394 Chapter 7 • The Riemann Integral
If we define the step function CJ^1 on [a, b] by
{CT(x) on [a, c - 8], }
CJ'(x) = L' on (c - 8, c),
f(c) on[c,b],then CJ^1 satisfies the required c:-condition on [a, c]. Thus, c E A. Therefore,
A= [a,c].
We shall conclude the proof of Part 1 by showing that c = b. Suppose c i:-b.
Then a< c < b so f has a limit from the right at c. Let L" = lim f(x). Then
x--+c+
3 b > 0 3 c + b <band c <:: x < c + b ==?-lf(x) - L"I < c:. If we define
{CJ
1
(x) on [a, c], }
T(x) = f(c) at x = c,
L" on ( c, b],then T is a step function satisfying the required c:-condition on [a, c + b]. So
c + b E A. But c + b > c = sup A. Contradiction. Therefore, c = b.
P art 2 ( <=): Suppose a< band f : [a , b] --) JR is such that Ve:> 0, 3 step
function CJ: [a, b]--) JR 3 Vx E [a, b], lf(x) - CT(x)I < c:. We shall prove that f is
regulated.
First, consider any x 0 E [a , b). We shall prove that lim f(x) exists. Let
X--+xci
c: > 0. By hypothesis , 3 step function Cle : [a, b] --) JR 3
Vx E [a,b], lf(x ) - Cle(x)I < c:/2.Since Cle is a step function, 3 b > 0 3 Cle is constant on (x 0 , x 0 + b). Thus,
Xo < x, y < Xo + b * Cle(x) = Cle(Y)
* lf(x ) - f(y)I:::; lf(x ) - Cle(x)I + ICle(x) - Cle(Y)I + h(y) - f(y)I
< c:/2 + 0 + c:/2 = c.Thus, by the Cauchy criterion for one-sided limits^10 a t x 0 , lim f(x ) exists.
x--+xci
Next, consider any x 0 E (a, b]. An argument similar to the one just given
will prove that lim_ f(x) exists (Exercise 16). Therefore, f is regulated. •
x--+x 0
- See Exercise 4 .2.22.