EXISTENCE OF THE INTEGRAL

The following lemma is the starting point for our study of the integrability of a bounded functionf on a closed intervalŒa; b.

Lemma 3.2.1

Suppose that

jf .x/j M; axb; (1)

and letP⁰be a partition ofŒa; bobtained by addingrpoints to a partitionP D fx0; x1; : : : ; xng ofŒa; b:Then

S.P /S.P⁰/S.P / 2M rkPk (2)

and

s.P /s.P⁰/s.P /C2M rkPk: (3)

Proof

We will prove (2) and leave the proof of (3) to you (Exercise 1). First suppose thatr D1, soP⁰is obtained by adding one pointcto the partitionP D fx0; x1; : : : ; xng; thenxi 1 < c < xi for somei inf1; 2; : : : ; ng. Ifj ¤ i, the productMj.xj xj 1/ appears in bothS.P /andS.P⁰/and cancels out of the differenceS.P / S.P⁰/. Therefore, if

Mi1D sup

xi 1xc

f .x/ and Mi 2D sup

cxxi

f .x/;

then

S.P / S.P⁰/DMi.xi xi 1/ Mi1.c xi 1/ Mi 2.xi c/

D.Mi Mi1/.c xi 1/C.Mi Mi 2/.xi c/: (4) Since (1) implies that

0Mi Mi r 2M; rD1; 2;

(4) implies that

0S.P / S.P⁰/2M.xi xi 1/2MkPk: This proves (2) forrD1.

Now suppose thatr > 1andP⁰is obtained by adding pointsc1,c2, . . . ,cr toP. Let P^.0/ D P and, forj 1, let P^{.j /} be the partition ofŒa; bobtained by addingcj to P^{.j 1/}. Then the result just proved implies that

0S.P^{.j 1/}/ S.P^{.j /}/2MkP^{.j 1/}k; 1j r:

Adding these inequalities and taking account of cancellations yields

0S.P^.0// S.P^{.r /}/2M.kP^.0/k C kP^.1/k C C kP^{.r 1/}k/: (5) SinceP^.0/DP,P^{.r /}DP⁰, andkP^.k/k kP^{.k 1/}kfor1kr 1, (5) implies that

0S.P / S.P⁰/2M rkPk; which is equivalent to (2).

Theorem 3.2.2

Iff is bounded onŒa; b;then Z b

Proof

Suppose thatP1 andP2 are partitions ofŒa; bandP⁰is a refinement of both.

LettingP DP1in (3) andP DP2in (2) shows that

s.P1/s.P⁰/ and S.P⁰/S.P2/:

Sinces.P⁰/S.P⁰/, this implies thats.P1/S.P2/. Thus, every lower sum is a lower bound for the set of all upper sums. SinceRb

af .x/ dxis the infimum of this set, it follows of all lower sums. SinceRb

af .x/ dxis the supremum of this set, this implies (6).

Theorem 3.2.3

Iff is integrable onŒa; b;then Z b

Proof

We prove thatRb

af .x/ dxDRb

the triangle inequality implies that

Now suppose that > 0. From Definition 3.1.3, there is a partitionP0ofŒa; bsuch that Z b

From Definition 3.1.1, there is aı > 0such that ˇˇˇ by Lemma 3.2.1, (8) implies that

Z b

for every Riemann sum off overP. SinceS.P / is the supremum of these Riemann sums (Theorem 3.1.4), we may chooseso that

jS.P / j< Sinceis an arbitrary positive number, it follows that

Z b

Lemma 3.2.4

Iff is bounded onŒa; band > 0;there is aı > 0such that Z b

a

f .x/ dxS.P / <

Z b

a

f .x/ dxC (12)

and Z b

a

f .x/ dxs.P / >

Z b

a

f .x/ dx ifkPk< ı.

Proof

We show that (12) holds ifkPkis sufficiently small, and leave the rest of the proof to you (Exercise 3).

The first inequality in (12) follows immediately from Definition 3.1.3. To establish the second inequality, suppose thatjf .x/j Kifax b. From Definition 3.1.3, there is a partitionP0D fx0; x1; : : : ; xrC1gofŒa; bsuch that

S.P0/ <

Z b

a

f .x/ dxC

2: (13)

IfP is any partition ofŒa; b, letP⁰be constructed from the partition points ofP0andP. Then

S.P⁰/S.P0/; (14)

by Lemma 3.2.1. SinceP⁰ is obtained by adding at most r points toP, Lemma 3.2.1 implies that

S.P⁰/S.P / 2KrkPk: (15)

Now (13), (14), and (15) imply that

S.P /S.P⁰/C2KrkPk S.P0/C2KrkPk

<

Z b

a

f .x/ dxC

2 C2KrkPk: Therefore, (12) holds if

kPk< ıD 4Kr:

Theorem 3.2.5

Iff is bounded onŒa; band

Z b

a

f .x/ dx D Z b

a

f .x/ dxDL; (16)

thenf is integrable onŒa; band Z b

a

f .x/ dxDL: (17)

Proof

If > 0, there is aı > 0such that Z b

a

f .x/ dx < s.P /S.P / <

Z b

a

f .x/ dxC (18)

ifkPk< ı(Lemma 3.2.4). Ifis a Riemann sum off overP, then s.P /S.P /;

so (16) and (18) imply that

L < < LC ifkPk< ı. Now Definition 3.1.1 implies (17).

Theorems 3.2.3 and 3.2.5 imply the following theorem.

Theorem 3.2.6

A bounded functionf is integrable onŒa; bif and only if Z b

a

f .x/ dx D Z b

a

f .x/ dx:

The next theorem translates this into a test that can be conveniently applied.

Theorem 3.2.7

Iff is bounded onŒa; b;thenf is integrable onŒa; bif and only if for each > 0there is a partitionP ofŒa; bfor which

S.P / s.P / < : (19)

Proof

We leave it to you (Exercise 4) to show that ifRb

a f .x/ dxexists, then (19) holds forkPksufficiently small. This implies that the stated condition is necessary for integra-bility. To show that it is sufficient, we observe that since

s.P / Z b

a

f .x/ dx Z b

a

f .x/ dxS.P / for allP, (19) implies that

0 Z b

a

f .x/ dx Z b

a

f .x/ dx < : Sincecan be any positive number, this implies that

Z b

a

f .x/ dx D Z b

a

f .x/ dx:

Therefore,Rb

a f .x/ dxexists, by Theorem 3.2.5.

The next two theorems are important applications of Theorem 3.2.7.

Theorem 3.2.8

Iff is continuous onŒa; b;thenf is integrable onŒa; b.

Proof

LetP D fx0; x1; : : : ; xngbe a partition ofŒa; b. Sincef is continuous onŒa; b, there are pointscj andc_j⁰ inŒxj 1; xjsuch that

f .cj/DMj D sup

xj 1xxj

f .x/

and

f .c_j⁰/Dmj D inf

xj 1xxj

f .x/

(Theorem 2.2.9). Therefore, S.P / s.P /D

Xn

jD1

f .cj/ f .c⁰_j/

.xj xj 1/: (20)

Sincef is uniformly continuous onŒa; b(Theorem 2.2.12), there is for each > 0aı > 0 such that

jf .x⁰/ f .x/j<

b a

ifxandx⁰are inŒa; bandjx x⁰j< ı. IfkPk< ı, thenjcj c⁰_jj< ıand, from (20), S.P / s.P / <

b a

Xn

jD1

.xj xj 1/D: Hence,f is integrable onŒa; b, by Theorem 3.2.7.

Theorem 3.2.9

Iff is monotonic onŒa; b;thenf is integrable onŒa; b.

Proof

LetP D fx0; x1; : : : ; xngbe a partition ofŒa; b. Sincef is nondecreasing, f .xj/DMj D sup

xj 1xxj

f .x/

and

f .xj 1/Dmj D inf

xj 1xxj

f .x/:

Hence,

S.P / s.P /D Xn

jD1

.f .xj/ f .xj 1//.xj xj 1/:

Since0 < xj xj 1 kPkandf .xj/ f .xj 1/0, S.P / s.P / kPk

Xn

jD1

.f .xj/ f .xj 1//

D kPk.f .b/ f .a//:

Therefore,

S.P / s.P / < if kPk.f .b/ f .a// < ; sof is integrable onŒa; b, by Theorem 3.2.7.

The proof for nonincreasingf is similar.

We will also use Theorem 3.2.7 in the next section to establish properties of the integral.

In Section 3.5 we will study more general conditions for integrability.

3.2 Exercises

1.

Complete the proof of Lemma 3.2.1 by verifying Eqn. (3).

2.

Show that iff is integrable onŒa; b, then sufficiently small. HINT:Use Theorem3:1:4:

5.

Suppose thatf is integrable andgis bounded onŒa; b, andgdiffers fromf only at points in a setHwith the following property: For each > 0,H can be covered by a finite number of closed subintervals ofŒa; b, the sum of whose lengths is less than. Show thatgis integrable onŒa; band that

Z b

HINT:Use Exercise3:1:3:

6.

Suppose thatgis bounded onŒ˛; ˇ, and letQW˛Dv0 < v1 < < vLDˇbe a fixed partition ofŒ˛; ˇ. Prove:

(a)

7.

A functionf isof bounded variation on Œa; bif there is a numberKsuch that Xn

jD1

ˇˇf .aj/ f .aj 1/ˇˇK

whenevera Da0 < a1 < < an Db. (The smallest number with this property is thetotal variation off onŒa; b.)

(a)

Prove: Iff is of bounded variation onŒa; b, thenf is bounded onŒa; b.

(b)

Prove: Iff is of bounded variation onŒa; b, thenf is integrable onŒa; b.

HINT:Use Theorems3:1:4and3:2:7:

8.

LetP D fx0; x1; : : : ; xngbe a partition ofŒa; b,c0 D x0 D a,cnC1 DxnD b, andxj 1cj xj,j D1,2, . . . ,n. Verify that

Xn

jD1

g.cj/Œf .xj/ f .xj 1/Dg.b/f .b/ g.a/f .a/

Xn

jD0

f .xj/Œg.cjC1/ g.cj/:

Use this to prove that ifRb

a f .x/ dg.x/exists, then so doesRb

a g.x/ df .x/, and Z b

a

g.x/ df .x/Df .b/g.b/ f .a/g.a/

Z b

a

f .x/ dg.x/:

(This is theintegration by parts formulafor Riemann–Stieltjes integrals.)

9.

Letf be continuous andgbe of bounded variation (Exercise 7) onŒa; b.

(a)

Show that if > 0, there is aı > 0such thatj ⁰j < =2if and⁰ are Riemann–Stieltjes sums off with respect togover partitionsP andP⁰ ofŒa; b, whereP⁰ is a refinement ofP and kPk < ı. HINT:Use Theo-rem2:2:12:

(b)

Letı be as chosen in

(a)

. Suppose that1 and2 are Riemann–Stieltjes sums off with respect togover any partitionsP1andP2ofŒa; bwith norm less thanı. Show thatj1 2j< .

(c)

Ifı > 0, letL.ı/be the supremum of all Riemann–Stieltjes sums off with respect togover partitions ofŒa; bwith norms less thanı. Show thatL.ı/is finite. Then show thatLDlimı!0CL.ı/exists. HINT:Use Theorem2:1:9:

(d)

Show thatRb

a f .x/ dg.x/DL.

10.

Show thatRb

a f .x/ dg.x/exists iff is of bounded variation andgis continuous on Œa; b. HINT:See Exercises8and9:

Im Dokument INTRODUCTION TO REAL ANALYSIS (Seite 136-143)