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 fPkg of [a; b] such that lim (U(Pk; f) − L(Pk; f)) = 0: k!1 Show that f is Riemann integrable and that Z b f = lim U(Pk; f) = lim L(Pk; f): a k!1 k!1 Proof. Given " > 0, by assumption there exists a N 2 N such that when k ≥ N, U(Pk; f) − L(Pk; f) < ". Specifically, U(PN ; f) − L(PN ; f) < ". Then, Z b Z b 0 ≤ f − f ≤ U(PN ; f) − L(PN ; f) < " a a Since " > 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 " > 0 be given. Choose N 2 N a k!1 such that when k ≥ N, U(Pk; f) − L(Pk; f) < ". 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!1 Z b Finally, since lim (U(Pk; f) − L(Pk; f)) = 0 and lim U(Pk; f) = f, k!1 k!1 a lim L(Pk; f) = lim [U(P; k) − (U(Pk; f) − L(Pk; f))] k!1 k!1 = lim U(P; k) − lim (U(Pk; f) − L(Pk; f)) k!1 k!1 Z b = f − 0 a Z b = f: a Week 8 Solutions Page 2 Exercise (5.1.6). Let c 2 (a; b) and let d 2 R. Define f :[a; b] ! R as ( d if x = c; f(x) := 0 if x 6= c: R b Prove that f 2 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 " > 0 be given. Let 0 < β < minfc − a; b − c; "=2dg. Let P := fa; c − β; c + β; bg. 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 < ": Therefore, by the Cauchy criterion, f is integrable. Next, for any " > 0, if we define P as above, we have Z b 0 = L(P; f) ≤ f ≤ U(P; f) < ": a R b Since " 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 2 Ag. Week 8 Solutions Page 3 Let P = fx0; x1; x2; : : : ; xn−1; xng 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)k2N such that limk!1(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!1 k!1 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 = supfL(P; −f): P is a partition of [a; b]g = sup{−U(P; f): P is a partition of [a; b]g = − inffU(P; f): P is a partition of [a; b]g 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 2 [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 2 (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 2 [a; b] such that f(c) = 0. R b Proof. Apply Exercise 5.2.4 with a f = 0 to obtain the conclusion..