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,
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