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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