1.7. The Riemann Integral 17

Then if we can define

b

a

f(x) dx in a way that satisfies monotonicity

it must also satisfy

(1.7.1)

b

a

f(x) dx ≤ inf

b

a

u(x) dx u ∈ U(f) .

The infimum exists because all of the step functions in U(f) are

bounded below by a lower bound for the function f.

Similarly, we define L(f) to be the set of all step functions v(x)

such that v(x) ≤ f(x) for all x ∈ I. Again, if we can define

b

a

f(x) dx

in such a way that it satisfies monotonicity it must also satisfy

(1.7.2) sup

b

a

v(x) dx v ∈ L(f) ≤

b

a

f(x) dx.

The supremum exists because an upper bound for the function f is

an upper bound for all of the step functions in U(f).

Putting inequalities (1.7.1) and (1.7.2) together, we see that if

V is any vector space of bounded functions which contains the step

functions and we manage to define the integral of functions in V in a

way that satisfies monotonicity, then this integral must satisfy

sup

b

a

v(x) dx v ∈ L(f) ≤

b

a

f(x) dx

≤ inf

b

a

u(x) dx u ∈ U(f)

for every f ∈ V. We next observe that even if we cannot define an

integral for f we still have the inequality relating the expressions at

the ends.

Proposition 1.7.1. Let f be any bounded function on the interval

I = [a.b]. Let U(f) denote the set of all step functions u(x) on I such

that f(x) ≤ u(x) for all x and let L(f) denote the set of all step

functions v(x) such that v(x) ≤ f(x) for all x. Then

sup

b

a

v(x) dx v ∈ L(f) ≤ inf

b

a

u(x) dx u ∈ U(f) .

Proof. If v ∈ L(f) and u ∈ U(f), then v(x) ≤ f(x) ≤ u(x) for all

x ∈ I, so monotonicity implies that

b

a

v(x) dx ≤

b

a

u(x) dx. Hence,