n=1

En

+

n=1

En

=

n=1

Gn

+

n=1

En

n=1

En

= (E1)

n=1

Gn

= (E1) lim

n(Gn)

= limn

(E1) (Gn) = limn

(En)

Since (E1)

(E2)

, we have limn (En) = infn (En).

3.5 Positive measurable functions

Proposition 3.5.1. Iff: R [0, ] is a (nonnegative extended real valued) measurable func-tion, then there is a monotone increasing sequence (fn)

n=1 of (nonnegative) measurable simple

functions fn : R [0, ) with

limn

fn(x) = f(x) (for each x R)

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Proof. For each n, define

En,j = f1j 12n , j2n 1 j n2n

and

En,1+n2n = f1([n, ]).

Note that each En,j is in L. Then put

fn =n2n+1j=1

j 12n

En,j .

Observe that fn(x)

f(x) always. For a given x where f(x) f(x) we find thatfn(x) =

1

2n2nf(x)

where y means the greatest integer y. Since 2n+1f(x) 22nf(x), it follows thatfn+1(x) fn(x) always.

Also as 2nf(x) 2nf(x) 1 we get fn(x) > f(x) 12n as soon as n > f(x). Sincefn(x) f(x) we get limn fn(x) = f(x) iff(x) < .

Finally, if f(x) = , then fn(x) = n for all n and so limn fn(x) = = f(x) in thiscase also.

Remark 3.5.2. It is tempting to defineR f d (for f: R [0, ] a measurable function) as

limnR

fn d (where fn is exactly as in the proof of the above proposition). But that makesit a little tricky to prove that the integral has the various obvious properties it has. So we use a

more complicated definition first and prove (as part of the monotone convergence theorem) that

we get the same value for the integral.

Another thing to watch out for is that we dont change the definition we already have when

f: R R is a measurable simple function.Definition 3.5.3. For f: R [0, ] a Lebesgue measurable function, we define

R

f d = supR

s d : s : R

[0,

) a measurable simple function with s

fFor E L, we define

E

f d =

R

Ef d

(noting that Ef is also a nonnegative measurable function).

Remark 3.5.4. By Corollary 3.4.8, when f: R R is itself a simple measurable function, thisgives the same value for

R

f d (andE

f d) as in the earlier definition (Definition 3.4.4).

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Lemma 3.5.5. Forf, g : R [0, ] Lebesgue measurable functions, we have(a) iff

g (pointwise) then

R

f d R

g d

(b) ifE F with E, F L, then E

f d F

f d

Proof. (a) iff g, then any simple function s : R [0, ) with s f also has s g and soR

s d R

g d. Taking the supremum over all such s we get the result.

(b) Since E F, we have Ef Ff and so

Ef d =

R Ef d R Ff d = F f d

by (a).

Theorem 3.5.6 (Monotone Convergence Theorem). Iffn : R [0, ] is a monotone increasingsequence of (Lebesgue) measurable functions with pointwise limitf, then

R

f d = limn

R

fn d

Proof. From fn fn+1 we have fj(x) limn fn(x) = f(x) for each j. HenceR

fj d R f d for each j by Lemma 3.5.5. Also R fn d R fn+1 d and so the sequence of integralsR fn d is monotone increasing. That means limn R fn d makes sense in [0, ]. We have

limn

R

fn d = supn1

R

fn d R

f d.

It remains to prove the reverse inequality.

Fix a simple function s : R Rwith s(x) f(x) for all x R. We will show limnR

fn d R

s d and that will be enough because of the wayR

f d is defined.To show this, fix (0, 1) and consider

En, = {x R : fn(x) s(x)}

Notice thatEn,

sd En,

fn d R

fn d (1)

Because fn+1 fn we know En, En+1,. For s(x) = 0, x En, and for s(x) >0, limn fn(x) = f(x) s(x) > s(x) implies x En, when n is large enough. So

n=1 En, = R and Proposition 3.4.10 (with Proposition 3.4.9) tells us thatR

sd = s(R) = s

n=1

En,

= lim

ns(En,) lim

n

R

fn d

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(using (1) at the last step). That gives us

R s d =

R sd limnR fn d.

As that is true for each (0, 1), it follows that R

s d limnR

fn d, and soR

f d = sup

R

s d : s simple measureable, s f

limn

R

fn d,

completing the proof.

Corollary 3.5.7. Iff, g : R [0, ] are (Lebesgue) measurable functions and 0 then(a) R f d = R f d(b)

R

(f + g) d =R

f d +R

g d

Proof. (a) By Proposition 3.5.1, there is a monotone increasing sequence (fn)n=1 of measurable

simple functions that converges pointwise to f. By the Monotone Convergence Theorem(3.5.6) we know that

R

f d = limnR

fn d. But also (fn)n=1 is a sequence of mea-

surable simple functions that converges pointwise to f and so we also haveR

f d =limn

R

fn d. Now from Proposition 3.4.6 we have

R

f d = limn

R

fn d = limn

R

fn d =

R

f d

(b) Again by Proposition 3.5.1, there are monotone increasing sequences (fn)n=1 and (gn)n=1of measurable simple functions with limn fn(x) = f(x) and limn gn(x) = g(x) foreach x R. By the Monotone Convergence Theorem (3.5.6) we know that

Rf d =

limnR

fn d andR

g d = limnR

gn d. But also (fn + gn)n=1 is a monotone

increasing sequence of measurable simple functions that converges pointwise to f + g andso we also have

R

(f + g) d = limnR

(fn + gn) d. From Proposition 3.4.6 we haveR

(f + g) d = limn

R

(fn + gn) d

= limn

R

fn d + R

fn d=

R

f d +

R

g d

Corollary 3.5.8. Iff, g : R [0, ] are (Lebesgue) measurable functions, 0 andE L,then

E

(f + g) d =

E

f d +

E

g d

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Proof. Apply Corollary 3.5.7 to E(f + g) = Ef + Eg to get

E(f + g) d =

R E(f + g) d

=

R

(Ef + Eg) d

=

R

Ef d +

R

Eg d

=

E

f d +

E

g d

3.6 Integrable functions

Definition 3.6.1. Iff: R [, ] is measurable, then we say that f is Lebesgue integrableif

R

f+ d < andR

f d < .

For integrable f, we define R

f d =

R

f+ d R

f d

Lemma 3.6.2. Iff: R [, ] is measurable, then f is Lebesgue integrable if and only ifR

|f| d <

Proof. Iff is Lebesgue integrable, then |f| = f+ + f is (Lebesgue) measurable andR

|f| d =R

f+ d +

R

f d < .

Conversely, iff is Lebesgue integrable andR|f| d < , then f+ |f| (and we know f+

is measurable) and so

R

f+ d

R |f| d < .Similarly

R

f d < .Exercise 3.6.3. Iff: R [, ] is (Lebesgue) integrable, show that

R

f d

R

|f| d

(an integral version of the triangle inequality).

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Lemma 3.6.4. Ifg, h : R [0, ] are (Lebesgue) measurable functions with both R

g d < and

Rh d < , and if

{x R : g(x) = }{x R : h(x) = } = ,

then f = g h is integrable andR

f d =

R

g d R

hd.

Proof. We know that f(x) = g(x) h(x) is defined everywhere and f is measurable (by Propo-sition 3.2.12). Also |f(x)| g(x) + h(x) and so (via Lemma 3.5.5)

R |

f|

d R(g + h) d = R g d + R h d < .

So f is integrable.From f = g h = f+ f we have f + g = f+ + h and so by Corollary 3.5.7 we get

R

f d +

R

g d =

R

(f + g) d =

R

(f+ + h) d =

R

f+ d +

R

h d

which gives us R

f d =

R

f+ d R

f d =

R

g d R

hd.

Theorem 3.6.5 (Linearity of the integral). If f, g : R [, ] are (Lebesgue) integrablefunctions and R then(a) f is integrable and

R

f d = R

f d

(b) iff(x) + g(x) is defined for each x R, then f + g is integrable and R

(f + g) d =R

f d +R

g d

Proof. (a) If = 0, the result is easy. For = 0, f is measurable (by Proposition 3.2.5) andR|f| d = ||

R|f| d < so that f is integrable. If > 0 then (f)+ = f+,

(f) = f, which gives (using Corollary 3.5.7 (a))R

f d =

R

f+ d R

f d

=

R

f+ d R

f d

=

R

f+ d R

f d

=

R

f d

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If < 0, then (f)+ = f = ||f, (f) = f+ = ||f+ and so f = (f)+ (f) = ||f ||f+. Using Corollary 3.5.7 (a) we have

R

f d =R

||f d R

||f+ d

= ||

R

f d R

f+ d

=

R

f+ d R

f d

=

R

f d

(b) f + g is measurable (by Proposition 3.2.5). We know R f+ d < , R f d 0, for all n sufficiently large

supx[a,b]

|f(x) sn(x)| = supx[a,b]

f(x) sn(x) <

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and

0 < b

a

f(x) dx Ln(f) < (b a).

So we deduce that

limn

sn(x) = F(x) =

f(x) ifx [a, b]0 ifx / [a, b]

and

limn

Ln(f) =

ba

f(x) dx

But by the monotone convergence theorem (remember we assumed f(x) 0)

limn

Ln(f) = limn

R

sn d = R

F d = [a,b]

f d

which shows that the two integrals are the same.

R. Timoney March 14, 2012

MA2224-ch3 (1)

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