The Riemann Integral

24 1. The Riemann Integral
Proposition 1.39. Suppose that f : [a,b] → R is bounded and integrable on [a,r]
for every a < r < b. Then f is integrable on [a,b] and
∫ b
a
f = lim
r→b − ∫ r
Proof. Since f is bounded, |f| ≤ M on [a,b] for some M > 0. Given ǫ > 0, let
r = b− ǫ
4M
(where we assume ǫ is sufficiently small that r > a). Since f is integrable on [a,r],
there is a partition Q of [a,r] such that
a
f.
U(f;Q)−L(f;Q) < ǫ 2 .
ThenP = Q∪{b}isapartitionof[a,b]whoselastintervalis[r,b]. The boundedness
of f implies that
supf − inf f ≤ 2M.
[r,b] [r,b]
Therefore
(
U(f;P)−L(f;P) = U(f;Q)−L(f;Q)+
sup
[r,b]
)
f − inf f ·(b−r)
[r,b]
< ǫ +2M ·(b−r) = ǫ,
2
so f is integrable on [a,b] by Theorem 1.14. Moreover, using the additivity of the
integral, we get
∫ b ∫ ∣
r
∣∣∣∣ ∫ b
f − f
∣ ∣ = f
∣ ≤ M ·(b−r) → 0 as r → b− .
a
a
r
An obvious analogous result holds for the left endpoint.
Example 1.40. Define f : [0,1] → R by
{
sin(1/x) if 0 < x ≤ 1,
f(x) =
0 if x = 0.
Then f is bounded on [0,1]. Furthemore, f is continuous and therefore integrable
on [r,1] for every 0 < r < 1. It follows from Proposition 1.39 that f is integrable
on [0,1].
The assumption in Proposition 1.39 that f is bounded on [a,b] is essential.
Example 1.41. The function f : [0,1] → R defined by
{
1/x for 0 < x ≤ 1,
f(x) =
0 for x = 0,
is continuous and therefore integrable on [r,1] for every 0 < r < 1, but it’s unbounded
and therefore not integrable on [0,1].
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