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.