Week 8 Solutions Page 1

Exercise (5.1.3). Let f :[a, b] → R be a bounded function. Suppose that there exists a sequence of partitions {Pk} of [a, b] such that

lim (U(Pk, f) − L(Pk, f)) = 0. k→∞ Show that f is Riemann integrable and that Z b f = lim U(Pk, f) = lim L(Pk, f). a k→∞ k→∞

Proof. Given ε > 0, by assumption there exists a N ∈ N such that when k ≥ N, U(Pk, f) − L(Pk, f) < ε. Specifically, U(PN , f) − L(PN , f) < ε. Then, Z b Z b 0 ≤ f − f ≤ U(PN , f) − L(PN , f) < ε a a Since ε > 0 was arbitrary, we must have Z b Z b Z b Z b f − f = 0 or f = f. a a a a Therefore, f is Riemann integrable. R b Next,we show that f = lim U(Pk, f). Let ε > 0 be given. Choose N ∈ N a k→∞ such that when k ≥ N, U(Pk, f) − L(Pk, f) < ε. Then, when k ≥ N

Z b Z b

U(Pk, f) − f = U(Pk, f) − f a a Z b = U(Pk, f) − f a

≤ U(Pk, f) − L(Pk, f) < .

R b Hence, we have that f = lim U(Pk, f). a k→∞ Z b Finally, since lim (U(Pk, f) − L(Pk, f)) = 0 and lim U(Pk, f) = f, k→∞ k→∞ a

lim L(Pk, f) = lim [U(P, k) − (U(Pk, f) − L(Pk, f))] k→∞ k→∞

= lim U(P, k) − lim (U(Pk, f) − L(Pk, f)) k→∞ k→∞ Z b = f − 0 a Z b = f. a Week 8 Solutions Page 2

Exercise (5.1.6). Let c ∈ (a, b) and let d ∈ R. Define f :[a, b] → R as ( d if x = c, f(x) := 0 if x 6= c.

R b Prove that f ∈ R[a, b] and compute a f using the definition of the integral (and propositions of the section).

Proof. Without loss of generality, assume that d > 0. Let ε > 0 be given. Let 0 < β < min{c − a, b − c, ε/2d}. Let P := {a, c − β, c + β, b}. Then,

3 X U(P, f) − L(P, f) = (Mj − mj)∆xj j=1

= (0 − 0)∆x1 + (d − 0)∆x2 + (0 − 0)∆x3 = d(2β) = 2dβ < ε.

Therefore, by the Cauchy criterion, f is integrable. Next, for any ε > 0, if we define P as above, we have

Z b 0 = L(P, f) ≤ f ≤ U(P, f) < ε. a

R b Since ε was arbitrary, this gives that a f = 0.

Exercise (5.2.1). Let f be in R[a, b]. Prove that −f is in R[a, b] and

Z b Z b −f(x) dx = − f(x) dx. a a

Proof. Note that for any set A ⊆ R, sup(−A) = − inf A and inf(−A) = − sup A, where −A = {−a : a ∈ A}. Week 8 Solutions Page 3

Let P = {x0, x1, x2, . . . , xn−1, xn} be any partition of [a, b], and let Ik = [xk−1, xk] denote the kth subinterval created by the partition. Then, using the above fact, we have that for k = 1, . . . , n,

n X U(P, −f) = sup(−f(x))∆xk I k=1 k n X = − inf f(x)∆xk Ik k=1 n X = − inf f(x)∆xk Ik k=1 = −L(P, f)

Similarly, L(P, −f) = −U(P, f). Hence,

U(P, −f) − L(P, −f) = −L(P, f) − (−U(P, f)) = U(P, f) − L(P, f)

Since f is integrable, from problem 5.1.3, there is a sequence of partitions

(Pk)k∈N such that limk→∞(U(Pk, f) − L(Pk, f)) = 0. Then, for the same se- quence,

lim (U(Pk, −f) − L(Pk, −f)) = lim (U(Pk, f) − L(Pk, f)) = 0 k→∞ k→∞ and so −f is integrable if and only if f is integrable. Finally, since f and −f are both integrable we have that

Z b Z b −f(x) dx = − f(x) dx a a = sup{L(P, −f): P is a partition of [a, b]} = sup{−U(P, f): P is a partition of [a, b]} = − inf{U(P, f): P is a partition of [a, b]} Z b = − f(x) dx a Z b = − f(x) dx. a Week 8 Solutions Page 4

Exercise (5.2.4). Prove the mean value theorem for integrals. That is, prove that if f :[a, b] → R is continuous, then there exists a c ∈ [a, b] such that R b a f = f(c)(b − a).

Proof. By the Min-Max Theorem, f achieves both an absolute maximum M and an absolute minimum m in [a, b], say at points c1 and c2. Without loss of generality, c1 < c2. Then we have that

Z b m(b − a) ≤ f ≤ M(b − a). a or 1 Z b m ≤ f ≤ M. (b − a) a

By the Intermediate Value Theorem, there must be a point c ∈ (c1, c2) ⊂ [a, b] such that 1 Z b f(c) = f. (b − a) a Therefore, for this c, Z b f = f(c)(b − a). a

R b Exercise (5.2.6). Suppose f :[a, b] → R is a continuous function and a f = 0. Prove that there exists a c ∈ [a, b] such that f(c) = 0.

R b Proof. Apply Exercise 5.2.4 with a f = 0 to obtain the conclusion.