7.4 Basic Existence and Additivity Theorems 389. Since f is integrable on [a+t:, b-t:], the middle terms of the right-hand sides
of these two equations are equal. Thus, when we subtract these two equations
we get
1:1 -1:1 = (1:+
0
1 - l:+o 1) + (1Lo1 -lLJ)::::; (M - m)e + (M - m)e = 2(M - m)t:,
where M =sup f[a, b] and m =inf f[a, b] (see Exercise 7.2.3).
Thus, by the generalized forcing principle (7.1.6), l: f -l: f ::::; 0. Therefore
l: f ::::; l: f , from which we conclude that f is integrable on [a, b].
(b) Since f is bounded on [a,b], 3M> 0 3 \:/x E [a,b], lf(x)I::::; M. Let
0 < h < b - a. By Theorem 7.4.2, f is integrable on [a, a+ h] and [a+ h, b], and
l
b la+h lb
f= !+ f.
a a a+hThen,
11 : -l:+h f I = ll:+h JI ::::; Mh. (See Exercise 6.)
€
Let e > 0. Choose 5 = M. Then0<h<8=?11:+hf-l: JI::::; Mh < M8 = M ~
=? 11:+hf-l: 11 < €.
Therefore, h-+O+ lim t+h a f = t a f. This proves the first equality in (b).
The remaining equalities in (b) are proved similarly. (Exercise 7) •Corollary 7.4.8 (Irrelevance of Endpoint Values) Suppose f, g: [a, b] --t
JR are bounded on [a, b] and \:/x E (a, b), f(x) = g(x). Then if f is integrable on
[a,b] so is g, and l: f = l: g.
Proof. Suppose f, g : [a, b] --t JR are bounded on [a, b], f is integrable
on [a, b], and \:/x E (a, b), f(x) = g(x). Then g is integrable on every closed
subinterval of (a, b), so by Theorem 7.4.7, l: g exists. Further,
t g = lim t+-~ g by 7.4.7 (b)
a h-+O+ a
= lim t+-~ f since f(x) = g(x) on [a+ h, b - h]
h-+O+ a
= l: f by 7.4.7 (b). •