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 =

1/2n

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|

δn.

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

function.

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.