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### Transcript of Week 7 Riemann Stieltjes Integration: Lectures 19-21 ars/week7-8.pdf  Week 7 Riemann Stieltjes...

• Week 7 Riemann Stieltjes

Integration:

Lectures 19-21

Lecture 19

Throughout this section will denote a monotonically increasing func-

tion on an interval [a, b].

Let f be a bounded function on [a, b].

Let P = {a = a0 < a1, , an = b} be a partition of [a, b]. Puti = (ai) (ai1).Mi = sup{f(x) : ai1 x ai}.mi = inf{f(x) : ai1 x ai}.

U(P, f) =ni=1

Mii; L(P, f) =ni=1

mii. ba

fd = inf{U(P, f) : P}; ba

fd = sup{L(P, f) : P}.

Definition 1 If

ba

fd =

ba

fd, then we say f is Riemann-Stieltjes

(R-S) integrable w.r.t. to and denote this common value by ba

fd :=

ba

f(x)d(x) :=

ba

fd =

ba

fd.

0

• Let R() denote the class of all R-S integrable functions on [a, b].

Definition 2 A partition P of [a, b] is called a refinement of another

partition P of [a, b] if, points of P are all present in P . We then write

P P .

Lemma 1 If P P then L(P ) L(P ) and U(P ) U(P ).

Enough to do this under the assumption that P has one extra point

than P. And then it is obvious because if a < b < c then

inf{f(x) : a x c} min{inf{f(x) : a x b}, inf{f(x) : b x c}

etc.

Theorem 1

ba

fd ba

fd.

For first of all, because for every partition P we have U(P, f) L(P, f).Let P and Q be any two partitions of [a, b]. By taking a common

refinement T = P Q, and applying the above lemma we get

U(P ; f) U(T ; f) L(T ; f) L(Q; f)

Now varying Q over all possible partitions and taking the supremum,

we get

U(P ) ba

fd.

Now varying P over all partitions of [a, b] and taking the infimum, we

get the theorem.

Theorem 2 Let f be a bounded function and be a monotonically

increasing function. Then the following are equivalent.

(i) f R().(ii) Given > 0, there exists a partition P of [a, b] such that

U(P, f) L(P, f) < .

1

• (iii) Given > 0, there exists a partition P of [a, b] such that for all

refinements of Q of P we have

U(Q, f) L(Q, f) < .

(iv) Given > 0, there exists a partition P = {a0 < a1 < cdots < an}of [a, b] such that for arbitrary points ti, si [ai1, ai] we have

ni=1

|f(si) f(ti)|i < .

(v) There exists a real number such that for every > 0, there exists

a partition P = {a0 < a1, < < an} of [a, b] such that for arbitrarypoints ti [ai1, ai], we have |

ni=1 f(ti)i | < .

Proof: (i) = (ii): By definition of the upper and lower integrals,there exist partitions Q, T such that

U(Q) b

a

fd < /2;

ba

fd L(T ) < /2.

Take a common refinement P to Q and T and replace Q, T by P in

the above inequalities, and then add the two inequalities and use the

hypothesis (i) to conclude (ii).

(ii) = (i): Since L(P ) bafd

bafd U(P ) the conclusion

follows.

(ii) = (iii): This follows from the previous theorem for if P P then

L(P ) L(P ) U(P ) U(P ).

(iii) = (ii): Obvious.(iii) = (iv): Note that |f(si) f(ti)| Mi mi. Therefore,

i

|f(si) f(ti)|i i

(Mi mi)i = U(P, f) L(P, f) < .

2

• (iv) = (iii): Choose points ti, si [ai1, ai] such that

|mi f(si)| 0 choose a partition as in (v) with replacedby /2.

3

• Lecture 20

Fundamental Properties of the Riemann-Stieltjes Integral

Theorem 3 Let f, g be bounded functions and be an increasing func-

tion on an interval [a, b].

(a) Linearity in f : This just means that if f, g R(), , R thenf + g R(). Moreover, b

a

(f + g) =

ba

fd +

ba

fd.

(b) Semi-Linearity in . This just means if f R(j), j = 1, 2 j > 0then f R(11 + 22) and moreover, b

a

fd(11 + 22) =

ba

fd1 +

ba

fd2.

(c) Let a < c < b. Then f R() on [a, b] if f R() on [a, c] as wellas on [c, b]. Moreover we have b

a

fd =

ca

fd +

bc

fd.

(d) f1 f2 on [a, b] and fi R() then baf1d

baf2d.

(e) If f R() and |f(x)| M then ba

fd

M [(b) (a)].(f) If f is continuous on [a, b] then f R().(g) f : [a, b] [c, d] is in R() and : [c, d] R is continuous then f R().(h) If f R() then f 2 R().(i) If, f, g R() then fg R().(j) If f R() then |f | R() and b

a

fd

ba

|f |d.

4

• Proof: (a) Put h = f + g. Given > 0, choose partitions P,Q of [a, b]

such that

U(P, f) L(P, f) < /2, P (Q, g) L(Q, g) < /2

and replace these partitions by their common refinement T and then

appeal to

L(T, f) + L(T, g) L(T, h) U(T, h) U(T, f) + U(T, g).

For a constant since

U(P, f) = U(P, f); L(P, f) = L(P.f)

it follows that

ba

fd =

ba

fd. Combining these two we get the

proof of (a).

(b) This is easier: In any partition P we have

(11 + 22) = 11 + 22

from which the conclusion follows.

(c) All that we do is to stick to those partitions of [a, b] which contain

the point c.

(d) This is easy and

(e) is a consequence of (d).

(f) Given > 0, put 1 =

(b)(a) . Then by uniform continuity of f,

there exists a > 0 such that |f(t) f(s)| < 1 whenever t, s [a, b]and |t s| < . Choose a partition P such that i < for all i. Thenit follows that Mi mi < 1 and hence U(P ) L(P ) < .(g) Given > 0 by uniform continuity of , we get > > 0 such that

|(t)(s)| < for all t, s [c, d] with |t s| < . There is a partitionP of [a, b] such that

U(P, f) L(P, f) < 2.

5

• The differences Mimi may behave in two different ways: Accordinglylet us define

A = {1 i n : Mi mi < }, B = {1, 2, . . . , n} \ A.

Put h = f. It follows that

Mi(h)mi(h) < , i A.

Therefore we have

(iB

i) iB

(Mi mi)i < U(P, f) L(P, f) < 2.

Therefore we have

iB i < . Now let K be a bound for |(t)| on[c, d]. Then

U(P, h) L(P, h) =

i(Mi(h)mi(h))i=

iA(Mi(h)mi(h))i +

iB(Mi(h)mi(h))i ((b) (a)) + 2K < ((b) (a) + 2K).Since > 0 is arbitrary, we are done.

(h) Follows from (g) by taking (t) = t2.

(i) Write fg = 14[(f + g)2 (f g)2].

(j) Take (t) = |t| and apply (g) to see that |f | R(). Now let = 1so that

bafd 0. Then ba

fd

= ba

fd =

ba

fd ba

|f |d.

This completes the proof of the theorem.

Theorem 4 Suppose f is monotonic and is continuous and mono-

tonically increasing. Then f R().

Proof: Given > 0, by uniform continuity of we can find a partition

P such that each i < .

6

• Now if f is increasing, then we have Mi = f(ai),mi = f(ai1).

Therefore,

U(P ) L(P ) =i

[f(ai) f(ai1)]i < f(b) f(a)).

Since > 0 is arbitrary, we are done.

Lecture 21

Theorem 5 Let f be a bounded function on [a, b] with finitely many

discontinuities. Suppose is continuous at every point where f is dis-

continuous. Then f R().

Proof: Because of (c) of theorem 3, it is enough to prove this for the

case when c [a, b] is the only discontinuity of f. Put K = sup|f(t)|.Given > 0, we can find 1 > 0 such that (c + 1) (c 1) < .By uniform continuity of f on [a, b] \ (c , c + ) we can find 2 > 0such that |x y| < 2 implies |f(x) f(y)| < . Given any partitionP of [a, b] choose a partition Q which contains the points c and whose

mesh is less than min{1, 2}. It follows that U(Q)L(Q) < ((b)(a)) + 2K. Since > 0 is arbitrary this implies f R().

Remark 1 The above result leads one to the following question. As-

suming that is continuous on the whole of [a, b], how large can be

the set of discontinuities of a function f such that f R()? The an-swer is not within R-S theory. Lebesgue has to invent a new powerful

theory which not only answers this and several such questions raised

by Riemann integration theory but also provides a sound foundation

to the theory of probability.

Example 1 We shall denote the unit step function at 0 by U which

7

• is defined as follows:

U(x) =

{0, x 0;1, x > 0.

By shifting the origin at other points we can get other unit step func-

tion. For example, suppose c [a, b]. Consider (x) = U(x c), x [a, b]. For any bounded function f : [a, b] R, let us try to compute bafd. Consider any partition P of [a, b] in which c = ak. The only

non zero i is k = 1. Therefore U(P ) L(P ) = Mk(f)mk(f).Now assume that f is continuous at c. Then by choosing ak+1 close

to ak = c, we can make Mk mk 0. This means that f R().Indeed, it follows that Mk f(c) and mk f(c). Therefore, b

a

fd = f(c).

Now suppose f has a discontinuity at c of the first kind i.e, in

particular, f(c+) exists. It then follows that |Mkmk| |f(c)f(c+)|.Therefore, f R() iff f(c+) = f(c).

Thus, we see that it is possible to destroy integrability by just dis-

turbing the value of the function at one single point where itself is

discontinuous.

In particular, take f = . It follows that 6 R() on [a, b].

We shall now prove a partial converse to (c) of Theorem 3.

Theorem 6 Let f be a bounded function and an increasing function

on [a, b]. Let c [a, b] at which (at least) f or is continuous. Iff R() on [a, b] then f R() on both [a, c] and [c, b]; moreover, inthat case, b

a

fd =

ca

fd +

bc

fd.

8

• Proof: Assume is