The Riemann Integral

Transcription

1 The Riemnn Integrl July 24, pper nd lower sums A prtition of closed intervl [, b] is subset P = {x 0, x 1,...,x n } of [, b] with = x 0 < x 1 < < x n = b. A prtition divides the intervl [, b] into subintervls I j = [x j 1, x j ], for j = 1,..., n. We ll cll I j the jth subintervl for the prtition. Now let f be bounded function on the intervl [, b], nd let P = {x 0,...,x n } be prtition of [, b]. Let m j = inf{f(x) : x I j }, M j = sup{f(x) : x I j } nd define the upper nd lower sums for the function f with respect to the prtition P by L(f; P) = m j (x j x j 1 ) nd (f; P) = M j (x j x j 1 ). Since m j M j for every j, it follows tht L(f; P) (f; P). We next sk how the upper nd lower sums will be ffected if we dd single new point x to the prtition P, to obtin new prtition P. Since x is not equl to ny of the originl prtition points, it s in the interior of one of the subintervls, sy x j 1 < x < x j. 1

2 Now most of the terms in the new upper nd lower sums for the prtition P re the sme s the corresponding terms in the originl sums. The only difference is tht the term corresponding to the subintervl I j is split into two terms for the subintervls [x j 1, x ] nd [x, x j ]. Thus, in the upper sum, the term M j (x j x j 1) gets replced by the two terms where M (x x j 1) + M (x j x ) M = sup{f(x) : x [x j 1, x ]}, M = sup{f(x) : x [x, x j ]}. Notice tht M nd M re both less thn or equl to M j, so M (x x j 1)+M (x j x ) M j (x x j 1)+M j (x j x ) = M j (x j x j 1). Therefore, we hve shown tht (f; P ) (f; P), nd similr resoning for the lower sums gives We hve proved L(f; P ) L(f; P). Lemm 1.1. If the prtition P is obtined from P by dding single point, then L(f; P) L(f; P ) (f; P ) (f; P). We next sk wht hppens to the upper nd lower sums if you dd mny points to given prtition. Let s cll prtition P refinement of the prtition P if P P. In this cse, either P = P, or you cn get from P to P in severl steps, dding one point t ech step. By Lemm 1.1, ech time you dd point, the lower sums get bigger, nd the upper sums get smller. We ve proved Theorem 1.2. Let f be ny bounded function on [, b] nd let P nd P be prtitions of [, b] such tht P is refinement of P. Then L(f; P) L(f; P ) (f; P ) (f; P). 2

3 Theorem 1.2 gives us wy of compring upper nd lower sums for prtitions tht re completely independent of ech other. Let P nd Q be ny two prtitions of [, b]. Then the prtition P Q is refinement of both P nd Q, so Theorem 1.2 gives We therefore hve L(f; P) L(f; P Q) (f; P Q) (f; Q). Corollry 1.3. For ny bounded function f on [, b] nd ny prtitions P nd Q of [, b] we hve L(f; P) (f; Q) 2 Integrbility A consequence of Corollry 1.3 is tht the set of ll lower sums is bounded bove, while the set of ll upper sums is bounded below. We define the lower nd upper integrls of bounded function f by L f = sup{l(f; P)} f = inf{(f; P)} where the supremum nd infimum re tken over ll prtitions P of [, b]. Corollry 1.3 ensures tht both the lower nd upper integrls re finite, nd tht L f It s possible tht the bove inequlity is strict. In this cse, we sy tht the function f is not integrble. We ll sy f is integrble if the upper nd lower integrls re equl, nd it this cse, we define f to be the common vlue of the upper nd lower integrls. Exmple 2.1 (Constnt functions). Let f hve the constnt vlue c on the intervl [, b]. Then for ny prtition P, we hve m j = M j = c 3

4 on ech subintervl, so so nd similrly L(f; P) = Therefore, f is integrble, nd L = = c m j (x j x j 1 ) c(x j x j 1 ) (x j x j 1 ) = c(x n x 0 ) = c(b ) f = c(b ), f = c(b ). f = c(b ). Exmple 2.2 (A non-integrble function). Define f on [0, 1] by f(x) = 0 if x is rtionl, nd f(x) = 1 if x is irrtionl. Let P be ny prtition of [0, 1]. Since the jth subintervl contins both rtionl nd irrtionl numbers, we hve m j = 0 nd M j = 1. Therefore L(f; P) = 0(x j x j 1 ) = 0 nd (f; P) = 1(x j x j 1 ) = 1 so the lower integrl is 0 nd the upper integrl is 1. Therefore f is not integrble 4

5 3 A Cuchy criterion for integrbility Theorem 3.1 (Cuchy Criterion). Let f be bounded function on [, b]. Then f is integrble if nd only if for every ε > 0 there is prtition P such tht (f; P) L(f; P) ε. Proo Suppose f is integrble. Let ε > 0 be given. By definition of integrbility, we cn pproximte the integrl to within ε from bove nd below 2 by upper nd lower sums respectively. Thus, there re prtitions P 1 nd P 2 such tht nd (f; P 1 ) Adding these two inequlities gives f < ε 2 f L(f; P 2 ) < ε 2. (f; P 1 ) L(f; P 2 ) < ε. Let P = P 1 P 2. Then P is refinement of P 1 nd P 2, so Theorem 1.2 gives nd so L(f; P 1 ) L(f; P) (f; P) (f; P 2 ) (f; P) L(f; P) (f; P 1 ) L(f; P 2 ) < ε. For the converse, let ε > 0 be rbitrry, nd let P be s in the sttement of the theorem. Then nd since ε > 0 is rbitrry, f (f; P) L(f; P) + ε L f L f + ε Since the reverse inequlity lwys holds, the upper nd lower integrls re equl, so f is integrble. 5

6 4 Integrbility of monotone functions Theorem 4.1. If f is monotone on [, b], then f is integrble on [, b]. Proo We ll prove the theorem under the ssumption tht f is incresing. The proof for decresing functions is similr. For ny positive integer n, let P n be the prtition tht divides [, b] into n subintervls of equl length b. To be explicit, the prtition points re n x j = + j b n. Since f is incresing, the infimum nd supremum of f over the jth subintervl [x j 1, x j ] re m j = f(x j 1 ), M j = f(x j ). The upper sum is (f; P n ) = M j (x j x j 1 ) = b n f(x j ) nd the lower sum is so L(f; P n ) = b n (f; P n ) L(f; P n ) = b n f(x j 1 ) (f(x j ) f(x j 1 )) = b n (f(x n) f(x 0 )) = b (f(b) f()). n Now let ε > 0 be rbitrry. Choose n sufficiently lrge tht (b )(f(b) f()) n < ε. Then (f; P n ) L(f; P n ) < ε. By the Cuchy criterion for integrbility, f is integrble. 6

7 Exmple 4.2. We will evlute the integrl 1 x. Note first tht, since the 0 function f(x) = x is monotone, integrbility is gurnteed by the previous theorem. Fix nturl number n, nd let P n be the prtition of [0, 1] with prtition points x j = j/n. Then nd similrly (f; P n ) = = x j (x j x j 1 ) j 1 n n = 1 j n 2 n(n + 1) = 2n 2 = n L(f; P n ) = n. Since the integrl is greter thn or equl to every lower sum nd less thn or equl to every upper sum, we hve n 1 0 x n. Since this holds for every positive integer n, it follows tht 1 0 x = Integrbility of continuous functions Theorem 5.1. If f is continuous on [, b], then f is integrble on [, b]. Proo First, the Boundedness Theorem ensures tht f is bounded, so we cn check for integrbility using the Cuchy Criterion. Let ε > 0 be rbitrry. 7

8 By the niform Continuity Theorem, f is uniformly continuous, so there is δ > 0 such tht f(x) f(x ) < ε b for every x, x [, b] with x x < δ. Let P be ny prtition such tht ech subintervl hs length less thn δ. Since ech subintervl I j is closed, the Mx Min Theorem ensures tht the restriction of f to I j chieves mximum nd minimum vlue, so m j = f(s j ) nd M j = f(t j ) for some s j, t j I j, nd since I j hs length less then δ, we hve t j s j < δ. By the wy we chose δ, so M j m j = f(t j ) f(s j ) < ε b (f; P) L(f; P) = M j (x j x j 1 ) m j (x j x j 1 ) = (M j m j )(x j x j 1 ) < ε = ε b (xj x j 1 ) = By the Cuchy Criterion, f is integrble. 6 Properties of integrls Theorem 6.1 (Linerity). is integrble, nd b (x j x j 1 ) ε (b ) = ε b 1. If f nd g re integrble on [, b], then f+g (f + g) = f + 2. If f is integrble on [, b] nd c R, then cf is integrble nd cf = c Proo We ll only prove the first ssertion. The second is similr, but slightly esier, nd will be left s n exercise. Let ε > 0, nd choose prtitions P 1 nd P 2 such tht L(f; P 1 ) > 8 f ε 2 g.

9 nd L(g; P 2 ) > g ε 2. Let P = P 1 P 2. Since, by Theorem 1.2, pssing to refinement will increse the lower sum, the bove inequlities continue to hold when P 1 nd P 2 re replced by P. Therefore L Since ε > 0 is rbitrry, A similr rgument gives so it follows tht (f + g) L(f + g; P) L(f; P) + L(g; P) L L L(f; P 1 ) + L(g; P 2 ) > (f + g) = f + (f + g) (f + g) g ε. f + f + (f + g) = g g f + g. Theorem 6.2 (Monotonicity). If f nd g re integrble on [, b], nd if f(x) g(x) for every x [, b], then f The proof is elementry, nd is left s n exercise. Theorem 6.3 (Additivity). Let f be integrble on [, b], nd let < c < b. Then f is integrble on [, c] nd [c, b], nd f = c f + g. c 9

10 Proo Let ε > 0. Since f is integrble on [, b], there is prtition P of [, b] such tht (f; P) < f + ε. By Lemm 1.1, we my ssume tht P contins c. (If not, dding it will mke the left side even smller.) Let P nd P be the prtitions of [, c] nd [c, b] obtined by intersecting P with [, c] nd [c, b] respectively. Then c f + Since ε > 0 is rbitrry, c f (f; P ) + (f; P ) = (f; P) < c f + c f A similr rgument using lower sums shows so c f + c f L f c f + L f L c c f f + L c f + ε. Since the right side is lwys less thn or equl to the left side, we hve equlity throughout, nd the result follows. Theorem 6.4 (Tringle Inequlity). If f is integrble on [, b], then f is lso integrble, nd b f f. Proo It s esy to check tht for ny prtition P, ( f ; P) L( f ; P) (f; P) L(f; P) so it follows from the Cuchy Criterion tht f is integrble. Also, since f(x) f(x) f(x) 10

11 for every x, monotonicity gives or equivlently f(x) f(x) f f. f(x) 7 The Fundmentl Theorem of Clculus Theorem 7.1 (Fundmentl Theorem of Clculus I). Let f be n integrble function on [, b], nd let F be continuous function on [, b] such tht F (x) = f(x) for every x (, b). Then f = F(b) F(). Proo Let P = {x 0,...,x n } be prtition of [, b]. Applying the Men Vlue Theorem on ech subintervl gives c j (x j 1, x j ) such tht Clerly F(b) F() = (F(x j ) F(x j 1 )) = F (c j )(x j x j 1 ) = f(c j )(x j x j 1 ) m j f(c j ) M j where m j nd M j re the infimum nd supremum of f(x) on the jth subintervl. Therefore L(f; P) F(b) F() (f; P). Since f cn be pproximted to ny prescribed ccurcy by upper nd lower sums, it follows tht f = F(b) F(). 11

12 Theorem 7.2 (Fundmentl Theorem of Clculus II). Let f be integrble on [, b], nd let F(x) = Then F is Lipschitz continuous, nd if f is continuous t x 0, then F is differentible t x 0 nd F (x 0 ) = f(x 0 ). Proo By dditivity, we hve x F(x) F(x 0 ) = x x 0 Since f is integrble, it s bounded, sy f M. Therefore if x 0 < x, the Tringle Inequlity nd monotonicity give x x x F(x) F(x 0 ) = f f M = M( ) x 0 x 0 x 0 which estblishes Lipschitz continuity. If f is continuous t x 0, then for x > x 0 so F(x) F(x 0 ) f(x 0 ) = F(x) F(x 0 ) = 1 = x x 0 f 1 x f 1 x f(x 0 ) x 0 x 0 1 x (f f(x 0 )) x 0 1 x f f(x 0 ) x 0 Let ε > 0. Since f is continuous t x 0, there is δ > 0 such tht f(x) f(x 0 ) < ε whenever < δ. Therefore, for x 0 < x < x 0 + δ, F(x) F(x 0 ) f(x 0 ) 1 x ε = ε x 0 nd therefore lim x x + 0 F(x) F(x 0 ) = f(x 0 ). We leve s n exercise tht the left hnd limit is lso equl to f(x 0 ). 12