1549901369-Elements_of_Real_Analysis__Denlinger_

7.4 Basic Existence and Additivity Theorems 393

Examples 7.4.17

(a) The function f(x) = x - 1

1

x ' is regulated on [O, 5], and

{

Ix - 11 .f -I 1 .}

0 if x = 1

w v Xo E [O ) 5 J ' J "(f ) Xo ) = { 0 if xo -j. 1; }.

2 if Xo = 1

(b) Th e D me .. hl et f unct10n. f ( ) x = {^1 if x is rational; } is not regulated

0 if x is irrational

on [a, b] if a< b. (See Exercise 4 .1.4.)

(c) Every monotone function f: [a,b] ---t IR is regulated on [a,b]. (See

Theorems 5.2.17 and 5.2.18.)

(d) Thomae's function T defined in Example 5.1.12 is regulated on any

[a,b]. (See Exercise 5.1.30.) Note that Vx E IR, j(T,x) = T(x). D

We shall find the following theorem useful in showing that regulated func-

tions are integrable.

Theorem 7.4.18 A function f : [a, b] , IR is regulated {::} Ve > 0, :3 step
function CT: [a, b] ,IR 3 Vx E [a, b], lf(x) - CT(x)I < c.

*Proof. Part 1 (::::} ): Suppose a < b and f : [a, b] ___, IR is regulated. Let
c > 0. Define

A= {x E [a,b] : :3 step function CT on [a,b] 3 Vt E [a,x], lf(t) -CT(t)I < c}.

The (constant) step function CT(x) = f(a) shows that a EA, so A is a nonempty
bounded set. Hence, :3 c = sup A. First, we show that c > a. Since f is regulated
on [a,b], lim f(x) = L exists, so :3 6 > 0 3 a+6 < b, and a< x < a+6::::}
x-+a+
lf(x) - LI < c. Then the step function CT(a) = f(a), CT(x) = L when x > a,
satisfies IJ(x) - CT(x)I <con [a, a+~], so a+~ EA. Thus, c >a.

Note that A is an interval since Vx E A, if a ::::; x' < x, then any step
function satisfying the given c-condition on [a, x] must satisfy that condition
on [a, x'], so x' E A. Thus,

A= [a,c) or A= [a,c].

Since f is regulated on [a, b], f has a limit from the left at c, say L' =
lim f(x). Then :3 6 > 0 3 a+ 6 < c and c - 6 < x < c ::::} lf(x) - L'I < c.
X-+C-
NOW c - 6 EA since a< c - 6 < c. So :3 step function CT on [a, b] such that

Vt E [a, c - 6], lf(t) - CT(t)I < c.