Lecture 1: Introduction to integration theory and bounded variation

Transcription

1 Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You took clss clled Mth 1 nd lerned something like this: Given bounded rel-vlued function f on n intervl [, b], we wnt to define f(x)dx. set with So wht we do is tht we prtition the intervl [, b]. A prtition of [, b] is finite = {x 0, x 1,..., x n } = x 0 < x 1 <... < x n = b. For every prtition, we define upper nd lower Riemnn sums: U (f) = n (x j x j 1 ) sup x [x j 1,x j ] f(x), nd L (f) = n (x j x j 1 ) inf f(x), x [x j 1,x j ] We tht prtition Q refines prtition if Q nd if Q refines we hve the inequlities U Q (f) U (f), nd L (f) L Q (f). (Why?) Then we define upper nd lower integrls by U b (f) = inf U (f), nd L b (f) = sup L (f). 1

2 If the two numbers, U b (f) nd L b (f) mtch, then we sy tht f is integrble nd denote this number by Otherwise, we sy tht f is not integrble. f(x)dx. The integrl bove is clled the Riemnn integrl. Are there ny drwbcks to this set-up. One thing we could worry bout is tht we hven t entirely worked out which functions re Riemnn integrble nd which re not. [We could do tht this qurter]. But we know something bout it. For instnce: roposition 1 A function f which is continuous on [, b] is Riemnn integrble. roof: Since [, b] is compct, we know tht f is uniformly continuous on [, b]. Hence for every ɛ > 0, there is δ > 0 so tht whenever x, y [, b] nd x y < δ, it must be tht f(x) f(y) < ɛ. Then if is prtition so tht x j x j 1 < δ for every j, we must hve (Why?) Thus for every ɛ > 0, we hve U (f) L (f) < ɛ(b ). U b (f) L b (f) < ɛ(b ). We conclude the difference must be zero, hence f is Riemnn integrble. But continuity is not relly necessry condition. For instnce, bounded function f with just finitely mny discontinuities is still Riemnn integrble. (Why?) A slightly worse f is given by f(x) = { 1 if x Q; 0 if x / Q. This is disster. Since the rtionls, Q, re dense in the rels, every point is point of discontinuity of f, nd in ny subintervl of [, b], we find point where f is 1 nd point where f is 0. We get U b (f) = b, while L b (f) = 0. 2

3 Thus the function f is not integrble. Wht s relly wrong with tht f? The rtionls Q re relly firly smll set. They re countble wheres the rels re not. Everywhere but on the rtionls, the function f is zero. Why should the rtionls hve so much weight in determining the integrl. (Answer: Becuse it s Riemnn integrl, loction, loction, loction.) A more prcticl minded person might view such question s nonsense. A function like f cn t relly pper in nture. How cn you even mesure whether number is rtionl or irrtionl? These prcticl people might think tht we should be interested only in much nicer functions nd wht identities we cn discover bout their integrls. For instnce, if we tke f(x) insted to be continuous function on [, b] nd let F (x) be its ntiderivtive then we hve f(x)dx = F (b) F (), the fundmentl theorem of clculus. (Secretly, this is some identity bout telescoping sums.) We cn get still fncier identities out of this. To wit: roposition 2 (Integrtion by prts) Let f nd g be once continuously differentible functions on [, b]. Then f(x)g (x)dx = f (x)g(x)dx + f(b)g(b) f()g(). roof We use the fundmentl theorem of clculus nd the product rule for derivtives to get [f(x)g (x) + f (x)g(x)]dx = f(b)g(b) f()g(). Integrtion by prts is definitely cool. It is somewht deeper sttement bout telescoping sums thn the fundmentl theorem nd lets us derive lots of identities nd estimtes. Is continuity of the derivtive relly necessry for integrtion by prts to work? This is in fct wht we ll be focusing on for the first two weeks of the course. But why do we cre? Is it of ny prcticl importnce to know which functions integrtion by prts works for? Wht this course is lrgely bout is the question of wht estimtes (inequlities) depend on. When we prove the integrtion by prts identity, we re ctully proving inequlities becuse both sides of the eqution re limits. When we hve theorem, we know the limits re equl, but we might cre how fst they re converging to 3

4 one nother. Wht does tht depend on? Continuity is complicted thing to quntify. Even ccepting tht the continuity of derivtives hs to be uniform on [, b], the input into ny one of our proofs is relly how δ depends on ɛ. We might like everything to be bsed on simple quntity insted. rt of wht we re going to see over the course of this term (nd studying nlysis in generl) is tht there s perfect universl definition of integrl tht is best for every sitution. Rther, we should introduce different integrls for different purposes nd usully there s nice clss of functions ech one works for. We should try to see how everything depends on the definition of the clss. But the simpler nd more generl the clss of functions is, the better. We begin the course by introducing clss of functions for which we ll see tht integrtion by prts works very well. These re the functions of bounded vrition. We proceed to define them: Given f : [, b] R, nd = {x), x 1,..., x n } prtition of [, b] s before, we define the vrition of f with respect to the prtition by V (f, ) = N f(x j ) f(x j 1 ). It is esy to see tht if Q refines then V (f, Q) V (f, ). We define the totl vrition of f on [, b] by V b = sup V (f, ). If V b (f) <, we sy tht f is of bounded vrition on [, b], or sometimes write f BV ([, b]). To work in the clss of functions of bounded vrition relly mens to use these definitions, tht is to mke ll estimtes depend only on totl vritions. I ll close this lecture with some exmples. Exmple 1: Let f be monotone incresing function on [, b]. Tht is, suppose tht if x < y, we hve f(x) f(y). Then the sum defining V (f, ) telescopes so tht we hve V (f, ) = f(b) f(), 4

5 regrdless of. Then this quntity lso gives the totl vrition V b f. Exmple 2: You re probbly dying to know wht hppens with nice functions. Let f be once continuously differentible on [, b]. Then V b f = f (x) dx. How do we know. ick prtition. Then by the men vlue theorem for ech prt [x j 1, x j ], there is point c j in the intervl so tht f (c j )(x j x j 1 ) = f(x j ) f(x j 1 ). Thus the vrition V (f, ) is Riemnn sum of f on the prtition. As we tke the sup of vritions we rrive t the integrl. Thus using the theory of bounded vrition on just nice functions mens tying one hnd behind our bcks nd mking everything just depend on totl vritions, tht is on integrls of bsolute vlues of derivtives. Finlly: Exmple 3 Let s recll our old enemy f(x) = { 1 if x Q; 0 if x / Q. We cn pick prtition where the x j s lternte between being rtionl nd irrtionl. This mkes the vrition rbitrrily lrge. We see tht this function is not of bounded vrition, which is just the tip of the iceberg of wht is wrong with it. 5