12 1. The Regulated and Riemann Integrals
Proof. A continuous function f(x) defined on a closed interval [a, b] is
uniformly continuous (see Theorem A.8.2). That is, given ε 0 there
is a corresponding δ 0 such that |f(x)−f(y)| ε whenever |x−y|
δ. Let εn =
and let δn be the corresponding δ guaranteed by
uniform continuity.
Fix a value of n and choose a partition x0 = a x1 x2 ···
xm = b with xi xi−1 δn. For example, we could choose m so large
that if we define ∆x = (b a)/m, then ∆x δn and then we could
define xi to be a + i∆x. Next we define a step function fn by
fn(x) = f(xi) for all x [xi−1, xi).
That is, on each half open interval [xi−1, xi) we define fn to be the
constant function whose value is the value of f at the left endpoint
of the interval. The value of fn(b) is defined to be f(b).
Clearly, fn(x) is a step function with the given partition. We
must estimate its distance from f. Let x be an arbitrary point of
[a, b]. It must lie in one of the open intervals of the partition or be
an endpoint of one of them; say x [xi−1, xi). Then since fn(x) =
fn(xi−1) = f(xi−1) we may conclude that
|f(x) fn(x)| |f(x) f(xi−1)| εn
because of the uniform continuity of f and the fact that |x xi−1|
Thus we have constructed a step function fn with the property
that for all x [a, b],
|f(x) fn(x)| εn.
So the sequence {fn} converges uniformly to f and f is a regulated
Exercise 1.5.6.
(1) Show that the continuous function f(x) = 1/x on the open
interval (0, 1) is not regulated, i.e., it cannot be uniformly
approximated by step functions.
(2) Give an example of a bounded continuous function on the
open interval (0, 1) which is not regulated.
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