7.2 Riemann Integrable Functions


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1 7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous on [, b] except t finite number of points {c 1,..., c m }, then f R[, b]. Theorem 4. If f : [, b] R is monotone on [, b], then f R[, b]. Proof. Either f is incresing or decresing. Assume f is incresing nd f(b) > f(). (Wht hppens if f() = f(b)?) Let ɛ > 0. Consider prtition P = {x 0,..., x n } of [, b] with subintervls of equl length, tht is, ech subintervl [x k 1, x k ] hs length x = x k x k 1 = (b )/n. The infimum of f occurs t the left endpoint f(x k 1 ) nd the supremum of f occurs t the right endpoint f(x k ). Then L(f; P ) = f(x k 1 ) x nd U(f; P ) = f(x k ) x. Choose n sufficiently lrge such tht the subintervls in the prtition P ɛ hve length x < ɛ/[f(b) f()]. Then U(f; P ɛ ) L(f; P ɛ ) = (f(x k ) f(x k 1 )) x = x Therefore, by the Integrbility Criterion, f is Riemnn integrble. (f(x k ) f(x k 1 )) = x(f(b) f()) < ɛ. How cn the preceding proof be modified to show decresing bounded function is Riemnn integrble? Theorem 5 (Additivity Theorem). Let f : [, b] R be bounded function nd c [, b]. Then f R[, b] iff its restrictions to [, c] nd [c, b] re Riemnn integrble. In prticulr f = A corollry to this theorem extends the result to set of points = c 0 < c 1 < < c m = b. c f + c f. f = m ci c i 1 f. Definition: If f R[, b], then f = b f nd f = 0. Properties of Integrls: If f, g R[, b] nd c is ny rel number, then (f + g) = f + g If f(x) g(x) for x [, b], then cf = c f f g.
2 Exmples 1. If f : [, b] R tkes only finite number of distinct vlues, is f step function? 2. Give n exmple of Riemnn integrble function f such tht f = 0 but f is not the zero function. 3. Define f on [0, 1] s follows: f(x) = 1/2 for x [0, 1/2), f(x) = 2/3 for x [1/2, 2/3) nd in generl f(x) = n/(n + 1) for x [(n 1)/n, n/(n + 1)). Also, f(1) = 1. The function f is not step function. () How mny discontinuities does f hve? (b) Is f continuous t x = 1? (c) Is f Riemnn integrble on [0, 1]? 4. Evlute f if c [, b], c f = 2 nd c b f = 5.
3 7.3 The Fundmentl Theorem Theorem 6 (Fundmentl Theorem of Clculus (First Form)). Let E = {c 1,..., c m } [, b] nd f, F : [, b] R such tht () F is continuous on [, b], (b) F (x) = f(x) for ll x [, b] \ E, (c) f R[, b]. Then f(x) dx = F (b) F (). (1) Proof. Let ɛ > 0. Assume the set E hs m distinct points. Let P ɛ be prtition contining the set E, P ɛ = {x 0,..., x n } such tht the Integrbility Criterion holds for f (by prt (c)), U(f; P ɛ ) L(f; P ɛ ) < ɛ. (2) By prts () nd (b), F is continuous on the closed subintervl [x i 1, x i ] of [, b] nd differentible on the open intervl (x i 1, x i ) (ech point of E is some endpoint). Apply the MVT to F on [x i 1, x i ]. Tht is, there exists u i (x i 1, x i ) such tht F (x i ) F (x i 1 ) = F (u i )(x i x i 1 ) = f(u i )(x i x i 1 ), i = 1,..., n. Note tht u i / E. Summing the previous identity from i = 1 to n we obtin nother identity: F (b) F () = [F (x i ) F (x i 1 )] = f(u i )(x i x i 1 ). (3) Since m i f(u i ) M i, where m i nd M i re the infimum nd supremum of f on [x i 1, x i ], the following inequlity holds: L(f; P ɛ ) = m i (x i x i 1 ) f(u i )(x i x i 1 ) M i (x i x i 1 ) = U(f; P ɛ ). Applying (2) nd (3), L(f; P ɛ ) Since ɛ is rbitrry, the conclusion (1) holds. f U(f; P ɛ ) nd L(f; P ɛ ) F (b) F () U(f; P ɛ ). Therefore, F (b) F () f(x) dx < ɛ. Definition: The indefinite integrl of f R[, b] is F (x) = Theorem 7. The indefinite integrl F is continuous on [, b]. x f = x f(t) dt for x [, b] Theorem 8 (Fundmentl Theorem of Clculus (Second Form)). Let f R[, b] nd f be continuous t point c [, b]. Then the indefinite integrl F on [, b] is differentible t c nd F (c) = f(c). Theorem 9. If f is continuous on [, b], then the indefinite integrl F is differentible on [, b] nd F (x) = f(x) for ll x [, b] (F (x) = [ x f(t) dt] = f(x)). Definition: If f is continuous on [, b], then the indefinite integrl F is clled n ntiderivtive of f (F (x) = f(x) for x [, b]). Theorem 10 (Substitution Theorem). Let u : [α, β] R hve continuous derivtive on [α, β] nd u([α, β]) [, b]. If f : [, b] R is continuous, then for x = u(t), β α f(u(t)) u (t) dt = u(β) All functions re continuous in the Substitution Theorem. u(α) f(x) dx.
4 Exmples 5. Show there does not exist continuously differentible function F on [0, 2] such tht F (0) = 1 nd F (2) = 4 nd F (x) 2 for x [0, 2]. (Hint: Assume there does exist function F nd pply the Fundmentl Theorem of Clculus (First Form) to get contrdiction.) 6. Compute the derivtive of F (x) = 0 x (1 + t 2 ) 2 dt. 7. If f is continuous on R, compute the derivtive of G(x) = sin(x) x f(t) dt. 8. Let f : [ 1, 1] R where f(x) = 1 for x [ 1, 0), f(x) = 1 for x (0, 1], nd f(0) = 0. Evlute the indefinite integrl F (x) = x 1 f(t) dt for x < 0 nd for x > 0. Sketch F nd show it is continuous. The function F is NOT { n ntiderivtive of f becuse F is not differentible t x = 0. (See definition of x 1, x < 0, ntiderivtive.) F (x) = x 1, x > Evlute the indefinite integrl 4 1 The Substitution Theorem does not pply to the indefinite integrl not differentible t t = 0. cos( t) dt by pplying the Substitution Theorem with u(t) = t. t 4 0 cos( t) t dt becuse u(t) = t is
5 Lebesgue s Integrbility Condition Definition: Set A is sid to be null set or set of mesure zero in R if for every ɛ > 0 there exists countble collection of open intervls {( k, b k )} tht contin A nd the sum of their lengths is less thn or equl to ɛ. Tht is, A ( k, b k ) nd (b k k ) ɛ. Exmples: (1) Any finite set of rel numbers is null set. (2) The set of rtionls Q is null set. Proof of (2): The rtionls re countble. Therefore, there exists n enumertion of the rtionls, Q = {r 1, r 2,..., r n,...} = {r k}. Ech rtionl cn be contined in smll open intervl, nd the sum of their lengths stisfies Q ( ) 1 2ɛ 2 k+1 ( r k ɛ 2 k+1, r k + ɛ ) 2 k+1, = ɛ 1 2 k = ɛ 1/2 1 1/2 = ɛ. Definition: If A is null set nd sttement holds for every x B \ A, then it is sid to hold for lmost every x in B or it holds lmost everywhere in B, tht is, the sttement holds.e in B. Theorem 11 (Lebesgue s Integrbility Condition). A bounded function f : [, b] R is Riemnn integrble iff f is continuous lmost everywhere on [, b]. Theorem 12 (Composition Theorem). Let f : [, b] R be Riemnn integrble nd f([, b]) [c, d]. h : [c, d] R is continuous, then the composition h f : [, b] R is Riemnn integrble. If Theorem 13 (Product Theorem). If f, g : [, b] R nd both f nd g re Riemnn integrble, then fg is Riemnn integrble. Proof. Apply the Composition theorem. The function h(x) = x 2 is continuous on ny finite intervl. Then h f = h(f) = f 2 nd h g = h(g) = g 2 re Riemnn integrble. Also (f + g) 2 is Riemnn integrble (why?). Therefore, fg = 1 2 [(f + g)2 f 2 g 2 ] is Riemnn integrble Theorem 14 (Integrtion by Prts). If F, G re differentible on [, b], then F G = F G b GF. The preceding formul is the wellknown integrtion by prts formul: udv = uv b vdu.
6 Exmples: 10. Show tht ny countbly infinite collection of rel numbers {x 1, x 2,..., x n,...} is null set. 11. Thome s function, f : [, b] R, restricted to finite intervl, with 0 < < b < (Exmple (h), pp ), is bounded function, continuous on the irrtionls in [, b] nd discontinuous on the rtionls in [, b]. In prticulr, the function is zero on the irrtionls nd positive on the rtionls. Thus, Thome s function f is continuous on [, b] \ Q. Is f R[, b]? If yes, wht is the vlue of f? 12. Let A = [0, 1] \ {1/2, 1/3, 1/4,..., 1/n,...}. Consider the chrcteristic function X A : [0, 1] R. Is X A R[, b]?
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