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Unformatted text preview: \\:;ii//>Integrabilitx g: bounded Eiecewise—monotonic functions. The definition of "piecewisemonotonic" is given on p.
77 of the text. Lemma. If f is bounded on [a,b] and monotonic on (a,b), then f is integrable on (a,bJ. (Note that we need to assume f is bounded in the hypo— thesis of this lemma. The function l/x for 0 < x s 1 f(x) ll
0 0 for x is monotonic on (0,1), but it is not bounded.) Proof. Choose M so that M < f(x) < M. We apply the Riemann condition.
Given 8 > 0, let us choose numbers c and d (close to a and b respectively), such that a < c < d < b and such that c  a < s/M and b  d < s/M. Now f is monotonic on [c,d] so it is integrable on [c,d].
Therefore we can find step functions 5 and t defined on
[c,d] such that s < f < t on [c,d], and such that I: t * f2 3 <'e. Extend t to a step function tl defined on [a,b] by setting M for a < x < c,
tl(x) = t(x) for c < x < d,
M for d < x < b. Similarly, extend .s to a step function 5 defined on. [a,b] l
by setting M for a < x < c,
sl(x) = t(x) for c < x < d,
M for d<X<b.
Then 51 < f < tl on all of [a,b]. Furthermore,
[b b c d b
t  J s = I (t s ) + I (t s ) + I (t s )
a l a l a l l l l d l l
d
= 2M(c—a) + I (tl—sl) + 2M(db)
c < 28 + e + 25 58.
Since a is arbitrary, the Riemann condition is satisfied.
Theorem. lg f is bounded and piecewise—monotonic on [a,b], then f is integrable 93 [a,b]. Proof. By hypothesis, there is a partition x0 < x1 < ... < xn of [a,b] such that f is monotonic on each open interval (xi_l,xi). By the preceding lemma, f is integrable on [xi_l,xi] for each .i. By the additivity _property of integrals (Theorem on p.‘D.2) , it follows that f is integrable on [a,b]. Exercise l. Suppose f is bounded on [a,b]. Suppose also that f
is integrable on 33951 closed interval [c,d] contained
in the open interval (a,b). Show that f is integrable on [a,b]. ...
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 Spring '09
 BRUBAKER
 Topology, Order theory, Metric space, Closed set

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