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.