Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous 2 hours of lectures by prctising with the concepts. Your work will be grded nd count 20% towrds the finl grde (with the remining 80% determined by the exm). See the web site http://www.mths.tcd.ie/ richrdt/ma2224/ for useful informtion. Contents 0.1 Wht is the best theory for integrtion?....................... 1 0.2 A look bck t the Riemnn integrl........................ 2 0.3 Chnged pproch for the Lebesgue integrl.................... 4 0.4 The Lebesgue integrl................................ 5 0.1 Wht is the best theory for integrtion? You ve seen the Riemnn integrl nd you know tht b f(x) dx mkes sense when f : [, b] R is continuous function on finite closed intervl in R. There is theory for wht the integrl mens bsed on Riemnn sums, or on upper nd lower sums, nd fter certin mount of bother we rrive t point where we cn just use integrls. Then we don t seem to hve to worry bout the theory ny more. Yet the point of wht we re going to do here is to do ll this theory gin in different wy. Why would we do it gin? Well it is not so much tht there ws nything wrong with the Riemnn integrl. It gives the right nswer nd is quite stisfctory for nice (continuous) functions on finite closed intervls. There re then vrious extensions of the notion of the integrl to include vrious improper integrls, like 1 e x dx = lim b b 1 e x dx. There re plenty of functions where the Riemnn integrl mkes sense but the integrnd is not continuous, like functions with finite number of finite jump discontinuities. Sy 4 f(x) dx 0 where 1 x < 1 f(x) = x 2 1 x 2 5e x 2 < x
2 2017 18 Mthemtics MA2224 From prcticl point of view, there is no rel bother with the Riemnn integrl for while, but eventully it runs out of stem. For exmple, in Fourier series things work out very nicely if we work in the Hilbert spce of squre-integrble functions nd we get into bother with tht unless we go beyond the Riemnn integrl. In differentil equtions nd prtil differentil equtions, similr limittions come into ply. In probbility theory, the Riemnn integrl is bsiclly indequte. There re rivls, but the Lebesgue theory of the integrl is lmost universlly used, nd our im is to try nd explin wht it is bout. We will stick to the one vrible cse lot, becuse we cn drw pictures nd grphs more esily, nd some of the ides seem more concrete there, but ctully most of wht we do cn be dpted rther esily to very generl settings. 0.2 A look bck t the Riemnn integrl One strts with the picture tht the integrl b f(x) dx should be the totl re between the grph y = f(x) nd the x-xis where f(x) > 0 minus the re of the region between the grph nd the x-xis where f(x) < 0. The bother comes in trying to explin tht precisely, nd indeed from the fct tht we cn t llow relly bd integrnds f(x). In the Riemnn theory we prtition the intervl [, b] into (mny smll) subintervls. We tke division points 1 < 2 < < n 1 where < 1 nd n 1 < b. To mke our life more esy we write 0 = nd n = b. We get picture like this where the verticl lines re t the positions x = i (i = 0, 1, 2,..., n).
Introduction 3 25 20 15 10 5 0.5 1 1.5 2 We squre off the verticl strips ccording to one of these strtegies: (i) Riemnn sum. We pick vlue x i where i 1 x i i (i = 1, 2,..., n) nd squre off the strips t level y i = f(x i ). Then the Riemnn sum for this setup is f(x i )( i i 1 ) = y i ( i i 1 ) (We define the integrl then s the limit of these sums s the mesh size mx 1 i n ( i i 1 ) goes to zero provided there is limit. Using uniform continuity of continuous functions on finite closed intervls, one cn show tht if the integrnd f(x) is continuous on [, b] nd provided the mesh size is smll enough, then ll the Riemnn sums will be close to hving the sme vlue, nd hence tht the limit exists by invoking Cuchy sequence rgument.) (ii) Upper sum. We squre off ech strip t the highest point on the grph within i 1 x i, or to be more precise t the lowest horizontl level y = y i tht is bove ll the points of the grph in i 1 x i. We tke nd then tke the upper sum y i = sup{f(x) : x [ i 1, i [} y i ( i i 1 ) These upper sums re ll too big, or bigger thn wht we wnt the integrl to be. (iii) Lower sum. We squre off ech strip t the lowest point on the grph within i 1 x i, or to be more precise t the highest horizontl level y = y i tht is below ll the points of the grph in i 1 x i. We tke y i = inf{f(x) : x [ i 1, i ]} nd then tke the lower sum y i ( i i 1 )
4 2017 18 Mthemtics MA2224 These lower sums re ll too smll, or smller thn wht we wnt the integrl to be. One cn prove tht upper sums get smller if we refine the prtition of [, b], tht is if we dd in more division points into the list = 0 < 1 < 2 < < n 1 < n = b. On the other hnd lower sums get bigger when we refine the prtitions nd it is possible to prove tht ll lower sums re smller thn ll upper sums. We cn sy tht the integrl b f(x) dx exists if we cn get the upper sums nd lower sums s close together s we like nd the integrl is the unique number tht is trpped between ll lower sums nd ll upper sums. lower sum b f(x) dx upper sum [This pproch is followed through in the notes from MA2321 2016 17, http://www. mths.tcd.ie/ dwilkins/courses/ma2321/ (Section 3) by Professor Wilkins, where it is lso discussed for more generl bounded functions on finite closed intervl [, b], not just continuous functions. One of the results is tht b f(x) dx mkes sense for continuous functions f : [, b] R. He uses the term Drboux integrl for this pproch to the Riemnn integrl.] 0.3 Chnged pproch for the Lebesgue integrl Here is rough guide to wht we will do in this course. There re lots of detils to be filled in lter. Insted of prtitioning the intervl [, b], the domin over which we re integrting, we insted prtition the rnge. So we drw horizontl lines t (sy) spcing 1/N prt, mybe t y = 0, 1/N, 2/N,... nd lso t y = 1/N, 2/N,.... As usul there is trde off between ccurcy nd the mount of effort, but we should hve N very big s we re going to mke n pproximtion, nd the pproximtion will be better if N is big. We then tke the prts of the grph between the levels we hve chosen. Here is picture for 3π f(x) dx = 3π sin x dx. π/2 π/2
Introduction 5 For ech horizontl level, sy 0.5 y < 0.75 s in this picture, we replce the grph by constnt nd integrte tht constnt. (We might tke the constnt vlue to be 0.5, perhps. Of course, we hve not divided stuff up into nrrow enough bnds here to get ny kind of good result. We should use more nd nrrower bnds.) We might sy then tht the integrl of those prts of the grph should contribute f(x) dx = 0.5 dx x {x:0.5 f(x)<0.75} x {x:0.5 f(x)<0.75} nd tht should be the totl length of {x : 0.5 f(x) < 0.75} times the height 0.5. We dd up the result of the sme clcultion over ll the different bnds to get n pproximte vlue for 3π f(x) dx nd then we tke limit s the bnds get nrrower to define the integrl. π/2 In this cse you my be ble to see tht the nswer we get from the pproximtion is wrong by t most 0.25 (the heights of the bnds we used) times the totl width 3π π/2. Now wht re the obstcles to crrying out this strtegy? One is tht the sets we del with, like {x : 0.5 f(x) < 0.75}, cn get complicted. Then we need to be sure we know wht we men by the lengths of these sets. Well, in the exmple, we just got 3 little intervls nd we re on sfe enough ground dding up the 3 lengths, but there re more complicted exmples. If you took 1 sin(1/x) dx the set {x : 0.5 sin(1/x) < 0.75} would hve infinitely mny 0 pieces. Tht mens we need to be ble to dd up the infinitely mny lengths to get the totl length of {x (0, 1] : 0.5 sin(1/x) < 0.75} (nd of other similr sets). So we will need to be ble to find lengths of rther complicted sets nd this does end up being rther tricky. In fct this concept of mesuring the length of (possibly) complicted set is probbly more thn hlf the difficulty to be sorted out in order to mke progress on the Lebesgue theory of the integrl. 0.4 The Lebesgue integrl In fct you my not see so clerly tht this is the strtegy becuse of the following:
6 2017 18 Mthemtics MA2224 (i) The difficulty mentioned just now of mesuring the lengths of complicted sets tkes bit of getting round. There re quite few detils in tht. We strt with rther esy sets nd build up to more complicted ones. We lso hve to llow infinite lengths (sy to del with the length of the whole rel line, or n intervl tht hs ± s one endpoint). And we need to be creful bout not llowing sets tht re too wild otherwise things won t work. (ii) For positive (or never negtive) functions we follow more or less the strtegy we hve outlined bove, though it will be expressed in terms of tking limits of sequences of integrls of simple functions. Agin we cn end up with s the vlue of the integrl of positive function. But for functions tht cn be negtive we dopt more restrictive pproch, not llowing ± s the integrl. Wht we will do is essentilly integrte the function where it is positive seprtely from where it is negtive (using the pproch for positive functions) nd then combine these vlues (provided both re finite). Perhps it will help to refer bck to this little sketch now nd gin so s to see how fr we hve progressed. R. Timoney Februry 19, 2018