1.7. The Riemann Integral 21
Putting together the last two inequalities with Proposition 1.7.1 we
obtain
0 sup
v∈L(f )
1
0
v(x) dx inf
u∈U(f )
1
0
u(x) dx 0.
So all of these inequalities are equalities and by definition, f is Rie-
mann integrable with integral 0.
To see that f is not regulated suppose that g is an approximating
step function with partition x0 = 0 x1 ··· xm = 1 and
satisfying |f(x) g(x)| ε for some ε 0. Then g is constant, say
with value c1 on the open interval (0, x1).
There exist points a1, a2 (0, x1) with f(a1) = 0 and f(a2) = 1.
So
|c1| = |c1 0| = |g(a1) f(a1)| ε
and
|1 c1| = |f(a2) g(a2)| ε.
But
|c1| + |1 c1| |c1 + 1 c1| = 1,
so at least one of |c1| and |1 c1| must be 1/2. This implies that
ε 1/2. That is, f cannot be uniformly approximated by any step
function to within ε if ε 1/2. So f is not regulated.
Theorem 1.7.5. (Regulated implies Riemann integrable). Ev-
ery regulated function f is Riemann integrable and the regulated in-
tegral of f is equal to its Riemann integral.
Proof. If f is a regulated function on the interval I = [a, b], then,
for any ε 0, it can be uniformly approximated within ε by a step
function. In particular, if εn = 1/2n, there is a step function gn(x)
such that |f(x) gn(x)| εn for all x I. The regulated integral
b
a
f(x) dx was defined to be lim
b
a
gn(x) dx.
We define two other approximating sequences of step functions for
f. Let un(x) =
gn(x)+1/2n
and vn(x) = gn(x)
−1/2n.
Then un(x)
f(x) for all x I because un(x)−f(x) =
1/2n
+gn(x)−f(x) 0 since
|gn(x) f(x)|
1/2n.
Similarly, vn(x) f(x) for all x I because
f(x) vn(x) =
1/2n
+ f(x) gn(x) 0 since |f(x) gn(x)|
1/2n.
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