Riemann is the Mann! (But Lebesgue may besgue to differ.)


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1 Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ > 0 there is some δ ɛ > 0 such tht mesh(p ) < δ ɛ = U f (P ) L f (P ) < ɛ for prtitions P of [, b]. By the Infrstructure of Integrtion hndout this gurntees tht sup { L f (P ) P is prtition of [, b] } = inf { U f (P ) P is prtition of [, b] } (1) with the common vlue denoted by b f, nd furthermore [mesh(p n )] 0 = lim n L f (P n ) = b f = lim n U f (P n ). (2) We strt by introducing definition of lower nd upper sums (corresponding to prtition of [,b]) for functions which re not necessrily continuous on [, b]. With these definitions in hnd the centrl question is this: which other functions on [, b] (besides the continuous ones) re nice enough so tht eqution (1) nd the impliction (2) hold? Before I tell you bout RiemnnLebesgue theorem which nswers tht question, let us introduce the definitions of lower nd upper sums s promised. 1 Definition. Given bounded function g : [, b] R nd prtition P = {, x 1, x 2,..., x m 1, b} of [, b], define n ( ) L g (P ) = inf g i [x i 1,x i ] nd where i is the length of [x i 1, x i ]. U g (P ) = i=1 ( ) n sup g [x i 1,x i ] i=1 1 i
2 2 Observtion. The key difference between this new definition nd the one used for continuous functions is tht we use inf [xi 1,x i ] g insted of min [xi 1,x i ] g, nd sup [xi 1,x i ] g insted of mx [xi 1,x i ] g. Extreme Vlue theorem which gurntees the existence of miniml/mximl vlue requires continuity, nd when the function hs no minimum/mximum but is bounded, it hs excellent substitutes: the inf nd the sup! Of course you lredy know tht when min exists it is the inf, nd similrly for mx. 3 Definition. A bounded function g : [, b] R is sid to be Riemnnintegrble on [, b] if sup { L g (P ) P is prtition of [, b] } = inf { U g (P ) P is prtition of [, b] }. In this cse the common vlue is denoted by b g nd is clled the Riemnn integrl of the function. When = b, we sy tht ny function on [, b] is Riemnnintegrble nd f = 0. 4 Observtion. Note tht to demonstrte tht bounded function g : [, b] R is Riemnnintegrble on [, b] ll one hs to do is show tht for ny ɛ > 0 there exist prtitions P nd Q such tht 0 U g (Q) L g (P ) < ɛ. This is consequence of Theorem 1 on Infrstructure of Integrtion hndout. 1 Problem. Show tht ny incresing bounded function g : [, b] R is Riemnnintegrble. Do the sme for decresing bounded functions. Wht bout impliction (2), you sy? Well, believe it or not, it comes for free, but the proof of this remrkble result, which we stte below, belongs to Rel Anlysis course, nd is too involved to be given here. 5 Theorem. (Riemnn s Theorem) If bounded function g : [, b] R is Riemnnintegrble on [, b] then 2 Riemnn sums [mesh(p n )] 0 = lim n L g(p n ) = b g = lim n U g(p n ). 6 Definition. Given prtition P = {, x 1, x 2,..., x n 1, b} of [, b], suppose one picks n evlution point z i in ech of the n closed intervls [x i 1, x i ] generted by the prtition. The object thus obtined is pointed prtition P of [, b]. Given bounded function g : [, b] R nd pointed prtition P of [, b], the Riemnn sum R g (P ) for g corresponding to P is defined s follows n R g (P ) = g(z i ) i i=1 2
3 7 Observtion. The key difference between the definition of Riemnn sum nd the definition of upper/lower sums is tht we use g(z i ) insted of sup [xi 1,x i ] g (nd inf [xi 1,x i ] g). In other words we build rectngles on the subintervls of [, b] by using vlues of the function g t evlution points s heights, rther thn using best upper/lower estimtes s heights. When g is continuous on [, b], nd prtition P is given, the Extreme Vlue theorem gurntees the existence of choices of evlution points which will mke the corresponding Riemnn sum equl the upper sum (similrly, the lower sum). Do you see how? The following corollry to Theorem 8 shows the relevnce of Riemnn sums. 8 Theorem. (Riemnn Sum Theorem) If bounded function g : [, b] R is Riemnnintegrble on [, b] then [mesh(p n )] 0 = lim n R g(p n) = Proof. This is trivil by squeeze theorem nd Theorem 8, since clerly L g (P ) R g (P ) U g (P ) no mtter which evlution points re chosen. 9 Observtion. Wht this theorem sttes is tht once one knows tht given function is Riemnnintegrble, one does not need to bother to use cumbersome upper/lower sums to pproximte the integrl. All one hs to do is pick ny sequence of prtitions whose mesh goes to zero, nd pick one s fvorite evlution points for ech prtition, nd then the corresponding Riemnn sums (re esy to clculte nd) converge to the integrl. b g. 3 Mesure for Mesure A big question remins: Which bounded functions on [, b] re Riemnn integrble? The complete nswer to it is given by the following stunning theorem which my be considered to be genesis of whole subject clled Mesure Theory. 10 Theorem. (RiemnnLebesgue Theorem) A bounded function g : [, b] R is Riemnnintegrble on [, b] if nd only if the set of discontinuities of g cn be covered by sequence of open intervls with totl length s smll s desired. In other words, ɛ > 0 there is sequence I 1, I 2, I 3,... of open intervls such tht 1. All points where g is not continuous lie in the set I 1 I 2 I The sum of the lengths of I 1, I 2, I 3,... (treted s limit of n infinite series, of course!) is less thn ɛ. 3
4 11 Definition. Subsets of R tht cn be covered by sequence of open intervls with totl length s smll s desired re sid to hve mesure zero. In prticulr, empty set hs mesure zero. 2 Problem. 1. Show tht subset of set of mesure zero hs mesure zero lso. 2. Show by induction tht every finite subset of R hs mesure zero. 3. Show tht every infinite countble subset of R hs mesure zero. (Recll tht n infinite set is countble if there is bijection between it t N; in other words, if there is sequence which contins exctly the elements of the set ppering without repetitions.) (Hint: if length of I k is α 2 k, wht is the sum of lengths of I 1, I 2, I 3,...?) 12 Corollry. If bounded function g : [, b] R is continuous t every point of [, b] with the exception of finitely mny points, then it is Riemnnintegrble on [, b]. The sme is true if if the set of discontinuities of g is infinite but countble. 13 Theorem. Chnging the vlues of bounded function g : [, b] R t finite set of points does not ffect Riemnnintegrbility (or the lck of it) of the function. In the cse when the function is Riemnnintegrble, such chnge does not chnge the vlue of the Riemnn integrl b g itself. Proof. The proof I know for this theorem is not hrd, but is rther technicl. I will show you the ide behind it in clss, but will not present it here. 14 Corollry. The concept of Riemnnintegrbility cn be defined for bounded functions on ll finite intervls, including the intervls tht re open or hlfopen. Proof. The key point is tht given such function h on n intervl with endpoints nd b, we cn mke it into bounded function on [, b] by defining it ny wy we like t nd t b. If by doing it one wy we get Riemnnintegrble function g on [, b] with b g, then doing it in ny other wy we get Riemnnintegrble functions with exctly the sme vlue of the integrl, by Theorem 13 of course! 15 Exmple. Recll tht the function 1 n, if x = m n in reduced form f(x) = 0, otherwise is continuous t every irrtionl number nd zero, nd is discontinuous t every nonzero rtionl number. Still, this function is good enough to be Riemnnintegrble to ny [, b]. 4
5 The following theorem is not s trivil s it my seem. In fct it is so nontrivil tht we omit the proof nd suggest tht you tke Rel Anlysis to see it! 16 Theorem. No intervl of positive length hs mesure zero. 3 Problem. 1. Prove tht union of two sets of mesure zero hs mesure zero. 2. Give n exmple of fmily of sets { S r r R } ech of which hs mesure zero, such tht the union r R S r does not hve mesure zero. 3. (Extr Credit) Suppose tht S 1, S 2, S 3,... re sets ll of which hve mesure zero. Prove tht their union lso hs mesure zero. This is usully expressed by sying tht countble union of sets of mesure zero hs mesure zero. 17 Corollry. The set of irrtionl numbers in ny intervl of nonzero length does not hve mesure zero. Proof. The set of rtionl numbers in n intervl J of nonzero length is countble nd so hs mesure zero. If the set of irrtionls in J lso hd mesure zero, so would their union (by Problem 3) which equls ll of J, nd does not hve mesure zero by Theorem Exmple. Recll tht the function f(x) = { 1, if x Q 0, otherwise is discontinuous t every point of R. Hence it is not Riemnnintegrble on ny intervl of positive length. 4 For rectngles in R 2 Perhps it should not come s complete surprise tht the results of the first section cn be boosted to higher dimensions with little trouble. Here is wht one hs to do to formulte the corresponding results in 2dimensions. 1. Replce bounded closed/open intervls with bounded closed/open rectngles respectively. 19 Definition. Given intervls I nd J in R, the set { (x, y) x I, y J } is clled rectngle nd is denoted by I J. Rectngle I J is bounded/closed/open if both I nd J re bounded/closed/open (respectively). 5
6 2. Lengths of intervls get replced by res of rectngles. 3. Functions considered re bounded functions of the form g : K R, where K is bounded (most often closed) rectngle in R The integrl of such function is denoted by K g. The results of Problems 2 nd 3 extend to higher dimensions wordforword. So should their proofs, fter you hve performed the terminology switch described bove. The following twodimensionl theorem is new. 20 Theorem. Let f : [, b] R be continuous function. Then the grph of f hs mesure zero in R 2. Recll tht the grph of f is the set { (x, f(x) ) x [, b] }. Proof. We know tht since f is continuous on [,b], it is Riemnnintegrble on [, b]. In prticulr for every ɛ > 0 there is prtition P of [, b] such tht U f (P ) L f (P ) < ɛ. Yet it is esy to see tht U f (P ) L f (P ) is the sum of the res of finitely mny closed rectngles whose union contins the grph of f (do you see how?). By enlrging ech rectngle bit in both direction (mking sure its re does not double) we cn drop off the boundries of the rectngles nd see tht the grph of f cn be covered by union of finitely mny open rectngles with the sum of res equl t most 2ɛ. This is good enough, since ɛ ws n unspecified positive number. 6
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