Answers & Hints for Selected Exercises 681
(a) f is bounded and continuous, hence integrable, on every [c, d] ~ (0, 1),
so by Thm. 7.4.7, f is integrable on [O, l]. The same is true on [-1, OJ. Thus,
by Thm. 7.4.5, f is integrable on [-1, l].
(b) The sequence {g(~+~nn)} shows that g is unbounded on [-1,1],
hence is not integrable there.
(a) (^10) (b) 5/2 (c) 10
(a) For x E (xi-1, xi ), O"(x) = m i and T(x) = Mi, so O" and T are step
functions on [a, b].
(b) For x E [a, b], either x = b or there is a unique i 3 x E [xi-l, xi). In the
first case, O"(x ) = f(x ) = T(x). In the second case, O"(x) =mi ::::; f(x) ::::; Mi=
T(x). Thus, \:/x E [a, b], O"(x) ::::; f(x ) ::::; T(x).
b k ( X; ) k
(c) By (7.4.3), (7.4.8), and (7 .2.9), fa O" = i~ fx, 1 O" = i~ m i (Xi -
k k
Xi-1) = S..(f, P) and Ja rb T = L (Jx x' T ) = L Mi(Xi - Xi-1) = -S(f, P).
i=l '^1 i=l
- The function f(x) = l/x if x I 0, 0 if x = 0, is continuous (hence inte-
grable) on every proper closed subinterval [c, d] of [O, 1], but is not bounded on
[O, 1] so is not integrable on [O, 1] This does not contradict Thm. 7.4.7 because
that theorem pertains only to bounded functions. - The function f(x) = sin~ if x I 0, 0 if x = 0, is continuous (hence
integrable) on every [c, d] ~ (0, 1), so by Thm. 7.4.7 it is integrable on [O, l]. As
shown in (4.1.12), lim f(x) does not exist, so f is not regulated on [O, 1].
x->O+ - Let a < b. By Thm. 5.7.3, 3 bounded, monotone increasing f : [a, b] ->IR
having Qn [a, b] as its set of discontinuities. Since f is monotone, it is integrable
on [a, b]. Since Q is dense in IR, Q n [a, b] is dense in [a, b]. By Thm. 5.2.17, f is
regulated on [a, b].
EXERCISE SET 7.5- (a ) Given P = {x1,X2, ... ,xn} , define m i(!)= inf{f(x) : Xi-l ::::; x::::;
xi}. Define m i (g) and m i(! + g) similarly. Then \:/x E [xi-l, xi], m i(!) +
n
mi(g) ::::; f(x ) + g(x), and so mi(!)+ mi(g) ::::; mi(!+ g). Thus, I: mi(f)6i +
n n i=l
I: m i(g)6i::::; I: mi(!+ g)6i. That is ,
i=l i=l
S..(f, P) + S..(g, P)::::; S..(f + g, P)::::; J:(f + g). (3)