1.7. The Riemann Integral 21
Putting together the last two inequalities with Proposition 1.7.1 we
0 ≤ sup
v(x) dx ≤ inf
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.
|c1| = |c1 − 0| = |g(a1) − f(a1)| ≤ ε
|1 − c1| = |f(a2) − g(a2)| ≤ ε.
|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
f(x) dx was defined to be lim
We define two other approximating sequences of step functions for
f. Let un(x) =
and vn(x) = gn(x)
Then un(x) ≥
f(x) for all x ∈ I because un(x)−f(x) =
+gn(x)−f(x) ≥ 0 since
|gn(x) − f(x)|
Similarly, vn(x) ≤ f(x) for all x ∈ I because
f(x) − vn(x) =
+ f(x) − gn(x) ≥ 0 since |f(x) − gn(x)|