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# 476 Class Note for MATH M0070 at Purdue

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MA544 LECTURE NOTES7 FALL 2009 ANTONIO SA BARRETO 1 INTRODUCTION These are my lecture notes for MA 544 fall 2009 They were actually typed in the fall 2005 and have not been changed even though it obviously contains several typos But I guess they are helpful Here is hw 1 for the course Give me a list of typos There are no required text books 1 will use a bit of each of the following references which have been placed on reserve in the library I A Friedman Foundations of Modern Analysis Dover Edition 1982 W Rudin Principles of Mathematical Analysis 3rd Ed McGraw Hill 1986 W Rudin Real and Complex Analysis 3rd Ed McGraw Hill 1986 HIL Royden Real Analysis PrenticeHall 1988 A Torchinsky Real Variables AddisonWesley 1988 2 PRELIMINARIES PROPERTIES OF R COMPACT SETS CONTINUOUS FUNCTIONS ETC We review some properties of the n dimensional Euclidean space R and its topology De nition 21 We denote R the set of real numbers The Euclidean space R is de ned as R z 11 zj E R As we all know R is a normed vector space One norm of an element 1 11 I In E R is HIM m my The distance between two points zy E R is ay HI yll We then de ne open and closed subsets De nition 22 A subset X C R is open iffor every I E X there exists an e gt 0 such that Bze y E R lt e C X A subset F C R is closed 1R F is open Exercise 21 If Um a E A is a family of open subsets of R show that U0 Ua is open Similarly show that if 1f Fm a E A is a family of closed subsets of R show that at Fa is closed The family of all open sets of R is described as a topology on R and R equipped with this family of open sets is called a topological space A subset U C R can be equipped with what is called the relative topology We say that a subset O C U is open with respect to the relative topology if 0 X N U where X C R is open 1 2 SA BARRETO De nition 23 A family Um a E A of open subsets of R is said to be an open cover of a subset K C R if K c U Ua aeA An open cover Um a E A of K is said to have a nite subcover if there exists Ua1Ua2 UaN with a1aN E A such that n K c U Ha j1 Theorem 21 Let K C R The following properties ofK are equivalent 1 K is bounded and closed 2 Every open cover of K has a nite subcover 5 Every sequence of points ofK has a subsequence that converges to a point of K Proof 1 gt 2 Since K is closed R K is open Since K is bounded then there exist ab E R a lt b lt 00 such that K C Q with Q ab gtlt ab gtlt gtlt ab If Um a E A is an open cover of K then Rn K U Um a E A is an open cover of Thus we need to prove that an arbitrary open cover of Q has a nite subcover Suppose by contradiction that this is not true ie there exists an open cover U04 of Q which does not have a nite subcover that covers Let us introduce some important notation Let U C R we de ne the diameter of U as diamU supHx 7 xy E U Then d diamQ b 7 an Let c b 7 a2 The intervals ac and bc determine 2 cubes Qj with diamQj d2 At least one of these sets let s say Q1 cannot be covered by a nite subfamily of U0 Then we divide Q1 in 2 cubes of diameter d4 and repeat the process We obtain a family of cubes B1 j E N such that d d1amBj a B1 3 B2 3 B3 Bj cannot be covered by any nite subfamily of U0 j 12 We then need the following Lemma 21 Let Bj a1jb1jgtlt a2jb2j gtlt anjbnj B1 3 B2 3 B3 then B y 0 j1 Proof Let us suppose rst that n 1 Denote al v aj bl v bj and E ah j E N be the set of lower endpoints of the intervals Let x sup E If mn E N then an S amn S bmn S bm Therefore x S bm for every m Since it is obvious that x 2 am we have x E Ij for all j E N In the general case for each h E 1n there exists xk E 1akjbkj Then x1xn E 00 j1 Bj B Let xquot E Bj Since U04 covers Bj j E N there exists 10 such that x E U040 Since UDO is open there exists 6 gt 0 such that x E Bx6 y E R Z i lt 6 C U040 On the other hand since MA544 LECTURE NOTES FALL 2009 3 xquot E Bj j 12 and diamBj d2j we can pickj large enough such that Bj C Uaol Therefore Bj is covered by the set UDO and this is a contradiction This ends the proof of the rst implication 2 gt 1 Suppose every open cover of K has a nite subcover First we show that K is bounded Let 6 gt 0 andfor each x E K let Bx6 y E R 1 lt 6 The family Bx6 x E K is an open cover ofK and therefore has a nite subcover Bxk6 1 S h S N Hence K C Ug1Bxk6 It is then easy to see that K is bounded Now we want to show that K is closed Let p E R K For each 4 E K let Bq rq be the ball centered in q with radius rq lt Hp7qH4l This gives an open cover ofK Since K is compact there exists on M qN such that K C U lBqjrqj W Now let Bprqj be the hall centered at p with radius rqj It follows from the de nition of rq that Bprq Bqrq 0 Therefore if Bprqj V then V is open p E V and V N W 0 So in particular V N K 0 This proves the implication 2 gt 3 Let yj j E N be sequence contained in Kl For 6 gt 0 the family Bx6 x E K is an open cover of Kl Hence there exists xk 1 S h S N and a subsequence yjm C such that Hij 7ka lt 6 Repeating this process we deduce that has a Cauchy subsequence yjm Therefore it converges Since K is closed its limit belongs to Kl This proves the implication 3 gt 1 IfK is not bounded one can construct a sequence of elements ofK that does not converge 5 also clearly implies that K is closed This ends the proof of the Theorem There is one result that we will need later and it is important to review it Theorem 22 Let Km 1 E A be a family of compacts subset of R such that for any nite subcol lection K0 1 g j g N m K0 y 0 Then may K0 y 0 Proof Suppose by contradiction that aeA K0 0 Fix an element K01 of the family By assumption K041 QweA D ad K0 0 This is equivalent to saying that if GD R Kw Then K01 C UWEAX aa1Ga Since G0 is open and K01 is compact there exist 12 M aN such that K01 C G0 But this implies that K0 ll which is a contradiction D Now we discuss continuity De nition 24 Let X C R We say that a function f z X 7 R is continuous at a point x0 6 X if for every 6 gt 0 there exists a 6 gt 0 6 depending on e and x0 such that 21 for all x E X with 7 xOH lt 6 gt 7 fx0l lt e We say that f is continuous at a subest X C R f is continuous at every point of Xl Exercise 22 Show that a function f z X 7 R is continuous everywhere if and only if for every open subset V C R f 1V is an relatively open subset of Xl Notice that 21 in the de nition of continuity involves two inequalities that is f is continuous at x0 if and only if Hx 7 xOH lt 6 gt fx0 7 e lt lt fx0 6 This motivates the following de nition De nition 25 Let X C R A function f z X 7 R is upper semicontinuous at x0 6 X for any 6 gt 0 there exists 6 gt 0 such that 22 x E X 7x011 lt 6 implies lt fx0 5 4 SA BARRETO Similarly a function f z X 7 R is lower semicontinuous at 10 E X for any 6 gt 0 there exists 6 gt 0 such that 23 16X HI7IOH lt6 implies fzo7eltfzl A function is upper or lower semicontinous in X if it is upper or lower semicontinous at every 10 E X Theorem 23 Let X C R A function f z X 7 R is lower semicontinuous on X if and only if f 1hoo is a relative open subset of X for all h E R Similarly f is upper semicontinuous if and only if f 17oo k is a relative open subset ofX for all h E R Proof Suppose f is lower semicontinuous in X Let h E R and let 10 E f 1hoo Then fzo gt kl Let 6 fzo 7 h and let 6 gt 0 be such that 22 holds Then for 7 Ioll lt 5 gt fzo 7 e kl So I z 7 roll lt 6 C f 1hool Conversely suppose that for every h f 1hoo is open for every h E R In particular for e gt 0 and 10 E X the set f 1fzo 7 600 9 10 is open Hence there exists 6 gt 0 such that for 7 roll lt 6 z E f 1fzo 7 600 Therefore gt fzo 7 e D The following is a very important result which is a direct consequence of Theorem 23 Theorem 24 Let fun a E A be a collection of lower semicontinuous functions Then supX is lower semicontinuous as Similarly if fun a E A be a collection of upper semicontinuous functions Then is upper semicontinuous Proof One just needs to show that if M 7 23 was then f 1k700 7 Ufglltltkoogtgt if fltzgt7gggfaltzgt then f1ltlt7ookgtgt7Ufglltlt7oo7kgtl Let us consider the rst statement If f supaEA fa and z E f 1hoo then and gt kl This implies that gt h for some 1 Otherwise S kl Therefore I E Ua f1h Reciprocally if z E Ua f1hoo then gt h for some 1 In this case gt h and hence I E f 1hool D Next we study the notion of oscillation of a function De nition 26 Let X C R and let f z X 7 R be a bounded function The oscillation of f over X is de ned to be 010 X suplf1 7 fyl 17y 6 X Exercise 23 Let X C R and let f z X 7 R be a bounded function on X C R show that wf7X 13 fI gggf l Next we localize to a point the notion of oscillation of a function De nition 27 Let X C R be an open subset let f z X 7 R be a bounded function and let 10 E a by Let Bzo6 y E R 7 roll lt 6 We de ne the oscillation of f at 10 as were 7 gmmm MA544 LECTURE NOTES FALL 2009 5 Notice that if 61 lt 62 then wfBzo61 S wfBzo62 so w6 wfBzo6 is a bounded nondecreasing function Hence the limit exists Theorem 25 Let X C R be open and let f z X 7 R be a bounded function and let 10 E X Then f is continuous at 10 and only if wfzo 0 Proof If f is continuous at 10 then for every 6 gt 0 there exists 6 gt 0 such that 7 roll lt 6 implies fzo 7 e lt lt fzo 6 This says that supfzz E Bzo6 lt fzo e and inffzz E Bzo6 gt fzo 7 6 Thus wfBzo6 lt 26 Hence wf Io 0 Conversely suppose that wfzo 0 Then ginvd BQo 6 0 That is for e gt 0 there exists 60 gt 0 such that for 6 lt 60 wfBzo6 lt 6 But wfBzo6 suplfz 7 fylzy E Bzo6 In particular for 7 roll lt 5 We have 7 fzol lt e D Theorem 26 Let X C R be open and let f z X 7 R be bounded and let 10 E X The function wfz is upper semicontinuous on X That is xed 10 E X then for all e gt 0 there exists 6 gt 0 such that z E X and lz 7 10 lt 6 implies that wfz lt wfzo 6 Proof Since wfzo lim5n0wf Bzo6 it follows that for any 6 gt 0 there exists 6 gt 0 such that wf7BI075 lt 010510 6 On the other hand recalling that if61 lt 62 then wf Bzo 61 S wf Bzo 62 wf z lt wf Bzo 6 z E Bzo6 This proves the theorem D We deduce from Theorem 26 and Theorem 23 that Corollary 21 Let X C R be open and let f z X 7 R be bounded Then for every 1 gt 0 the set Ea z 6 ab wfz 2 a is closed Theorem 27 Let X C R be open and let f z X 7 R be bounded Let D be the set of points of ab where f is not continuous For 6 gt 0 let E5 z 6 ab wfz 2 6 Then D U135 UEln 6gt0 nEN Proof We know that f is not continuous at 10 if and only ifwf 10 gt 0 But it is easy to see that mm gt0 7 Unzww 2a 04gt0 Since E5 C E7 for 6 gt 7 D U Eln D nEN The structure of the set D is important and deserves a name De nition 28 A set F C R is an E if 00 F U Fn F7 closed 711 5 SA BARRETO AsetGClR isaG5 if G a G7 Gn open 711 So theorem 2 implies that D is an F01 3 SETS OF MEASURE ZERO AND CANTOR SETS We want to make sense of a way to characterize when a subset X C R is small For now we will restrict ourselves to one dimension De nition 31 A subset X C R has content zero and write cX 0 if for an arbitrary e gt 0 there exist a nite collection of open intervals 1112111IN such that N XC U 1 and 11 12M 1m lt e j1 Here Ij denotes the length of the interval Iji We say that X has Lebesgue measure equal to zero and denote mX 0 if for every 6 gt 0 there is a countable family of open intervals 1 j E N such that 00 00 X c U 1 and Em lt 6 j1 j1 In particular cX 0 then mX 0 Exercise 31 Show that one can take the intervals to be closed in the de nition 31 Theorem 31 Sets of Lebesgue measure zero have the following properties 1 IfmX 0 andY C X then mY 0 2 IfX is compact and mX 0 then cX 0 5 fY Xj and mXj 0 for all j E N then mY 0 We leave the proof as an exercise Next we give examples of some 77weird77 sets which will be a rich source of examples and counter examples during the course and therefore it is useful to review themi Let C be the subset of 01 de ned in the following way Let r 6 01 Let E0 01i Let E1 be the set obtained by removing the open interval of length r centered at That is 1 1 E10 lt17 rgtiuiilt1rgt71i So E1 consists of two intervals of length 1 7 r Therefore the sum of the lengths of the intervals that makeup E1 is 1 7 r These can be thought of two copies of 01scaled by the factor 1 7 r Then repeat the process taking into account the scaling factor So let E2 be the closed set obtained from E1 by removing the open interval centered at the middle of each interval of E1 of length 1 7 rr and so on So E2 will consist of four intervals of length i1 7 r In the general case if n gt 1 the set En is obtained from En1 by removing 2 1 open intervals of length 7 rn71 ri So En consists of 2 intervals of length 7 Notice that the sum of the lengths of the intervals that make up the set En is 1 7 r MA544 LECTURE NOTES FALL 2009 7 So we have E03E13E23M3En The Cantor set is de ned by Theorem 32 C is compact and its content is equal to zero Proof C is the intersection of closed sets so it is closed Since it is contained in 01 it must be compacti Given 6 gt 0 pick N such that 1 7 7 N lt 62 Since EN 3 E one can cover E by nitely many intervals with sum of their lengths less than 6 D Now we will construct a similar set but with measure not equal to zero As before let F0 01i Let F1 be the set obtained by removing the middle open interval of length Ti That is F1 7 050 7 TU1T711 So the sum of the lengths of the intervals that makeup E1 is 1 7 7 Now instead of removing the interval of length 1 7 7 7 from each interval we remove lessi Lets say we remove 1 7 7 7 2 from Fli So let F2 be the closed set obtained from F1 by removing the middle open interval of length 1 7 7 7 2 of each interval of Fli So F2 consists of four intervals of length l2 a1 7 7 1 7 7 2 eachi Let F3 be the set obtained from F2 by removing the middle open interval of length l2quot3 from each interval of F2 So F3 consists of 8 intervals of length 2 31 7 7 1 7 7 21 7 T3 In general if we let ln denote the length of each interval of F then Fn1 is the set obtained by removing the middle open interval of length lnrn1 from Fni Lemma 31 For n 2 1 Fn consists of 2 intervals of length n 31 Zn 2 H17r1i j2 Proof By de nition F1 0 1 7 T U 7 1 So the claim holds for n 1i Supposes 31 holds for it holds and we want to show it holds for n 1 Since we have 2n1 intervals of the same size we work with the one which contains 0iTake the interval 0 ln ln01 and remove its middle open interval of length lnin1i So we get two intervals of the same length and the one which contains zero is ln 0 1 7 TWA 0 ln1739rn1i But ln1 l 1 7 TWA This ends the proof of the lemma D As above the modi ed Cantor set is de ned by K 61F Unlike the rst Cantor set C the sums of the lengths of intervals of Fn do not go to zero as n 7gt 00 In this case the sum of the intervals that makeup Fn is Mn 2nLn H 1 7T1 j1 3 SA BARRETO But we know that 14139 0 2 1 J Theorem 33 The Cantor set K is compact has empty inten39m but it does not have measure zero Proof K is closed and contained in 01 Therefore it is compact Since Fn consists of 2 disjoint intervals of length In 2 H1l 7 Tj It is easy to see that K has empty interiori Otherwise K and hence Fn n 12M would contain an interval I of size 6 Since the intervals that makeup Fn are disjoint I would have to be contained entirely inside one of its intervalsi Just pick n so that 1 lt 6 If K had measure zero then for any 6 gt 0 there would be a family In n E N of open intervals such that K C Ur In and 221 lInl lt 6 Since K is compact this collection of intervals can be assumed to be nite so K would have content zeroi Then N KC 11 Mg um UIN Elm lt 5 j1 Pick 6 lt H1l 7 T1 Let U I1 U12 U UINi We claim that there exists m E N such that U D Fm and hence U D for allj 2 mi Suppose this is not true That is for all m E N Gm Fm U 3e 0 Since U is open Gm is closed and therefore compact and Gj D Gj1i By theorem 2 Gmlt FmgtU l mEN mEN That is absurd So U D Fm for some mi Therefore 6 would have to be bigger than or equal to the sum of the lengths of the intervals of Fmi That is e 2 1111 7 Tj gt 1111 7 T1 That is not possible in view of the choice of e D 4 THE RIEMANN INTEGRAL IN R In this section we will discuss the Riemann integrali For simplicity we restrict ourselves to R but all the results proved here can be easily extended to higher dimensions I believe most students in this class have studied this before Our goal is to do a brief review of the methods and to emphasize their limitations A partition 73 of an interval ab a b lt 00 is a nite collection of points 1011 C ab with a Io zj S 1 and b zni We say that a partition 731 is a re nement of if C 731 Let f ab 7 R be a bounded function and let 73 1112 be a partition of abi We de ne the upper and lower sums with respect to a xed partition 73 E1 gt171l U737f 2M1 1H1 11 M1 SUP f17 j1 L73fa ij 1 7 zj mj inf j1 1106 Obviously M737 a S U037 a MA544 LECTURE NOTES FALL 2009 9 Proposition 41 Let W and W1 be partitions of ab Suppose W C W17 ie W1 is a re nement of W Let f a z 7 R be bounded and let M supab andm infab Then mba S M7371 E 1473171 E U731f S UW7f S Ma a Proof Since Mj S M and mj 2 m7 UWf S M02 7 a and 7211 7 a S LW7 We use induction to prove the other inequalities Suppose that W1 contains one more point than W Let W1 W and suppose xquot E xj7xj11ll Let Mj1 511 f17 Mm SUP 757 l i wwil m inf in inf 17W 81744 Recall that Mj suphwidd x and mj inf 7 71fxl It is then clear that Mj1 S My M12 S My mil 2 my mm 2 mi Then UW17 f U037 f Mj11 11 Mj2rj1 1 MjIj1 11 S 0 M7317 f M73716 mj11 11 mj2rj1 1 mjIj1 Ir 2 0 If W1 W consists of h points7 one just needs to repeat this argument h times D De nition 41 The upper and lower integrals of f in a7 12 are respectively 1 b A fdzing73f Handyman De nition 42 We say that f is integrable and denote f E R if Zfdzabfdz fabfdz It is very important to characterize the class of functions Which are integrable in the sense of Riemannl The concept of oscillation of a function over a set is an important one7 not just for the theory of integration Theorem 41 Let f a7 12 7 R be a bounded function The following are equivalent 1 f e R 2 For every 6 gt 0 there exist partitions W and Q of a z such that UfW 7 Lf7 Q lt 6 5 For every 6 gt 0 there exists a partition W of a z such that UfW 7 LfW lt 6 4 For every 6 gt 0 there exists a partition W x07 M xn of a z such that Zw rnl 7 11 lt 67 W wf7lrj7rj1ll j1 Proof The implications 1 ltgt 2 follow almost directly from the de ntion It is obvious that 3 gt 2 just take W Ql If 2 holds7 let W1 W U Ql From Proposition 417 LO Q S Lf7731 S 1105731 Uf773A Then UfW1 7LfW1 lt e The implications 4 ltgt 3 follow from exercise 23 D 10 SA BARRETO Corollary 41 ff a7b 7 R is continuous then f E R Proof Since a7b is compact7 and f is continuous7 f is bounded The compactness of ab also implies that f is uniformly continuous So for any 6 gt 07 we can pick 6 gt 0 such that for any I y 6 a7 b with z 7m lt 6 was 7 fyl lt eltb7agt Let 73 10 be a partition of a7 b such that lzj 7 zj1l lt 6 Hence wjf wf7 zjzj1 lt eb 7 a Then 201195 11 lt 6b i a ZltIj1 11 6 j1 j1 In View of Theorem 417 f E R D The goal of this section is to prove Theorem 42 Let f ab 7 R be a bounded function and let D C a7b be the set of points where f is not continuous Then f E R if and only if mD 0 In View of Theorem 317 Corollary 21 and Theorem 27 Theorem 42 follows directly from Theorem 43 Let f a7 b 7 R be a bounded function Then f E R if and only iffor every 6 gt 0 the set E z 6 a7 b wf7 I 2 6 has content equal to zero lncleecl7 suppose Theorem 43 has been proved Suppose rst that mD 0 Since D UneN Eln7 we have mE1n 07 for all n E N Since E C E77 for 6 gt 7 it follows from Theorem 31 that mE5 0 for all 6 gt 0 Since E5 is compact7 by Corollary 217 Theorem 31 guarantees that cE5 0 Conversely7 if cE5 0 for all 6 gt 07 then in particular cE1n mE1n 0 for all n E N Then Theorem 31 implies that mD 0 Now we prove Theorem 43 Proof The implication f E R gt cE5 07 for all 6 gt 07 is easier and we prove it rst Suppose that f E R and x 6 gt 0 We want to show that cE5 0 Let 6 gt 0 Since f E R7 we know that there exists a partition 73 10771n of ab such that ijzj1 7 zj lt 66 j1 Let j zjzj1 and let A j l S j S n7 and 1 E5 Since for every I E E N 1k 6 S wfz S wfIk it follows that 5 Z lel 5 Emu 11 S 2011411 11 S ZWltI11 11 lt 65 jeA jeA jeA j1 Hence 2 HA lt e jEA The intervals j cover E5 it follows that cE5 0 Now we prove the converse We begin with the following Lemma 41 Suppose thatwf7 I lt 6 for all z E 174 Then there exists a partition 73 1011771N of p7 q such that 41 Mj7mjlt67 jl2N7 Mj sup mj inf x 744 7 gt171 MA544 LECTURE NOTES FALL 2009 11 Proof Extend for z lt p and for z gt L For each I E q there exists an open interval LE 9 I such that supxet 7 infxer lt 6 where E is the closure of LB This gives an open cover of 10 q and hence it has a nite subcover 1I N Now take a partition 73 of 10 4 such that an interval of 73 is contained in the closure of one of the intervals LE Then 41 holds D Suppose cE5 0 for all 6 We want to show that f E R Let 6 gt 0 Since cE5 0 there exist open intervals 11IN such that E C I and that llll lINl lt 6 We begin with a partition 73 of a 12 consisting of intervals that fall in two categories 731 and 732 de ned as follows J E 731 if J C for some j and J E 732 if J N E 0 Let J be an interval in 732 Since wfz lt 6 for all z E J Lemma 41 shows that there exists a partition 73 of J such that sup 7 infgcgj lt 6 Doing this for every interval in 732 we get a re nement 73 of 73 such that each interval of 73 falls in two classes de ned as above which we denote by 731 and 735 Hence U05 110573 2 MJ mJNJl 2 MJ mJNJL M1 supfr7 mj supf Jep JePg J J Since f is bounded S M and hence M 7 m S 2M for any J E 73 For J 673 M 7m lt 6 Hence U027 7Lf73 2 mm 2 am lt 2Mb7 a6 JePl JePz Since 6 gt 0 is arbitrary this shows that f E R D 41 Limitations of the Riemann Integral Theorem 42 shows that the class of functions integrable in the sense of Riemann is small The srt limitations is that one can have two functions which are equal outside a set of measure zero with one integrable and the other not integrable Consider the following function 1 if IEQ XQW 0 if zgo Exercise 41 Show that wow I 1 Therefore XQ is discontinuous everywhere and hence not Riemann integrable On the other hand Fz 0 is Riemann integrable and Fz XQI I g So they are equal outside a set of measure zero one is integrable and the other one is not Proposition 42 Let X0 and XK be the characteristic functions of the Cantor sets de ned above Then the set of discontinuities of X0 and XK are C and K respectively Proof Let us take the case of X0 The other one is identical Since C is closed R C is open Since X0 0 if I g C X0 is continuous in R C Since C has empty interior for every point of C there is a sequence of points of R C converging to it Hence the set of discontinuities of X0 is exactly equal to C D Corollary 42 X0 6 R but XK g R Another limitation of the Riemann integral is in its relation with limit operations Before we discuss these examples we will review some concepts about convergence of functions in the next section 12 SA BARRETO 5 METRIC SPACES Several or perhaps all topics discussed in this section are not new for most students in MA 544 However these concepts are an important part of the course and worth reviewing De nition 51 A metric space is a pair M d consisting of a set M and a function d M X M 8 000 satisfying the following properties 1 dzy dyz for all zy E M 2 d17y dy 2 2 dz 2 for all zy2 E M triangle inequality 5 dzy 0 if and only y In what follows we denote a metric space either by M or M d we want to emphasize its metric Example 1 M R n 21 and dzy 11 7 y12 In 7 yn2 Exercise 51 Verify that R d is a metric space Example 2 Let C01 M denote the set of real valued functions de ned in 01 Let d129 Epl WI 7 WW Properties 1 and 3 are obvious One just needs to verify property 2 Let fgh E C01 Since by the standard triangle inequality lf1 MIN S lf1 9Il 191 h1l7 we immediately see that df7h S df79 d97h De nition 52 Let X be a vector space over R or mJ A norm on X is a function NXgt0oo such that Nz 0 ltgt z 0 Nz lAlNz A E R or C NIySNINy One often uses the notation Nz Exercise 52 Let X N be a normed vector space Show that Who MI 7 y is a metric on X MA544 LECTURE NOTES FALL 2009 13 51 Sequences and their convergence A sequence of elements of a metric space M is a countable subset xj j E N of elements xj E M De nition 53 A sequence converges to x E M for every 6 gt 0 there exists N E N which depends on e such that for allj gt N dxjx lt e In this case we say that lim xjx or xjax jam An important concept is that of a Cauchy sequence De nition 54 A sequence is Cauchy if for every 6 gt 0 there exists N E N which depends on e such that for all jh gt N dxjxk lt 6 Theorem 51 Every convergent sequence is Cauchy Proof This is an easy consequence of the triangle inequality Let xj E M With xj A x For 6 gt 0 pick N such that dxjx lt 62 for all j gt N Then for j k gt N dxjxk S dxjx dxkx lt 6 Therefore xj is Cauchy D It is not the case that all Cauchy sequences in a metric space M d converge to an element x E M Example 51 Let M Q denote the set of rational numbers With the metric dpq lpi ql Q is not complete To see that pick a sequence qj E Q Which converges to qj is Cauchy as it converges in R but it does not converge in De nition 55 A metric space M d is complete for every Cauchy sequence C M there exists x E M such that converges to x 52 Completion of metric spaces One wonders if given a metric space there is a complete metric space that contains it De nition 56 Let M d and be two metric spaces A map i M a M is an isometry if 490749 d 9 Notice thati is injective In this case we call i an embedding ofM into fi is onto we say that M and M are isomorphic Theorem 52 Let M d be a metric space Then there is a metric space with the following properties 1 is complete 2 There is an embedding i M a M 5 is dense in 4 If X p is another metric space satisfying 12 and 5 then and X p are isomorphic Proof Will be in the homework assignment D 14 SA BARRETO 6 SPACES OF FUNCTIONS The most important examples of metric spaces in MA 544 are certain spaces of functions We begin With the following fundamental examples7 Which are the motivation for the introduction of the Lebesgue measure Exercise 61 Let M CO7 1 be the space of continuous real valued functions in 01 Let Hflloo sup WIN xe01 1 117 llfllp lfzlde 7 p 2 1 This is the Riemann integral 0 Show that Hpr l S p S 00 are norms Theorem 61 The metric space C0717dp7 d1f7 g 7 ng is incomplete ifl S p lt 00 and is complete if p 00 Proof It is easier to prove that C0717dp7 l S p lt 007 is incomplete Let 1 if z e 0 g MI 1Iif16l7l 0 if IE1 Let 1 if mpg x 0 if zeg Then 0 if IE0 fz7fnz 17zi if z 7 0 if IE1 Hence a 01lfav79avlpdavATalighvi Vdacsi 41 l 2 So dpfnf A 071 S p lt 007 as n A 00 But f g C071 Now we prove that the metric space C0717dm is complete Let be a Cauchy sequence in C0717dm Then for any 6 gt 0 there exists N E N such that 6 1 7 fmzHDo lt e for all mm 6 N Fix 1 E 01 Then is a Cauchy sequence in R and therefore it converges Let def In view of 61 fn converges to f uniformly on 01 Since fn is continuous and 01 is compact7 f E CO7 This ends the proof of the Theorem D MA544 LECTURE NOTES FALL 2009 15 We know from Theorem 52 that C0 1 dp 1 S p lt 00 can be completed The question is what is its completion This is a very important question in analysis Let fg E RGO 1 ie f and g are Riemann integrable in 01 We will say that f and g are equivalent if there exists a set F with mF 0 such that f1917 V I 610711FA This is an equivalence relation Theorem 62 Let Lp01R denote the space of equivalence classes offunctions which are Riemann integrable and that for each f E Lp01R let Hpr 7 Mimosa Then Lp01R is a normed vector space but it is not complete Proof We will take p 1 but the general case is the same Since the set of discontinuities of is contained in the set of discontinuities of f and f is Riemann integrable then also is Riemann integrable Recall that by assumption if f E R then f is bounded So there is no convergence problem with the integral It is clear that does not depend on the choice of the representative of the class of The properties of the norm are clear with the exception of its nondegeneracy We need to show that if f E R and 0 then f E 0 This follows from Lemma 61 Ifgz 2 0 for all z in 01 M and folgzdz 0 then 91 0 for all I such that g is continuous at 1 Proof If 910 0 and g is continuous at 10 then there exists 6 gt 0 such that if z 7 10 lt 6 then 91 gt gzo2 This implies that folgzdz gt 69zo2 gt 0 D If 0 then 0 for all z in the set of points where is continuous Since f is Riemann integrable it follows that f 0 except on a set of measure zero Thus f E 0 To prove that L101R is incomplete we recall the construction of the Cantor set of nonzero measure Let fn XFW be the characteristic function of the set Fn de ned in Lecture 1 fn is obviously Riemann integrable Moreover fn fn1 XFFW1I By the construction Fn Fn1 consists of 2 disjoint intervals of length 2 H11 7 Tj7 1 Therefore 1 n fn1gtd1 H17 Tj39r 1 g Tn1 0 F1 If N gt 1 we write N N fn fnN S Zfnj fnj71 Zfnj fnj7l j1 j1 Hence n N nj T7 Hf gammaer 17 15 SA BARRETO Hence fn is a Cauchy sequence in L1017Ri But it obviously converges to the characteristic function of the Cantor set K Which is not Riemann integrable D

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