1549901369-Elements_of_Real_Analysis__Denlinger_

Answers & Hints for Selected Exercises 6 77

Thus, s_u, Pn) = i~ l(xi)6.i = i~ ( 1 - (-1 + ~t)

3

) · ~

= l ~ [2 _ ~ + 27i^2 _ 27i^3 ] = 6- 27 n(n+l) + 81 n(n+l)(2n+l) 81 n^2 (n+l)^2

n iL, =l n ""1i2 ~ Ti2 2 ri3 6 10 4

= 6 -^2 ; (1 + ~) +^2 ; (1 + ~) (2 + ~) -^841 (1 + ~)

2

--> -~. Similarly,

S(f, Pn) = i~ l(Xi-1)6.i = ~ i~ [1-(-1 +
3
(i~l))

3

] = ~ f: [1-(-1 + ~t)

3

]

= 6 (1 - ~) -^2 ; (1 - ~) +^2 ; (1 - ~) (2 - ~) -^841 (1 - ~)

2

--> -~.

13. For I , g satisfying the hypotheses, and V partitions P of [a, b], s_(f, P) :::;

• S(f, P), SO J.b a I= _!!_I J.b 5 J.b ag = J.b a g.

15. (a) For I and {Pn} satisfying the hypotheses, {s(f, Pn)} and {S(f, Pn)}
    are monotone increasing and monotone decreasing sequences, respectively, by
    Thm. 7.2.4. By Thm. 7.2.5, {s(f, Pn)} is bounded above by S(f, Pi) and
    {S(f, Pn)} is bounded below by s(f, P 1 ). Thus, by the monotone convergence
    theorem (2.5.3) both of these sequences converge. Since '<In E N, s(f, Pn) 5
    S(f, Pn), and since limits preserve inequalities, lim s(f, Pn) 5 lim S(f, Pn)·
    n--+ oo n--+ oo
    (b) By the monotone convergence thm, lim s(f, Pn) =sup {s_(f, Pn) : n EN}
    n-+oo

(^5) £ t I and n--+oo lim S(f, Pn) =inf {S(f, Pn) : n E N} ;?.'. t a f. Thus, n-+oo lim s(f, Pn) 5
J.__g b I^5 t a I^5 n--+oo lim S(f, Pn)· So, if n-+oo lim s(f, Pn) = n--+oo lim S(f, Pn) = L, then
l:I = l:I = L.
(c) Let l(x) = x on [O, 2]. Then I is integrable on [O, 2]. For the partition
Pn = {O, 2 ~, 22 n, 23 n,. · · , ~=, 2}, take K = 2n. Then s_(f, Pn) = 1+ ~(1-:k)-->
~, while S(f, Pn) = 2 + ~(1 + :k)--> ~-

17. Trivial. Go through the proof, replacing "<" by "5", and see that it works.


18. The proof of Case 2 is just like that of Case 1 except that this time mi =
    n
    l(xi) and Mi = l(xii). Then S(f, P) - s(f, Q) = b-;;a I: [!(xi) - l(xi_i)J =
    i=l
    b-;;a [l(a) - f(b)] < (b -a) [!(a) - l(b)] c. Apply Riemann's criterion.

[

k-1

21. s(f, Q)-s(f, P) = i~ mi(xi -Xi-1) + mk,1(x'k - xk-1) + mk,2(xk - x'k)

+ I: n mi(xi -Xi-1) l - [k-1 I: mi(Xi - Xi-1) + mk(Xk - Xk-1)

i=k+l i=l

+ f mi(Xi - Xi-1)] = mk,1(x'k-Xk-1)+mk,2(xk-xk)-mk(Xk-Xk-1).

i =k+l

Similarly, S(f, Q)-S(f, P) = Mk,1(x'k-Xk-1)+Mk,2(xk-xk)-Mk(Xk-Xk_i).