Lebesgue Measure and Integration: An Introduction (Pure and Applied Mathematics: A Wiley Series of Texts, Monographs and Tracts)

rp(x) = LckXEk(x), k=l

IS

called a simple function. A simple function assumes a finite number of

' taej

LEBESGUE "EASUAAIILE FUNCTIONS

· teaJ values and assumes each of these on a measurable set:
;·oJermine if the following functions are "simple" on [0, 1):

i.
.

2. rp(x) =

I,

x = 1/ n, n = 1,2, .. .

-I ,

xf.J f n.

{I,

x irrational

0, x rational.

~- rp(x) = [x) (the greatest integer function). Cantor function (Appendix A).

4.

j4.3':3

Problema

i. If Eisa measurable set, then the characteristic function onE, X£, is i. l

a measurable function on E. If

'

. ~OXIMAnON THEORDI 4.2 Let f be a measurable function defined jon~ measurable set E. Then there exists a sequence ofsimple functions ('Pt) ;qn )/:,so that

(fmite or infmite) lfof1al/ x E E. Iff is bounded on E. then lim
(unif)

~11 E.lff is nonnegative, the sequence ('l'k) maybe constructed so that it is a (mobotonically increasing sequence.

f roof' Suppose that f is nonnegative on E. We will construct a .mopotonically increasing sequence ('l't) with lim 'l'k =f. The idea is

4,3 APPROXIMATING AIEASURABL.E FUNCTIONS

131

essentially Lebesgue's: Divide the range off and approximate by level curves. Since f (E) C (0, ooJ, we partition (0, ooJ: (O,oo] = (0, I) U [l , oo] = (0, 1/ 2) U (1/ 2, I) U (l , oo). Define E11 =F ((0, 1/ 2)), E,z =F1 ((1 / 2, 1)), E, =F 1([1 ,oo]), and '1'1 = 0 · X e, + 1/ 2 · Xe11 + 1 · Xe, · Clearly 'l't ~!on E. Step I.

1

Step 2. (O,oo] = (0, I) u (1 , 2)u[2,ooj = (0, I/ 4)U [l / 4, I/2)U [l / 2, 3/ 4) u [3/ 4, 1) u [I, 5/ 4) u (5/ 4, 6/ 4) u (6/ 4, 7/ 4) U [7 / 4, 8/ 4) U (2,oo]. We have decomposed {0, ooj into 22 + 22 + I subintervals at the 2nd step. Fonn inverse images:

~. =r'([o.~>) .

-t(rl4'! 2)I) ,... ,

E22 =f

~s =r' (r~ . ~>) . Define

M

.,., 2

I 7 = 0·Xe11 +4 · Xen +· ··+ -4 ·Xe11 + 2X,U2 1 or

Note: Et; =:'

Ew-t U Elli for i =

I, 2.

Step k. [0, oo) = [0, I) U [I, 2) U · · · U [k - I, k ) U [k, oo] and partition into 2k + ~ + · + ~ + I subintervals = k · 2k + I disjoint subintervals and form inverse images. Thus, k-2" . I "\;""'1 'l'k = L..-ki=l 2

X e.,+ kXe,.

Note that E~r; = Ek+t 21 _ 1 U Ek+l 21 . To construct l"k+ l • divide the intervals [(i - l )/2k , i/ 2k) in half, and then l"k to l"k +J at those x's where 'l't changes.

140

LEBESGUE MEASURABLE FUNCTIONS

Certainly the 'Pk are nonnegative simple functions. We must show 'Pk < 'Pk+l and lim 'Pk = f on E. If f(xo) = oo, then 'Pk(x0 ) = k Yk and lim 'Pk(x0 ) = oo. If f(x 0 ) < oo, then fork> f(xo), 0 (2k·2k)/2+ 1 = k = 'Pk(x0 ), or x 0 Ej- 1 ([k+ l,oo]), and then 'Pk+ 1 (x 0 ) = k+ 1 > k = c.pk(x0 ). We have shown that for f > 0, 0

< 'PI < ... < 'Pk < 'Pk+l <

and lim 'Pk = f on E. Iff is nonnegative and bounded onE, say, 0 < f < M onE, then for all k > M, 0
4.3.4

Problems

Determine r.p 1, c.p 2 , c.p 3 for the following functions: 1. f(x)=x 2 ,

2. f(x)

={

O<x
~'

O<x
oo,

X=

0.

3. Cantor function (Appendix A). 4. f(x)

={

1,

x irrational

0,

x rational,

0 < x < 1.

4.4

4.3.5

ALMOST UNIFORM CONVERGENCE

141

Summary

1.

Lebesgue

Riemann Integrable Functions

Measurable Functions

2. If (fk) are measurable onE and iff is defined onE with f = Iimfk almost everywhere on E, then f is measurable on E. 3. Every nonnegative measurable function on E is the limit of a nondecreasing sequence of simple functions on E.

4.4

ALMOST UNIFORM CONVERGENCE

In this section we prove a remarkable theorem due to Egoroff: If we have pointwise convergence of a sequence of measurable functions on a set of finite measure, then we have uniform convergence on a "large" subset of that set. An example to illustrate this result is useful.

Example4: fk(x) = xk, 0 < x < 1 andf(x) vergence is not uniform on [0, 1] sup I fk(x)- f(x) ( O<;x-<;1 but is uniform on [0, I-E], 0 sup 0-<;x-<;1-E

I=

= 0,

0 < x < 1. The con-

1),

< E < 1, since

lfk(x) -f(x)

I= (1-

Et

(Egoroff, 1911) Let E be a measurable set ofreal numbers with finite measure. If (fk) is a sequence of measurable functions which converge to a real-valued function f almost everywhere onE, then, given E > 0, there exists a measurable subset E, of E such that J.L(E- E,) < E and the sequence (fd converges uniformly to f onE,.

THEOREM 4.2

142

LEBESGUE MEASURABLE FUNCTIONS

Proof Using the sequence (fk), we will construct a monotonically decreasing sequence of nonnegative measurable functions, (gk). This new sequence will be shown to converge uniformly to zero on £ 0 from which it will immediately follow that limfk = f(unif) on Ee

Let A denote the subset of E where f is not the limit of the sequence (fk)· Because A has measure zero, J..L(E) = J..L(E- A), and limfk = f on E-A. Since the limit of a sequence of measurable functions is measurable (Theorem 4.1), f is a measurable function on E-A. Defining gk=sup{ifk-fl,lfk+l-fl, ... } for k=1,2, ... , we have from Propositions 4.5, 4.6, and Theorem 4.1 again, that gk is measurable on E-A. Furthermore, 0 < gk+l < gk and limgk = 0 on E-A. In other words, the sequence (gk) is a monotone decreasing sequence of nonnegative measurable functions converging to zero on E-A. We will show that, except for a "small" subset of E- A, this convergence is uniform. That the original sequence (fk) would converge uniformly to f on this same set is apparent. The technical aspect of the argument begins: Let E > 0 be given. Stage 1: We first construct an increasing sequence of measurable subsets of E-A that "fill" E-A: Ek {xEE-AI gk(x)<1}. Clearly E l c E ~ c · · ·, U£ k = E- A, E k measurable (why?). It follows from Theorem 3.4 that limJ..L(Ek) = {t(E- A), so fork sufficiently large, say K1, 0 < J..L(E- A) - J..L(E k1 ) < E/2. The set E k1 is "almost" E- A and on this set, 0 < gk < 1 for all k > K1 • But why the sets E k? Everything certainly "works as advertised", but ... ?

Pick u E E-A. Since limgk(u) = 0, we have fork sufficiently large, say, k(u), that 0 < gk(u) < 1 for all k > k(u). Pick vEE- A. Since limgk(v) = 0, we have fork sufficiently large, say, k(v), that 0 < gk(v) < 1 for all k > k(v). Pick IV E E-A. Since limgk(w) = 0, we have fork sufficiently large, say, k(w), that 0 < gk(w) < I for all k > k(w).

Think about what we are doing: With each point of E - A we are associating a natural number, looking for a "magical" natural number that "works" for all points of E-A. (Idea: How about with each natural number associating a set of points, a subset of E ··-A.) For example, with the number one, associate the points of E- A such that g 1 < I. With the

4.4

ALMOST UNIFORM CONVERGENCE

143

number two, associate the points of E - A such that g 2 < I. Maybe in this way we can find a "not so magical" natural number that "works" for not all, but "most" of E-A. By now you realize that the subset of E- A associated with the number one is E l, the subset of E- A associated with the number two is E ~, ... , the subset of E - A associated with k is ELand the "so-called" "not so magical" natural number K 1 "works" on "most" of E- A: E So, after all this, 0 < gk < I for all k > K 1 onE k1 •

k.

Stage 2: Now things become quick and straightforward (thankfully). Understanding Stage 1, we define another sequence of measurable sets in the following manner: E~ {x E E- A I gk(x) < 1/2}. Again, ET c E~ c · · · , UE~ = E- A, lim ~t(E~) = ~t(E- A), and thus for k 2 sufficiently large, say K2 , 0 < ~t(E- A)- ~t(EkJ < E/2 . Also, 0 < gk < 1/2 f~r k > K 2 on Ek2 •

Stage n:

We have measurable sets

UEk

=

£7 c E2 c · · · ,

E- A, lim~t(Ek)

k

=

~t(E- A),

k

and fork sufficiently large, say Km 0 < ~t(E- A) - ~t(E'KJ < E/2n with 0 < gk < 1/n fork> Kn on E'Kn· Each of the sets E kI ,Ek2 , ... , E'k, ... is "almost" E-A. We will n show that

is "almost" E- A and that we have uniform convergence on Ef.

~t( (E- A)- nE'Kn)

=

~t(U( (E- A)- E'KJ)

<

L IL ( (E -

=

L(~t(E- A) -~t(E'kJ)

A) - E

'KJ

144

LEBESGUE MEASURABLE FUNCTIONS

Only uniform convergence on

n

EKn for k remams. Let 8 > 0 be given. We show 0 < gk < 8 on Recall 1/N < 8. that so N Choose sufficiently large. E'iN = {x E E- A I gKN < 1/N}. But E'iN :J nEKn and gk < gKN for all k > KN, that is, 0 < gk < 8 for all k > KN and all X

E

nEKn· n

•

The proof is complete.

By the way, where in the above argument did we use the assumption J..L(E) < oo? The example, fk = X[k,k+IJ with f = 0 shows J..L(E) < oo is necessary. 4.4.1

Problem

Using Example 4, the reader should repeat the argument of Egoroff's Theorem with fk(x) = x\ 0 < x < 1 and f(x) = 0, 0 < x < 1: gk(x) = xk, 0 < x < 1 and limgk = 0 on [0, 1). Show E k = [0, 1) for all k and E kr = E l. ii. Show E~ = [0, 1/2(!/k)) and K 2 > 277 if, ior example,£= .01. 111. Show E'k = [0, 1/n(l/k)) and (1 - Ej2n)K" < ljn.

1.

(1' 1)

1

n

E~

0

I

~----~~~~1

less than _e ·

2n

X

4.4

ALMOST UNIFORM CONVERGENCE

145

In 1912 Lusin showed that measurable functions are "almost" continuous. We use Egoroff's Theorem to establish this result.

4.3 (Lusin, 1912) Iff is a real-valued measurable function, defined on a set E offinite measure, then we may construct a closed subset Ec of E so that !-L(E- Ec)
THEOREM

Proof" The idea here is to approximate f with a sequence of simple functions (¢k) (Approximation Theorem 4.2), each being continuous except possibly at a finite set of points, and thus the set of discontinuities of all members, being a countable set, has measure zero. Cover this set of discontinuities, with a sequence of open intervals (h) so that /-L(Uh) < E/2. On the closed set E- (Uh), the simple functions are continuous and converge to f. Applying Egoroff, we have Ec so that EcCE- (Uh), !-L((E- (Uh))- Ec) < E/2, ¢k~f uniformly on Ec. Since the uniform limit of a sequence of continuous functions is continuous, the proof is complete. • 4.4.2

Problem

f(x) =

1'

x irrational

{ 0,

x rational,

Looking at the proof ofLusin's Theorem, what will the set Ec be? How about (¢k)? Even though measurable functions are "almost continuous," the reader should realize that "almost continuous" may in fact be nowhere continuous, as problem 4.1.3 indicates. The life of the individual only has meaning in so far as it aids in making the life of every living thing nobler and beautiful. -A. Einstein ... simplicity in mathematics, like simplicity of character, is an ideal to be achieved only by unremitting toil. -G. Temple Great spirits have always encountered violent opposition from mediocre minds. -A. Einstein

~ (4n)! 1103+26390n 9801 f:Q (n!) 4 396 4 n

.!_= v's 7r

-Ramanujan

5 Lebesgue Integration

We introduce the Lebesgue integral, in some respects a generalization of the primitive notion of "area under a curve". This new integral is defined for a class of functions that contains the Riemann integrable functions and permits certain limiting operations to be handled much more easily. The development proceeds as follows: Section 5.1 reviews the development of the Riemann integral via step functions. The same approach, via simple functions, yields the Lebesgue integral for bounded functions on Lebesgue measurable sets of finite measure. These efforts comprise Section 5.2. Of course, we want to remove the restrictions "bounded" and "finite measure." This is done in Sections 5.3 and 5.4. Finally, Section 5.5 is concerned with applications of lim J fn = Jlimfn· It is the so called "convergence theorems," which demonstrate the power and usefulness of the Lebesgue integral, a raison d'etre. Caution: Again, "measurable set" means "Lebesgue measurable set of real numbers,·' ''measurable function·' means ''Lebesgue measurable function''. All functions will have domains as subsets of R, with ranges in R or Re. 5.1

THE RIEMANN INTEGRAL

We define the Riemann integral of a step function in the obvious way. The extension to more general bounded functions f on [a, b] is via approximation from above and below by step functions. 147

148

LEBESGUE INTEGRATION

5.1.1

Definition

A real-valued function¢ with domain [a, b] is called a step function if there is a partition a

= Xo

<

<

XJ

0

0

0

<

Xn

= b

of the interval such that¢ is constant on each subinterval h that is,

cp(x) = ck for x with cp(xk) =db k

=

0, 1,

0

0

0,

E

h,

C1

---

no

• •

-0

• -------o

Cn o - - - - - ------o

• a=

- ---- X0

X1

+- Xn-

c2 o-----o

1

Xn =

•

• 5.1.2

(xk-i, xk);

k=l, .. o,n,

• 0------

=

Definition

Let¢ be a step function on [a, b]: k = 1, 2,

Xk-i <X< XkJ

k = 0, 1,

0

0

0,

0

0

0,

n

n,

The Riemann integral of¢ on [a, b], denoted by

J: cp(x) dx, is

•

•

•

• -o

• a= X 0

0·

_______ -----{)

EB

EB X2

X1

Xn-

8

•

•

1

b

5.1

5.1.3

149

THE RIEMANN INTEGRAL

Comments

1. We could write¢= L ]ck X(xk-J,Xk)

+L

odk X{xk}' and J% cj;(x) dx = Lk=1Ck!-L( (xk-1,xk)) + Lk=Odk~-t({dk}) = Lk=1ck(xk- xk-1)·

The step function's values at the endpoints of the subintervals have no bearing on the existence or value of the Riemann integral of a step function (dk does not appear in the definition of the integral):

¢I (x)

=

1'

O<x
3,

x= 1

2,

l<x<2

1, and

cP2(x)

¢I= ¢ 2 forO< x < 1 and 1 < x < 2, ¢I(l)

la ¢ (x) dx = 2

1

=

I

{ 2,

¢ 2 (1), but

la ¢ (x) dx. 2

1· 1+2 · 1=

2

2. The value of the Riemann integral of a step function is independent of the choice of the partition of [a, b] as long as the step function is constant on the open subintervals of the partition, for example,

¢(x) =

1' { 2,

Partition: {0, 1, 2}, J~ cj;(x) dx = 1 · 1 + 2 · 1 = 3. Partition: {0, 1/2, 1, 7/4, 2}, J~ ¢(x) dx = 1 · 1/2 + 1 · 1/2 + 2 · 3/4

2. 1/4

= 3.

+

More formally, the Riemann integral of a step function is well defined; it is independent of the particular representation of¢. For example, if ¢(x) = cb xk-1 < x < xk and we add another partition point x*, xk-1 < x* < xb we have

PROPOSITION

5.1

If¢ and '1/J are step functions on [a, b], and k is any real

number, then 1. (k¢) is a step function on [a,b], and J:(k¢)(x)dx

(homogeneous);

= k J%¢(x)dx

150

LEBESGUE INTEGRATION

2. (¢

+ '1/J)

is a step function on [a, b], and

1b (¢ + '1/J)(x) dx = 1b ¢(x) dx + 1b '1/J(x) dx

(additive);

3. J: ¢(x) dx < J: 'ljJ(x) dx if¢< 'ljJ on [a, b] (monotone); 4. If a< c < b, the integrals J~ ¢(x) dx, fcb ¢(x) dx exist, and J~ ¢(x) dx + J: ¢(x) dx = J: ¢(x) dx (additive on the domain).

Proof The arguments are straightforward: ¢ and 'ljJ each have an associated partition. The union of these partitions will also be a partition of [a, b] for¢, '1/J, (k¢), and (¢ + 'ljJ). For part 4, adjoin" c" if need be to the partition associated with¢. The reader may provide "rigor." • We now define the Riemann integral for more general functions f on [a, b]. Since we will be approximating f from above and below by step functions it is imperative that f be a bounded function on [a, b]. 5.1.4

Definitions

Let f be a bounded function on [a,b], say, o
J!(x)dx

=

sup{1b ¢(x)dx I ¢ <J, ¢a step function}. -b

The upper Riemann integral off on [a, b], fa f(x) dx, is defined by -b.

b

J/(x)dx=inf{1 '1/J(x)dxl f<'ljJ, 'ljJ a step function}.

5.1.5

Comments

1. Since o
lh

¢(x) dx < 1b (3 dx

= (3(b-

a),

5.1

THE RIEMANN INTEGRAL

151

the set is bounded above. The least upper bound is a real number: The lower Riemann integral for a bounded function on a closed bounded interval is well-defined ("The" least upper bound, not "a" least upper bound). Similarly for the upper Riemann integral. 2. Since cjJ < 1/J, cjy(x) dx < 1/J(x) dx by monotonicity for step functions. Because cjJ is arbitrary, we may interpret this inequality as saying 1/J(x) dx is an upper bound for the set

J:

J:

J:

{1b cjy(x) dx I

c/J


c/J

a step function}.

But _a Jb f(x) dx is the smallest upper bound. We have

Again, interpret this last inequality as saying the lower Riemann integral, f(x) dx, is a lower bound for the set _a

t

{1b 1/J(x) dx I f < 1/J, Since the upper Riemann integral, b

1jJ a step function} .

J! f(x) dx, is the greatest lower bound, -b

j !(x)dx < J/(x)dx. It follows that a bounded function

f on [a, b] satisfies

for any step functions cjJ
152

LEBESGUE INTEGRATION

5.1.6

Definition

A bounded function f on [a, b] is Riemann integrable on [a, b] whenever -b Jb f(x) dx =fa f(x) dx.

_a b

=

f'J(x) dx

5.1. 7

b

faf(x) dx

b

Denote the common value by fa f(x) dx;

fa

-

= J(x) dx.

Comment

The definition of the Riemann integral for a bounded function f on [a, b] agrees with our original definition 5.1.2 when f is a step function, say, A

b ,-.,

A

h

¢ = f = '1/J: fa ¢(x) dx E {fa ¢(x) dx monotoncity,

A

¢ < ¢, ¢a step function} and by

J: ¢(x) dx < J: ¢(x) dx,b i.e., J: ¢(x)b dx is an upperb bound

and a member of this set. Hence - b

I

b

J f(x) dx = fa ¢(x) dx = fa '1/J(x) dx. _a A

b

A

A

A

Similarly, fa f(x) dx = fa '1/J(x) dx =fa ¢(x) dx. We have defined what it means for a bounded function f on [a, b] to be Riemann integrable; a common value for the lower and upper Riemann integrals. An equivalent condition, that is frequently easier to apply, is given by the next result. 5.2 A bounded function f on [a, b] is Riemann integrable iff for every E > 0, we have step functions¢ and '1/J, ¢ < f < 'ljJ on [a, b]. so that PROPOSITION

0 < 1b '1/J(x) dx -1b ¢(x) dx

= 1b['I/J(x)- ¢(x)]dx <E .

.___ ___. a= X0

Xj

-

-

-

--

-~

-- -

~-

.______

Xn-

1

x, = b

5.1

THE RIEMANN INTEGRAL

153

1b[ 1/;(x)- ¢(x) ]dx ='--i_

Proof Assume the bounded function f is Riemann integrable on [a, b] and let E > 0 be given. From 5.1.5 and the definitions of le~st upp~r bound and greatest lower bound we have step functions ¢ and 1/J, J < f < {/;, so that bf(x) dx--E = Jb f(x) dx- E < 1b J(x) dx < Jb f(x) dx 2 2 _a a _a 1

rb

E

. ' 2 b' b' b' Thus 0 < fa 1/;(x) dx- fa c/J(x) dx = fa [1/;(x) - ¢(x)] dx < E. Now let E > 0 be given and assume we have step functions ¢and 1/J, ¢
A

0

< 1b 1/J(x) dx- 1b c/J(x) dx = 1b [1/;(x) - cb(x)] dx <

E.

But, for any bounded function f on [a, b], b 1

b -b b ¢(x) dx < J !(x) dx < J /(x) dx < 11/J(x) dx. -b

b

We conclude that 0
j that is,

b !(x) dx

-b

=

j /(x) dx =

f, is Riemann integrable on

L

By the arbitrary

b 1 f(x) dx,

[a, b].

Note: When the bounded function f is Riemann integrable on [a, b], w~ have ¢ < f < 1/J on [a, b], f; ¢(x) dx < f; f(x) dx < f; 1/;(x) dx, and fa [1/;(x) - ¢(x)] dx < E, for some step functions¢ and 1/J. • The next problems are applications of Proposition 5.2, a revisitation of Cavalieri, Fermat, and so on.

154

5.1 .8

LEBESGUE INTEGRATION

Problems

I. Show f(x) = x 2 , 0 < x < 1 is Riemann integrable on [0, 1]. 1. Form the partition 0 < lin< 2ln < · · · < (n- I)ln < nln.

J(x)

(k 1)

2

n

=

,

otherwise,

f,

~(x) =

(~) 2' otherwise.

f,

jJ

n. Show ¢(x) dx = 113- 1l2n + 1l6n2 113 < J~x dx. 1

1

' 1 1 1 'lj;(x)dx=-+-+-2 3 2n 6n

and

thus

and, thus,

(See Problem 2.4.5.) 1 2 -1 2 n1. You could conclude 113 < J 0 x dx < J 0 x dx < 113 and then 2 JJ x dx exists and has value 113. Please show

1~(x) -1 1

0

<

dx

1

¢(x) dx

= ~.

Thus JJ x 2 dx exists by Proposition 5.2. What is its value? 2. Show that JJ xP dx exists, p any natural number. Is it difficult to evaluate this integral? Look at the next problem. 3. Show JJ x(pfq) exists and JJ x(p/q) = 1I(pI q + 1), p, q natural numbers. 1. Form these partitions:

... ' ~}, { 0,~,~' n n n

{o, (!)q, (~)q, ... , (~)q}, {o,(~)p,G)p,

... , (:)p}·

5.1

155

THE RIEMANN INTEGRAL

Then

1 (k- 1)p q(k- 1)q- ·-I l:q (k- 1)p+q- ·-=2: k=1

1

1

n

n

n

n

n

n

_ (k- 1)p [(k- 1)q- +···+ (k- 1)q- ]·-1 -2: 1

n

1

n

n

n

q terms

2:c: r[ Gr-c: ~rJ 1

=

<

l:(~Y[(~r-c: ~rJ

(k)p+q-1 ·-,I
k=1

and

1 L:q (k- 1)p+q-l 1 q L:q ( -k)p+q-J ·-n n n n n

·-=-.

Interchange p and q to conclude that Jd xp+q-l dx,

Jd xP/qdx, and Jd pxp+q-l dx =

n.

Jd xq/p dx exist (Proposition 5.2) and that Jd xqfpdx, Jd qxP+q- 1dx = Jd xp/qdx. Since (p + q) Jd xP+q- 1dx = Jd xq/pdx + Jd xP/qdx, you need only show Jd xPiqdx + Jd xqfpdx = 1, for then Jd xp+q- 1dx = 1/(p + q) and Jd xp/qdx = 1/(1 + pjq).

156

LEBESGUE INTEGRATION

This is obvious from the following figure: lf-----

l plq'd JoX X

1

0

or the relationship,

iii. Complete problem 2. Problems like the last three are highly motivational. We desperately seek an alternative method. Generally, we abstract to try and determine what is essential and what is not. We follow in Cauchy's footsteps: Specific functions are to be replaced with more general classifications, such as continuous, monotone, and so forth. 5.1.9

Problem

Every continuous function 1.

f on [a, b] is Riemann integrable.

f is bounded on [a, b] so

defined.

J_ab f(x) dx

-b

and fa f(x) dx are well-

5.1

u.

THE RIEMANN INTEGRAL

157

f is uniformly continuous on [a, b] by Theorem 2.8. Given t. > 0, choose 15 > 0, so that

lf(x)- f(y)

111.

E

I< b _a

whenever I x - y I < 15, x, y E [a, b]. Take any partition {x 0 ,x 1 , ... ,xn} of [a,b] so that xk- xk-J < 15, k = 1, 2, ... , n and, thus, E

- b _a

E

_a for

x,y E (xk_ 1 ,xk),

that is, f(x)
1v. Define¢= infj, ;J; = sup/ on (xk-J, xk) and f, otherwise. b b v. Show fa 7/J(x) dx- fa ¢(x) dx < (t./(b- a))(b- a), and apply Proposition 5.2. A

A

then

So continuous functions are Riemann integrable (Is existence in Problem 5.1.8 a little easier now?). Can discontinuous functions be Riemann integrable? A little thought should convince you that a finite number of discontinuities will not cause any problems, and if you recall Riemann's example of Chapter 1, in that case, a countable dense set of discontinuities did not cause any difficulties. 5.1.1 0

Problem

Let

f(x) =

1, { 0,

x irrational,

O<x
x rational,

O<x< I.

Show J~f(x) dx = 0, J ~ f(x) dx = 1 and conclude f is not Riemann integrable. Of course f is nowhere continuous. Maybe "countable" is what we are looking for in the way of a new criteria for Riemann integrability.

158

LEBESG UE INTEGR ATION

5.1.11

Probl em

A nondecreasing or nonincreasing real-valued function is said to be monotone.

I. Show every mono tone function f on [a, b] is Riemann integrable. 1. A pictur e provides the idea. Assume f is nondecreasing. Then f(a) <j(x )
11.

f

--- ..

" flb)- flax

'

---

-

---

~-- . - -

b-a -n

-

-·--~----'-

a

be: q n

b~(l

-

..

-

- - - -- - · - - -

n

- - - _ L __

__._

b-g, b n

2. Every mono tone function f on [a, b] has a countable set of discontinuities. Suppose f is nondecreasing. 1. If a < c < b, show lim_ f(x) = sup f = f(c-) exists,

X-->C

a<Sx
lim f(x) = inf f = f(c+) exists,

X->c+

c<x<Sb

and f(c-)
5.1

VI.

THE RIEMANN INTEGRAL

159

f has a countable set of discontinuities.

Countability of the set of discontinuities is looking better and better.

5.1.12

Problem

Define a function f on the Cantor set (Appendix A) by 1,

f(x)

1.

Let Let

=

{ 0,

x is a member of the Cantor set otherwise. 1, 0 < x < 1. Then ¢;0 < f < ~0 on [0, I]. = I on [0, 1/3], [2/3, I], and zero otherwise: [0, I]. Continue the obvious way. Then I], and

¢;0 = 0 and ~ 0 =

¢; 1 = 0 and ~ 1 ¢; 1 < f < ~J, on

Jn
J

1

0<

I/(x) /

dx <

/(x) dx < 1-

(I.3 + 92+ ···)

i.e., 0 = J~f(x) dx = J~f(x) dx. This function is Riemann integrable even though f is discontinuous at each point of the Cantor set, an uncountable set. "Uncountable" is okay sometimes. Finally, Lebesgue realized that instead of "measuring" discontinuities by "counting" we should be "measuring" discontinuities by "lengths of covers." The next result characterizes Riemann integrability! THEOREM 5.1 (Lebesgue, 1902) A bounded function on a closed bounded interval is Riemann integrable iff the function is continuous almost everywhere. (The set of discontinuities is limited to a set of Lebesgue measure zero.)

Proof' We will use the Riemann integrability criteria (Proposition 5.2): A bounded function f on [a, b] is Riemann integrable iff we can approximate f from below and above by step functions whose integrals can be made arbitrarily close to each other: If we have step functions ¢ < f < ~, the difference ~ - ¢ is a measure of how much f

160

LEBESGUE INTEGRATION

may vary on any subinterval of a partition. The standard means of measuring this variation of f on an interval J is by calculating sup{f(x) I X E J n [a, b]}- inf{j(x) I X E J n [a, b]}, traditionally denoted by w1 (J). This real number w1 (J) (f is bounded on [a, b]) is nonnegative. It is natural then to measure the variation off at a point x 0 E [a,b],denotedbywr(x0 ),asw1 (x 0 ) =inf{w1 (I) I Ianyopeninterval containing x 0 }. We would expect that for a function continuous at x 0 , w1 (x 0 ) = 0. In fact, for a bounded function f on [a, b], f is continuous at x0 in [a, b] iff w1 (x 0 ) = 0. Here is a rough sketch of the argument to establish this result. Suppose f is continuous at x 0 E [a, b] and E > 0 is given. We have 8 > 0 so that for X E (xo- 8, Xo + 8) n [a, b], -E f(xo)- E. Subtracting, w1 ((x 0 - 8,x0 + 8)) < 2c Then w1 (x 0 ) < w1 ((x0 - 8,x0 + 8)) < 2E for every E > 0. So w1 (x 0 ) = 0. Conversely, suppose w1 (x 0 ) = 0 for some x 0 E [a, b] and let E > 0. Then inf{ w1 (I) I I any open interval containing x 0 } = 0 implies we have an open interval I* containing x 0 so that 0 < w1 (l*) < E. Since I* is open, choose 8 > 0 so that (x 0 - 8, x 0 + 8) c I*. Then X E (xo- 8,xo + 8) n [a,b]===?- E 0. This implies that the set of discontinuities off, say, D, may be written as D = U{x E [a,b]l w1 (x) > 1/n}. n

What do we know about the sets Dn = {x E [a,b]l w1 (x) > 1/n}, n= 1,2, ... ? We claim Dn is a closed set in [a,b], or, equivalently, D~ = {x E [a, b]l w1 (x) < 1/n} is an open set in [a, b]. Let xED~. We will determine a 8 > 0 so that (x- 8, x + 8) n [a, b] c D~. The argument is very close to what we did above. Since w1 (x) = inf{w1 (I) I I any open interval containing x} < 1/n, we have an open interval I* containing x so that wr(I*) < 1jn. Because I* is open, there exists a 8 > 0 so that (x- 8, x + 8) n [a, b] c I* n [a, b]. We are done if we can show (x- 8, X+ 8) n [a, b] c D~. Let z E (x- 8, X+ 8) n [a, b]. Then w1 (z) <w1 ((x-8,x+8)) <w1 (l*) < 1/n;zED~. The preliminaries have been lengthy, but the proof of the main theorem is now fairly easily constructed if we keep this idea in mind: We can cover the discontinuities off with a set of small measure. On this "small" set, f

5.1

THE RIEMANN INTEGRAL

161

may be badly behaved, but f is bounded. On the rest of the interval, a "large" set, f is well-behaved (continuous). Integrability is determined by the behavior off on the "large" set. It's time to begin in earnest. We first suppose f is continuous a.e. on [a, b], that is, the set of discontinuities off, D, has measure zero, and let If I < Bon [a, b]. We will show f is Riemann integrable on [a, b]. Because Dn = {x E [a,bJI w1 (x) > 1/n} is a subset of D, Dn has measure zero. Thus we have Dn c Uh, h open intervals, L:: /(h) < Ej4B. But Dn is a closed and bounded subset of [a, b], hence compact. By Heine-Bore!, we have a finite subcover of Dn by the open intervals, h, that is, m

Dn C

Uik,h ' ' i=l

open intervals. Then the set [a, b] - (hi U h 2 U · · · U hJ is a .finite union of closed intervals 1 1, 1 2 , ••• , h· That is,

Recall that all points of Dn are in hi U h 2 U · · · U Ikm· Thus wf(x) < 1/n on J 1 U J 2 U · · · U h. Can we now define step functions ¢, 1/J in the obvious way: ¢(x) = inf/ on the intervals J1, ... , h, hi, ... , lkm,

1/J(x) =sup/ on the intervals J1, ... , h, hi, .. ·, hm? The problem here is that we need some kind of estimate for supf(x) - infj(x), J;

1;

having only the information that w1 (x) < 1/n on each h Does "pointwise" small, w1 (x) < 1/n, imply "global" small, w1 (Ji) < 1/n, or something like this? Again, we call on Heine-Borel. Since w1 ( x) < 1/ n for each x E h we have an open interval containing lx so that w1 (Ix) < ljn. But then the collection {lx} is an open cover of the closed, bounded, and hence compact set, h A finite subcover must cover h We have a partition of Ji, and the "sup of/"-"inf off" over any of the subintervals of this partition is less than 1jn. This is exactly what we need! (Clearly this does

162

LEBESGUE INTEGRATION

not imply w1 (JJ < 1/n!) Do this for each J,., i = 1,2, ... , L. We have a finite collection of subintervals whose union is J 1 U h u · · · U J L' and on any one of these subintervals, say J*, w1 (J*) = sup{f(x)

I

x E J* n [a, b]} - inf{j(x)

Now define step functions

I

x E J* n [a, b]}


¢, {/; in the obvious way (finite number of

subintervals is critical to being a step function): A

¢ = infj on the subintervals of J1 ,Jz, ... , h and hi, ... ,lkm, 1/J =supf on the subintervals of J1, ]z, ... , h and hi, ... , hm. Then

1

b [1/J(x) A

- <j>(x)]dx <-1 "" L-J /(subintervals of J,.) A

n

< ~ (b - a) + 2B (

+ 2B "" L-J l(hJ

4~)

< E for n sufficiently large. So

f is Riemann integrable on [a, b] (Proposition 5.2).

We next suppose f is Riemann integrable on [a, b]. We want to show the bounded function f is continuous a.e. on [a, b]. Because the set of discontinuities of j, D, is given by D = UDn = U{x E [a, b]l w1 (x) > 1/n}, if we can show J-L(Dn) = 0, we would be done, since a countable union of sets of measure zero is a set of measure zero. Fix ann, say, N, and consider the set DN

= {x

E

[a, b]l w1 (x) > 1/ N},

with

E

> 0.

Because f is Riemann integrable on [a, b] by assumption, we have step b functions ¢(x)]dx<,_E/N. ,Let a=xo< x 1 < · · · < xn = b be a partition associated with ¢ and '1/J, and split the collection {(x 0 , x 1), ••• , (xn~!, xn)} into two subcol!ections V 1 and V2, as follows: If(xk~J,xk) n DN #- 0, put it in V 1 . Otherwise, put it in V2. Then A

A

A

A

5.1

THE RIEMANN INTEGRAL

every point of D N is in 'D 1 or a member of the set {a, x 1 , x2, So

... , Xn-1,

163

b}

For I:v I , some point of DN is in each subinterval, i.e., each subinterval ' ' of'D 1 containsapointxwithw1 (x) > 1/N.Butthenw-¢> 1/Nonthis subinterval. Conseq uen tl y,

~>

L

>

~L

(lengths of subintervals that contain points of DN ),

D1

so I:v 1 (xk- xk_ 1) <E. The intervals of 'Dh along with the finite set {a, x 1 , ... , Xn_ 1 , b} contain all points of DN. So J..L(DN) < E, and the proof of this theorem is complete. Congratulations! • We have characterized Riemann integrable functions. We now show that the integral properties that hold for step functions (Proposition 5.1) remain valid for Riemann integrable functions. This theme, step functions first (easiest), and then more general functions as combinations or limits of easier functions, is repeated throughout this chapter. 5.2 If bounded functions f and g are Riemann integrable on [a, b], and k is any real number, then

THEOREM

1. (kf) is Riemann integrable on [a, b], and

1b

(kf)(x) dx =k

1b

f(x) dx

(homogeneous);

2. (f +g) is Riemann integrable on [a, b], and

(additive);

3.

1b

f(x) dx <

1b

g(x) dx iff< g on [a, b]

(monotone);

164

LEBESGUE INTEGRATION

4. If a < c < b, f is Riemann integrable on [a, c] and [c, b], and (additive on the domain);

5. !fa
1b

f(x) dx < (J(b- a)

(mean value).

Proof' Theorem 5.1 quickly yields integrability of the appropriate functions. A more direct argument, based on Proposition 5.2, is of interest.

1. If k = 0, obvious. Assume k > 0 and f is Riemann integrable on [a, b]. Let E > 0 be given. From Proposition 5.2 we have step functions ¢, ,J; on [a, b] so that ¢ < f < ,J;,

and

b

'

b'

b

bA

Since fa (k¢)(x)dx = k fa ¢(x)dx and fa (k'!jJ)(x)dx = k fa 1/J(x)dx (Proposition 5.1),

1b

(k¢) (x)dx = k


A

1b 1b 1b ~(x)dx ¢(x)dx

f(x)dx

5.1

THE RIEMANN INTEGRAL

165

and

But, A

A

k¢ < kf < k'lj; on [a, b]. That is, we have step functions (k/fy), (k(/;) bracketing (kf) so that f:(k(/;)(x)dx- f:(k/fy)(x)dx <E. But then (kf) is Riemann integrable on [a, b] by Proposition 5.2. Since k f: f(x)dx b b b and fa (kf)(x)dx lie between k fa ¢(x)dx and k fa 1/J(x)dx, they are equal by the arbitrary nature of E. The reader may argue the case when k < 0. 2. From Proposition 5.2, A

A

and and and Adding, ¢1 + Jg < f tion 5.1 yields

1b ¢1 (x)dx

+ g < (/;1 + (/;g,

and application of Proposi-

+ 1b /fyg(x)dx = 1b ( ¢1 + Jg) (x)dx < jb ( f + g)(x)dx _a

J

-b

<

a(f + g)(x)dx

< 1b((J;1 +(J;g)(x)dx = 1b (/;1 (x)dx + 1b (/;g(x)dx.

166

LEBESGUE INTEGRATION

Application of Proposition 5.2 again yields existence of Jt(f + g)(x)dx, and it along with Jtf(x)dx + Jt g(x)dx, lies bA bA bA bA between fa 'l/Jt(x)dx +fa 'l/Jg(x)dx and fa r/Jt(x)dx +fa f/Jg(x)dx. 3.

1b f(x)dx = sup{1b ¢(x)dx I ¢

<J,

< sup{1b ¢(x)dx I ¢ < g, since

¢ a step function} ¢ a step function},

f
= 1b g(x)dx. 4. Let a< c
by Proposition 5.2. But

and 0 < 1c ;j;(x)dx- 1c /f;(x)dx

= 1c(;j;- /f;)(x)dx < 1b(;j;- /f;)(x)dx <E by Proposition 5.1 and thus J~f(x)dx makes sense by Proposition

5.1

THE RIEMANN INTEGRAL

167

5.2. An analogous arguments yields existence of Jt f(x)dx. Thus

and

This completes the argument for 4. 5. The constant functions o: and (3 are step functions satisfying o: < f < (3 on [a, b]. From part 3,

o:(b- a)=

1b

o: dx

<

1b

f(x)dx <

1b

(3 dx

=

(3(b- a).

•

We conclude our brief treatment of the Riemann integral with some examples, a convergence theorem pertaining to:

b 1

?

(limfk)(x)dx · lim

1b a

fk(x)dx,

and The Fundamental Theorem of Calculus. Example 1:

1. f(x)

=

x rational

1, { 0,

x irrational

enumeration of the rationals in [0, 1], we may define a uniformly bounded, monotonic sequence of functions (fk) as follows: 1,

fk(x)

=

{

0,

otherwise

Iimfk = f, Jd fk(x)dx = 0, and as a consequence, limfdfk(x)dx = 0. However, Jd(Iimfk)(x)dx is not even defined! Then

2. f(x)

= 0, 0 < x < 1, fk(x) =

< 1I k

k 2 X,

0<

-k2 (x- 2/k),

1/k < x < 2/k

0,

2/k <X< 1.

X

168

LEBESGUE INTEGRATION

Then limfk = f(not uniformly), the sequence (fk) is not uniformly bounded, Jdfk(x)dx = 1, and

Uniform boundedness is not enough, nor monotonicity. The concept we are looking for with the Riemann integral is uniform convergence: The uniform limit of a sequence of Riemann integrable functions is Riemann integrable and the integral of the limit is the limit of the integrals.

5.3 Suppose we have a sequence (fk) of Riemann integrable functions on [a, b]. If limfk = f ( unif) on [a, b] then THEOREM

1. (limfk) is Riemann integrable on [a, b]; 2. lim J:fk(x)dx = J:f(x)dx = J:(limfk)(x)dx. Proof

1.

f is Riemann integrable on [a, b]. Let E > 0. From uniform convergence of (fk) on [a, b], we have a natural number K so that E

E

fk(x)- 4(b _a)
+ 4(b _a)

(1)

for all x E [a, b], k > K. So f is bounded on [a, b], and because fk is Riemann integrable on [a, b], we have step functions ¢k and 'lj;b ¢k K. But then, £

f

¢k(x)- 4(b- a)

< 'lj;k(x) + 4(b- a)

5.1

THE RIEMANN INTEGRA L

169

on [a,b], k > K: We have step functions ¢k- (E/4(b - a)) and '1/Jk + (E/4(b - a)), bracketing f, and

1b [

('1/Jk(x)

+ 4(b ~a))=

k > K. Thus

1b

(¢k(x )- 4(b

~ aJ] dx

['1/Jk(x) - ¢k(x)] dx + ~ < E,

f is Riemann integrable on [a, b].

2. We may integrate (l) by Propos ition 5.2:

b 1

fk(x)d x-

4E <

1b a

f(x)dx <

1b a

fk(x)d x

+ 4E ,

k > K, that is,

k > K, and the proof is complete. 5.1.13

•

Problem

Suppose we have a sequence of Riema nn integrable functions (fk) on [a, b]. If f(x) = "L.'b.dk(x), a< x < b, the series converging uniformly on [a, b] to f, then

l (f;,

tk(x)) dF

l

f(x)dx

~ f;, J.' fk(x)dx

In other words, the series may be integrated term by term.

=

Hint: Fn(x) 2:. k=l fk(x). Then Fn(x) is Riemann integrable on [a, b], lim Fn = f uniformly on [a, b], and application of Theorem 5.3 completes the argument. We may weaken some of these requirements of Theorem 5.3 when we deal with the Lebesgue integral. And now, one of the most celebrated theorems in mathematics: "The Funda menta l Theorem of Calculus"!

1 FU

LEBESGUE INTEGRATION

5.4 The Fundamental Theorem of Calculus: Let f be a bounded function on [a, b]. If f is Riemann integrable on [a, b] and F(x) - J: f(t)dt, then F is continuous on [a, b]. Additionally, iff is continuous at x 0 E (a, b), then F is differentiable at x 0 and F'(x0 ) =f(xo). THEOREM

Proof We show that F is continuous with the assumption off being Riemann integrable. Equivalently, limx~x0 [F(x)- F(x 0 )] = 0 for x 0 E [a, b]. We begin: Since f is bounded on [a, b] -B
a< t
We may integrate because f is Riemann integrable by assumption, and thus J:0 -Edt< J:J(t) dt < J:0 Bdt, a< x 0 < x < b (Theorem 5.2), that is, -B(x- x 0 ) < F(x)- F(x0 ) < B(x- xo),

a< x 0 < x
\ \ \ \ \ \ \ \ \ ~

0

We have limx_,x+ [ F(x)- F(x 0 )] = 0. A similar argument shows 0 limx_,x-0 [F(x) - F(x0 )] = 0. F is continuous on [a, b]. Now, assume f is continuous at x 0 , a< x 0 Xo

X - Xo

Since f is continuous at x 0 , given f(xo)-

E


+ E,

E

x 0 E (a,b).

> 0, we have 8 > 0 so that t E (xo - 8, Xo

+ 8) n [a, b].

We may integrate (f is continuous; the Riemann integrable exists by

5.1

THE RIEMANN INTEGRAL

171

Theorem 5.2 or Problem 5.1.9)

X E [xo, Xo

+ 8) n [a, b]. That is,

[f(x 0 ) - E](x- x 0 ) < F(x)- F(xo) < [f(xo) + E](x- xo), X

E

[xo, Xo -E

+ 8) n [a, b]. Thus

< F(x) - F(xo) - f(xo) < X- Xo

E,

X E (xo,Xo

+ 8) n [a,b].

Hence limx_,x+[(F(x)- F(x 0 ))j(x- xo)- f(x 0 )] = 0. The reader may 0 . • argue 1lmx_,x-. 0 5.1.14

Comment

Iff is continuous on [a, b],

a<x
5.1.15

Problems

-1,

-l<x
0,

O<x
1,

l<x<2

1. Suppose f(x) =

Calculate jxf(t)dt, -1

Graph f.

rf(t)dt and graph.

}o

Investigate the continuity of these three functions at x x = 1 using the definition. Calculate! (

1·~/(t)dt),

:x: ( lo'f(t)dt)

and graph. Compare with the graph of f.

= 0 and

where appropriate,

172

LEBESGUE INTEGRATION

X, 2. Suppose f(x)

=

0 <X< 1

Graph f.

1'

3-x

'

Calculate and graph Jt f(t)dt, ft; 2 f(t)dt. Investigate the continuity and differentiability of these three functions at x = 1 and x = 2 using the definition. Calculate :x (lox f(t)dt),

~(

1:

2

f(t)dt)

where appropriate, and graph. Compare with the graph of f. The "other?" Fundamental Theorem of Calculus is what we frequently use for evaluating integrals.

5.5 Iff is continuous on [a, b], differentiable on (a, b), and if the derivative, f 1 , is Riemann integrable on [a, b], then

THEOREM

lb

f'(x) dx

= f(b)- f(a).

Proof Since f' is Riemann integrable on [a, b] by assumption, we have step functions¢ < f' < 'ljJ on [a, b] with ¢(x) dx < f' (x)dx < '!j!(x) dx, and '!j!(x) dx¢(x) dx < E for any E > 0. Take the common partition formed by ¢and '!j!, say {a, x 1, x 2 , ... , Xn-i, b}. Suppose¢= 2:::~ 1 c;X(x,_"x,) and 'ljJ = 2:::7= 1 d;X(x,_ 1,x;)· Thus C; < f' < d; on (xi-!, xJ

J:

J:

J:

J:

Since n

f(b)- f(a) = 'L)f(x;)- f(x;_J)J i=l

n

=L

f'(e;)(x;- x;-J),

i=l

by the Mean Value Theorem (Theorem 2.10), we have

J:

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

that is, J: ¢(x)dx
173

and

1/(b) -f(a) -1bf'(x)dx I <E.

•

The argument is complete.

Looking back at Problem 5.1.8, existence is via Problem 5.1.9 and evaluation follows from this last result, Theorem 5.5. Isn't mathematics powerful: For it is unworthy of excellent men to lose hours like slaves in the labor of computation. -Gottfried W. Leibnitz

This concludes our treatment of the Riemann integral. 5.2 THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

We develop the Lebesgue integral for bounded functions f on a set E of finite Lebesgue measure. The treatment parallels that of the Riemann integral, replacing step functions with simple functions. The reader might review 5.1 at this time. 5.2.1

Definition

A function ¢, whose domain is a measurable set E with finite measure, is called a simple function, provided n

¢(x)

=L

ck XEk (x),

k=i

where the ck are real numbers (not necessarily distinct) and UEk mutually disjoint measurable subsets of E. (We may assume n

n

UEk = E since E

0

I

=

=E-

UEk I

is a measurable set and 0 · p,(E0 ) = 0.) Every step function is a simple function. The converse is false.

E, Ek

174

LEBESGUE INTEGRATION

Example 2:

A simple function that is not a step function:

¢(x) ¢

=

{

l)

x irrational

0,

x rational.

= 1 · X{irrationals} + 0 · X{ rationals}

= X{irrationals} ·

A simple function is "simply" a finite linear combination of characteristic functions and is easily shown to be a measurable function (Problem 4.3.3).

5.2.2

Comment

A simple function may have many representations.

¢(x)

Let £ 1 = (0, 1], £ 2 D3

= (1, 2], D 4 =

= (1,2] =

O<x< 1

2,

1<x<2

0,

2

=

E3 (2, 3]. Then

¢

-1,

-1XEI

<X<

3.

= (2,3] and D 1 = (0, 1/2),

= [1/2, 1],

+ 2XE2 + OXE]

= -1Xn 1

+ -1Xn2 + 2Xn + 0Xn

=

+ -1Xn + 2Xn

-1Xn 1

D2

3

2

3

4

•

We define the Lebesgue integral of a simple function¢ on E.

5.2.3

Definition

Suppose ¢ is a simple function defined on a measurable set E, that is, 11

¢(x)

=

I::Ck XEk(x), k=l

n

with

U Ek = E, Ek mutually disjoint, f.L(E) < oc, ck real. 1

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

The Lebesgue integral of cp on E,

175

JE cp, is defined as

-

-

E3

5.2.4

~~Ek~

Comments

1. If cp is a step function on [a, b], cp is a simple function, and

The Lebesgue integral of a step function agrees with the Riemann integral of a step function. 2. Because of the many possible representations of cp we must check to see that our definition is not ambiguous. Suppose

E

=

n

m

i=l

}=!

UE; = UFJ·

Somehow we want to refine both of these decompositions of E into a "common" decomposition. Each set in this "common" decom-

176

LEBESGUE INTEGRATION

position would be a subset of E; and Fj. "And" suggests intersection: E =

UU(E; n Fj) = UU(E; n Fj). i

j

j

i

Now suppose¢= "E;c;XE; = "E1 d1 XFJ, {E;} and {F1} mutually disjoint collections of Lebesgue measurable sets, J.t(E) < oo. Then we claim

L cu·L(E;) = L d1 tJ(F}), i

j

the Lebesgue integral is independent of the representation. We know the nonempty E; n Fj are measurable and mutually disjoint. Because n

m

i= I

j= 1

UE; = E = UFJ, L c; tJ(E;) = L c; L tJ(E; n Fj) = L L c;tJ(E; n Fj) =

i

J

i

j

L L d1tJ(E; n Fj) = L ~ L tJ(E; n Fj) j

i

}

i

since if E; n Fj # 0, then c; = c;XE; = ¢ = d1XFJ = d1, and the argument is complete. Example 3: We calculate some Lebesgue integrals of simple functions. 1. ¢(x) =

-1,

O<x< 1

2,

l<x<2

0

2<x<3. ¢

r

}(0,3]

=

-lX(o,I]

+ 2X(l,2] 3

¢ = -1 . 1 + 2. 1 = 1 = { cj;(x) dx

Jo

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

2. ¢(x) =

{

0,

x rational, 0 < x < 1

1'

x irrational, 0 < x < 1 .

¢=

f

177

X[O,I]n{irrationals}

¢ = lJ.L([O, 1] n {irrationals})= 1.

}[O,l]

3. ¢(x)=

Then

1,

X

2,

xECn{irrationals},

3,

X

fro, I]¢ =

E C n {rationals}

E

1·0

whereCistheCantorset.

[0, 1] - C

+ 2 · 0 + 3 · 1 = 3 because J.L( C) = 0.

5.3 If¢, 'ljJ are simple functions defined on a set E with .finite measure, and k is any real number, then PROPOSITION

1. (k¢) is a simple function onE, and JE(k¢) = k JE ¢ (homogeneous); 2. (¢ + 'lj;) is a simple function onE, and JE(¢ + 'l/J) = JE¢ + JE'l/J (additive); 3. JE ¢ < JE 'ljJ if¢ < 'ljJ on E (monotone); 4. If E 1 and E 2 are disjoint measurable subsets of E with E = E1 U £2, the integrals f£ 1 'ljJ and f£ 2 'l/J exist, and JE 'l/J = 1 'l/J + f£ 2 'l/J (additive on the domain).

J£

E =

n

m

i= I

j= I

UEi = UFJ,

{E;} and { fj} are mutually disjoint collections of measurable subsets of E.

178

LEBESGUE INTEGRATION

I. Jdk¢) = L7=I(kc;)XE, =k:L7=1C;XE, =kfE¢· 2. Au E; n Fj. The nonempty sets in the collection of A,1, 1 < i < n, 1 < j < m, are mutually disjoint measurable sets whose union is E.

Then n

(¢

m

+'1/J) =I: 2)c; +~)XA;J' i= }=I

and

I

n

m

m

n

=I: I: c;J-L(E n Fj) +I: L d1J-L(E, n Fj) i= j= 1

I j= I

n

I i= I

m

m

n

=I: c I: J-L(E n Fj) +I: d1 I: J-L(E; n Fj) }=I }=I 1

1

i=l

i=l

n

m

i=l

J=l

=L C;J-L(E;) +I: ~J-L(Fj)

'1/J,

3. If¢ < then 'ljJ- ¢is a nonnegative simple function onE, whose integral will be nonnegative by definition of the integral, and then from parts 1 and 2 we have

0

<

1 ('ljJ- ¢) = 1 'ljJ + 1-¢ = 1 'ljJ- 1¢ 1¢<1'1/J·

and, thus,

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

m

=

179

m

L d J.L(Fj n E + L d J.L(Fj n £ 1

1)

}=I

1

2 ).

}=I

But {Fj n EJ}, {Fj n E 2 } are collections of mutually disjoint measurable subsets of E 1 , E 2 , respectively, with m

m

U(Fj n EJ),

E1 =

E2=U(FjnE2),

J=l

and, smce

J=l

the integral is independent of the "decomposition" used, we have

and thus

• We have completed the preliminaries for extending the definition of the Lebesgue integral from simple functions to bounded functions.

5.2.5

Definitions

Suppose f is a bounded function defined on a measurable set E with finite measure; say a < f < (3 on E, J.L( E) < oo. Let ¢ and 7./J denote simple functions such that¢ < f < 1j; on E. The lower Lebesgue integral off on E, JEf, is given by

j j =sup{1 ¢ I ¢
¢ a simple function}.

180

LEBESGUE INTEGRATION

The upper Lebesgue integral off on E,

j j- inf{h 1/J I f < 1/J, 5.2.6

JE f, is

1/J a simple function} .

Comments

1. The constant simple functions a:, (3 assure us that the lower and upper Lebesgue integrals are well-defined, since the appropriate sets will be nonempty and bounded above and below.

-------+------·----,-·-·- ------r------·-

CXf.l(E)

f Ef

f3 f.1 (E)

----+-------··-··-·--'----~----·-

a f.1 (E)

f3 f.1 (E)

2. If ¢ < f < 1/J, ¢ and 1/J simple functions, then by monotonicity (Proposition 5.3),

Because 1/J is arbitrary, we interpret this inequality as JE ¢is ajower bound for the set {IE 1/J I f < 1/J, 1/J aJiimple function}. Because IE f is the greatest Jower bound, IE¢ < IE f: Similarly, IE¢ < IE f for ¢ < f; view IE f as an upper bound. Since I Ef is the least upper bound, we may conclude: I Ef
5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

181

measure has lower and upper integrals, satisfying

for any simple functions ¢ < f < 'ljJ on E.

5.2.7

Definition

A bounded function f, defined on a measurable set E with finite measure, is Lebesgue integrable on E whenever the lower and upper Lebesgue integrals are the same. Denote the common value by IE f:

In this case, ¢ < f < 'ljJ onE implies IE¢ < IEf < IE 'ljJ for any simple functions ¢, '1/J.

5.2.8

Comment

The definition of the Lebesgue integral given in 5.2.7 agrees with the original Definition 5.2.3 when f is a simple function: Iff= J a simple functio~, then IE J E {IE¢ I ¢ < ¢} and by monotonicity IE¢ < IEAJ, i.e., IE¢ is an up~er bound and a memb~r of the set {IE¢ I ¢ < ¢ }. Hence I Ef =IE¢. Similarly IE f = IE¢, and the upper and lower Lebesgue integrals have the common value IE¢. The reader should recall that the Riemann integral of a step function agrees with the Lebesgue integral of the same step function (5.2.4) and that every step function is a simple function. The next theorem goes much, much farther. It tells us that the Lebesgue integral exists for any Riemann integrable function and, in fact, the Lebesgue integral and Riemann integral have the same value. The next theorem will also be used in conjunction with Theorem 5.5 to calculate Lebesgue integrals in Sections 5.3, 5.4, and 5.5. THEOREM

5.6

Let f be a bounded function on the interval [a, b]. Iff is

182

LEBESGUE INTEGRATION

Riemann integrable on [a, b], then

f

is Lebesgue integrable on [a, b], and

rb f(x) dx = r f. }a }[a,b] Proof' rjJ a step function}

f

=sup{

lra,bJ

¢ I ¢ <J,

<sup{ f. ¢ I ¢
lra.hJ

=/

_[a,b]

<j

[a.b]

rjJ a step function}

rjJ a simple function}

f f

= inf{ f. 1/J

lra.hJ

< inf{ f. 1/J

lra.bJ

1

1

J<

1/J,

?jJ a simple function}

J<

1/J,

?jJ a step function}

= inf{lb ?j;(x) dx I f

< ?/;,

1/J a step function}

-b

= J/(x)dx. Since f is Riemann integrable, J:f(x) dx = J! f(x) dx and the conclusion follows. • So Riemann integrability implies Lebesgue integrability. That the Lebesgue integral generalizes the Riemann integral, that there are func-

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

183

tions that we can integrate in the Lebesgue sense and not in the Riemann sense, is revealed by

Example 4:

f(x)

=

x rational, 0 < x < 1

1, { 0,

x irrational, 0 < x < 1 .

J~f(x)dx = 0, J~f(x)dx =

1. ThusfisnotRiemannintegrable. However it is trivally Lebesgue integrable. It is a simple function after all, and

r

}[O,l]

f =

1 JL({rationals} n [0, 1])

= 0.

Lebesgue characterized the types of bounded functions that are Riemann integrable (Theorem 5.1): continuous almost everywhere. We will eventually show that bounded Lebesgue measurable functions play an analogous role for Lebesgue integration. Before beginning that undertaking, we have a criteria for Lebesgue integrability similar to Proposition 5.2:

5.4 A bounded function f, defined on a set E with finite measure, is Lebesgue integrable if.!for every E > 0 we have simple functions ¢and '1/J, ¢ < f < 'ljJ on E, so that PROPOSITION

184

LEBESGUE INTEGRATION

The reader should compare this figure with that of Proposition 5.2. In particular, £ 1 and £ 2 are not necessarily intervals. Proof Assume the bounded function f is Lebesgue integrable on the measurable set E, J.L(E) < oo, and let E > 0. From the definitions of greatest lower bound and least upper bound we have simple functions ¢ and -J;, ¢ < f < -J; on E, so that

< Thus 0 <

fI

<

f

fe-0 < I+~ =

h1 + ~ ·

IE -0 - IE¢ = IE( -0 - ¢) < E.

For the other direction, let E > 0 be given along with simple functions¢ and '1/J, ¢ < f < '1/J, so that 0 < IE 'ljJ- IE¢ = IE( 'ljJ- ¢) < E. Then again, from greatest lower bound and least upper bound properties (5.2.6)

Hence 0 < IE fnature of E.

L:J <

E and

the conclusion follows from the arbitrary

Note: When the bounded function f is Lebesgue integrable on E, J.L(E) < oo, we have ¢ < f < 'ljJ on E, IE¢ < IE!< IE '1/J, and IE( 'ljJ- ¢) < E, for some simple functions ¢and '1/J. •

We now develop some examples, using the Fundamental Idea of Lebesgue: Partition the range. Then the domain off will be decomposed, hopefully, into measurable sets iff is not too badly behaved. In these examples IE!= I:J(x)dx. Calculate first the Riemann integral and then the Lebesgue integral as indicated. Example 5:

1. Let

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

f(x) = A

¢(x) =

yx, k- 1 n

2

'

( k n 1) < x < (kn)2,

l
2

k

A

185

(kn1) < x< (kn)2,

~(x)=-,

n

I < k < n, (/;(I) = I.

Also, 2

Le: r·~< 1

2:e: (~r-e: rl 2:(n [(~r-e: rl 2 L(~Y-~. 1

1 ) [

L(k-l )sk- ) =

1

1

<

<

Thus 23 = 2 .

t

)o

x2

dx

=

r vx.

)[o,t]

2.

¢(x) = 1 + kn A

A

k

'1/J( X) = 1 + -

n

,

1, In (I + k-n 1) < x < In (I + k) n ,

/ k- 1) <

In ( 1 +

n

x < In

( + k) 1

n ,

f

otherwise;

f otherwise.

186

LEBESGUE INTEGRATION

Then 1

A

A

( 1jJ - ¢) = - (In 2 - 0)

1r

n

[O,In 2]

=

l 1 -ln2 <-. n n

Thus fto, 1n2] ex exists. Show

< n In ( 1 + ~)

1

1-

n

+1

n

< (n + 1) In ( 1 + ~) < l + ~ I.e., 1-

l

n+

J

<

1

ex < -

_[O,ln 2]

and conclude fto,ln 2] ex = 1 =

J

ex < 1 + -l

n

[O,ln2]

'

J bn 2 ex dx.

3.

f(x)

4. f(x) =

=

{

2'

{ 0,

x 1/3,

1]

x E [l

1,

U [l l] U ... 4' 3

otherwise

0

<x < 1

.

1 <X< 2

2,

Let Ek ={xI (k- 1)/n
fro,

5. f(x) = {

x, 2-x

O<x<1 -

Ek as in 4.

l<x<2

'

l

E

2 I"' ('lj;- ¢) = - ~J-L(Ek) =-. A

A

n

n

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

187

6. Suppose f is continuous on [a, b]. Thus m
¢(x) = (k-1) ~

1/J(x)

=k

M-m , n

and

M-m , n

Then

1

[a,b]

~

~

(1/;-¢)=

M-m"'\:"' n

~J.L(Ek)=

M-m n

(b-a).

Compare with 5.1.9. We are now ready to prove the Lebesgue integral counterpart of Theorem 5.1; a characterization of Lebesgue integrability in terms of Lebesgue measurable functions. 5.7 Let f be a bounded function on a set E with finite measure. Then f is Lebesgue integrable onE iff f is measurable on E.

THEOREM

Let If I< M onE and assume f is measurable on E. We will s_how f is ~Lebesgue integrable by constructing simple functions ¢ and -0, ¢
o
188

LEBESGUE INTEGRATION

Now we assume f is Lebesgue integrable and bounded on the set E, J-L(E) < oo. We will show f is a measurable function onE by showing that f equals almost everywhere the "inf" of a sequence of simple functions. Then the result will follow by Theorem 4.1. Because f is bounded and Lebesgue integrable on E, we have simple functions cPn and ¢n so that cPn < f < '1/Jn onE, IE cPn < IE!< IE '1/Jn, and n=1,2,3, ... Define two measurable functions (Theorem 4.1 ): and Certainly cPn < ¢* < f < 'ljJ* < '1/Jn onE for all n > 1. We want to show ¢* = 'lj;* almost everywhere on E and thus conclude f = 'ljJ* a.e. By Proposition 4.3 f will be measurable on E. Consider the set

{x E E I 'lj;*(x)- ¢*(x) > 0}

=

C

V{ x V{ X

E E

E

I '1/J*(x)- ¢*(x) >

~}

E I '1/Jn(x)- cPn(x) > ~}

for all n > 1. If we show {x E E I 'lj;*(x)- ¢*(x) > 1/m} has measure zero we would be finished. We know this set is measurable because 'ljJ* - ¢* is a measurable function. By construction, Id'l/Jn- cPn) < 1/n. But then 1/n > ]E('l/Jn- cPn) > (1/m)f.l( {xI '1/Jn(x)- ¢n(x) > 1/m}) by application of Proposition 5.3 with £1 ={xI '1/Jn- cPn > 1/m} and E2 ={xI '1/Jn- cPn < 1/m }. Thus J-l ({xI

¢n(x)- cPn(x) > __!_}) < m m

n

for all n > 1 .

i.e., J-L({x I 'lj;*(x)- ¢*(x) > 1/m}) = 0 and the proof is completed by recalling that a countable union of sets of measure zero is a measurable set of measure zero. • We have characterized Lebesgue integrability. We now show that the Lebesgue integral has the important properties of linearity and monotonicity.

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

189

5.8

If the bounded functions f and g are measurable on the set E, J.L(E) < oo, and k is any real number, then f and g are Lebesgue integrable onE and

THEOREM

1. (kf) is Lebesgue integrable on E, and JE(kf) = k JEf (homogeneous); 2. (f +g) is Lebesgue integrable on E, and JEif +g) = JEf + JE g (additive); 3. JEJ < fEg iff< gonE (monotone); 4. If E 1 and E 2 are disjoint measurable subsets of E withE = £ 1 U £ 2 , f is Lebesgue integrable on E 1 and E 2 , and (additive on the domain); (mean value). Proof Existence of Lebesgue integrability for the various parts 1 through 5 follows immediately from the previous theorem. However, the following arguments are of interest:

1. Let k > 0. Because f is Lebesgue integrable, we have simple functions ¢ and '1/J so that ¢ < f < '1/J, JE ¢ < fEf < JE '1/J and f£('1/J- ¢) < Ejk. But then k¢ < kf < k?jJ, k JE ¢ < k fEf < k JE '1/J and k f£('1/J- ¢) = JE(k?jJ- k¢) <E. The last inequality implies kf is Lebesgue integrable. Thus k JE ¢ = JE k¢ < JE kf < JE k?jJ = k JE '1/J. So k JEf and JE kf are between JE k¢ and JE k?jJ, with JE k( '1/J - ¢) < E. The reader may complete the argument when k = 0 and k < 0. 2. f is Lebesgue integrable: ¢1 < f < ?/J1 , JE ¢J < JEf < JE ?/J1 with

1(

'1/Jf - ¢J) <

~.

g is Lebesgue integrable: ¢g < g < '1/Jg, JE ¢g < JEf < JE '1/Jg with JE( '1/Jg - ¢g) < E/2. Adding, we have ¢1 + ¢g < f + g < ?/J1 + '1/Jg, JE(¢J +¢g)< JEf + fEg < f£('1/Jj + '1/Jg) With f£[('1/Jj + '1/Jg)(¢J + ¢g)] < E. Again this last inequality implies f + g is Lebesgue integrable on E and

190

LEBESGUE INTEGRATION

Thus IEf + IEg and IEif +g) are between IE(¢1 +¢g) and IE(1/Jt + 7/Jg), where IE[(1/Jt + 1/Jg)- (¢J +¢g)]< f. 3. IEg- IEf = jE(g- f) by parts 1 and 2. Sinceg- f > 0, let J = 0. Then 0 =IE¢< IE(g- f). 4. Because f is Lebesgue integrable on E, we have simple functions ¢ < f < 1j; on E so that

I£

Since IE¢= IE 1 ¢ + I£2 ¢ and IE 1j; = 1 1j; + I£2 1/J (Proposition 5.3), we conclude ¢ < f < 1j; on £ 1 and ( 1j;- ¢) < f and 1 ¢ < f < 1/J on E 2 and I£2 ( 1/J - ¢) < f. Thus f is Lebesgue integrable on £ 1 and £ 2 and

I£

Again, IEJ + IEJ and IE! are between IE¢ and IE 1/;. Thus IE! = IEJ + IEJ by the arbitrary nature of f. 5. Immediate from part 3:

•

The argument is complete.

The next result shows one of the important advantages of the Lebesgue integral over the Riemann integral: If we change the values of a Riemann integrable function on a set of measure zero we may destroy integrability. This is illustrated by the functions f(x) = 0, 0 < x < 1, g(x)

Id

1'

={ 0,

x rational x irrational.

Here f(x) dx = 0, g differs from f on a countable set (rationals, measure zero), but g(x) dx does not exist (Problem 5.1.10). On the other hand, Lebesgue integrability is independent of unpleasant behavior on a set of measure zero.

Id

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

191

Iff is a bounded, Lebesgue integrable function on a set E of finite measure, and g is a bounded function onE such that g = f a.e. onE, then g is Lebesgue integrable on E and THEOREM 5.9

Proof The function f is measurable by Theorem 5. 7 and application of Proposition 4.3 yields measurability for g, and thus integrability for g (Theorem 5.7). Let A ={xI f(x) -j:. g(x)}. The set A has measure zero, thus E- A is measurable and, by Theorem 5.8,

Example 6:

1. x rational

2x,

g(x)

=

{

I-x

O<x
x irrational

'

and f(x) = I - x, 0 < x < I. Thus {

lro,IJ

f

= {

lro,IJ

(I - x)

=

t

(I - x)dx = _21 = { g. }o lro,IJ

Jd g(x)dx does not exist since g is continuous at only one point, X=

1/3.

2.

g(x) =

x 2,

xECn{irrationals}

x,

x E C n {rationals}

I,

xrf_C

= 1,

C the Cantor set, and f(x) integrable, and

1

[0,1]

f=

] lo

0 < x < I. Then

Idx=I=

0

O<x< I,

1

[0,1]

g.

f is Riemann

192

LEBESGUE INTEGRATION

3. l

g(x)

=

x

q'

{ 0,

=

p / q;

p, q integers, (p, q)

1

=

otherwise,

and f(x) = 0, 0 < x < 1. Thus

{

f= tf(x)dx=O= {

}o

}[o,J]

g=

}[o,J]

f

}o

1

g(x)dx

since g is continuous a.e. and thus Riemann integrable. Sometimes Riemann integrals are easier to calculate because of Theorem 5.9. 5.2.9

SUMMARY

Riemann Integral 1. ¢, '1/J step functions on [a, b]:

k

=

1, 2, ... , n

k = 0, 1, ... , n, where

b

2.

I

¢(x)dx

~ ck(xk- xk_ 1)

a

3. ¢

< f < 'lj;; ¢, '1/J step functions, f bounded:

and

I

-b

/(x)dx

b

= inf{l

'lj;(x)dx I f <

'lj;}.

4. If Jb f(x)dx = J!J(x)dxfortheboundedfunctionf ,fisRiemann integrable on [a, b]. Denoted by J%f(x)dx.

5.2

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS ON SETS OF FINITE MEASURE

193

5. A bounded function f is Riemann integrable on [a, b] iff f is continuous a.e. on [a, b]. Lebesgue Integral 1. ¢, 1j; simple functions onE, J.l(E)

< oo:

n

¢(x)

=I: ckXEk(x) I

where E =

U Eb

Ek mutually disjoint Lebesgue measurable sets.

3. ¢ < f < 1);; ¢, 1j; simple functions,

f bounded:

and

jj =

inf{ 11); I

f < 1);} .

4. If JE f = JE f for the bounded function j, f is Lebesgue integrable on E. Denoted by fEf· 5. A bounded function f is Lebesgue integrable onE, J.l(E) < oo, iff f is Lebesgue measurable on E.

Riemann Integrable Functions

Lebesgue Integrable Functions

We now remove the restrictions that 11(£) < oo and f be bounded.

194

LEBESGUE INTEGRATION

5.3 THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

This section is the starting point for many treatments of the Lebesgue integral, and the reader who has omitted the previous two sections may treat it as such. For those who have studied the previous two sections, this material may be viewed as a natural generalization in the following progression, or evolution, of ideas: Riemann integral for bounded functions on closed, bounded intervals via step functions ------* Lebesgue integral for bounded functions on sets of finite measure via simple functions ------* Lebesgue integral for nonnegative measurable functions on measurable sets via approximation by nonnegative monotone sequences of simple functions ------* Lebesgue integral for measurable functions via: f = f+ - f-, with j+ and f- nonnegative measurable functions; Define fEf = fEf+ - fEf- · Examples will clarify some difficulties and problems faced when dealing with measurable (sometimes unbounded) functions that are both positive and negative.

Example 7: O<x
1/x, 1. f(x) = { 1/(x- 2),

f is unbounded (positive

l<x<2.

and negative). How to define J(o, 2 )f? Maybe

k k 1 (0,2)

f .?

!+

E--->0

.

?

(L2)

(OJ]

. __:?__ hm

1'Ill e--->0

f

iJ

-1 dx

E

1

X

• J2-E 1 dx + hm E--->0

1+ I1'l l e--->0 (d] X

I

1

X -

1

(1,2-e) X -

= oo- oo (not defined).

2

2

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

-1/2

2. f(x) =

'

0<x<1

1/(x-2),

1<x<2.

X {

i

11

f ?· lim

(0,2)

E-->0

I x-2dx +lim

12-i

c-->0

E

I

1 X -

2

195

dx

= 2-00 = -()().

-1/2

3. f(x) =

X {

O<x
'

-(2- x)-1/2,

l<x<2.

i i J (0,2)

!+

f .?

(0,1]

?

· lim c-->0

f

(1,2)

11

I x -2dx + lim

c

c-->0

12-£ I

-1

V2 -

dx X

=2-2=0. 4. f(x)

= x- 112 , 0 < x < 1. f is nonnegative and unbounded, and

5. f(x) = Ijx 2 , x > l. f is nonnegative and bounded on a set of infinite measure, with

1

- 12 ?· lim [Loc) x R-->x

6. f(x) =

(-It(n+ 1)- 1, { 0,

r

mr

1R 1

- 12 dx x

= I.

< x < (n + I )7r

n

= 0, 1, 2, ....

otherwise

?( 1 - 21+13 -14 + .. . )

}[o,x/ .

7r

= 7r In 2 ·

196

LEBESGUE INTEGRATION

7. f(x)

={

sin(x)jx,

x:;iO x=O.

1'

{

j

lim {R sin(x) dx =

?

}[o,x)

R->x

}o

X

rx sin(x) dx }o

= 7r •

X

2

In these examples, the functions, being unbounded, causes problems with integral interpretations, since the Riemann integral is only defined for bounded functions, as has been the Lebesgue integral (so far). The functions, being both positive and negative, add further complications. One thing at a time. Recall (Proposition 4.6) that any measurable function can be written as the difference of two nonnegative measurable functions: f = f+ -f-. As a result of this decomposition, it is natural to define the Lebesgue integral off, whether f is positive, negative, or both, as

and this will be meaningful if we have meaning for JEf+ and JEf- in some sense. This will be our approach then: restrict our attention to nonnegative measurable functions whose domains need not have finite measure, and then in the next section remove the condition that f be nonnegative. Example 8:

f nonnegative and unbounded: 1

1. f(t) =

{


t- ,

0

1

(2- t)- 112 ,

I < t < 2.

i h J f ?·

f+

(0,2)

2. f(t) = Ijt,

(0,1]

t

1

> I.

f · lim ?

[I ,oo)

3. f(t)

=

I/t

2

f ?· oo+2=oo.

(1,2)

R->x

1

-1 __:?__ lim [l.R] t R->oo

lR 1

-1 dt = oo . t

,

1

[l,oo)

1

f · lim R->oo

1

-12 = lim

[l,Rjf

R->oo

lR 1

-I2 dt = I . t

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

197

t > 0.

4. f(t) = 1/t,

i

• -1 ?· hm

(O,oo) l

e--->0+

. 1'Im

?

E--->0+

1] l~ 1. -+ j e

1 . -dt+ hm t e->O+

(e,l)

1 t

1'Im

e->O+

I

(!,~)

1 -dt t

-1

t

?

· oo+oo=oo.

5. f(x) =

t {

t

-1/2

_

2

0
'

t> 1

'

?l l f ·

(o,oo)

r 2I +

(0,1)

=

1

t- 2

[!,oo)

2 + 1.

How do we define the Lebesgue integral for a nonnegative, not necessarily bounded, measurable function? That is the problem! The idea that "works", is inherent in the Approximation Theorem 4.2: We can approximate any nonnegative measurable function f with a monotone increasing sequence of nonnegative simple functions; 0 <
IE

IE

Roughly speaking, we must make sense of rPm that is, the Lebesgue integral of a nonnegative simple function. But that is easy to do (if you've read the previous section some of the material that follows will be repetitious), and since simple functions are by definition bounded, boundedness will cause no difficulties. We begin with simple functions as linear combinations of characteristic functions.

198

5.3.1

LEBESGUE INTEGRATION

Definition

Let ¢ be a nonnegative simple function on R, that is, n

¢(x) = :~:::>kxEk (x), k=l

where Ek are mutually disjoint Lebesgue measurable subsets of R,

and ck are nonnegative real numbers (Note: Nothing lost in assuming

for if not, then n

E0

= R- UEk k=l

is a measurable set and n

R= UEk.)· k=O

Thus p,(Ek) will be infinite for at least one k, 1 < k < n. Obviously a simple function ¢ has many representations as linear combinations of characteristic functions. 5.3.2

Definition

The Lebesgue integral of a nonnegative simple function ¢, on a measurable set E, written JE ¢, is defined by

where¢=

n

n

1

I

'L::>kxEk' Ek mutually disjoint, R = U Ek, ck > 0.

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

199

We employ the usual convention of defining JE ¢ = 0 whenever¢= 0 on E, even if J..L(E) = oo (0. oo = 0). Also, if J..L(E) = 0, then J..L(E n Ek) = 0 for all k and JE ¢ = 0. The problem, oo- oo, does not arise because ck > 0. 5.3.3

Comments

1. For nonnegative simple functions on E, J..L(E) < oo, we are in agreement with our previous Definition 5.2.3. 2. Even though a given simple function ¢ has many representations, the Lebesgue integral of¢ is well-defined: Suppose n

¢ = :L.:.:CkxEk' ck > 0 I

and m

¢ = LdJXDJ' dj > 0 I

with R

n

m

I

I

= UEk = UDJ,

Ek and D1 mutually disjoint measurable subsets of R. We show that n

m

I

I

L ckJ..L(E n Ek) = L ~J..L(E n D1), that is, the integral as we have defined it is independent of the representation of¢. Note that

and

200

LEBESGUE INTEGRATION

Thus

t

Cki'(E n £,)

=

t

=

t

Cki'( En

( Q(E, n

cki'(Q(E n E;

n

fj)))

n fj))

m

= L ck~= ~-t(E n Ek n Fj) I

n

I

m

= LLd1 ~-t(E n Ek n F)) =

I

I

m

n

L

L I

m

I

d1 ~-t(E n Ek n Fj·) n

= L d1 L ~-t( E n Ek n F)) I

I

m

=

:Ld1 ~-t(EnFJ) I

since, for Ek n En F) =f. 0, ck = ckXEk = ¢ = ~XFJ = ~' and if Ek n En F1 = 0, no contribution because ~-t(0) = 0. 3. The Lebesgue integral of a nonnegative simple function is a nonnegative real number or oo. PROPOSITION 5.5

If¢, 'ljJ are nonnegative simple functions on R, if E is any

measurable subset of R, and k is any nonnegative real number, then 1. (k¢) is a nonnegative simple function onE, and (homogeneous);

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

201

2. (cp + 'lj.;) is a nonnegative simple function on E, and (additive);

4.

J

j 'lj.; ifO < ¢ < 'lj.; onE

E

E

¢<

3.

(monotone);

If E 1 and E 2 are disjoint measurable subsets of E with E = £ 1 U £ 2 , the integrals

J£

1

'lj.; and

f£

2

'lj.; exist, and

(additive on the domain).

Proof" 1. Suppose cp

=

'2: I ciXE;· Then kcp = '2: kciXE; and

2. Let cp = '2:ickXEk and 'lj.; the n · m sets;

=

'2:Td;XF;, 0 < ckld;· The idea is to form

E1 nF1, £ 1 nF2, ... , E, nFm E2 n F1, E2 n F2, ... , E2 n Fm

If Ek n Fj =I= 0, define cp + 'lj.; as ck + d1. The nonempty Ek n Fj are mutually disjoint measurable subsets of R,

R

=

U(Ek n Fj), k,j

and ¢

+ 7f.; =

2)ck + d;)XEknF;· k,j

202

LEBESGUE INTEGRATION

Thus

n

m

L L(ck + d1)tL(Ek n Fj

=

n E)

k=l J=l n

=

m

m

L:ck LM(Ek k=l

nEJ nE) + L~ LM(Ek nFi n E)

J=l

J=l

n

=

n

k=l

m

L cktL(Ek n E)+ L d1tL(Fj n E) k=l

J=l

3. Suppose¢= I: k=l ckXEk' Ek mutually disjoint, and '1/J = Fj mutually disjoint, where n

2:::}: 1~XF, 1

m

UEk=R=UFJ· I

I

Since 0 < ¢ < '1/J, 0 < ck <

m

~

on nonempty Ek n Fj and thus

n

m

< Ld1 LM(Ek n Fj n E)= L~M(F} n E) J=l

k=l

J=l

4.

h= '1/J

=

L d1tL(F} n E) = L d1tL(F} n (E 1 u E2)) L ~ [M(F} n E1)

+ tL(F} n E2)]

= L d1tL(F} n EJ) + L d1tL(F} n E2) =

hi + h2 '1/J

'1/J .

•

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

5.3

203

We now define the Lebesgue integral of a nonnegative measurable function. In fact, we give two commonly used definitions and show their equivalence. 5.3.4

Definitions

Definition A

Iff is a nonnegative, measurable function, defined on a measurable set E, the Lebesgue integral off over E, fEf, is given by ¢ nonnegative and simple} .

Definition B

If f is a nonnegative, measurable function, defined on a Lebesgue measurable set E, and ¢n is a nonnegative monotone sequence of simple functions, 0 < ¢n < ¢n+l onE, with lim ¢n(x)

=

(finite or infinite)

f(x)

for all x E E, the Lebesgue integral off over E, JEf, is given by

rf

}E

lim

r¢n = }Er(lim ¢n).

}E

Some comments are in order before we show the equivalence of these definitions, hereafter referred to as A and B. 5.3.5

Comments

1. Suppose

f is nonnegative, bounded and measurable on a set E of

finite measure. Does Definition A agree with Definition 5.2.5? Yes! Theorem 5.7 tell us that JEf = J E f (integrable by 5.2.5). But then,

j j=sup{h¢1 ¢
¢simple} d> simple and nonnegative},

since f is nonnegative, and this is 5.3.4.

204

LEBESGUE INTEGRATION

2. Because f is nonnegative, we always have simple functions below f(¢ = 0). Thus the set {IE¢ I ¢ < f} is nonempty and the "sup" is a nonnegative member of the extended reals. 3. Iff js nonnegative a?d simple, say, f =¢,A then IE¢ E {IE¢ I ¢<¢}and IE¢< IE¢ by Proposition 5.5; IE¢ is a member of and an upper bout?d of the set {IE¢ I ¢
on E.

The sequence (JE ¢m) is a nondecreasing sequence in the extended reals (Proposition 5.5), so the limit is defined in the extended reals: lim IE Jm is a nonnegative real number or oo. 5. IEf, as given by B, is well-defined: Suppose we have seql!ences, of simple functions (¢n),A (¢m), 0 < c/>n < c/>n+i and 0 < c/>m < c/>m+i onE with lim¢n = lim¢m =f. We claim lim

h

cf>n = lim

h

Jm .

The argument is as follows: Pick 0 < ¢n ¢n on E. We will show

Because ¢n is nonnegative and simple, we have

where E = U~=l (En Ek) and the nonempty En Ek are mutually disjoint measurable subsets of E. So we must show

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

205

But the integral is additive on the domain (Proposition 5.5). Thus

Our claim will be justified provided we can show lim { m

}EnEk

Jm > ckp,(E n Ek),

If ck = 0, immediate. Suppose ck > 0. Let a be any number between zero and one. (This idea of 0 < a < 1 has been attributed to W. Rudin.) We construct a sequence of sets (Bm) as follows:

Bm = {x

E EnEk I

¢m(x) > ack}.

Bm is measurable, Bm C Bm+l since Jm < ¢m+l> and En Ek = UBm (If p E E n Eb ¢n (p) = ck and consequently limm ¢m (p) > ck > acb i.e., ¢m(p) > ack form sufficiently large, in other words, p E Bm for m sufficiently large.). So the sequence Bm is an expanding sequence of measurable sets that "fills" En Ek. Then the sequence (p,(Bm)) is monotone increasing with (Theorem 3.4). But the sequence UEnEk ¢m) is also monotone increasing with

Therefore,

But this holds for any a between 0 and 1. Thus

and the argument is complete. The reverse inequality is obtained by interchanging ¢m and ¢n. 5.6 The Definitions A and B of the Lebesgue integral of a nonnegative measurable function are equivalent. PROPOSITION

206

LEBESGUE INTEGRATION

By the Approximation Theorem 4.2, we have a monotone sequence of nonnegative simple functions, 0 < ¢n < ¢n+I onE, with Proof

lim(/Jn =f onE, and

{

}E

Jn < }E{ ¢n+I·

We show

Suppose 0 < ¢*
1

¢*
1Jn

= f

> ¢* onE and the argument in

(finite or infinite).

In other words, lim JE ¢n is an upper bound for the set {fE ¢ I ¢ < f}. Therefore,

On the other hand, JE ¢n E {fE ¢ I ¢ < f} for all n, and the sequence (JE ¢n) is nondecreasing (Propo~ition 5.5). So lim JE (hn < sup{fE ¢ I ¢ < f}. Combining, lim JE cPn = sup{fE ¢ I ¢ < f}. • We have the familiar properties of the integral.

5.7

I f f and g are nonnegative measurable functions, defined on a measurable set E. and k is any nonnegative real number, then PROPOSITION

1. (kf) is nonnegative, measurable, and JE(kf) (homogeneous);

= k JEf

2. (f +g) is nonnegative, measurable, and JE(f +g)= JEf + fEg (additive); 3. JEf < JEg ifO
5.3

~----

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

207

Proof Measurability of the appropriate functions follows from Proposition 4.5. 1. From the Approximation Theorem 4.2, we have a sequence (/!Jn) of

simple functions satisfying 0 < ¢n < ¢n+i with lim ¢n =f. But then, 0 < k/!Jn < k/!Jn+i and limn(k/!Jn) = (kf). Using Definition B,

k hf = k lim

1

/!Jn = lim

1 1 k/!Jn =

kf.

2. lim
that is, JEf < JE g.

f£


4. JE
1

and nonnegative, possibly in the extended reals. The result follows by taking limits.

•

This completes the proof.

We showed (Approximation Theorem 4.2) that every nonnegative measurable function is the monotone limit of a sequence of nonnegative simple functions. It was very natural then to use this result to define the Lebesgue integral for a nonnegative (possibly unbounded) measurable function in terms of the Lebesgue integral for nonnegative simple functions. How does this work for specific functions? Example 9:

1. f(x)

=

l/x

2

,

x > 1. Calculate ftl,oo) 1/x 2 .

Define a sequence of simple functions as follows:

¢, (x) =

1, 1 22' 0,

X=l l<x<2 2 <X.

208

LEBESGUE INTEGRATION

Note:

{

)p ,oo)

= { ¢1 = 1 · ~)p ,2) 2

¢1

x=l

1'

1/G)2, 1/W2, ¢2(x) =

1/G)2, 1I(¥)

2

~<X<.§ 2 - 2

,

3 <X.

0, Note:

In general, x=1

1'

1/(1 +

k )

2

zn-1

0,

.1

With

[l,oo)

,

1 + (k-1) zn- 1 <

X

< 1+

k

zn-1 ,

n+I<x, ¢n = 2n-1 [

1 (2n-l+l) 2

+ ··· +

It is straightforward to show 0

1

((n+I)2n-1) 2

]

·

< ¢n < ¢n+l and lim ¢n

=

f on

[I,oo). With the observation 1/((k + I)k) < I/(k · k) < I/((k- 1) · k), the reader may show 2n-1 [

(2n-l

1

1

]

+ 2)(2n-l + 1) + ... + ((n + 1)2n-l + 1)(n + 1)2n-l <

r

}[l,oo)

¢n

n-1 [ 1 1 ] <2 2n- 1 (2n-l + 1) + ... + ((n + 1)2n-l- 1)(n + 1)2n-l '

that is, 1

1 -----,- < 1 n+ 1 + 2n-l

1 [l,oc)

¢n < 1 -

1

n+ 1

·

5.3

THE LEBESGUE INTEGRAL FOR NONNEGA TJVE MEASURABLE FUNCTIONS

209

Thus lim f[I,oo) cPn = 1. By Definition B,

~ - lim { cPn { }[!,oo) }[!,oo) X

=

1.

2. Calculate J(o,IJljy'X.

cP! (X)

=

f

1, 0 < X < 1.

}(o,I]

cP! = 1.

1~<x<~

~' ~<x
h

1[W2-o2]

(0, 1]
+

t[m

2

-W

2 ]

+ ~[m2-m2] 1+ 2

12

4' ll

4' 10

¢3(x)

=

4'

0

02 + :&) .

<X< ({i)2

(I4:i <X< (141)2 (1'\)2 <X< (1~)2

(i)2 < 4

4'

(~)2 <

X X

< (~)2 < (~)2 .

The reader should determine ¢n, show 0

< ¢n <

¢n+b lim ¢n = f,

210

LEBESGUE INTEGRATION

and that

· i (0,1]


1 1 + 2n-l ( = + (2n-1 n

1)2

1 oc) +···+----(n2n-!)2 .

Hence,

cPn = 1 + 1 = 2 ,

lim { }(o,I]

and

r

_I =2

}(O,l]

vfx

.

After a few calculations like the last two examples, we look for an easier way to calculate Lebesgue integrals. The reader has probably realized that in these examples we "really" approximate d f by a monotone sequence ifn) of nonnegative measurable functions and then approximated each fn by a sequence of simple functions. Specifically:

2 2 1*. For calculating ][I,x) 1lx , let fn(x) = llx , I< x < n, and 0 2 J[!.x)fn = ][J,n]fn = J71lx dx = I - Iln, Then otherwise. ~nd .lim][ 1 ,x~fn =.1. Woul·d·n't }t be "convenient " if hm J[!.oo)fn = )[!.x) hmfn = ][J,x) Ilx . Did you use Theorem 5.6? 2*. For calculating .[( 0. 11 1I yx, let

fn(x)

2 lln < x < 1

= { II yx,

0 < x < 1I n

n,

f(o.J]fn Then lim J(o.!]fn = 2.

=

Jd

1

2

n

2

.

ndx + J/;n2

11 yxdx = 2- 2ln

and

Did you use Theorem 5.6? In these calculations, many simplifications arise if lim Jfn = J(limfn) is valid for monotone sequences of nonnegative measurable functions, not just monotone sequences of nonnegative simple functions. Again, Definition B requires approximati on of f by monotone sequences of simple functions. Possible, but technically cumbersome.

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

211

Recall that the limit of a sequence of measurable functions is again a measurable function. Wouldn't it be "nice" if simple functions could be replaced by measurable functions, i.e., since ¢n r f, J ¢n r J f, ¢n simple, can we show fn I f, f fn I fl, ln measurable? This is the essence of the next theorem. 5.10 Lebesgue Monotone Convergence Theorem ( LMCT), Beppo Levi, 1906 Let (fk) be a monotone increasing sequence of nonnegative measurable functions on a measurable set £: 0 <Ji
THEOREM

Proof We give two arguments. The first is based on Definition A for the integral of a nonnegative measurable function. To that end, we observe that (limfk) is nonnegative (0 < fk) and measurable on E (The limit of a sequence of measurable functions is measurable by Theorem 4.1.). Thus

1

(limfk)

= sup{1 ¢I ¢ < (limfk),

¢ nonnegative and simple}.

Since the integral preserves monotonicity for nonnegative measurable functions, and 0 < lk < lk+I < · · · < (limlk) on E, we have 0 JE ¢, then this would say lim JEfk is an upper bound for the set {fE ¢ I ¢ < (limfk), ¢ nonnegative and simple}. But the least upper bound of this set, JE(limlk), would be less than or equal the upper bound, lim JElk, and the conclusion would follow. We now show lim JEfk > JE ¢, ¢ < (limfA:), ¢ nonnegative and simple. Since ¢ is nonnegative and simple, JE ¢ = I:f Cift(E n Ei), ci > 0 and N

U(£ n Ei) = E where En Ei are mutually disjoint measurable subsets of I

E. Because JElk is additive onE for nonnegative measurable functions (Proposition 5. 7), it is sufficient to show lim k

(t { I

j EnE;

lk) >

t

1

ciJ..L(E n Ei),

212

LEBESGUE INTEGRATION

and this will be accomplished, if we show

r

lim fk > C;p,(E n E;) . k }EnE; If c; = 0, done. Assume c; > 0. The idea is to construct an increasing sequence of measurable sets (Bk) which "fill" En E;, on each of which fk is "close" to c;. The ingenious idea that follows is apparently due to W. Rudin. Let 0 ac;}. These sets are measurable, Bk C Bk+h since fk < fk+h and UBk = En E; because (limfk) > ¢ = c; on En E;. The sequence (p(Bk)) is nondecreasing, and limp(Bk) =p(EnE;) by Theorem 3.4. But fEnE/k > JBkfk > ac;p(Bk). Therefore, lim { fk > ac; lim p,(Bk) = ac;p,(E n E;), k }EnE; k Thus limk fEnE/k Definition A.

O
> c;p(E n E;), and the theorem

IS

proved usmg

The second argument is based on Definition B, the Approximation Theorem 4.2, and extensive use of Proposition 5.7. From the Approximation Theorem we have: 0 S cPtt <

< · ·· S ¢tn <···Sit.

limn¢tn =It onE,

and limnfE¢tn =fElt;

0 < ¢21 < ¢22 S ... S ¢zn < ... Slz,

limn¢2n =12 onE,

and limnfE¢2n = JEfz;

0 ::; cPkt ::; ¢k2 < ... < cPkn ::; ... ::; Jk,

limn ¢kn = lk on E,

and limn IE cPkn = I Elk;

!/!12

etc.

Construct a new sequence of simple functions, lim (/Jk = (limfk) as follows:

¢,=¢,1;

¢2 =max{ ¢12, ¢22} > ¢12 > ¢11 = ¢,;

etc.

((/Jk),

with

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

213

The Jk are simple and 0 < ¢, < (/J 2 < · · · < Jk < .. · , with lim Jk = (limfk). By Definition B, lim JE Jk = JE(Iimfk). However, 0 < Jk
1

(limfk).

Recalling JE(limfk) = lim JE Jk from above, the conclusion follows. 5.3.6

•

Comments

1. What if we initially have a monotone increasing sequence of measurable functions (gk), not necessarily nonnegative: g 1 < g 2 < g 3 < · · · ? We could try subtracting g 1 : 0 < g2- gl < g 3 - g 1 < · · · (oo- oo?). The sequence (gk- g 1) satisfies the hypotheses of the LMCT. So,

We would have to be able to subtract. No problem if all integrals are finite, and we would need JE(f- g) = JEf- JE g. These conditions may not be available to us. Consider this example: -1

gk(x)

=

{ 0,

k<x '

otherwise

2. It's natural to consider monotone decreasing sequences. Can we draw any useful conclusions in this setting. Suppose It > / 2 > · · · . Then -/1 < -h < · · ·, or 0
k<x otherwise

214

LEBESGUE INTEGRATION

5.3. 7 Problems

Approximate unbounded measurable functions (whose domains have finite measure) with bounded measurable functions via "truncation" of the range. Note the frequent use of Theorem 5.6. t- 112 .

1. Calculate { }(0,1]

Hint:

fk(t)

o < r < 1;e

0, =

{

_

t

112

'

{', -1
Calculate { }(0,1]

2. Calculate

r (1- P)-

1 2 / .

}[0,1)

Hint:

(1 -

fk(t) =

t2)-1/2

{ 0,

Show 0
'

0 < t < 1 - 1/k 1 - 1/k < t < 1.

fro. 1)(1- t

2

r

112

= limfro,l)fk =lim

1

3. Suppose f(t) = {

(2- t)-1 12,

0, -1

Hint:

fk(t)

=

Calculate {

t- '

t ' 0,

}(0.2)

0 < t < 1/k 1/k < t < 1- 1/k 1 - 1/k < t < 1 + l/k

(2- t)-1/2,

1 + 1/k <

0,

2-ljk 2
1

< 2- 1;e

f.

'

5.3

Thus r f J(0,2)

5.3.8

215

THE LEBESGUE tN;fEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

. J(0,2)J r rk = hm . [f1-1/k -1d = 1tm 1/k t t + J2-1/k( 1+1/k 2 -

t

)-112d] t

=

oo.

Problems

All but number 9 deal with bounded measurable functions on sets of infinite measure. Essentially, "truncate" the domain, and notice how often Theorem 5.6 is used. 1. Calculate (

e- 1•

}[o,co)

Hint:

fk(t)={e-r, 0,

O
Show 0
j

2. Calculate

(1

( -oc,oc)

fk(t) =

3. Calculate (

Jro,cc) e-r

+ t 2 )- 1. + t2)-l

(1 Hint:

thus

It I< k It I> k

'

{ 0,

t- 1 •

}[Lac) 1

Hint:

fk(t) = { t0,

Calculate (

l
k
t-a, a > 1.

}[l,oc)

4. Show (

-tlnt

}(O.I]

= !. 4

-tlnt, Hint:

fk(t) =

{ 0,

ljk
=

Jro,oo) limfk =

216

LEBESGUE INTEGRATION

5. fk(t) =

Show

O
1, { 0,

fro,oo)

k < t.

Iimfk

Show limfk

= lim fro,oc)fk·

Does LMCT apply?

= 0 uniformly on [0, oo ). Iimfk and lim (

Calculate ( }[O,oo)

fk· Are they the same? What is

}[o,oo)

the problem? 0,

0


k < - t< -

e

Show limA = 0 uniformly on [0, oo).

Iimfk and lim (

Calculate ( }[o,oo)

8. fk(t)

0
0, =

-

{ I,

k < t

.

Can we use LMCT? Why?

limfk and lim

Calculate {

fk· Any problem?

}[o,oo)

}[O,oo)

f

fk·

}[O,oo)

9. An unbounded measurable function on a domain of infinite measure whose integral may be evaluated by a combination of techniques used above.

f(t)

={

t

-1/2

O
'

-2

1< t

t '

Calculate

f

}(O,oo)

f.

THE LEBESGUE /NiTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

5.3

217

5.8 If (gk) is a sequence of nonnegative measurable functions defined on a measurable set E, then

PROPOSITION

Let fn

Proof

=

g 1 + · · · + gn and use LMCT:

• 5.3.9

Problems

Show Lxn(l- x)

={

0

1 ' 0,

O<x<1 X=

1.

Is the convergence uniform on [0, 1]? Calculate

r xn(l L }[o,I]

x).

0

2.

r

2:x(1-xt?

}[O,l]

2: r

x(l-xt.

O }[O,l]

O

Show Lx(l-

xt

0

1 ={ ' 0,

O<x
0.

Is the convergence uniform on [0, 1]? Calculate L 3

.

k, ~ :Xr ~ k,I] (} :Xr. (}

I]

Show

"' " L...t 0

r

0 }[o,I]

x(1- x)n.

?

X

(1+xf

=

{

0,

x+1,

x=O O<x< 1.

Is the convergence uniform on (0, 1]? Calculate

L r .(1 +Xr· X

O }[O,lj

218

LEBESGUE INTEGRATION

L( 2r = o 1+x x2

Show

{

0,

x=O

2 1+x,

O<x< 1.

Is the convergence uniform on [0, 1]? Calculate 3/2

5 '

0

312

k.1] ~ (1: x r ~ k.1] (1: x r· 7

2

L lc

[0,1]

x2

2

n·

(1 +X )

2

x=O

0<x<1. Is the convergence uniform on [0, 1]? Calculate

L lc 0

6. Consider the integral

[0,1]

x3/2 2 (1 +x)

n·

JE e-x:

Ji

Suppose E = [0, 1]. Of course 0, 11 e-x= J~ e-xdx = 1- e- 1• On the other hand, e-x= 2::(( -x)k jk!), -oo < x < oo, and the question arises, does lio, 1] e-·~~ = 0, 11 2::(( -x)k jk!) equal I:lio,!]((-x)k/k!)?Thatis, 1-e- 1 · 2::((-1l+ 1/(k+ 1)!). But is Proposition 5.8 applicable? What if we define fn(x) = 1+((-x)/1!)+((-x) 2 /2!)+ .. ·+((-xr/n!) for O<x 0 on [0, 1]? Is 0 <j~
f!

-x/

r e-x=lim r. (1-kx)k =lim }o{

}[0,1]

1

}[0,1j

1) k+l)

=

k ( 1- ( 1- k lim k + 1

=

1 -e -1 .

(t-kx)kdx

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

219

Suppose now E = [O,oo). We cannot use uniform convergence on this set E. We can, however, truncate the domain:

f(x), fk(x)

=

O<x
{ 0,

k <X.

Then 0
Note:

If fk(x)

{(I -

O<x
xjk)k, 0,

=

k<x

then 0 < fk < fk+l and using LM CT,

r

e-X = lim

}[o,oo)

r

fk = lim

}[o,oo)

=lim

r fk = lim }or (I - kx) kdx

}[o,k]

k = 1. k+l

7.

Show

1.

r

2)e-nx- e-2nx) =

}[O,oo)

I

L

r

I }[O,oo)

(e-nx- e-2nx) =

L _I =

00.

2n

n. Does

111.

5.3.1 0

Is Proposition 5.8 applicable? Show

Comment

A very interesting application of the LMCT is the following: Suppose f is any nonnegative Lebesgue measurable function on R. Then for any

220

LEBESGUE INTEGRATION

Lebesgue measurable set£, we may calculate JEf· That is, on the sigma algebra M of Lebesgue measurable sets, we have a set function, say A: EEM,

A(E)

~f.

What properties, if any, does this set function A have in common with the set function p,, the Lebesgue measure set function? Review Proposition 5. 7 at this time. 1. 11.

Clearly ..\(0) = 0. If A and Bare Lebesgue measurable sets, A (monotone). But,

.X(B) =

ff = f f + ff }s ls-A }A

>

C B,

then p(B) > p,(A)

f f = .X(A). }A

The set function A is also monotone. In particular, A( E) > 0 for all EEM. 111.

If 00

E

=

UEk,

Ei n £1 =

0

I

when i -1- j, then

The Lebesgue measure p, is countably additive. Can we show

Actually, Proposition 5. 7 and induction yields

Finite additivity is not a problem. Countable additivity will follow from the LM CT.

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

221

We define a new sequence of functions onE as follows:

f(x), gl (x) =

{ 0, f(x),

gz(x)

=

{ 0,

x E E- (£1 U £ 2 ) .

f(x),

gn(x)

=

X

{ 0,

X

U7 Ek E E- U7 Ek. E

We have a nonnegative, monotone increasing sequence of Lebesgue measurable functions (gn) onE with limgn = f on E. Application of the LMCT yields

=

rf = JErlim

}E

gn

= lim

r

}E

gn

= lim

r f Jul Ek n

In summary, .A. is a nonnegative extended real-valued set function defined on the sigma algebra of Lebesgue measurable sets, M, such that

.A.(0) = 0; n. .A.(£) > 0 for all E

1.

IlL

.A. (

0

Ek) =

E

M;

~ .A.(Ek), Ei n £1 = 0, i -::j: j

and EkE M.

These three properties are the most important properties of any "measure" and they are the defining properties for measures in general. Given a measure, we can construct another measure relatively easily. For sequences that are not monotonic but nonnegative we have the following theorem;

222

LEBESGUE INTEGRATION

(Fatou, 1906) If (fk) is a sequence of nonnegative measurable functions defined on a measurable set E, then

THEOREM

5.11

The idea here is that "lim inf" yields monotone increasing sequences that are then amenable to the LMCT. We begin: f 1 - inf{ / 1,/2, ... }. f 1 is measurable, 0 < f 1 < fn for all n, and JEf 1 < JEfn for all n > 1. That is, Proof

In other words, JEf 1 is a lower bound for the set JEfn, .. . }. We have

hL

< h/1

= inf{h!J,

h/2, ... ,

{f£/1,

JEf2, ... ,

hfn, . ·}·

In general, if

then 0 < / 1 < / 2 < · · · < £n < · · ·, fElm < JEfm = inf{JEfm, JEfm+l, .. .}. The sequences (JE[_m), (JEfm) are nonnegative, monotone increasing sequences of perhaps extended-real numbers, and hence have limits in the extended reals: lim m

r

r

f
But, application of the LMCT tells us that lim

rf

m } E _m

=

r(lim/ m) r(lim inffk).

}E

m -

=

}E

5.3

THE LEBESGUE INTEGRAL FOR NONNEGATIVE MEASURABLE FUNCTIONS

223

Thus

r

}E

(liminffk)
rfm

}E

= liminf

rfkl

}E

•

and this is what we wanted to show.

Example 10: 1. fn = 1/nX[o,nj·limJo,x)fn = 1/= 0 = fro,cc) limfnBut fro,CXJ) lim inffn = Jro,CXJ) limfn = 0 < 1 = lim inf Jro,oo)fn· 2. gn = n Xp;n,2/n]· lim gn = 0, and

111.

limfro,CXJ)gn = 1/= J[o,x) limgn gn -+ 0 on [0, oo ); LMCT? (not monotone);

IV.

Jro,oc) lim inffn

1. 11.

=0

= 0;

< 1 = lim inf Jro,oc)fn-

v. Whatifgn = -nX[!/n,2/n]? 3. Nonnegative is necessary, even with uniform convergence!

fn

= -1/nX[o,n]·

fn-+ 0 (unit) on [0, oo).

lim

f fn

}[O,oc)

=

-1

I= f

}[O,oc)

limfn

= 0,

but

Je lim inf fn = 0 > -1 = lim inf fe!n4. fn = X[n,n+l]

Je lim inf fn = 0 < 1 = lim inf

J

fn .

(Inequality may be strict.) This concludes our treatment of the Lebesgue integral for nonnegative measurable functions defined on arbitrary measurable sets of real numbers.

224

LEBESGUE INTEGRATION

5.4

THE LEBESGUE INTEGRAL AND LEBESGUE INTEGRABILITY

Men pass away but their deeds abide. ~Cauchy

The requirement that f be nonnegative is eliminated. We discuss measurable functions defined on any measurable set of real numbers. 5.4.1

Definitions

Let f be a measurable function defined on a measurable set E. We know that f can be written as the difference of nonnegative measurable functions: f = j+ - f- (Proposition 4.6). Calculate JEf+, JEf- according to Definition A or B. If both !£!+ and JEf- are oo, we say that the Lebesgue integral off on E is not defined (oo - oo is not defined in Re). If either JEf+ or JEf- (but not both) are finite, we define the Lebesgue integral off by

If both JEf+ and JEf- are finite, we say f is Lebesgue integrable on E and

In this case, JEf E R. Caution:

Hereafter, "Lebesgue integrable" will be shortened to "integr-

able". 5.4.2

Comments

1. How does the new definition (5.4.1), applied to a nonnegative,

measurable function f, compare with the previous definition for nonnegative, measurable functions (5.3.4)? Because f is nonnegative, f+ = f, f- = 0, and JEf- = 0. The Lebesgue integral off on E exists (5.4.1) and JEf = JEf+. We have consistency. 2. How does the new definition (5.4.1), applied to a bounded, measur-

5.4

THE LEBESGUE INTEGRAL AND LEBESGUE INTEGRABILITY

225

able function f on a set of finite measure E, compare with our original definition for bounded functions on a set of finite measure (5.2.5), or an equivalent condition, Proposition 5.4? Because f is bounded, measurable, and f = f+ - f-, the functions f+, f- are nonnegative, bounded, measurable functions on a set of finite measure E. Integrability follows from Theorem 5.7, that Is, f Ef- = f Ef- and f Ef+ = J Ef+. But,

j j+

=

sup{k ¢I¢
= sup{k ¢I¢
¢ simple and nonnegative}

= kf+,

and similarly for f-. We have: f = f+ - f-, f bounded and measurable on a set of finite measure E, f+ and f- are integrable 5.2.5, and J Ef+ = fEf+ = fEf+, JEJ- = fEf- = JEf-. We want to show f Ef = f E f = fEf+- JEf-. We apply Proposition 5.4 to f+ and f-. Let E > 0 be given. We have ¢+
to

the

"old"

Definition

Proposition 5.3. So, f is integrable according to the old definition (5.2.5) by Proposition 5.4. 3. When we say f is integrable on E, it is understood that f is measurable on E. 4. Iff is integrable onE, fEf = JE(f+- f-)= fEf+- JEf- · 5. Suppose f is integrable on E and f = / 1 - j 2 , where / 1 and / 2 are nonnegative measurable functions such that JE f 1 and JEf2 are both finite. Is it true that JE f = JEf1 - f£12? Because f is integrable, JE f = fEf+- JEf-. Since f+ - f- = f = / 1 - /2 , we have f+ + !2 = /1 + f-, and because(/++ / 2 ) and (/1 +f-) are nonnegative measurable functions, application of Proposition 5. 7 yields

226

LEBESGUE INTEGRATION

We may subtract since all terms are finite and conclude

In other words, iff is integrable, and we have any representation for f as the difference of two nonnegative measurable functions having finite integrals, the difference of their integrals may be used to calculate the integral of f.

Recall Example 6.

Example 11:

1/x, 1. f(x) = { 1/(x- 2),

~

f+

=

0<x<1 1<x<2.

-1 = oo and ~

~

(0,1 J X

(0,2)

f-

=

1

-1

(I ,2) X -

(0,2)

2

oo

=

'

The Lebesgue integral off on (0, 2) is not defined, and we can't have integrability without an integral.

2. f(x)

={

~

X

-1/2

0<x<1

'

1/(x- 2),

f+

(0,2)

=

~

l<x<2.

x- 2I = 2 and

~

(0,1]

=

f-

(0,2)

1

(1,2)

-1 = oo. X- 2

The Lebesgue integral off is defined, we may write f(o, 2 )f = -oo, but f is not integrable on (0, 2). -1/2

3. f(x) =

~

(0,2)

f

X

0<x<1

'

{

-(2- x)-1/2,

+

=~ (0,1 J

1<x<2.

x- 2I = 2 and

l

(0,2)

f-

=

1

(2 -

I

X)-;;:

= 2.

(1,2)

Lebesgue integral of f is defined, we may write Jro, 2 )f = 2- 2 = 0, and thus f is integrable on (0, 2).

The

5.4

4. f(x)

={

X

-1/2

THE LEBESGUE INTEGRAL AND LEBESGUE INTEGRABILITY

227

0<x<1

'

1jx,

X> 1.

by Proposition 5.7. The Lebesgue integral off is defined, f(o,oo)f = oo, and f is not integrable on (0, oo).

5. f(x) =

-x -1/2 {

-2 X

r

}(O,oo)

'

X> 1.

'

f+ =

0<x<1

r

}(!,oo)

x- 2 = 1 and

r

}(O,oo)

f--

r

}(0,1]

x-!- 2 -

.

The Lebesgue integral off is defined, f(o,oo)f = 1 - 2 = -1, and f is integrable on (0, oo ). The usual properties (homogeneous, monotonic, etc.) for integrable functions and their integrals are valid (Proposition 5.12), and could be proven now. We defer these arguments at this time. We would rather discuss relationships between fEf, JE If I, and JE g when g = f almost everywhere on E. 5.9 Suppose f is a measurable function defined on a measurable set E. Then f is integrable on E iff If I is integrable on E. PROPOSITION

Furthermore,

Assume f is integrable on E. We want to show If I is measurable and IE If I+, IE If 1- < oo. But since f is measurable, If I is measurable (Proposition 4.6), IE If 1-= 0, and IE If I+= IE If I= IE(f+ +f-) = IEf+ + IEf- < oo because f is integrable on E. Thus If I is integrable on E. Assume If Iis integrable on E. We show f is integrable on E. Since f is measurable by hypothesis, and IEf+ + IEf- = IE(f+ +f-) = IE If I = IE If I+< oo, the nonnegative integrals IEf+, IEf- are both Proof'

228

LEBESGUE INTEGRATION

finite. Consequently, f is integrable on E.

~f

=

~f+ - ~f-

<

~f+

+ 1f-

= 1f+ + 1f- = 1(f+ +f-)= 1 l f I

5.4.3

Comments

I. This proposition is false if we are dealing with the Riemann integral: Let f(x)

=

1' { -1

x rational,

'

x irrational,

0< x < I 0 < x < I.

Then f is measurable on [0, 1], Id lf(x) I dx = 1, but I~f(x)dx = -1 -1
for all n. We have a contradiction f+(x) = +oo}) = 0. The reader may f-(x) = +oo}) = 0.

unless show

11.( {x E E I J-L({X E E I

Note: f(x) = Ijx, 0 < x < 1. Then f is real-valued everywhere, but is not integrable. 3. If f, g are integrable on E, then If I, I g I are integrable on E, and since lf+gl
5.4

5.4.4

THE LEBESGUE INTEGRAL AND LEBESGUE INTEGRABILITY

229

Problems

1. Suppose f(x)

={

(-l)n+1n.49 1

(n+ 1) -1/2 <x
0,

otherwise.

Show f is integrable on [0, 1]. 2. Suppose (n+ 1)-1/2 <x
(-l)n+1n5, f(x) =

{ 0,

otherwise.

Show f is not integrable on [0, 1]. ii. Does an "improper" Riemann integral on [0, 1] exist?

1.

3. Suppose f(x)

=

-~cos (;2) + 2x sin (;2}

O<x< 1.

i. Show f is not integrable over (0, 1]. Hint:

lf(x)

I> 2/x I cos(1/x 2 ) I -2x > 1/x- 2x on the

interval ((2n + 1/3);r)- 1/ 2 < x < ((2n- 1/3);r)- 112 .

4. Suppose f(x) =

-I COS

x

(;r) x2

. - -x Sill

;r

(;r) , x2

0,

0 <X< 1, X= 0.

Show f is not integrable on [0, 1]. u. Show an "improper" Riemann integral exists on [0, 1].

1.

lll.

Let F(x) =

. - -1 X 2 Sill

2;r

0,

(Jr) x -

2

'

0 <X< I, X

=0.

Then F'(x) =f(x) and the Newton-Leibnitz integral ts F(I)- F(O) = 0. 5. Suppose f(x)

(-1t(n

={ 0,

+ l)- 2,

n;r < x < (n

otherwise.

+ I );r,

n = 0, 1, 2, ...

230

LEBESGUE INTEGRATION

Show f is integrable on [0, oo). In fact,

6. Suppose

f(x) =

(-lf(n+ 1)-1, {0

mr

<

x

<

(n

+ 1)1r,

n = 0, 1, 2, ...

otherwise.

Show f is not integrable on [0, oo). ii. Show an "improper" Riemann integral off makes sense: In fact,

1.

00

fo f(x)dx

=

1r In 2. sin(x)

7. Suppose f(x) =

x

{

1'

'

0

< x,

x=O.

Show f is not integrable on [0, oo). ii. An "improper" Riemann integral off makes sense: In fact, it 00 can be shown that J0 sin(x)jxdx = 1rj2. 1.

5.10

Iff is a measurable function defined on a measurable set E, and g is integrable onE with If I < I g I, then JE If I < JE I g I and f is integrable on E.

PROPOSITION

We have JE If I < JE I g I < oo from Propositions 5. 7 and 5.9. To show f is integrable onE requires: f measurable onE (given) and showing JEf+, JEf- are both finite. But 0 < fEf+ + JEf- = JE If I < oo, and the argument is complete. Proof

•

5.4.5

Problems

1. f(x) =

1 2'

x=O

1 - cos(x) x2

0 <X.

Show 0
Let fk(x)

231

THE LEBESGUE INTEGRAL AND LEBESGUE INTEGRABILITY

5.4

0 < x < 2k7r

f(x),

={ 0,

x > 2k7r

.

Then 0
r

lro,CXJ)

f =

r

lro.x)

= lim

limfk

lc

fk = lim

. {cos(x) - 1 = 11m X

2

k1r

0

1

1 - cos(x)

0

[0,2k1r]

[O,x)

. 1 =1m

lc

fk = lim

2krr

+1 2

k1r

X

2

dx

sin(x) d }

0

X

X

2k-rr . ( )

1

sm x d

X.

X

1 - cos(x) = rXJ sin(x) dx. Thus { x }o x2 lro,oo)

Is (sin(x)/x? integrable on [0, oo)? Calculate fro,oo)(sin(x)/x) Hint:

r

2

(sin(x)) = lim

}[o,c:o)

1 - cos(2x) dx r" 2x }o 2

x =

.1

2 k"

hm

1 - cos(x)

0

Note:

X

2

dx

=

lox 0

sin(x) X

From problems 1 and 2,

r

1 - COS(x) = {DC (Sin(x)) {X sin(x) dx = x }o x2 lro,x) x Jo that is,

2

dx.

2

.

232

LEBESGUE INTEGRA T/ON

Isn't this amazing! sin(x). X

here

1

0

37r

27r

I

On closer inspection, show

Oh,

that is,

1

1 =

0

Then

0

5.4

THE LEBESGUE INTEGRAL AND LEBESGUE INTEGRABILITY

233

and now it is reasonable that

3. Are the following functions integrable on the sets E? 1.

tan- 1 (x) x 312 on(O,oo). lnx

11.

m.

xl+a

on (1, oo),

1 - cos(x) x 2+a

sin(x) tv.

xl+a

a> 0.

on(O,oo),O
on (0, oo), 0
Sets of measure zero do not affect Lebesgue integrability. 5.11 Iff= g a.e. on a measurable set E, and if g is integrable on E, then f is integrable on E and

PROPOSITION

Proof' The function g is measurable onE by the assumption of being Lebesgue integrable on E. Since f is equal almost everywhere to a measurable function g, f is measurable on E (Proposition 4.3). Application of Proposition 4.6 yields measurability off+ and on E. Let A= {x E E I f(x) -1- g(x)}. Then f+ = g+ and = g- on E- A, and IE-Af+ =IE-A g+ and IE-Af- =IE-A g-, that is, f is measurable on E- A, IE-Af+, IE-A < oo: f is integrable on E-A. Because A is a measurable subset of E, f is measurable on A (Problem 4.1.4), J.L(A) = 0, and hence IAf+, IAf- = 0. But then IEf+ = IE-Af+ + IAf+ < oo and IEf- = IE-Af- + IAf- < oc: The function f is integrable on E.

.r-

.r-

.r-

234

LEBESGUE INTEGRATION

Then

Example 12: 1. f(x)

g

=

= 0. f

2. f(x)

x rational

1, { 0,

x irrational

= g a.e. and

( -oc,oc)

f =

1

x rational,

x,

={ 1-x '

g(x) = 1 -X, 0

1

( -oo,oo)

0

x irrational, <X

<

}[O,l]

x irrational

1 !=1 (-oo,oo)

4. f(x) = {

r

}[O,l]

(-oo,oc)

<x <1

g =

t}o

'

g(x)

= e-lxl.

(1 - x)dx

=

1 2

g=2.

~'

x=-,

0,

x irrational

p

q

r g = o. 1 = }[O,l]

0

}[O,l]

x rational

3. f(x) = { -lxl e ,

<x<1

1.

f = g a.e. on [0, I] and { f = { x2,

g = 0.

O<x
g(x) _ 0.

5.4

THE LEBESGUE INTEGRAL AND LEBESGUE INTEGRABILITY

xz

5. f(x) =

r

}[o,rr]

f =

' { sin(x),

r

x in the Cantor set

g(x)

235

= sin(x).

otherwise

g = 2.

}[o.rr]

5.12 lff,g are integrable on a measurable set E, and k is any real number, then PROPOSITION

1. (kf) is integrable onE, and IE(kf) = k IEf (homogeneous); 2. (f +g) is integrable onE, and IE(f +g) = IEf +IE g (additive); 3. IEf < IE g iff < g on E (monotone); 4. If E 1 and E 2 are disjoint measurable subsets of E withE = E 1 U E 2 , f is integrable on E 1 and E 2 , and (additive on the domain).

Proof 1. If k > 0, IE (kft =IE kf+ = k IEf+ < x and IE(kf)~ = k IE f~ < oo because kf+, kf~ are nonnegative measurable functions (Proposition 5. 7). By definition, (kf) is integrable on E.

Furthermore, IE(kf) = Idkft - IE(kf)~ = k IEf+- k IEf~ = k IE f, where the last equality is the definition off being integrable on E. If k < 0, (kft = (-k)f~, (kft = (-k)f+, IE (kft = -k IEf~ < oo, and IE (kfr = -k IEf+ < oo, that is, (kf) is integrable on E. Again,

2. Since f, g are integrable on E, (Proposition 5.9). Because IE

If I , I g I are integrable on E If I= JE If I+< oo, fE I g I=

236

LEBESGUE INTEGRATION

fEigl+
=

~!+ + ~g+- ~~-- ~g­

= ~!+~g. 3. Since f
Because all terms are finite, we may subtract: JEf < JE g. 4.

(Proposition 5. 7) In the next section we prove the major convergence theorem of Lebesgue integration, the so-called Lebesgue Dominated Convergence Theorem (LDCT).

5.5

5.5

CONVERGENCE THEOREMS

237

CONVERGENCE THEOREMS

We continue our discussion on the question:

The reader may recall the results that have been established: 5.3 If (fk) is a sequence of Riemann integrable functions on [a, b], and if limfk = f uniformly on [a, b], then

THEOREM

1. (lim fk) is Riemann integrable on [a, b]; 2. lim fk(x)dx = J:(lim fk)(x)dx.

J:

5.10 (LMCT) If (fk) is a monotone increasing sequence of nonnegative measurable functions on a measurable set E, then

THEOREM

5.11 (Fatou) If(fk) is a sequence of nonnegative measurable functions on a measurable set E, then THEOREM

5.8 If (gk) is a sequence of nonnegative measurable functions defined on a measurable set E, then

PROPOSITION

Example 13: 1,

1. f(x) =

{ 0,

x rational x irrational.

If r 1 , r 2 , ..• is any enumeration of the rationals, define

238

LEBESGUE INTEGRATION

1, fk(x) =

{ 0,

x=r 1 ,r2 , ..• ,rk

otherwise.

Then 0 < fk < fk+h limfk = f, fro.x) f = = 0 via LMCT.

f[o,x) limfk

=lim f[o,x) fk

2. 1.

~ lc

3

2x = 2 [O,IJ (1 + x t

lc

[0.1]

3

~(I+ t 2x x2

=

lc

[O.!J

(I +X 2 )2x = -3

2.

11.

111.

O<x
~ k.t] [I+ (n- ~x][I + nx] k.t] ~[I+ (n- ~x][l + nx] =

~ j, ~:

O<x
1] {

-- 1.

x=O

3.

via Theorem 5.3. Can we use Proposition 5.8? The reader may have noticed the frequent occurence of "monotone"

5.5

CONVERGENCE THEOREMS

239

and "nonnegative" in our "limiting" discussions. It is these restrictions that we seek to modify or eliminate, although, other requirements must be imposed. The new requirements do not severely restrict applications of the Lebesgue integral, and in fact result in a very powerful tool for analysis, the so-called Lebesgue Dominated Convergence Theorem (LDCT). 5.12 (Lebesgue's Dominated Convergence Theorem, 1910) Let ( fd be a sequence of measurable functions, defined on a measurable set E, such that THEOREM

limfk = f a.e. on E. Suppose we have an integrable function g on E such that Ifk Then f is integrable on E and

I < g on

E.

Proof Since sets of measure zero do not affect Lebesgue integrability, and J.L( {x E E I g(x) = oo}) = 0 because g is integrable on E, we may as well assume limfk = f (real-valued) on E. Furthermore, f, as the limit of a sequence of measurable functions, is measurable (Theorem 4.1 ), and the reader may argue (Proposition 5.10) that f, and for that matter, all other functions appearing below, are integrable on E.

Our first argument will be based on the Lebesgue Monotone Convergence Theorem: monotone increasing and nonnegative. Monotonicity is not a problem: Recall that for a sequence of functions (fk) we may construct two related mo_~_wtone sequences (£) and (fk) where £=inf{fk,fk+J, ... } andfk=sup{fk,fk+ 1 , ••• } respectively. So, we have the following relationships between g,f, and terms of the sequences (£)and (fk): -

-

-g <£ <£+1
(1)

Everything follows from this string of inequalities. Initially, all functions being integrable, along with monotonicity, yield (Proposition 5.12)

240

LEBESGUE INTEGRATION

We want to use the "squeezing" principle:

-oo< rfk+l< rfk+l' rf< rfk+t<+oo }E-

}E

}E

}E

I

I I

I I I I I

t

? Common Limit?

If we can show a common limit, that is, lim fEf k = lim JEf b then lim JEfk exists and lim JEfk = JEf = JE(limfk), which is the conclusion we want. Returning to (I), we have -

-

0 < g + [_k < g + f k+l < 2g and 0 < g- fk < g- fk+l < 2g. The sequences (g +f k) and (g- fk) are nonnegative, monotone increasing, and have limits g +f and g- f respectively. Application of the monotone convergence theorem yields

r g + lim r f k = r lim(g + f k) = r (g +f) = r g + r f }E }E}E }E }E }E and

that is,

and this argument is complete. The next argument is more direct; an application of Fatou's Theorem. "Fatou" for (g + fk):

leg+ lef

=

le(g +f)= leliminf(g + fk) < liminf le(g + fk)

=

liminf(leg + hfk) =leg+ liminf lefk,

5.5

CONVERGENCE THEOREMS

241

so fEf < lim inf JEfk· "Fatou" for (g- fk): 1g - 1 ! = 1(g- f)= 1liminf(g- fk) < liminf 1(g- fk)

= liminf(hg -1fk) = hg -lim sup hfk,

Comments

5.5.1

1. We need "domination" onE: If fk = kX(O,l/k)• then limfk = 0, and lim fro,l]fk = 1 i= 0 = fro,i](limfk). 2. We need "integrable domination" on E. If fk = X[k, 2kJ• then limfk = 0 and lim JRfk = oo i= 0 = JR(Iimfk). The following problems illustrate some applications of the convergence theorems that have appeared in this course. They partially answer "Why Lebesgue?"

Problems

5.5.2

(Theorem 5.6 should be used quite often.) 1. 1.

kx2 2 = 0. Show lim { }[O,l]l + k X 2 2

2 We see that fro,!] (kx)/(1 + k x ) = Jd (kx)/(1 + k x )dx = 2 thus and (In (I+ k 2x 2 ) )/(2k) !6 = (In (1 + k ) )/(2k), 2 2 lim fro,!] (kx)/(1 + k x ) = lim(ln(l + e))/(2k) = 0. No problem, but, in anticipation of exercises to come, we try other 2 2 approaches. For example, lim(kx)/(1 + k x ) = 0, 0 < x < 1. But we do not have uniform convergence on [0, 1] : fk(1/k) = 1/2. How about 0
LEBESG UE INTEGR ATION

242

Evalu ate lim

ii

k3j2

1

+k

[0,1]1

: X

2

two ways.

0.

g(x)

Hint:

=

_

{

X

112

'

2. 1.

e-kx

Evaluate lim { }[O,oo)

g(x) =e-x .

Hint:

e-kx

n. Evalu ate lim {

2 •

}[o,oo)

k 312 xe-kx.

iii. Evalu ate lim { }[o,oo)

(sin(x))k

3. Show lim {

= 0.

}[O;rr /2]

.1

4. Show hm

=

[-2,2]1

+X2

lim(x 2k/(1

Hint:

5. fk(x)

x2k

k =

2.

0,

0 <X< 1

1/2,

x

1,

1<x <2

+ x 2k)) =

1 k, 0 < x. Show limfk (x) 1+x2

Evalu ate lim {

)[O,IJ

6. Evalu ate

Jk,

r L xnn

)[o,I]

lim

0

1 Jk, (l,x)

lim

=1

1,

O<x <1

1/2,

x

0,

1 <X

=

r

}[O,cc)

Jk.

=1

243

CONVERGENCE THEOREMS

5.5

7. 1.

xe-kX,

Evaluate {

k

=

1, 2, ....

}(O,oc)

xe -kx

=

gn(x)

Hint:

{ 0,

0< X< n

'

-

-

,

n< x

0 < gn < gn+J, and use LMCT to show J(o,oo) xe-kx = 2 lim fro,n] gn = limn xe -kx dx = 1/ k , where the last equality follows by integration by parts for the Riemann integral.

H

11.

Show

r

X

lro,oo) ex- 1

= ""-;. ~k

= x(l:::J e-kx) = l:1 xe-kx, x > 0, and J(o,oo) x/(ex- 1) = Jo,oo) 2::: 1 xe-kx, where

x(ex- 1)- 1

Hint:

Jro,oo) xj(ex- 1) = the first equality follows from, "sets of measure zero do not affect the Lebesgue integral".

m. Show

tv. Show

x2

lc

[O,oo) ex - 1

h

1 . L 3 2 k 1

= -

1

x"'

--=((a+ 1)r(a+ 1),

[O,oo) ex - 1

a> 0.

(Zeta-Gamma)

Recall tan- 1 (x) =

lo

x

1

1+

t.2

dt

-1
=1

JI: = I: J ? O
g(t)

=

1,

244

LEBESGUE INTEGRATION

and LDCT. l

Another approach:

1+ t

= 2

{ l::o(t4n _ t4n+1),

l /2,

O
Then t 4n - t 4n+ 2 > 0, 0 < t < l, and we may conclude

-7r =

tan~

4

1(1) = lol 1 +1 t dt = 1 1 +1 2

0

=

r (t 4n '"""" lro,IJ La

[0,1]

14n+ 2)

=

t

2

=

1

[0,1)

1

1 + t2

'"""" ( 1 4n + 1 La

1 ) 4n + 3

without constructing a dominating function.

9. Show that ln 2 = 1 Recallln 2 =

.. H znt.

1

2

1

+ - ··· .

3

1 dt = r {2:: 1 r lro,IJ 1/2, }o + t 0

I

, fi()-{2::~(-tt k t 1/2,

(- t

t'

O
O
Again, we could observe that O
since t 2n -

pn+ 1

0 > ' -

g(t) = 1, 0 < t < 1.

5.5

CONVERGENCE THEOREMS

Then ln2 -

-

10. Suppose f(t)

=

t-

1

,

~

t

1 1 ) ( 2n+ 1 - 2n+ 2

> 0 and fk(t) =

cl+l /k {

t

O< t
' -1-1/ k

1 < t.

'

Show

and k(x llk- 1), k(x 11k- 1) xl/k

1 <X.

Then

=

In X lnx

11. If f(t)

0 _< t

r ! = }[Ir limfk = r fk .

}[I ,x]

I

,x]

limk (x 11k -1),

r

J[O,x )

={ <

l

lim

}[I ,x]

0 <X.

= "£ akl, -oo < t < oo, ak > 0, then show

. H znt:

gn(t)

7

r! dt = }I

k e -t'"' L.., akt =

le- 1, 0 < t < n 0,

oo, an d

n


fr[O,oo) t k e

-1

'

'"'

L..,

r ) t k e -t , ak J[O,oo

0 < gn < gn+h

fr

= 1'Imn [O,n] t k e - f .

245

246

LEBESGUE INTEGRATION

120 1.

12

Evaluate {

lro,XJ)

e- cos(t)o

12

e- cos(t) =limn l:::o((-I)k/(2k)!)t 2ke-

Hint:

2

1

limfn(t) = e- cos(t), 0 < t < oo, lfn(t) I< e-

1

2 0

12 ,

e1

0

Apply LDCT:

r elro,XJ)

12

cos(t) =lim

0

=

hm

lc

( ~

~ -1 )k

[O,n]

n

2k -t2

(2k)' t e

0

.

0

~ (-Ill

2k -t

hm ~ (2k) 1 [O,n] t e n 0 0

=

r fn(t) lro,oo) 2

o

=

hm 10

1

(

n

22

V:

= e- 114 0 11.

o

(See problem 170)0

130 Show {

s!nat ="" 2 a 2' e - 1 ~ k +a

}(o,oo)

l

(O,oo)

2 /

4 0

2

}(O,XJ)

Let fn (t) =

0

e- 12 cos(xt) = y7i 1r e-x

Show {

Hint:

o

sinat = 1

e - 1

h

a> 00

sm at(e- 1 + e- 21 + 0)0 0

(O,CXJ)

2.:::7 e-kl sin at, and define g(t)

= {

?, -1 2e '

t) lc

1 - + ( -1 + 4 1! 2 2! 22nn!

O
[O,oo)

e -1

2

CONVERGENCE THEOREMS

5.5

247

Apply LDCT to conclude

1

smot = 1 (O,oo) e - 1

1 (O.:xo)

r.

=lim

te-kt }(o.x) k=!

n

=lim

ic

e

-kt .

Q

sm at = 0

[O,oo)

gn(t)

Hint:

14 .

1.

Show

f

Hint:

n. Show

f

+ k2

1- I

n

In(!) = I

Evaluate

Hint:

I

> 0.

7r .

6

y

t-

1

O
= ?_ "'~ 6 L.k 71

(t- l)- 1 ln t =

t- 1 1nt(t -1)- 1 = t- 1 lnt:L:: 1 t-k, t

>

1.

Ll J(I,oo) t-(k+i) Int. gn(t)

={

ta-!

15. Show }(o,I) 1 + tf3 = a, (3

2

1 In(!)="' (1- t)k-J ' k t 1-t

t-(k+I)lnt

n J(!,oo)

1

-

< t.

r(k+I) ln t

=

1 2 , and complete the

k

1 0

'

0,

Use LMCT to show argument.

r

< t.

~ ~k =

}(l,oc)

Hint:

sinot.

.

0,

1

}(O,l)

2

e-kt

sin at,

e-kt

={

t r

sinot

k=! }(O,x)

n

Sh ow

. r 1lmJn

0

1

1

+ (3 + 0 + 213 - a+ 313 + ·. · where

248

LEBESGUE INTEGRATION

Hint:

(1

+ t,B)- 1 =

(1-

t,B

+ t 2!3- · · ·), 0 < t <

1.

Form to:- 1(1- ti3) + to:- 1 (t 2!3- t 3!3) + to:- 1 (t 4!3- t 5!3) +···and let fk(t) = to:- 1(t 2ki3- t(lk+ 1)!3), 0 < t < 1 and k = 0, 1,2, · · · . Thus

r

~(0,1)

r fk :2:: ~(0,1)

2:1k =

=Let

1 + 2k;3-

a+

(2~ + 1);3) ·

Can you use LDCT? 16. 1.

Recall1 +

f < e 1k, (1 + f/ < e 1

n. Evaluate li_F J(o,oo) ( 1 +

1

,

-k < t, and lim(l

f/ e-o:r, a > 1.

1 , e(l-o:)r = Next show ( a- 1 ~[O,oo) e(I-o:)r

Hint:

1.

m. Show 1m Hint:

gk(t)

={

1 (1 [O,k]

'

Evaluate lim k

k < t.

0,

i

O
< t.

1

1

~[0,1]

Vl.

Show lim

a> 0.

(O,oc)

r (1 - k )\o:t if a < ~[O,k]

v. Evaluate lim (

e -t t o:-1 ,

-f) < e-t/k, 0 < t < k and define

(1- tjk)\ fk(t) = { 0 lV.

a> I.

O
- -t)k t o:-1 -k

Show (1

+ f)k = e1•

(1 + kt) -k-

.

1 = 1. r ~(O,oo) (1 + tjk )k

1.

5.5

CONVERGENCE THEOREMS

Look for a dominate function g: If 0 < t < I,

If I < t,

< [k(k- 1) -

2 4

<2' - t

0

!!_]-I k

k > 20

17. 2

e-x = foo }[O,oo) 2

Show { Let

0, and

1

fk

= Vk

[O,oo)

1fj2

1

cos2k+I (t)dt

0

= Vk

(2k)(2k- 2) (2) (2k + 1)(2k- I) (3) 0

0

0

o

0

0

249

250

LEBESGUE INTEGRATION

and

= Jk (2k+ 1)(2k-1)···(1). (2k + 2)(2k) ... (2) IV.

7r

2 . '

Since fk, gk are dominated by the integrable function e -r, t > 0, we have

f

lim

fk

=

}[o.oc)

f

Iimfk

}[o,k]

v.

(1

[O,oc)

e-t

2)2 = (lim =lim( -

4

VI.

1 )( 1 ) fk

lim

[O,oo)

r

}[o,oo)

gk

[O,oc)

fk·

r

}[o,oo)

gk)

Appendix A

Cantor's Set

When you have eliminated the impossible, whatever remains, however improbable, must be the truth. -Sir Arthur Conan Doyle

A.1

CANTOR'S SET

The German mathematician Georg Cantor (1845-1918) is regarded as the creator of set theory. Earlier (Chapter 2) we used the "Cantor diagonalization argument" to show the set of real numbers is uncountable. In this appendix, the beautiful Cantor set, a source of so many examples and counterexamples in mathematics, is developed. We begin with a constructive (geometrical) process applied to the interval [0, I]:

Step 1.

252

Divide [0, 1] into three equal parts and remove the "middle

A.1

CANTOR'S SET

253

third", the open interval (1/3, 2/3): 2 3

l

0

3

2 3

1

3

1

1

0

2 3

3

1

The closed set F 1 = [0, 1/3] U [2/3, I] is left. Step 2. Divide the two closed subintervals of F 1 into three equal parts and remove the "middle thirds," the open intervals (1/9, 2/9), and (7 /9, 8/9): 0

1

~

7

8

9

9

9

9

1

•

0

2

1

2

7

8

9

3

3

9

9

•

1 9

•

2 9

•1

•

2 3

3

~

1

•

I

9

•

§ 9·

F2

The closed set F 2

=

[0, 1/9] u [2/9, 1/3] u [2/3, 7/9] U [8/9, 1] is left.

Continuing in this way, after n- I steps we have deleted 1 + 2 1 + 22 + ... + 2n- 2 = 2n-i - I disjoint open intervals, and have left the closed set Fn-i consisting of 2n-i closed intervals, each of length I/3(n-l).

•1

254

APPENDIX A

Step n. Divide the remaining 2n-l closed intervals into three equal parts and remove the "middle thirds." The closed set that is left, Fn, consists of 2n closed subintervals, each of length 1j3n.

A The Cantor set, C, is what's left in [0, 1] after deleting "middle thirds," namely, DEFINITION

The constructive approach is complete. We now give an analytic definition of C. DEFINITION

B

The Cantor set, C, is the points of [0, 1] which have a base three (ternary) expansion without the digit 1. That is, for x E C, ' an= 0 or 2

0

Why two definitions? The idea here is that depending on the property to be discussed, Definition A will be preferable to Definition B and conversely. It's like the statement; Prove 0 = 1r- 1r3 /3! + 1r5 /5! - · · · . Who believes it, but when we recognize this is sin( 1r), with the "unit circle" interpretation of sine and the "power series" interpretation of sine-no problem. Our task at hand: reconcile Definition A and Definition B, the constructive versus descriptive interpretations of the Cantor set. Some discussion of ternary expansions is appropriate. We will show that every number in the interval (0, 1) has at least one, and at most two, ternary expansions. Recall x = .a 1a2 a 3 • • ·an · · · means an = 0, 1, or 2 and 3 X= I: an/3n. The integers 0, I' 2 are called the ternary digits and x = .a 1a2 ···an··· a ternary expansion of x.

r

3

Claim:

Every number in (0, I) has a ternary expansion.

Step 1. Since 0 < x < I, x will be in one of the "thirds": (0, I/3) or [1/3,2/3) or [2/3, 1). So 0 < 3x < 1 or I< 3x < 2 or 2 < 3x < 3. Let a 1 = [3x], where [ ] is the greatest integer function. Thus a 1 = 0 or I or 2

A.1

CANTOR'S SET

255

and a 1 < 3x < a 1 + 1, that is, a1

al

1

3 -

3

3

-~<x<-~+-t·

Since a 1 < 3x < a 1 + 1, 0 < 3x- a 1 < 1, and we have a number, 3x - a1 , between 0 and 1 again. Hence this number falls in one of the "thirds": [0, 1/3) or [1/3,2/3) or [2/3, 1), and 0 < 3(3x- aJ) < 3. Step 2. Let a2 = [3(3x- a 1) ]. a2 < 3(3x- a 1) < a2 + 1, that is,

Thus a2 = 0 or

1 or 2 and

Since a2 < 3(3x- a 1) < a2 + 1, 0 < 3(3x- aJ)- a2 < 1, and we have a number, 3(3x- a 1)- a2, between 0 and 1 again. Hence this number falls in one of the "thirds": [0, 1/3) or [1/3,2/3) or [2/3, 1), and 0 < 3(3(3x- a1 ) - a2 ) < 3. Step n. Let a,= [3"x- 3"- 1a 1 - · · ·- 3a,_J]. Thus a,= 0 or 1 or 2 and 1 an < - 3"x- 3"- a 1 - · · ·- 3a n- 1
Continuing the process, we have shown that, given 0 < x < 1, n

=

1,2, · · ·.

Every number in (0, 1) has a ternary expansion. Every number in (0, 1) has at most two ternary expansions. Suppose x = .a 1a2 · · ·a, · · · = .b 1b 2 · · · b,. · · · are two different ternary expansions ~f x and N is tfile first value of n where b, =1- a,, that is, x = .a1a2 · · · aN-laNaN+l · · · = .a 1a2 · · · aN_ 1bNbN+l · · · with aN< bN. Be~ause aN, bN = 0, 1, or 2, Jnd aN < bN, we have these possibilities: Claim:

256

APPENDIX A

aN= 0, bN

= 1 or 2, and aN= 1, in which case bN = 2. But

<'""" 00

b

-63kk I

=X.

Thus bN =aN+ 1, aN+J = aN+ 2 = · · · = 2 and bN+l = bN+ 2 = · · · = 0. Wehavex .a 1a2 ···aN-JaN222··· orx .a 1a2 ···aN_ 1(aN+l)OOO···. 3 3 Either the ternary expansion is unique or we have exactly two such representations, one ending in "2's" and the other in "O's", starting from the same location. We return our attention to the Cantor set. We show that Definition A implies Definition B; removal of "middle thirds" implies every point of the Cantor set has a ternary expansion without the digit 1. In Step 1 of the constructive process we removed (1/3, 2/3). What can we say about the ternary expansions .a 1a2 ···an··· of such numbers? We claim a 1 = 1 and among a 2 , a 3, · · · , some a; is not zero and some a; is not 2 2 3 two. If a 1 = 0, then x = 0/3 + a2 /3 + · · · < 2/3 + 2/3 + · · · = 1/3, but 1/3 < x < 2/3. If a 1 = 2, then x = 2/3 + a2 j3 2 + · · · > 2/3. We have a 1 = 1. If a2 = a 3 = · · · = 0, x = 1/3 and if a 2 = a 3 = · · · = 2, then x = 2/3. Thus the ternary expansion of every number in (1/3,2/3) starts with 1, and contains a nonzero entry and a "non" two entry. That is, every number in (1/3,2/3) must have a one in its ternary expansion. 1

-3 =3 .1- 00 . . . =3 .0222 ... Note: 2

-3 =3 .200 . . . = .1222 ... 3 -

A.1

CANTOR'S SET

257

The open interval (a 1 , b 1 ) removed at the first step can be written with a 1 = .l,b 1 = .2. In Step 2 of the constructive process we removed (1/9, 2/9) and (7 /9, 8/9). What can we say about the ternary expansions of such numbers? We claim a2 = 1 and among a3 , a4 , · · · , some ai is not zero and some ai is not two: If a2 = 0, then x = aJ/3 1 + 0/3 2 + · · · . Several cases must be considered: Suppose a 1 = 0. Then x = 0/3 1+ 4 0/3 2 + .. · < 2/3 3 + 2/3 + · · · = 1/9. Can't happen. Suppose a 1 = I. 2 2 3 Then 1/3 < x = 1/3 + 0/3 + .. · < 1/3 + 0/3 + 2/3 + .. · = 1/3+ 1/9. Can't happen. Suppose a 1 = 2. Then 2/3 < x = 2/3+ 0/3 2 + ... < 7/9. Can't happen. The other possibility is that a2 = 2, that is, x = a 1 j3 1 + 2/3 2 + · · · . Again, suppose a 1 = 0. Then 2/9 < x = 0/3 1 + 2/3 2 + · · · < 1/3. Can't happen. Suppose a 1 = 1. Then 1 2 5/9 < x = 1/3 + 2/3 + · · · = 6/9. Can't happen. Finally, what if a1 = 2. Then 8/9 < x = 2/3 1 + 2/3 2 + · · · = I. We have a2 = 1, that is, x=.a 1 1a3a4 •··. If a 3 =a4 =···=0, then x= 1/9 or 7/9 (4/9 was removed at step 1). If a 3 = a4 = · · · = 2, then x = 2/9 or 8/9 (5/9 was removed at step 1). Thus every number in (1/9, 2/9) and (7 /9, 8/9) has a ternary expansion that must have a 1 in the second entry, and contains at least one nonzero entry and at least one "non" two entry. Note:

I

-9 =3 .01000 ... = .00222 .... 3 ' 2

-9 =3 .02000 ... = .01222 .... 3 ' 7 9

- = .2100 ... = .20222 .... 8

-

9

3

=

3

-

3

-

'

.2200 ... = .2122 .... -

3

-

The open intervals (a 2 , b2 ) removed at the second stage can be written with a 2 = .(2o:t) I, b 2 = .(2o: 1)2, a 1 = 0 or I. 3 3 In general, at step n of the construction, the open interval (an, bn) is removed if an = .(2o:J) (2o: 2 ) .. · (2an- J) 1 and bn =. (2o: 1) (2o: 2 ) • · · 3 3 (2an_J)2. To see this, the idea is that the left-hand endpoint of an interval removed at the n'h step is "constructed" from a right-hand endpoint bn-l

258

APPENDIX A

of an interval removed at the preceding step (an = bn-l + 1/3n) or "constructed" from 0 (an= 0 + 1/3n). For example, at the first step we removed (1/3,2/3) = (0+ 1/3,0+2/3) = (.1,.2): "Constructed" from 3 0. At the second step, we removed (1/9, 2/9) = (0 + 1/9,0 + 2/9) = 3 (.01,.02), "Constructed" from 0, and (7/9,8/9)=(.2+1/3 2 ,.2+ 3 2/3 2 ) = (.21, .22), "Constructed" from right-hand endpoint, .2, of 3 ( .1, .2). Induction provides "rigor". In general, or

that is, or

1

and

We've shown that if xis removed in the geometric process, then x must have 1 in its ternary expansion and a nonzero and "non" two appearing thereafter. That is, Definition A implies Definition B. The converse remains: If x = .a 1a2 · · ·an · · · has a one in its ternary 3 expansion with a 0 and a 2 appearing thereafter, then it was removed by the geometrical process. Suppose x = .1 · · · 0 · · · 2 · · · . Then 3 1/3 < x < 2/3 and x was removed at step 1. Suppose x = .a 1 1 · · · 3 2 · · · 0 · · ·. Then a 1 = 0 or 2 (a 1 = 1 was removed at Step 1), and 1/9 < x < 2/9 or 7/9 < x < 8/9. Thus x was removed at Step 2. In general, x = .(2aJ) (2a 2) · · · (2an-d 1 · · · 2 · · · 0, ak = 0 or 1, that is, 3 .(2o:J)(2a2) · · · (2an-J)l < x < .(2aJ)(2a2) · · · (2o:n_J)2, and an< X< bw The number x was removed at the n1h step. (Induction provides "rigor".) We have completed the arguments for showing the equivalence of the constructive and descriptive definitions of the Cantor set. We now investigate some remarkable properties of the Cantor set. 1. The Cantor set is uncountable.

This statement may seem false. After all, the only obvious numbers are 0,1,1/3,2/3,1/9,2/9,7/9,8/9, ... , the countable set of endpoints of the open intervals that we removed. So, what now? We try some other points: l /4 = 1/3 - 1/9 + 1/27 - · · · =3 .02020 · · ·

A.1

CANTOR'S SET

259

3/4 .2020 · · · . At least we are convinced something other than 3 "endpoints" are in C. It is appropriate that the "Cantor diagonalization process", along with Definition B, shows uncountability: Assume Cis countable and make a "list":

with anm Let x

=

0 or 2.

= .a 1a2 a3 · · · where an = { 3

=2

0

if a12 n

2

if ann= 0

The number x does not appear on our list. 2. The Cantor set is measurable with measure zero. (After all, what is the probability that a ternary expansion does not have a "1 "?) Measurability is obvious. As for measure zero, given E > 0, choose n so large, say, N, such that (2/3)N < E. Recall F N consists of 2N closed intervals each of length 1/3N, that is, f.l(FN) < E. Since C = nF12 c F N• f.l( C) < E. From the arbitrary nature of E, f.l( C) = 0. Definition A works great! The Cantor set is an uncountable set of measure zero. 3. The Cantor set contains no open intervals: The Cantor set 1s nowhere dense. Since the length of the intervals removed is one ( 1/3 + 2(1/9) + 4(1/27) + · · · = 1), it is not surprising that C does not contain any intervals: In the construction of the Cantor set, intervals of the form ( (3k + 1)/3 12 , (3k + 2)/3 12 ), 0 < k < 3n-i- 1, n = 1, 2, ... , are removed, and thus given any interval (a,b) C (0, 1), we have ((3k+ l)/3n,(3k+2)/3n) C (a,b) for n sufficiently large. 4. Every point of the Cantor set is a limit of points of the Cantor set (perfect).

260

APPENDIX A

Let x 0 E C and (a, b) any open interval contammg x 0 . Since Xo E C = n Fm xo c Fn for all n. Let In be a closed interval of Fn that contains x 0 . Because the length of any In is 1j3n, we may choose n so large, say N, such that/N c (a, b). At least one of the endpoints of IN is different from x 0 . That is, given any open interval about x 0 , it contains a point of the Cantor set different from x 0 , and a point not in the Cantor set by part 3. Thus x 0 is a limit point of the points of C. A point in C is a limit point of points in C and points not in C. 5. It is possible to construct Cantor type sets (nowhere dense, perfect, etc.) of positive measure a, 0 < a < 1. These are the so-called, generalized Cantor sets. A very nice discussion may be found in Stromberg (1981) [St]. We will now discuss a function associated with the Cantor set, the socalled Cantor function, that has very surprising properties. Specifically, we construct a function such that: 1.

11.

is continuous and nondecreasing on [0, 1]; maps the Cantor set (a set of measure zero) onto [0, 1] continuously; = 0 for almost all x in [0, 1]: We start at elevation zero, end at elevation one, while moving horizontally most of the time!

I

111.

Curiouser and curiouser! -Lewis Carroll

We need some results regarding monotone functions. Recall a function f is said to be nondecreasing on (a, b) if a< x < y < b implies f(x)
A.l

Let f: (a, b)--+R be nondecreasing and suppose

a< c
A.1

261

CANTOR'S SET

Proof a< x < c implies f(x) 0 be given.

From the definition of "sup" we have a < x 1 < c with sup{f(x)

I

a< x < c}-

t:


I

a< x < c}.

Since f is nondecreasing, a < x 1 < x < c implies sup{f(x) I a< x < c}-

t:


That is, limx-.c- f(x) = sup{f(x) I a< x < c}. The reader may argue "inf" and "right-hand limit". If c < d, choose x so that c < x
A.2 Let

f : [a, b]

onto [c, d] be nondecreasing. Then

f

is

continuous on [a, b].

Let t: > 0 be given and suppose a< x 0
< f(xo) < f(xt).

We want to show f(xo) Suppose f(x 0 )
=

limx_,x-0 f(x)

=

f(xo).

d

f(xo)

f(xol

a

Then c
b

262

APPENDIX A

Something must, because f is onto! A similar argument shows f(xo) = f(xti), limx__,a+f(x) = f(a), and limx__,b- f(x) = f(b). • We are now in a position to start construction of the Cantor function . 1. First define a function ¢ that maps the Cantor set onto [0,

1].

Let x be any point in the Cantor set: x = .a 1a2 • • ·an · · · = 3 .a1a2 ···an··· (3), with ak = 0 or 2. We map such a point to the point .(a 1/2)(a 2 /2) · · · (an/2) · · · (2) (binary expansion), that is,

where an is either 0 or 2. For example,

¢(~)

=¢(.0222 ...

(3))=.011} .. .

¢ (~) = ¢(.2000...

(3)) = .1000 .. .

1

(2) =

2' 1

(2)

= 2'

¢G)=¢(~)=~·¢(~) =¢G)=~. ¢(

~) = ~ ,

¢

(!)

=

~,

etc .

Obviously¢ maps the Cantor set into [0, 1]. Is the mapping onto? Certainly any number in [0, 1] has a binary expansion, say .b1 b2 · · · bn · · ·( 2), bk = 0 or 1. Just "double" each digit. Then

and . (2b 1) (2b 2 ) · · · (2bn) · · · (3) is a member of the Cantor set. So, ¢ maps the Cantor set onto [0, 1]. 2. We claim¢ is a nondecreasing function on the Cantor set. Let x, y E C, x < y;

an= 0 or 2, bn

=0

or 2,

A.1

with an

CANTOR'S SET

263

= bn for n < Nand aN = 0, bN = 2, that is, x=.a 1a 2 ···aN_ 10aN+ 1 ···, 3

an= 0 or 2, bn = 0 or 2.

Then N-1

¢(x)

L 2~:1 + 2

=

0

x

N+1

+ L 2~:1 N+l

1

c;-

=

.(~1 ) (~) ...

<

.(~1) (~) ... c;-1)

=

¢(y).

1 )

(~) (~) ...

(2)

(D (b;+1)...

(2)

The function ¢ is nondecreasing on the Cantor set. 3. We extend ¢to [0, 1], and this extension, , will be our desired function. Here is the idea: If a point is not in the Cantor set, it is in an open interval that was removed during the construction, for example, take a point in (1/3, 2/3), say, 1/2. We already know ¢(1/3) = ¢(2/3) = 1/2, that is, (1/3) = (2/3) = 1/2, and is to be nondecreasing. Thus (1/2) should be 1/2. How about .4? Oh, ( A)= 1/2 would be reasonable. Apparently, we want

[~3'3~]

___!_.

{~} [~ 2] 2 ' 9'9

___!_.

{~} 4 '

[79'98] ___!_. {3} etc. 4 '

264

APPENDIX A

A nice way to encompass these ideas and see that we have an extension of¢ is to define as: (x) =sup{ ¢(y) I y < x,y E C}. If x E C, then (x) = ¢(x). If x rf. C, then xis in an interval removed, say x E (an, bn), where an E C. Then (x) = ¢(an) = ¢(bn)· For example, 4/27 = .010222 ... (3), and .002222 ... (3) < .010222 ... (3) < .02000 ... (3) and thus (4/27) = ¢(.002222 · · · (3)) = 1/4. maps [0, 1] onto [0, 1]. 4. is nondecreasing and continuous on [0, 1]. That is nondecreasing is an immediate consequence of the property that if A is a nonempty subset of B, then sup A < sup B. As for continuity, is a nondecreasing map of [0, 1] onto [0, 1], and thus continuous by the previous propositions. maps a set of measure zero, the Cantor set, continuously onto a set of measure 1, [0, 1]. 5. We claim is differentiable, with derivative zero, almost everywhere on [0, 1].

Recall that in the construction of the Cantor set, an open interval (an, bn) was removed iff

with

ak

= 0 or 1. But,

= .a 1 ... an_ 1 100... (2) = (.(2ai) ... (2an_I)2).

Thus is constant on the intervals removed; ' is zero on the intervals removed. The function is zero almost everywhere. I

A.1

CANTOR'S SET

265

A graph of is in order: 7 8 ..

1

3

4

3

4

5 8

1

2

1

3 8

2

1 1 4 1

4

1 8 ..

8

,----,------;--+-··

0

1

9

+

2 9

--+-~- +~---

1

3

By the way, what is { }[o, 1]

2

7

3

9


8

9

q,' ?

}[o, 11

An interesting article on this material [F 1] appears in the April 1994 issue of Mathematics Magazine.

1

Appendix B

A Lebesgue Nonmeasurable Set We are nature's unique experiment to make the rational intelligence prove itself sounder than the reflex. Knowledge is our destiny. -J. Bronowski

8.1

A LEBESGUE NONMEASURABLE SET

The first example of a Lebesgue nonmeasurable set of real numbers was discovered by Vitali in 1905. In the next few years several mathematicians (Van Vleck ( 1908) and F. Bernstein ( 1908) among others) discovered such sets. All of their constructions used the Axiom of Choice: for any nonempty collection C of sets there is a choice function f such that f(A) E A for each A E C. The natural question then became: Is it possible to construct a Lebesgue nonmeasurable set of real numbers without the Axiom of Choice? In 1970 Solovay showed that the Axiom of Choice was required; uncountably many choices are needed: It is not possible to construct a Lebesgue nonmeasurable set of real numbers without using the Axiom of Choice! We now construct a Lebesgue nonmeasurable set of real numbers. This construction involves the notions of "equivalent relations" and "equivalence classes". Examples serve to illustrate the concepts involved.

Example I. Let I be the set of integers: I= { ... , -3, -2, -1, 0, 1, 2, 3, ... }. Two integers are said to be equivalent if they differ by a multiple of 266

8.1

three: We write n 1 "' n2 provided n 1 - n2 do these "classes" look like? "Starting with 0":

267

= 3 · k, k = 0, ± 1, ±2, .... What

Add and subtract multiples of three: 10

"Starting with I":

A LEBESGUE NONMEASURABLE SET

= {... ,-6,-3,0,3,6, ... }.

Add and subtract multiples of three:

II={.

0

0

'-5, -2,1,4, 7,

0

0

.}

0

Continuing,

12

= { ...

,-4,-1,2_,5,8, ... },

1_3 = { ... '

-6, -_;-}_, 0, 3, 6,.

0

.},

I 5 = { ... ,-1,2,2_,8, ... },

and so forth. We notice that

I 0 = h = I_ 30 = ... = hk = ... ,

h = I-4 = ho2 = · · · = hk+2 = ... , and that I= I0 U I 1 U I2 with I; n I1 = 0 for i -:f: j, 0 < i,j < 2. We have decomposed the set of integers, with the relation, n 1 rv n2 provided 11 1 - 11 2 = 3k, into three disjoint sets ("equivalence classes"): Every integer belongs to one of the sets I 0 , I 1, or h; ii. If 11] rv 112, In 1 = In 2 ; 111. If In 1 n In 2 -:f: 0, In 1 = In 2 • 1.

Example 2. Let Q denote the set of rational numbers: Q = {plq I p, q integers, q -:f: 0}. Two rational numbers plq,rls are said to be equivalent, p 1q rv r1s, if ps = qr. What do these "classes"

268

APPENDIX B

look like? 0, ~,

0 , ~, 1

-~2 , ... }.

"Starting with 0":

Q0; 1 =

"Starting w1"th I"·.

Q l/l = { 1' -I '2' -2'...

{

1 -1 2 -2

}

.

Continuing,

=

Q1; 2

1 -1 2 -2 3 -3 } { l'-2'4'-4'6'-6'··· '

-3 3 -6 6 } = { 7'-7' 14 '-14'""" '

Q_ 317

and so forth. We notice that Ql/2

=

Q-500/-1000 = Q3/6

Q-3/4

=

Q-300/400

=

Q33/-44

= ... ' = ... '

and that

Q

= UQp/q.

We form the so-called equivalence class:

Qpjq =

{~ E

Q I ps = rq} .

The collection {Qp/q I pjq E Q} of all such equivalence classes is a countable collection of pairwise disjoint subsets of Q whose union is Q. In fact, the reader may show: Every rational number belongs to one of the sets Qp;q; ii. If pjq rjs, Qpjq = Qr;s; 111. If Qp/q n Q,;s # 0, Qp/q = Q,;s· 1.

"-J

Example 3.

Let R be the set of real numbers. Two real numbers x, yare said to be equivalent, x y, if their difference is a rational number: x y if x - y = r, r rational. What do these "classes" look like? "-J

"-J

8.1

A LEBESGUE NONMEASURABLE SET

269

"Starting with 0": Ro = {O+r I rational}= Q, R..J'i = { J2 + r I r rational},

R 314 = {

~+r I

r rational} = R 0 = Q,

Rrr = {1r + r I r rational},

and so forth. We notice that Ro = R3/4 = ... = Rp/q = ... = Q, Rrr = Rrr+n

r

rational,

and so forth. We haveR= URx. Form the so-called equivalence classes: R" = {y E R I x - y is rational } .

Then the reader should show: Every real number belongs to one of the sets Rx; 11. If x 1 "'x2 , then Rx 1 = Rx2 ; ii. If Rx 1 n Rx2 # 0, then Rx 1 = Rx2 •

1.

In fact, the collection {Rx I x E R} of all such equivalence classes is an uncountable collection of pairwise disjoint subsets of R, in fact, each is dense in R, whose union is R. We are ready to construct a Lebesgue nonmeasurable set of real numbers in the interval ( -1, 1). Our first task is to decompose this interval in essentially the same manner as R of Example 3: For x E (-1, 1), define Ax= {y E (-1, I) I y- x = r, r rational}. For A 1; 2 = {y E (-1, I) I y- 1/2 is rational}= {rationals in example, (-1, 1)}, A_ 11 ..J2 = {y E (-1, 1) I y+ 1/v'2 is rational}, and so forth.

270

APPENDIX 8

Again, A1;2 = A_3;4 =···={rationals in (-1, 1)} and (-1, 1) = UAx, The reader may show: Each set A, is countable; 11. If xis rational, A, is the set of rationals of ( -1, 1); m. If xis irrational, then each member of Ax is an irrational number; IV. If x 1 and x 2 are members of ( -1, 1) with x 1 - x 2 a rational number, then Ax 1 = Ax2 ; v. If x 1 and x 2 are members of ( -1, 1) with x 1 - x 2 an irrational number, then Ax 1 n Ax2 = 0. VI. The collection of distinct sets Ax is uncountable. 1.

In summary, we have decomposed the interval ( -1, 1) into an uncountable collection of pairwise disjoint sets; each of these sets is itself countable, one such set consisting of the rationals in ( -1, 1), and each of the others consisting only of irrational numbers. From each of these disjoint subsets Ax, pick a point (We have used the Axiom of Choice, picking a point from each of an uncountable number of disjoints sets). Call this setN. N is an uncountable set, a subset of ( -1, 1), and N n Ax is a single point. That N is nonmeasurable will result from the countable additivity and translation invariance requirements for Lebesgue measure. Enumerate the rationals in ( -2, 2) : r 1 , r 2 , r 3 ... . Define N + rn = {x + rn I x EN, -2 < rn < 2, rn rational}, and note that since Nc(-1,1), N+r 11 C(-3,3). We claim (N+rn)n(N+ rm) = 0 if r11 # rm: Suppose the intersection is nonempty; z E (N + rn) n (N + rm)· Then z = x + r11 = y + rm, or x- y is rational, with x, yEN. But this means that x andy are in the same equivalence class. However, N was constructed by taking points from mutually disjoint sets, so x must equal y. But then r 11 = r m• and this is a contradiction to r11 # 'm· So, the collection {N + r11 } forms a countable, mutually disjoint cover of ( -1, 1), each member of which is a subset of (-3, 3): ()C

(-1, 1) C U(N +rn) C (-3,3). n=l

If N is measurable, N + r 11 is measurable, and because Lebesgue measure IS translation invariant (1-l(N) = 1-1(N + rn) ), and countably additive

8.1

A LEBESGUE NONMEASURABLE SET

271

we have

2

~I'( (-I, 1)) <

I'(Q(N + r")) ~ ~J•(N + r,) ~ ~J•(N)

<JL((-3,3))=6. So, X

2 < LJL(N) < 6. n=i

The left-hand inequality, 2 < L-~ 1 JL(N), implies JL(N) > 0. The right-hand inequality, L- ~ 1 JL(N) < 6 implies JL(N) = 0. We can't have it both ways. The set N must be nonmeasurable. Summary of Construction:

I.

Partition (-I, I) into an uncountable collection of mutually disjoint countable sets; II. Use the Axiom of Choice to pick a point from each of these sets. This set N is the set we show to be nonmeasurable; III. Form a countable collection of mutually disjoint translates of N : {N + r n I r 1, r 2 , . .. an enumeration of the rational numbers of ( -2, 2)}; IV. Observe (-I, I) c u(N +rm) c (-3,3). Because Lebesgue measure is translation invariant and countably additive, measurability of N implies JL(N) = 0 and JL(N) > 0. What if A is any set of positive Lebesgue measure: JL(A) > 0? (Every subset of a set of Lebesgue measure zero is Lebesgue measurable and has Lebesgue measure zero.) Since 00

A c U(-n,n), JL(A n (-N,N)) > 0 n=i

for some natural number N, and thus An ( -N, N) must be uncountable since any countable set is Lebesgue measurable with measure zero. Select any countable subset of A n (- N, N). This countable set will play the role of the "rationals" in the proceeding construction. Mimic the previous arguments to establish that every set of real numbers with positive Lebesgue measure contains a Lebesgue nonmeasurable subset.

272

APPENDIX B

Over the years attempts having been made to enlarge the class of Lebesgue measurable sets [SS]. The "idea" would be to find a measure >. so that 1. >.(E) defined for all E c R (all subsets of Rare "measurable"); 2. >.([0, 1]) = 1; 3.

>-CQ En) =~>.(En), En mutually disjoint (countable additivity);

4. >.(E +a) =>.(E) if E measurable (translation invariant). As has been shown, Lebesgue measure satisfies parts 2,3, and 4, with requirements 3 and 4 forcing abandonment of 1. How about replacing 3 with

finite additivity, in an attempt to "measure" all subsets of R? We have the following results [Ba]. There exists a set function >.defined for all subsets of R such that 1'. 2'. 3'. 4'.

0 <>.(E) < oo for all E c R >.([0, 1]) = 1 >. is finitely additive >. is translation invariant.

Then Banach, Hausdorff showed the above result holds in Rn iff n = 1, 2. In 1933 Kolmogorov [Ko] developed probability theory from an axiomatic standpoint with countable additivity being of fundamental importance: We want countable additivity, even at the expense of not being able to "measure" all sets of real numbers.

Appendix C

Lebesgue, Not Borel

The unexamined life is not worth living. -Socrates

C.1

LEBESGUE, NOT BOREL

We show that there exist Lebesgue measurable sets that are not Borel sets: the class of Borel sets, B, is properly contained in the class of Lebesgue measurable sets, M. We will use the PROPOSITION

C.l Iff is a continuous, strictly increasing mapping of R

onto R, then f maps Borel sets to Borel sets. Proof The argument is indirect. Let C be the collection of all subsets of R whose image under f is a Borel set. That is, if A E C, then f(A) is a Borel set. We will show that this collection C is a sigma algebra that contains the closed intervals, i.e., B. We claim Cis a sigma algebra of subsets of R.

Suppose A E C. We claim AcE C. Since f is strictly increasing, f is 1 - 1, and consequently f(Ac) nj(A) = 0. Thus R = f(R) = f(A U A c)= f(A) Uf(Ac), that 273

?74

APPENDIX C

is, R- f(A) = f(Ac). Because A E C, f(A) is a Borel set. Thus R- f(A) = f(Ac) is a Borel set, that is, Ac E C. Next, suppose (Ak) is a sequence of sets in C. Since f(UAk) = Uj(Ak), and each f(Ak) is a Borel set (since by assumption, Ak E C), the countable union of Borel sets is a Borel set, that is, f(UAk) is a Borel set. In other words, UAk E C. Cis a sigma algebra. We now show closed intervals are in C. Suppose I is a closed interval. Then f(I) is an closed interval: f([a, b]) = [f(a),f(b)J (Intermediate Value Theorem, increasing). Thus C is a sigma algebra containing all closed intervals. Because 8 is the smallest such sigma algebra, 8 c C. Borel sets are mapped to Borel sets. We are ready to prove the

C.l There is a Lebesgue measurable set of real numbers that is not a Borel set.

THEOREM

Proof Let