RiemannStieltjes Integration

8/14/2019 RiemannStieltjes Integration

1/50

Chapter 7

Riemann-Stieltjes Integration

Calculus provides us with tools to study nicely behaved phenomena using small

discrete increments for information collection. The general idea is to (intelligently)

connect information obtained from examination of a phenomenon over a lot of tiny

discrete increments of some related quantity to close in on or approximate some-

thing that behaves in a controlled (i.e., bounded, continuous, etc.) way. The clos-

ing in on approach is useful only if we can get back to information concerning the

phenomena that was originally under study. The benet of this approach is most

beautifully illustrated with the elementary theory of integral calculus overU. It en-ables us to adapt some limiting formulas that relate quantities of physical interest

to study more realistic situations involving the quantities.

Consider three formulas that are encountered frequently in most standard phys-ical science and physics classes at the pre-college level:

A l * d r t m d l.Use the space that is provided to indicate what you know about these formulas.

Our use of these formulas is limited to situations where the quantities on theright are constant. The minute that we are given a shape that is not rectangular,

a velocity that varies as a function of time, or a density that is determined by our

position in (or on) an object, at rst, we appear to be out of luck. However, when

the quantities given are well enough behaved, we can obtain bounds on what we

275


2/50


3/50

7.1. RIEMANN SUMS AND INTEGRABILITY 277

subinterval, yields an estimate of

25 ft/min 15 min tt

25 94

uft/min

u 15 min 573

8ft.

Excursion 7.0.1 Find the estimate for the distance travelled taking increments of

one minute (which is not small for the purposes of calculus) and using the minimum

velocity achieved in each subinterval as the estimating velocity.

***Hopefully, you obtained 61 feet.***

Notice that none of the work done actually gave us the answer to the original

problem. Using Calculus, we can develop the appropriate tools to solve the problem

as an appropriate limit. This motivates the development of the very important and

useful theory of integration. We start with some formal denitions that enable us to

carry the closing in on process to its logical conclusion.

7.1 Riemann Sums and Integrability

Denition 7.1.1 Given a closed interval I [a b], a partition of I is any nitestrictly increasing sequence of points S x0x1 xn1xn such that a x0and b xn. The mesh of the partition x0x1 xn1xn is dened by

meshS max1njnn

bxj xj1

c

Each partition of I, x0x1 xn1xn, decomposes I into n subintervals Ij dxj1xj

e, j 1 2 n, such that Ij D Ik xj if and only if k j 1 and is

empty for k / j or k/ j 1. Each such decomposition of I into subintervals iscalled a subdivision ofI.

Notation 7.1.2 Given a partition S x0x1 xn1xn of an interval I [a b], the two notations xj and(

bIjc

will be used forbxj xj1

c, the length of

the j th subinterval in the partition. The symbol or I will be used to denote

an arbitrary subdivision of an interval I .


4/50

278 CHAPTER 7. RIEMANN-STIELTJES INTEGRATION

If f is a function whose domain contains the closed interval I and f is bounded

on the interval I, we know that f has both a least upper bound and a greatest lower

bound on I as well as on each interval of any subdivision of I.

Denition 7.1.3 Given a function f that is bounded and dened on the interval

I and a partition S x0x1 xn1xn of I, let I j dxj1xj

e, Mj

supx+Ij

f x and mj infx+Ij

f x for j 1 2 n. Then the upper Riemann sum off with respect to the partition S, denoted by US f, is dened by

U S f n

;j1Mj xj

and the lower Riemann sum of f with respect to the partition S , denoted by

L S f, is dened by

L S f n;

j1mj xj

where xj bxj xj1

c.

Notation 7.1.4 With the subdivision notation the upper and lower Riemann sums

for f are denoted by U f and L f, respectively.

Example 7.1.5 For f x 2x 1 in I [0 1] andS |

01

4

1

2

3

4 1

},

US f 14

t3

2 2 5

2 3

u 9

4and L S f 1

4

t1 3

2 2 5

2

u 7

4.

Example 7.1.6 For g x

0 , for x + TD [0 2]

1 , for x + TD [0 2]U I g 2 and L I g 0 for any subdivision of[0 2].To build on the motivation that constructed some Riemann sums to estimate a

distance travelled, we want to introduce the idea of rening or adding points to

partitions in an attempt to obtain better estimates.


5/50


Denition 7.1.7 For a partition Sk x0x1 xk1xk of an interval I [a

b] , let

kdenote to corresponding subdivision of [a

b]. If S

nandS

mare

partitions of[a b] having n 1 and m 1 points, respectively, andSn t Sm, thenSm is a renement ofSn orm is a renement ofn. If the partitions Sn andSmare independently chosen, then the partition Sn C Sm is a common renement ofSn andSm and the resulting Sn C Sm is called a common renement of n

andm .

Excursion 7.1.8 LetS|

01

2

3

4 1

}andS

|0

1

4

1

3

1

2

5

8

3

4 1

}.

(a) If and ` are the subdivisions of I [0 1] that correspondS andS ,

respectively, then |v0 12w v12 34w v34 1w}. Find`.

(b) Set I1 v

01

2

w, I2

v1

2

3

4

w, and I3

v3

4 1

w. For k 1 2 3, let k be

the subdivision of Ik

that consists of all the elements of`

that are contained

in Ik. Find k for k 1 2 and3.

(c) For f x x 2 and the notation established in parts (a) and (b), nd each ofthe following.

(i) m infx+I

f x


6/50


(ii) mj infx+Ij

f x for j 1 2 3

(iii) m`j inf|

infx+J

f x : J + j}

(iv) M supx+I

f x

(v) Mj supx+Ij

f x for j 1 2 3

(vi) Mj sup|

supx+J

f x : J + j}

(d) Note how the values m, mj, m j , M, Mj , and Mj compare. What you ob-

served is a special case of the general situation. Let

S x0 ax1xn1xn bbe a partition of an interval I [a b], be the corresponding subdivisionof[a b] andS denote a renement ofSwith corresponding subdivision de-

noted by `. For k 1 2 n, let k be the subdivision of Ik consistingof the elements of ` that are contained in Ik. Justify each of the followingclaims for any function that is dened and bounded on I .

(i) If m

inf

x+If x and m j

inf

x+Ijf x , then, for j

1 2 n, m

nmj

and mj n infx+J

f x for J + j.


7/50


(ii) If M supx

+I

f x and Mj supx

+Ij

f x, then, for j 1 2 n, Mj n

M and Mj o supx+J

f x for J + j .

Our next result relates the Riemann sums taken over various subdivisions of an

interval.

Lemma 7.1.9 Suppose that f is a bounded function with domain I [a b]. Let be a subdivision of I, M sup

x+If x, and m inf

x+If x. Then

m b a n L f n U f n Mb a (7.1)

and

L f n L b` fc n Ub` fc n U f (7.2)for any renement ` of. Furthermore, if < andD are any two subdivisionsof I, then

L

b< f

cn UD f (7.3)

Excursion 7.1.10 Fill in what is missing to complete the following proofs.

Proof. Suppose that f is a bounded function with domain I [a b], M supx+I

f x, and m infx+I

f x. For Ik : k 1 2 n an arbitrary subdi-vision of I, let Mj sup

x+Ijf x and mj inf

x+Ijf x. Then Ij t I for each

j 1 2 n, we have that

m n mj n1

, for each j 1 2 n.

Because xj bxj xj1c o 0 for each j 1 2 n, it follows immediatelythat

2

mn;

j1

bxj xj1

c n n;j1

mj xj L f


8/50


and

n;j1

mj xj n n;j1

Mj xj U f n3

Mb a .

Therefore, m b a n L f n U f n Mb a as claimed in equation(7.1).

Let ` be a renement of and, for each k 1 2 n, let k be thesubdivision of Ik that consists of all the elements of

` that are contained in Ik.In view of the established conventions for the notation being used, we know that

1J J + ` " 2!k k + 1 2 n F J + k also, for each J + k,J

tIk

"mk

inf

x+Ikf x

ninf

x+Jf x and Mk

sup

x+Ikf x

osupx+J

f x. Thus,

mk( Ik n L k f and M k( Ik o U k f

from which it follows that

L f n;

j1mj (

bIjc n n;

j1L j f L b` fc

and

U f 4

on;

j1 U j f 5 .

From equation (7.1), L ` f n U ` f. Finally, combining the inequalitiesyields that

L f n L b` fc n Ub` fc n U fwhich completes the proof of equation (7.2).

Suppose that < and D are two subdivisions of I. Then < C D is

6

< and D. Because is a renement of< , by the

comparison of lower sums given in equation (7.2), Lb

< fc n L f. On the

other hand, from being a renement ofD, it follows that7

.

Combining the inequalities with equation (7.1) leads to equation (7.3).


9/50


***Acceptable responses are: (1) Mj n M, (2) m b a, (3) M3nj1 bxj xj1c,

(4)3n

j1 Mj (bIjc

, (5) U` f, (6) the common renement of, and(7) U f n U f D.***

If f is a bounded function with domain I [a b] and , ,[a b] is theset of all partitions of [a b], then the Lemma assures us that L f : + , isbounded above by b a sup

x+If x and U f : + , is bounded below by

b a infx+I

f x. Hence, by the least upper bound and greatest lower bound prop -

erties of the reals both sup L f : + , and infU f : + , exist tosee that they need not be equal, note thatfor the bounded function g given in Exam-

ple 7.1we have that sup L g : + , 0 while infU g : + , 2.

Denition 7.1.11 Suppose that f is a function onU that is dened and bounded onthe interval I [a b] and, ,[a b] is the set of all partitions of [a b]. Thenthe upper Riemann integraland the lower Riemann integralare dened by

=ba

f x dx inf+, U S f and=b

af x d x sup

+,L S f,

respectively. If5b

af x d x 5b

af x dx, then f is Riemann integrable , or just

integrable, on I , and the common value of the integral is denoted by

=ba

f x d x.

Excursion 7.1.12 Let f x

5x

3 , for x + T

0 , for x + T.

For each n + M , letn denote the subdivision of the interval [1 2] that con-sists of n segments of equal length. Use n : n + M to nd an upper bound for


10/50


521 f x dx. [Hint: Recall that3nk1 k

n n 12

.]

***Corresponding to each n you needed to nd a useful form for U n f. Your

work should have led you to a sequence for which the limit exists as n *. Forn + M, the partition that gives the desired n is

|1 1 1

n 1 2

n 1 n

n

}. Then

n I1 I2 In with Ij v

1 j 1n

1 jn

wand Mj 8

5j

nleads to

Un f 21

2 5

2n. Therefore, you should proved that

52

1 f x d x n21

2.***

It is a rather short jump from Lemma 7.1.9 to upper and lower bounds on theRiemann integrals. They are given by the next theorem.

Theorem 7.1.13 Suppose that f is dened on the interval I [a b] and m nf x n M for all x + I. Then

m b a n=b

a

f x d x n=

b

a

f x d x n Mb a . (7.4)

Furthermore, if f is Riemann integrable on I , then

m b a n=

b

a

f x d x n Mb a . (7.5)

Proof. Since equation (7.5), is an immediate consequence of the denition of

the Riemann integral, we will prove only equation (7.4). Let G denote the set of all

subdivisions of the interval [a b]. By Lemma 7.1.9, we have that, for `, + G,m b a n L bf `c n U f n Mb a .


11/50


Since is arbitrary,

m b a n L bf `c n inf+&

U f

and inf+&

U f n Mb a i.e.,

m b a n L bf `c n =ba

f x d x n Mb a .

Because ` is also arbitrary, m b a n sup +&

L f ` and sup +&

L f ` n

5ba

fx

d x

i.e.,

m b a n=b

a

f x d x n=

b

a

f x d x .

Combining the inequalities leads to equation (7.4).

Before getting into some of the general properties of upper and lower integrals,

we are going to make a slight transfer to a more general set -up. A re-examination

of the proof of Lemma 7.1.9 reveals that it relied only upon independent application

of properties of inmums and supremums in conjunction with the fact that, for any

partition x0x1 xn1xn, xj xj1 0 andn3

j1bxj xj1c xn x0. Now,

given any function : that is dened and strictly increasing on an interval [a b], for

any partition S a x0x1xn1xn b of [a b],

: S : a : x0 : x1 : xn1 : xn : b t : [a b] ,

:bxjc: bxj1c 0 and n3

j1

b:bxjc : bxj1cc : b : a. Consequently,

: S is a partition of

[: a : b] ? I : I [c d] F : S t I ,which is the smallest interval that contains : [a b]. The case : t t returnsus to the set-up for Riemann sums on the other hand, : [a b] need not be an

interval because : need not be continuous.


12/50


Example 7.1.14 Let I [0 3] and : t t2 JtK. Then : I [0 1 C

[2 5 C [6 11 C 12. For the partition S |0 12 1 54 2 83 3} of I, : S |0

1

4 2

41

16 6

82

9 12

}is a partition of[0 12] which contains : I.

Denition 7.1.15 Given a function f that is bounded and dened on the closed

interval I [a b], a function : that is dened and monotonically increasing onI , and a partition S x0x1 xn1xn of I with corresponding subdivision , let Mj sup

x+Ijf x and mj inf

x+Ijf x , for Ij

dxj1xj

e. Then the upper

Riemann-Stieltjes sum of f over : with respect to the partition S , denoted by

US f : or U f :, is dened by

US f : n;

j1Mj :j

and the lower Riemann-Stieltjes sum of f over : with respect to the partition S,

denoted by L S f : or L f :, is dened by

L S f : n;

j1mj :j

where :j b:

bxjc:

bxj1cc.Replacing xj with :

bxjc

in the proof of Lemma 7.1.9 and Theorem 7.1.13

yields the analogous results for Riemann-Stieltjes sums.

Lemma 7.1.16 Suppose that f is a bounded function with domain I [a b] and:is a function that is dened and monotonically increasing on I . LetSbe a partition

of I , M supx+I

f x, and m infx+I

f x. Then

m : b : a n L S f : n U S f : n M: b : a (7.6)and

L S

f : n L bS` f :c n UbS` f :c n U S f : (7.7)for any renementS ofS. Furthermore, if < andD are any two subdivisions

of I, then

Lb

< f :c n U D f : (7.8)


13/50


The bounds given by Lemma 7.1.16 with the greatest lower and least upper

bound properties of the reals the following denition.

Denition 7.1.17 Suppose that f is a function on U that is dened and boundedon the interval I [a b], , ,[a b] is the set of all partitions of [a b] , and: is a function that is dened and monotonically increasing on I . Then the upper

Riemann-Stieltjes integraland the lower Riemann-Stieltjes integralare dened by

=ba

f x d: x inf+,

U S f : and

=ba

f x d: x sup+,

L S f :,

respectively. If5ba f x d: x 5b

af x d: x , then f is Riemann-Stieltjes in-

tegrable , orintegrable with respect to : in the Riemann sense , on I, and the

common value of the integral is denoted by

=ba

f x d: x or

=ba

f d:.

Denition 7.1.18 Suppose that: is a function that is dened and monotonically in-

creasing on the interval I [a b]. Then the set of all functions that are integrablewith respect to : in the Riemann sense is denoted by 4 :.

Because the proof is essentially the same as what was done for the Riemann

upper and lower integrals, we offer the following theorem without proof.

Theorem 7.1.19 Suppose that f is a bounded function with domain I [a b], : isa function that is dened and monotonically increasing on I , and m n f x n M

for all x + I. Then

m : b : a n=b

a

f d: n=

b

a

f d: n M: b : a . (7.9)

Furthermore, if f is Riemann-Stieltjes integrable on I , then

m : b : a n =b

a

f x d: x n M: b : a . (7.10)

In elementary Calculus, we restricted our study to Riemann integrals of con -

tinuous functions. Even there we either glossed over the stringent requirement of

needing to check all possible partitions or limited ourselves to functions where some

trick could be used. Depending on how rigorous your course was, some examples of


14/50


nding the integral from the denition might have been based on taking partitions

of equal length and using some summation formulas (like was done in Excursion

7.1.12) or might have made use of a special bounding lemma that applied to x n for

each n + M.It is not worth our while to grind out some tedious processes in order to show

that special functions are integrable. Integrability will only be a useful concept if it

is veriable with a reasonable amount of effort. Towards this end, we want to seek

some properties of functions that would guarantee integrability.

Theorem 7.1.20 (Integrability Criterion) Suppose that f is a function that is bounded

on an interval I [a b] and: is monotonically increasing on I . Then f + 4 :on I if and only if for every > 0 there exists a partitionSof I such that

U S f : L S f : > (7.11)

Excursion 7.1.21 Fill in what is missing to complete the following proof.

Proof. Let f be a function that is bounded on an interval I [a b] and : bemonotonically increasing on I.

Suppose that for every > 0 there exists a partition Sof I such that

U S f : L S f : >. (*)From the denition of the Riemann-Stieltjes integral and Lemma 7.1.16, we have

that

L S f : n=b

a

f x d: x n1

n2

.

It follows immediately from (*) that

0n =

b

a

fx

d

: x

=b

a

fx

d

: x

.

Since was arbitrary and the upper and lower Riemann Stieltjes integrals are con-

stants, we conclude that5

b

af x d: x 5ba f x d: x i.e.,

3

.


15/50


Conversely, suppose that f + 4 : and let 0 be given. For ,

,[a b] the set of all partitions of [a b],5b

a f x d: x inf+, US f : and5ba

f x d: x sup+,

L S f :. Thus,

2 0 implies that there exists a S1 +

,[a b] such that5b

af x d: x U S1 f :

5ba

f x d: x 2

and there

exists S2 + ,[a b] such that5b

af x d: x

24

.

Therefore,

US1 f : =b

a

f x d: x

2and =

b

a

f x d: x L S2 f :

2.

(**)

Let S be the common renement ofS1 and S2. Lemma 7.1.16, equation (7.7)

applied to (**) yields that

5

=b

a

f x d: x

2and

=ba

f x d: x 6

2.

Thus

US f : L S f:r

U S f : 5ba

f x d: xs

7

.

***Acceptable responses are: (1)5

b

af x d: x, (2) U S f :, (3) f + 4 :,

(4) L S2 f : 5b

af x d: x, (5) US f :, (6) L S f :, and (7)5b

af x d: x L S f :.***Theorem 7.1.20 will be useful to us whenever we have a way of closing the gap

between functional values on the same intervals. The corollaries give us two big

classes of integrable functions.

Corollary 7.1.22 If f is a function that is continuous on the interval I [a b],then f is Riemann-Stieltjes integrable on [a b].


16/50


Proof. Let : be monotonically increasing on I and f be continuous on I. Sup-

pose that

0 is given. Then there exists an@

0 such that [:

b :

a

]@

.

By the Uniform Continuity Theorem, f is uniformly continuous in [a b] from

which it follows that there exists a = 0 such that

1u 1) du )+ I F u ) =" f u f ) e .Let S x0 ax1 xn1xn b be a partition of [a b] for which meshS=and, for each j , j 1 2 n, set Mj sup

xj

1nxnxjf x and mj inf

xj

1nxnxjf x.

Then Mj mj n @ and

US f : L S f : n;j1

bMj mj

c:j n @ n

;j1

:j @ [: b : a] .

Since 0 was arbitrary, we have that

1 0 " 2S S+ ,[a b] F US f : L S f : .

In view of the Integrability Criterion, f + 4 :. Because : was arbitrary, weconclude that f is Riemann-Stieltjes Integrable (with respect to any monotonically

increasing function on [a b]).

Corollary 7.1.23 If f is a function that is monotonic on the interval I [a b] and: is continuous and monotonically increasing on I , then f + 4:.

Proof. Suppose that f is a function that is monotonic on the interval I [a b]and : is continuous and monotonically increasing on I. For 0 given, let n + M,be such that

: b

: a

f b

f a

n.

Because : is continuous and monotonically increasing, we can choose a partition

S x0 ax1 xn1xn b of [a b] such that :j b

:bxjc : bxj1cc

: b : an

. If f is monotonically increasing in I, then, for each j + 1 2 n,


17/50


Mj supxj

1

nx

nxj

f x f bxjc and mj infxj 1nxnxj f x f bxj1c and

US f : L S f : n;

j1

bMj mj

c:j

: b : an

n;j1

bfbxjc f bxj1cc

: b : an

f b f a

while f monotonically decreasing yields that Mj f bxj1c, mj f bxjc andUS f : L S f : : b : a

n

n;j1

bfbxj1

c f bxjcc

: b : an

f a f b .

Since 0 was arbitrary, we have that

1 0 " 2S S+ ,[a b] F U S f : L S f : .

In view of the Integrability Criterion, f + 4:.Corollary 7.1.24 Suppose that f is bounded on [a b] , f has only nitely many

points of discontinuity in I [a b], and that the monotonically increasing function: is continuous at each point of discontinuity of f . Then f + 4 :.

Proof. Let 0 be given. Suppose that f is bounded on [a b] and continuous

on [a b] E where E ?1 ?2?k is the nonempty nite set of points ofdiscontinuity of f in [a b]. Suppose further that : is a monotonically increasing

function on [a b] that is continuous at each element of E. Because E is nite and

: is continuous at each ?j+

E, we can nd k pairwise disjoint intervals duj )je,j 1 2 k, such that

E tk>

j1

duj )j

e+ [a b] and

k;j1

b:b

)jc : bujcc `


18/50


for any ` 0 furthermore, the intervals can be chosen in such a way that eachpoint

?m +E

D a

b

is an element of the interior of the corresponding interval,

[um )m]. Let

K [a b] k>

j1

buj )j

c.

Then K is compact and f continuous on K implies that f is uniformly continuous

there. Thus, corresponding to ` 0, there exists a = 0 such that

1s 1t bs t + K F s t =" f s f t `c .Now, let S x0 ax1xn1xn b be a partition of [a b] satis-

fying the following conditions:

1j bj + 1 2 k " uj + SF )j + Sc, 1j bj + 1 2 k " buj )jc D S 3c, and 1p 1j dbp + 1 2 n F j + 1 2 k F xp1 / ujc " xp =e.Note that under the conditions established, xq1 uj implies that xq )j . If

M supx+I

f x, Mp supxp

1nxnxpf x and mp inf

xp

1nxnxpf x, then for each

p, Mp mp n 2M. Furthermore, Mp mp ` as long as xp1 / uj . Usingcommutativity to regroup the summation according to the available bounds yields

that

US f : L S f : n;

j1

bMj mj

c:j n [: b : a] ` 2M`

whenever `

2M [: b : a] . Since 0 was arbitrary, from the Inte-grability Criterion we conclude that f + 4:.

Remark 7.1.25 The three Corollaries correspond to Theorems 6.8, 6.9, and 6.10in our text.

As a fairly immediate consequence of Lemma 7.1.16 and the Integrability Cri-

terion we have the following Theorem which is Theorem 6.7 in our text.


19/50


Theorem 7.1.26 Suppose that f is bounded on [a b] and : is monotonically in-

creasing on [a

b].

(a) If there exists an 0 and a partition S of[a b] such that equation (7.11)

is satised, then equation (7.11) is satised for every renementS ofS .

(b) If equation (7.11) is satised for the partitionS x0 ax1 xn1xn band, for each j, j 1 2 n, sj and tj are arbitrary points in

dxj1xj

e,

then

n;j1

nn

fb

sjc f btjcn

n:j .

(c) If f + 4 :, equation (7.11) is satised for the partition

S x0 ax1 xn1xn b

and, for each j, j 1 2 n, tj is an arbitrary point indxj1xj

e, thennnnnn

n;j1

fb

tjc

:j =b

a

f x d: x

nnnnn .Remark 7.1.27 Recall the following denition of Riemann Integrals that you saw

in elementary calculus: Given a function f that is dened on an interval I x : a n x n b, the R sum for I1 I2 In a subdivision of I is given by

jn;j1

fb

Gjc

(bIjc

where Gj is any element of Ij . The point Gj is referred to as a sampling point.

To get the R integral we want to take the limit over such sums as the mesh

of the partitions associated with goes to 0. In particular, if the function f

is dened on I x : a n x n b and ,[a b] denotes the set of all partitions

x

0 ax

1 x

n1x

n b

of the interval I , then f is said to be R integrable

over I if and only if

limmesh[ab]0

jn;j1

fb

Gjc b

xj xj1c


20/50


exists for any choices ofGj + dxj1xje. The limit is called the R integral and isdenoted by

5ba f x d x .

Taking : t t in Theorem 7.1.26 justies that the old concept of an Rintegrability is equivalent to a Riemann integrability as introduced at the beginning

of this chapter.

The following theorem gives a sufcient condition for the composition of a

function with a Riemann-Stieltjes integrable function to be Riemann-Stieltjes inte-

grable.

Theorem 7.1.28 Suppose f + 4 : on [a b], m n f n M on [a b], M is con-tinuous on [m M] , and h x M f x for x + [a b]. Then h + 4 : on[a b].

Excursion 7.1.29 Fill in what is missing in order to complete the proof.

Proof. For f + 4 : on [a b] such that m n f n M on [a b] and M acontinuous function on [m M], let h x M f x for x + [a b]. Suppose that 0 is given. By the

1

, M is uniformly continuous on

[m M]. Hence, there exists a = 0 such that = and

1

s 1

t s t+

[m M]F

s

t

="

M s

M t

. (!)

Because2

, there exists a S x0 ax1xn b + ,[a b]

such that

U S f : L S f : =2. (!!)

For each j + 1 2 n, let Mj supxj

1nxnxjf x , mj inf

xj 1nxnxjf x, Mj

supxj 1nxnxj

h x, and m`j infxj

1nxnxjh x. From the Trichotomy Law, we know that

A jj : j + 1 2 n F bMj mjc =kand

B jj : j + 1 2 n F bMj mjc o =k


21/50


are disjoint.

If j + A, then u )+ dxj1xj e " f u f ) =. It follows from(!) that3

i.e., h u h ) . Hence, Mj m`j n . Since

B t 1 2 j, (!!) implies that

=;j+B

:j n;j+B

bMj mj

c:j n

4

=2.

Because = by choice, we conclude that3

j+B:j . Consequently, for

K supmntnM

M t, we have that rMj m`js n 2K for each j + 1 2 n and3j+B

rMj m`j

s:j 2K. Combining the bounds yields that

U S h : L S h :

n3j1

rMj m`j

s:j

3j+A

rMj m`j

s:j

3j+B

rMj m`j

s:j

n5

2K.

Since 0 was arbitrary, the Integrability Criterion allows us to conclude that

h + 4:.***Acceptable responses are: (1) Uniform Continuity Theorem, (2) f + 4:, (3)M f u M f ) , (4) US f :L S f :, (5) [: b : a].***

7.1.1 Properties of Riemann-Stieltjes Integrals

This section offers a list of properties of the various Riemann-Stieltjes integrals.

The rst lemma allows us to draw conclusions concerning the upper and lowerRiemann-Stieltjes sums of a constant times a bounded function in relationship to

the upper and lower Riemann-Stieltjes sums of the function.


22/50


Lemma 7.1.30 Suppose that f is a function that is bounded and dened on the

interval I

[a

b]. For k a nonzero real number and g

k f , we have

infx+I

g x

!!!!

k infx+I

f x , if k 0

k supx+I

f x , if k 0supx+I

g x

!!!!

k supx+I

f x , if k 0

k infx+I

f x , if k 0

.

Proof. We will prove two of the four equalities. For f a function that is dened

and bounded on the interval I [a b] and k a nonzero real number, let g x k f x.

Suppose that k 0 and that M supx+I

f x. Then f x n M for all x + I

and

g x k f x n k M for all x + IHence, k M is an upper bound for g x on the interval I. Ifk M is not the least upper

bound, then there exists an 0 such that g x n k M for all x + I. (Here, can be taken to be any positive real that is less than or equal to the distance between

k M and supx+I

g x.) By substitution, we have k f x n k M for all x + I. Sincek is positive, the latter is equivalent to

f x n M r

ks for all x + Iwhich contradicts that M is the supremum of f over I. Therefore,

supx+I

g x k M ksupx+I

f x .

Next, suppose that k 0 and that M supx+I

f x. Now, f x n M for allx + I implies that g x k f x o k M. Hence, k M is a lower bound for g xon I. If k M is not a greatest lower bound, then there exists an 0, such that

g x o k M for all x + I. But, from k f x o k M and k 0, we concludethat f x n M k for all x + I. Since k is negative, M k Mwhich gives us a contradiction to M being the sup

x+If x . Therefore,

infx+I

g x k M ksupx+I

f x .


23/50


Theorem 7.1.31 (Properties of Upper and Lower Riemann-Stieltjes Integrals)

Suppose that the functions f , f1 , and f

2are bounded and dened on the closed in-

terval I [a b] and: is a function that is dened and monotonically increasingin I.

(a) If g k f for k+ U0, then 5ba gd: !!

k5b

af x d: x , if k 0

k5b

af x d: x , if k 0

and5

b

agd:

!!

k5b

af x d: x , if k 0

k5b

af x d: x , if k 0

.

(b) If h f1 f2, then

(i)5b

ah x d: x o 5b

af1 x d: x

5ba

f2 x d: x , and

(ii)5b

ah x d: x n 5b

af1 x d: x

5ba

f2 x d: x .

(c) If f1 x n f2 x for all x + I, then

(i)5b

af1 x d: x n

5ba

f2 x d: x, and

(ii) 5ba f1 x d: x n 5b

af2 x d: x.

(d) If a b c and f is bounded on I x : a n x n c and: is monotoni-cally increasing on I , then

(i)5c

af x d: x 5b

af x d: x 5c

bf x d: x, and

(ii)5c

af x d: x 5b

af x d: x 5c

bf x d: x.

Excursion 7.1.32 Fill in what is missing in order to complete the following proof

of part d(i).

Proof. Suppose that a b c and that the function f is bounded in the

interval I [a c]. For any nite real numbers < and D, let G d 0 is given. Since=b

a

f x d: x sup+&[ab]

L f : and

=cb

f x d: x sup+&[bc]

L f : ,


24/50


25/50


Therefore,=b

a

f x d: x =

c

b

f x d: x o L b)) f :c o L b) f :c 4

.

Since 0 was arbitrary, we conclude that=ba

f x d: x =c

b

f x d: x o=c

a

f x d: x .

In view of the Trichotomy Law,5c

a f x d: x o 5ba f x d: x5cb f x d: xand

5ba

f x d: x 5cb

f x d: x o 5ca

f x d: x yields the desired result.

***Acceptable responses include: (1) L f :,

(2)5b

af x d: x5c

bf x d: x, (3) Lemma 7.1.16, (4) same as completion

for (2).***

Given Riemann-Stieltjes integrable functions, the results of Theorem 7.1.31

translate directly to some of the algebraic properties that are listed in the follow-

ing Theorem.

Theorem 7.1.33 (Algebraic Properties of Riemann-Stieltjes Integrals) Suppose

that the functions f , f1, f2 + 4: on the interval I [a b].(a) If g x k f x for all x + I, then g + 4: and=b

a

g x d: x k=b

a

f x d: x .

(b) If h f1 f2 , then f1 f2 + 4 : and=ba

h x d: x =b

a

f1 x d: x =b

a

f2 x d: x .

(c) If f1 x n f2 x for all x + I, then=ba

f1 x d: x n=b

a

f2 x d: x .


26/50


(d) If the function f + 4 : also on I x : b n x n c , then f is Riemann-Stieltjes integrable on I

CI`

and=ca

f x d: x =b

a

f x d: x =c

b

f x d: x .

(e) Iff x n M for x + I, thennnnn=b

a

f x d: x

nnnn n M[: b : a] .(f) If f

+ 4:` on I, then f

+ 4:

:` and=b

a

f db

: :`c =ba

f x d: x =b

a

f x d:` x .

(g) If c is any positive real constant, then f + 4c: and=ba

f dc: c=b

a

f x d: x .

Remark 7.1.34 As long as the integrals exist, the formula given in (d) of the Corol-

lary holds regardless of the location of b i.e., b need not be between a and c.

Remark 7.1.35 Since a point has no dimension (that is, has length 0), we note that=aa

f x d: x 0 for any function f .

Remark 7.1.36 If we think of the denition of the Riemann-Stieltjes integrals as

taking direction into account (for example, with5b

af x d: x we had a b and

took the sums over subdivisions as we were going from a to b), then it makes sense

to introduce the convention that=ab

f x d: x =b

a

f x d: x

for Riemann-Stieltjes integrable functions f .


27/50


The following result follows directly from the observation that corresponding

to each partition of an interval we can derive a partition of any subinterval and vice

versa.

Theorem 7.1.37 (Restrictions of Integrable Functions) If the function f is (Rie-

mann) integrable on an interval I , then f I is integrable on I for any subintervalI of I.

Choosing different continuous functions for M in Theorem 7.1.28 in combina-

tion with the basic properties of Riemann-Stieltjes integrals allows us to generate

a set of Riemann-Stieltjes integrable functions. For example, because M1 t t2,M2 t t, and M3 t < t D for any real constants < and D, are continuouson U, if f + 4 : on an interval [a b], then each of f

2

, f, and < f D willbe Riemann-Stieltjes integrable with respect to : on [a b]. The proof of the next

theorem makes nice use of this observation.

Theorem 7.1.38 If f + 4: and g + 4: on [a b] , then f g + 4:.

Proof. Suppose that f + 4 : and g + 4 : on [a b]. From the AlgebraicProperties of the Riemann-Stieltjes Integral, it follows that f g + 4 : on[a b] and f g + 4 : on [a b]. Taking M t t2 in Theorem 7.1.28 yieldsthat f g2 and f g2 are also Riemann-Stieltjes integrable with respect to :on [a b]. Finally, the difference

4 f g f g2 f g2 + 4: on [a b]

as claimed.

Theorem 7.1.39 If f + 4: on [a b], then f + 4: andnnnn=b

a

f x d: x

nnnn n=b

a

f x d: x .

Proof. Suppose f + 4 : on [a b]. Taking M t t in Theorem 7.1.28yields that f + 4 : on [a b]. Choose < 1, if 5f x d: x o 0 and< 1, if5f x d: x n 0. Thennnnn=b

a

f x d: x

nnnn

Date post:	30-May-2018
Category:	Documents
Upload:	hyd-arnes
View:	213 times
Download:	0 times

RiemannStieltjes Integration

Documents