1.7. The Riemann Integral 23
Suppose ε 0 is given. Since f is Riemann integrable there are
step functions uf and vf such that vf (x) f(x) uf (x) for x I
(i.e., uf U(f) and vf L(f)) and with the property that
b
a
uf (x) dx
b
a
vf (x) dx ε.
Similarly, there are ug U(g) and vg L(g) with the property that
b
a
ug(x) dx
b
a
vg (x) dx ε.
This implies that
b
a
(uf + ug)(x) dx
b
a
(vf + vg )(x) dx 2ε.
Since (uf + ug) U(f + g) and (vf + vg ) L(f + g), we may
conclude that
inf
u∈U(f +g)
b
a
u(x) dx sup
v∈L(f +g)
b
a
v(x) dx 2ε.
As ε 0 is arbitrary, we conclude that
inf
u∈U(f +g)
b
a
u(x) dx = sup
v∈L(f +g)
b
a
v(x) dx
and hence (f + g) R.
Exercise 1.7.7.
(1) At the beginning of this chapter we mentioned the function
f : [0, 1] R which has the value f(x) = 0 if x is rational
and 1 otherwise. Prove that for this function
sup
v∈L(f )
1
0
v(x) dx = 0
and
inf
u∈U(f )
1
0
u(x) dx = 1.
Hence, f is not Riemann integrable.
(2) Prove that the absolute value of a Riemann integrable func-
tion is Riemann integrable.
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