176 Class Note for MATH 501 at PSU
176 Class Note for MATH 501 at PSU
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Lectures on Analysis I Math 501 John Roe Fall 2005 Lecture 1 Real and Complex Numbers We shall assume that the basic properties of the real number system are known These are summarized by Proposition 11 R is an archimedean complete ordered fied What does this mean a A field basic properties of additional multiplication sub traction division commutative associative distributive laws b 00 ordered There is an order relations lt with the expected properties In particular x2 2 O for all 13 This is true in any ordered field The fact that every positive element is a square is special Archimedean no infinite or infinitesimal real numbers There is a unique injective homomorphism of rings Z gt R taking 1 to 1 Proof The archimedean property is that every element of R is smaller than the image of some element of Z complete no holes contrast Q There are various ways of expressing completeness The stan dard way is the Least Upper Bound Axiom Let S be any nonempty set of real numbers A number a e R is an upper bound for S if for all 1 e S 1 lt a Let US be the set of all upper bounds for S The least upper bound axiom says that if S and US are both nonempty then US has a least member This is called the least upper bound for S and written sup 8 Example 12 Let S x 6 R1332 lt 2 Then US 2 a E R a gt 061 2 2 US has a least member namely supS The corresponding statement with R replaced everywhere by Q would be false Once we have the notions of sequence and convergent sequence there are various other ways of expressing completeness For example Proposition 13 Bozano Weierstrass Every bounded sequence of real numbers has a convergent subsequence Every bounded monotonic sequence of real numbers is convergent A sequence of real numbers is called a Cauchy sequence if its members get closer and closer to one another Formally am is a Cauchy sequence if for every 6 gt 0 there is N such that xnl syn1 lt 6 whenever n n gt N Example 14 The successive truncations of an infinite decimal form a Cauchy sequence Proposition 15 Cauchy Every Cauchy sequence of real num bers is convergent Notice that this formulation of completeness does not make any use of the order relation It is therefore suitable for generalization to other situations eg C where no order relation is available Definition 16 The complex numbers C comprise all formal ex pressions a bi where a b E R Addition and subtraction of complex numbers are done compo nentwise Multiplication is done by forcing the distributive laws and i2 1 that is a bic di 2 ac bd ad bci It is then a theorem that C becomes a field under these oper ations To see that division is possible we define the complex conjugate of z a I bi to be 2 a bi and note that zZ is the positive real number a2 b2 z2 Thus we can put 1 aJa2 b2 Ina b2 Zlzl2 provided z 72 0 Remark 17 The complex numbers do not form an ordered field Remember in any ordered field squares are positive Proposition 18 Triangle inequality For complex numbers z and 111 we have z I w Proof Writing z a bi w c di we need to prove ac2bd2 lt va2b2vc2d22 Square both sides and cancel terms it s enough to show ac bd g la2 b2lC2 d2 Square and cancel again enough to show 20de g a2d2 b2c2 But right minus left is ad bc2 2 0 so this is true D Because satisfies the triangle inequality we can use it to define convergence for sequences of complex numbers just take the same definition as for real numbers It is then true that C is complete in the Cauchy sense Proposition 15 Exercise 11 Show that a bounded monotonic increasing se quence of real numbers converges to its least upper bound Exercise 12 Show that any sequence of real numbers either has an increasing subsequence or else a decreasing subsequence Deduce the Bolzano Weierstrass theorem 13 using the previ ous exercise Deduce Cauchy s version of completeness from Bolzano Weierstrass Exercise 13 Cantor s nested interval theorem Let ambn n 12 be a sequence of closed intervals in R that are nested in the sense that an1bn1 g ambn and whose lengths bn an tend to zero Show that the intersection nanbn contains one and only one real number c Exercise 14 Use the previous exercise to show that the real numbers are uncountable ie cannot be listed as c1c2 Hint Suppose such a listing is possible Pick an interval a1b1 of length at most 1 that doesn t contain c1 then a subinterval a2b2 of length at most that doesn t contain c2 and so on Apply the nested interval theorem What can you say about the limit point c Exercise 15 Let z and 111 be complex numbers of absolute value lt 1 Show that the absolute value of 2 111 1 1172 is also less than 1 Hint Expand the difference 1 v z2 z w2 in terms of complex conjugates and factorize the resulting mess Lecture 2 Metric spaces Much of the basic machinery of continuity convergence com pleteness and so on will function in a much broader context than that of R or C Definition 21 A metric space is a set X equipped with a func tion d X x X gt R called a metric or distance function such that i d1313 2 O for all xx moreover da13 0 if and only if 33513 ii d1313 da a for all 3333 symmetry iii d1313 lt d1313 I d13 13 for all x13 13 triangle inequal ity Motivating example dzw z w on C or R Familiar definitions can be reformulated for this new context Definition 22 Let 51 be a sequence in the metric space X We say xn converges to 1 e X if for every 6 gt 0 there is an integer N such that d1313n lt 6 whenever n gt N Similarly we can reformulate the definition of a Cauchy sequence Definition 23 Let xn be a sequence in the metric space X We say am is a Cauchy sequence if for every 6 gt 0 there is N such that ld 13n Bn lt 6 whenever n n gt N The metric space X is complete if every Cauchy sequence converges Definition 24 A function f X gt Y between metric spaces is continuous at 1 e X if for every 6 gt 0 there is 6 gt 0 such that df13f13 lt 6 whenever d1313 lt 6 It is continuous if it is continuous at every 13 E X Exercise 21 Show that a function f as above is continuous at 13 if and only if whenever mm is a sequence converging to 13 the sequence f13n converges to f13 Though this exercise is sometimes useful much more important is the characterization of continuity in terms of open sets Definition 25 A subset U of a metric space X is open if for every 1 e U there is e gt 0 such that the entire ball B13 e 1 513 E X d1313 lt e is contained in U The triangle inequality shows that every ball is open so there are plenty of open sets A set F whose complement X F is open is called closed Note carefully that closed does not mean the same as not open Many sets are neither open nor closed and some may be both Lemma 26 The union of any collection of open sets is open The intersection ofa finite collection of open sets is open The empty set Q and the entire metric space X are open and closed Theorem 27 Let X and Y be metric spaces Then f X gt Y is continuous iff for every open U g Y the inverse image f 1U 13 e X fx 6 U is open in X Proof Suppose that f is continuous and let U g Y be open Let 1 E f 1U then f13 e U so by definition of open there is e gt 0 such that Bf13e Q U By definition of continuous there is 6 gt 0 such that if x e B136 then f13 e Bf13e c U But this means that B136 g f 1U Thus the set f 1U is open Conversely let a E X e gt O and suppose that f satisfies the condition in the theorem In particular we may consider U Bf13e an open set such that 1 e f 1U Our hypothesis tells us that f 1U is open which means that there is a 6 gt 0 such that B136 g f 1U We have shown that whenever x E B136 f13 e Bf13e This gives us continuity D Here are some useful examples of metric spaces Example 28 Any set X can be made into a metric space by assigning it the discrete metric deny O a 513 1 x7 x In this metric all subsets of X are open and closed The only convergent sequences are those which are eventually constant Any function from a discrete metric space to another metric space is continuous Example 29 Here is a very significant example that we shall return to several times Consider the collection CO1 of all continuous functions 01 gt C We can define a metric by dfg suplft 9tl t 6 01 Convergence of a sequence of functions in this metric is called uniform convergence Exercise 22 Show that the sequence of functions on 01 de fined by Mt rite t does not converge in the metric space CO 1 even though for all t 6 01 fnt gt O as n gt 00 One says that fn converges pointwise but not uniformly Lecture 3 Closed sets There is an alternate characterization of closed sets in terms of limit points Definition 31 Let X be a metric space A g X A point a E X is a limit point of A if it is the limit of a sequence of distinct points of A Proposition 32 A is closed if and only if it contains all its limit points Proof A is closed if and only if B XA is open Suppose B is open Then for each b e B there is e gt 0 such that dba gt e for all a e A It follows that no sequence in A can converge to b so b is not a limit point of A Since this is true for all b e B A contains all its limit points Suppose B is not open Then there is some b e B such that every ball Bb e meets A Choose a1 6 A arbitrarily and by induction an E Bb en 0 A where en 2 mindba1 dban1 Then an is a sequence of distinct points of A converging to b g A D Definition 33 An isolated point of A is a point of A that is not a limit point of A Exercise 31 Show that a e A is isolated if and only if there is e gt 0 such that Ba 6 0A 2 a Definition 34 The closure A of a set A in a metric space is the union of the set and all of its limit points Exercise 32 Show that A is the collection of all limits of con vergent sequences in A whether or not the members of the sequence are distinct Proposition 35 The closure ofA is a closed set containing A Moreover it is the smallest set with these properties any closed set F that contains A also contains the closure of A Proof By definition A contains A Let 1 be a limit point of A Then there is a sequence am in A converging to 13 Each am either belongs to A or is a limit point of A in either case there is a gm 6 A with d13nyn lt 1n Then yn is a sequence in A and converges to 13 so 1 e A Thus A is closed Proposition 32 Suppose that F contains A Then every limit point of A is a limit point of F Thus if F is closed F contains all the limit points of A that is F contains A D Exercise 33 A closed ball in a metric space is a set Ba r 1 513 E X da13 lt 7 Show that a closed ball is closed Must every closed ball be the closure of the open ball of the same center and radius Exercise 34 Show that the intersection of any finite or infi nite collection of closed sets is closed Closed subsets can be very scattered The following is an im portant example of this Example 36 The Cantor set is the subset 0 g 01 defined by the following inductive process Let CO 01 01 0 U 21 02 0 U 3 U g U 1 and continue inductively in this way defining Cn1 by removing the open middle third of each of the 2quot closed intervals that make up On Let C 107 it is a closed set The figure below indicates the first few stages in the construction of the Cantor set Proposi on 37 The Cahtor set 5 Cosed uhcouhtabe has ho soated pomts aha cohtahs ho opeh hterva Moreover there 5 a cohthuous hohdecreaShg futhoh the Cantor function f 01 4 01 that maps 0 ohto 01 and maps the compe rheht ofC to a couhtaDe set Proof Outline Each element a e C is the intersection of a family of nested closed intervals F0 I F1F2 where each Fn is either the left third or the right third of Fn1 Assign to a the code sequence an where an 6 01 is 0 if Fn is the left third of Fn1 and 1 otherwise One sees that 00 2am a 13 71 Each code sequence represents a member of C and each mem ber of C has only one code sequence If two code sequences agree up to the n th stage the corre sponding members of C are at most distance 3 quot apart and conversely Thus any member of C is close to infinitely many other members and C has no isolated points Examining the construction we see that 01 contains no interval of length gt 02 contains no interval of length gt 5 and so on thus 0 contains no interval of positive length Define a function f C gt 01 by sending 00 2am 00 an a ngl W H fut nzl This function is increasing and it is continuous because two members of the Cantor set are close if and only if the initial seg ments of their code sequences are the same Moreover ifp q E C are two endpoints of an open interval pq removed in the con struction of the Cantor set one checks that fp fq One can therefore extend f to be constant on each complementary interval p q and the extended function will still be continuous and nondecreasing Since only countably many intervals were removed during the construction of the Cantor set fIC is countable Since f maps 0 onto an uncountable set C itself is uncountable D Exercise 35 Verify that if f A gt B is onto and B is uncount able then A is uncountable Exercise 36 Show that any open subset of R is a union of countably many disjoint open intervals Lindelof s theorem The example of the Cantor set shows that the closed analog of this statement is completely false Hint First show the result with out the countability restriction To get that note that each interval must contain a rational number A graph of the Cantor function f is shown below Lecture 4 Com pactness Definition 41 A metric space X is sequentially compact iff every sequence of points of X has a subsequence that converges in X The Bolzano Weierstrass theorem 13 tells us that every bounded closed subset of R or C considered as a metric space in its own right is sequentially compact Proposition 42 IfA is a subset of any metric space X and A is compact in its own right then A is bounded and closed in X Proof Let 1 e X be a limit point of A Then there is a sequence an in A converging to 13 But by compactness an has a subsequence converging in A Thus 1 e A and A is closed Suppose that A is not bounded Then one can construct by in duction a sequence an in A such that dana0 gt 1 I dan1a0 for n 2 1 The triangle inequality shows that danam gt 1 for n 72 m Clearly an has no convergent subsequence D Exercise 41 Let f X gt Y be continuous and surjective Show that if X is compact then Y is compact also What has this got to do with the familiar calculus principle that a continuous func tion on a closed bounded interval is itself bounded and attains its bounds Despite the above evidence it s not in general true that Closed and bounded equals Compact Counterexample We shall analyze this in detail Definition 43 Let X be a metric space An open cover u for X is a collection finite or infinite of open sets whose union is all of X A Lebesgue number for u is a number 6 gt 0 such that every open ball of radius 6 is a subset of some member of u Theorem 44 Lebesgue Every open cover ofa sequentialy compact metric space has a Lebesgue number Proof Suppose that u does not have a Lebesgue number Then for every n there is am e X such that B13n1n is contained in no member of u If X is compact the sequence am has a subsequence that converges say to 13 Now 1 belongs to some member U of u since u is a cover Thus there is e gt 0 such that B13 e Q U There is n gt 26 such that d13n13 lt 62 But then Ban 171 g B13n 62 g B13e Q U which is a contradiction D Exercise 42 Connectedness Show that R cannot be written as the disjoint union of two nonempty open sets Let the sets be U and V Then UV forms an open cover of each closed interval a b By considering a Lebesgue number for this cover show that a b lies entirely within one of the two sets Proposition 45 The following conditions on a metric space X are equivalent a Every sequence in X has a Cauchy subsequence b For every 6 gt 0 there is a finite cover ofX by balls of radius 6 In this case we say X is totally bounded some writers say pre compact Proof a implies b Choose x1 6 X and then inductively choose am e X such that d13nam 2 6 when m lt n so long as this is possible The process must terminate because if it didn t it would produce a sequence with no Cauchy subsequence When it does terminate with xN say it does so because the balls B13n6 n 1N cover X b implies a Notice the following implication of b given any 6 gt 0 any sequence in X has a subsequence all of whose members are separated by at most 6 call this a 6 close subse quence Let xn be any sequence in X Let x35 be a 2 1 close subse quence of mm let 32 be a 2 2 close subsequence of 513 and so on by induction Then x3 is a Cauchy subsequence of the original sequence D A metric space X is said to be covering compact if every open cover of X has a finite subcover Proposition 46 The following conditions on a metric space X are equivalent a X is sequentially compact b X is complete and totally bounded c X is covering compact Proof It is easy to see that a and b are equivalent Suppose a Let u be an open cover of X Let 6 gt 0 be a Lebesgue number for M which exists because of Theorem 44 Since X is totally bounded it has a finite cover by 6 balls But each such ball is a subset of a member of u so M has a finite subcoveh In the other direction suppose C Let xn be a sequence without convergent subsequence Then for each 1 e X there is some eac such that am g B13 ex for all but finitely many n The B13ex form a cover of X Picking a finite subcover we obtain the contradiction that am g X for all but finitely many n D Remark 47 For general topological spaces the notions cover ing compact and sequentially compact are not equivalent and covering compact is usually the most appropriate one Exercise 43 A map f X gt Y between metric spaces is uni formy continuous if for each 6 gt 0 there is 6 gt 0 such that dfafa lt 6 whenever d1313 lt 6 The extra information beyond ordinary continuity is that 6 does not depend on 13 Show that ifX is compact every continuous f is uniformly continuous Hint use Theorem 44 Lecture 5 Complete Spaces Rudin chapter 5 first few pages Definition 51 Let X be a compact metric space Then CX denotes the space of all continuous functions X gt C It is a metric space equipped with the metric dfg SUDfx 9x I a E X Compare Example 29 Compactness of X ensures that the supremum exists why We can also consider the space CRX of continuous real valued functions Proposition 52 For a compact X the metric spaces CX and CRX are complete Proof Most completeness proofs proceed in the same three stage way Given a Cauchy sequence identify a candidate for its limit show that the candidate is in the space in question and show that the sequence approaches the candidate limit in the metric of the space in question Let fn be a Cauchy sequence in CX Then for each a e X fn13 is a Cauchy sequence in C Since C is complete this sequence converges Denote its limit by f13 We show that f is a continuous function on X Fix x e X and let 6 gt 0 be given Because fn is Cauchy there is N such that for n n 2 N we have fna fn IZ lt 63 for all 13 Take 71 N and let n gt 00 to find that fN13 f13 lt 63 for all 13 Now fN is continuous at x so there is 6 gt 0 such that fN13 fN1 lt 63 whenever x 13 lt 6 It follows that whenever x x l lt 6 f fl lt DECK fN IfN I3 fN33 IfN13 f13 lt 6 This shows that f is continuous Finally to show that fn gt f in CX we must prove that for every 6 gt 0 there is N such that fn13 f13 lt e for all 1 whenever n gt N But in fact we already proved this in the previous paragraph D We shall encounter many other complete spaces of functions Definition 53 Let X be a metric space A mapping f X gt X is a strict contraction if there is a constant a lt 1 such that dfafa adaa for all 3333 6 X Note that a contraction must be continuous Theorem 54 Banach A contraction on a complete metric space has a unique fixed point a point 1 such that f13 13 Proof A fixed point is unique because if 33513 are two such then d1313 dfafa adaa which implies daa O To prove existence start with any x0 6 X and define x1 f130 5132 fa1 and SO on If da0a1 r then danxn1 lt an and so an 1 a which tends to O as n gt 00 Thus mm is a Cauchy sequence which converges to a point 13 We have d33n733nk lt an ank17 lt Iim imxn1 13 so 1 is a fixed point D Exercise 51 Banach s fixed point theorem can be extended as follows if f X gtX is a map and some power fNZfoof is a strict contraction then f has a unique fixed point Prove this Banach s fixed point theorem is one of the most important sources of existence theorems in analysis Another important source of existence theorems which also uses completeness is the Baire category theorem Definition 55 A subset of a metric space is dense if its closure is the whole space Example 56 The rational numbers Q are dense in R It is often important that R has a countable dense subset A space with this property is called separable Theorem 57 Baire In a complete metric space the intersec tion of countaby many dense open sets is dense Proof Let 1 e X and let 6 gt 0 Let Um n 12 be dense and open Choose x1 6 U1 Bae2 There is r1 such that Fx1r1 g U1 without loss of generality take r1 lt 64 Since U2 is dense there is x2 6 U2 0 B131r1 Proceed inductively in this way choosing am and m such that am 6 Un Ban1rn1 FCIZn 7 g Un rn lt rn12 Then mm is a Cauchy sequence say converging to a point x Since am e B13j rj for all j lt n the point x belongs to Fwy rj for every j and hence to Un Moreover d1313 lt 6 Since 1 and 6 were arbitrary Un is dense D Remark 58 The name category theorem comes from the fol lowing traditional terminology A set A is nowhere dense if the complement of its closure is dense A set is of first category if it is the countable union of nowhere dense sets Baire s theorem then says that in a complete metric space the complement of a set of first category sometimes called a residual set is dense Remark 59 Notice that this gives us another proof that R is uncountable compare Exercise 14 For the complement of a point in R is a dense open set Exercise 52 Consider the complete metric space CR0 1 of continuous real valued functions on 01 For ab 6 01 show that the set of functions f which are nondecreasing on the in terval a b is nowhere dense in CR01 Using Baire s theorem deduce that there exist functions f e CR0 1 that are nowhere monotonic that is monotonic on no subinterval of 01 Lecture 6 Convergent Series We define the convergence of a series 2301 fn of real or complex valued functions as the convergence of the sequence of its partial sums Proposition 61 Let X be compact and let an be a series in CX If there exist positive constants Mn such that 00 SUDfna 11 6 X g Mn and 2 Mn lt oo n21 then the series an converges in CX Remark 62 This sufficient condition for uniform convergence is traditionally known as Weierstrass M test It is not a necessary condition Exercise find an example that shows this Proof Let gN be the partial sum gN Zgzlfn Using the triangle inequality for N gt N we have NI NI NI lgNx gNal 2 mm lt Z Ifnalt 2 Mn nN1 nN1 nN1 and so NI nN1 Since ZMn converges the right hand side tends to zero as NN gt 00 Thus 9N is a Cauchy sequence in CX which is complete by Proposition 52 and we are done D Remark 63 A series Zan is called absolutey convergent if E an converges there is a similar definition of uniform absolute con vergence of a series of functions The Weierstrass M test in fact proves uniform absolute convergence Often need to consider series of functions on metric spaces that are not compact typically open subsets of CC Definition 64 Let X be any metric space A sequence of func tions on X converges uniformly on compact subsets on X if its restriction to every compact subset of X converges uniformly A series converges uniformly on compact subsets if its sequence of partial sums does Exercise 61 Let S2 be an open subset of CC Show that a sequence of functions converges uniformly on compact subsets of S2 if for every point z E Q there is a closed disk Dzrz g S2 on which the sequence converges uniformly Example 65 The series 1 0 22 converges uniformly on compact subsets on CCN Use Weier strass M test Later in the course we shall see that it converges to the familiar function 7tCOt7tz The most important examples are power series on C of the form 020 anzquot where the an are complex constants Proposition 66 For any power series 2330 anzn there is a num ber r nonnegative real number or 00 called the radius of convergence such that a 2330 anzn converges uniformly on compact subsets on the open disk DOr b 2330 anzquot diverges for each z with z gt r The behavior of the series on the circle of convergence z r is not specified by the proposition Balls in the complex plane are traditionally called disks and denoted D Proof Let U p 2 O the sequence anpquot is bounded and let 7 supU taken as I oo if U is not bounded above Then if z gt r the terms of the series 2300 anzn are unbounded and thus the series cannot converge This proves b To prove a we shall show that the series converges uniformly and absolutely on every closed disk 50 5 with s lt 7 Indeed there is p with s lt p lt r and the sequence anpquot bounded say by K Then for all z E FOs n anzn g Mn 2 K 5 p Since sp lt 1 ZMn is a convergent geometric series and thus 020 anzn converges uniformly on FOs by Proposition 61 D Exercise 62 Cauchy s formula for the radius of convergence is 17 lim sup an1quot VII gt00 Prove this You need to know the definition of the limit superior of a sequence of real numbers it is limsupbn lim sup bn VII gt00 m oongtm Notice that the limit on the right is the limit of a monotonic sequence so it always exists possibly infinite Easy to see that power series can be added and subtracted termwise within the disk of convergence They can also be mul tipled Proposition 67 Let f2 Zanzquot and gz anzquot be con vergent power series with radii of convergence r and 5 Put n on Z akbnk kO Then the power series chzquot has radius of convergence at least minrs and within the disk of this radius it converges to fzgz Proof This statement about power series follows from a general fact about absolutely convergent series if Zan and an con verge with at least one of them converging absolutely then Zen converges to the product 2 an 2 bn To prove that let An Bn and On be the sequences of partial sums It suffices to show that Aan Cn tends to zero But this difference is Z ajbk Z aj Z bk Z ajBn Bra 1 jkltnjkgtn j1 kn j1 j1 Assume that Zan converges absolutely Let M Zan and let K be a bound for the Bn Let 6 gt O and choose first m and then N large enough that 6 6 Z ajlt an Bnjllt fOFOltjltmngtN j m1 Write m n For n 2 N the first sum here is bounded by M e2M and the second is bounded by e4K 2K So the whole expression is bounded by e for sufficiently large n as required D Exercise 63 Let a e C be a constant The binomial series is the series 00 aln n faz Z Z n20 n where aim 2 aa 1a n 1 Show that the radius of convergence of the binomial series is at least 1 Exercise 64 Let faz denote the binomial series of the previous exercise Show that fabz fazfbz for z lt 1 Since f1z 1 z this justifies the notation 1 2 for faz Hint Start by proving by induction the formula n n L k rt k a l39b k0nkka b Remark 68 In the next lecture we shall need the following fact if a is real and positive the binomial series for 1 z converges uniformly and absolutely on the closed disk FO1 To prove this it is enough to see that Zaquotn lt 00 for these a which can be seen by showing that lan lt K n n5 for a suitable constant K and for any 3 with 1 lt B lt 1 a eg B 1 02 Exercise 65 Fill in the details of the above argument Show that for sufficiently large n the validity of the inequality for n implies its validity for n 1 Lecture 7 More about power series Rudin prologue This lecture will not be discussed in class It should be treated as a reading assignment Let Zanzquot be a power series with radius of convergence 7 and converging within DOr to a function fz The series Znanzn l obtained by formal differentiation also has radius of convergence 7 one can see this from Exercise 62 for example Moreover Proposition 71 With notation as above the function f is dif ferentiable on DOr and the series Znanzn l converges on DOr to the derivative h gtO h Proof Start with an algebraic fact if w z lt s then wn Zn inn 1011 1w z2sn2 This follows from the identity w Zn inn 1011 z w Z2wn 2 2wn 3Z n 1Zn 2 which is easy to check Let gz Znanzn l and fix 5 lt 7 Then using the identity above provided that zz h lt s we have fz h fz h and the sum on the right converges to a finite limit since 5 lt r In particular fz h fzh gt gz as h gt O and we are done D go lt an chn 1lanls 2 Remark 72 Repeating the argument shows that a function f de fined by power series is differentiable infinitely often smooth kAoreoven W fis de ned by the convergent pommu39senes Ejanzquot then am fltno7u VVarning if vve start mch ari arbhrary smooth function f R gt R and form the Maclaurin series with these coe hjentsthat senes need not converge to f See Ex ample 75 below The exponential function is defined by 00 n 2 Z 6 n20 n39 This series has infinite radius of convergence Using proposi tions 67 and 71 one verifies treat these as exercise a Addition law eZw 62611 b Differentiation law the function Z gt 62 is its own derivative The sine and cosine functions are defined in terms of the expo nential by I eiz 6 132 00 1 nz2n1 snnz Z 21 n20 2n 1 eiz i2 00 1 n 2n COSZ L 2 2 2 n20 271 The exponential sine and cosine functions are real valued for real arguments and we have em coszz39sinz for all z The addition law for the exponential function yields the corre sponding laws for sine and cosine sinz w sinzCOSw COSzSinw cosz w COSZCOSw sinzsinw In particular sian cos22 1 the special case 111 z of the second identity One sees by computation that cos has a positive real zero define 7r by letting 7r2 be the smallest real zero of cos We have COS7r2 O and sin7r2 1 The identities now give sinz 7r2 COSz cosz 7r2 sinz Iterating these we find that cos and sin are 27r periodic so the exponential function is 27ri periodic In particular we get the famous formulae Of course we are all familiar with the properties of the expo nential sine and cosine functions The point though is to show that these properties can be derived rigorously by purely analytic means Remark 73 Algebraically the mapping 1 l gt 27 is a homomor phism from the group R of real numbers under addition onto the group T of complex numbers of absolute value one under multiplication The kernel of this homomorphism is exactly the subgroup Z g R The exponential function grows fast Lemma 74 Exponentias kill polynomials For any constants ab gt 0 one has limlcgt00 xGe bw O Proof Let m gt a be an integer Then et gt tmm for 75 positive and thus xGe bx xaebx lt mlb mxa m tends to O as 1 gt 00 D Example 75 Let f13 O for 1 lt O f13 e lx for 1 gt 0 Then f is differentiable everywhere even at zero the limit which defines its derivative there is lim h le lh h gtO which vanishes because exponentials kill polynomials One can continue by induction to show that f is differentiable infinitely often and all its derivatives at the origin are zero Thus the Maclaurin series for the smooth function f does not converge to f in any neighborhood of the origin Remark 76 We shall see that this is a purely real variable phe nomenon Next semester we ll prove that differentiability for functions of a complex variable automatically ensures analyticity representability by power series This is a deep fact Lecture 8 The Weierstrass Approximation Theorem Recall that a subset of a metric space is dense if its closure is the whole space A classical method in analysis is to prove a result first for a dense subset of some space and then to extend it to the whole space by continuity How can we find dense subsets of CX Definition 81 A collection L of continuous real valued func tions on a set X is a lattice if whenever it contains functions f and 9 it also contains their pointwise maximum and minimum usually written f V9 and f g in this context Proposition 82 Stone Let L be a lattice of continuous func tions on a compact space X If for a 13513 6 X and any a a E R there is a function f e L having f13 a and f13 a then L is dense in CX The property appearing in the statement is called the two point interpolation property Proof Let h e CX be given and let 6 gt 0 We are going to approximate h within 6 by elements of L By hypothesis for each 213513 6 X there exists fm e L such that fmx h13 and Jim513 ha Fixing a for a moment let These sets are open why and x 6 Vin so they cover X Take a finite subcover and let gx be the pointwise maximum of the corresponding functions fm e L Because L is a lattice gas 6 L and by construction iKy 6 lt gxy Vy Ma 9x013 So we have approximated h from one side by members of L Now we play the same trick again from the other direction et Was y I gxy lt hy 6 Again these form an open cover of X take a finite subcover and let 9 be the pointwise minimum of the corresponding 953 Then gEL and by construction h eltglthe as required D There is a more classical formulation which makes use of alge braic rather than order theoretic operations Lemma 83 There is a sequence of polynomials pna on 11 converging uniformly to x Proof Writing x i1 x2 1 this follows from the fact that the binomial series for 1 012 converges uniformly for t 6 11 This was discussed in Remark 68 B One says that a subset of CX is a subagebra if it is closed under pointwise addition subtraction multiplication of functions and multiplication by scalars Lemma 84 A closed subalgebra of CRX that contains the constant functions is a lattice Proof Let A be the given subalgebra and g 6 A let h f g There is no loss of generality in assuming by rescaling that h g 1 everywhere Any polynomial in h belongs to A and hence so does h limanz by closure and lemma 83 But now fg h fg h f9 2 ll ng 2 ll belong to A as well D A subalgebra of CX is said to separate points if for every 33513 e X there is a function in the subalgebra taking different values at 1 and at x Theorem 85 Stone Weierstrass A subalgebra ofCRX which contains the constants and separates points is dense Proof The closure of the subalgebra is a closed subalgebra con tains the constants and separates points Using the algebraic operations it has the two point interpolation property The previous lemma shows that it is a lattice Hence by Stone s theorem 82 it is all of CX D The original result of Weierstrass was Corollary 86 Weierstrass Every continuous function on a closed bounded interval is the uniform limit of polynomials Proof The polynomials form an algebra which contains the con stants and separates points D Exercise 81 In the complex case the Stone Weierstrass the orem takes the following form a subalgebra of CX which contains the constants and separates points is dense where the condition means that the algebra is closed under pointwise complex conjugation Prove this Try to give an example to show that the complex theorem is not valid without the extra condition we shall discuss this in detail later Lecture 9 Compact Sets in CX Let X be compact Then CX is a complete metric space In this complete metric space closed bounded sets need not be compact Example 91 Consider the metric space CO 27r The functions fnt1 8in2nt n 12 belong to the closed unit ball of CO27r and dfnfm 2 1 if n 72 m Thus the sequence fn has no convergent subsequence so the closed unit ball is not compact Notice that the fn oscillate more and more rapidly We are going to prove a theorem which says that the only way compactness can fail is through this rapid oscillation Definition 92 Let f be a collection of functions from a metric space X to a metric space Y Say J is equicontinuous if for every 1 E X and e gt 0 there is 6 gt 0 such that Vf e f d113 lt 5 gt dfx fx lt 6 Example 93 The subset of CO 1 consisting of all functions f such that f t g 1 for all t 6 01 is an equicontinuous set Theorem 94 Arzea Ascoi Let X be compact metric A subset of CX is compact if it is closed bounded and equicon tinuous We need some technology in order to prove this Definition 95 The support Supportgigt of a continuous func tion gb is the closure of 13 x 72 0 Definition 96 Let u be an open cover of a metric space X A partition of unity subordinate to u is a collection gbUUEu of continuous functions X gt 01 such that a SupportgbU is contained in U b For each 1 e X ZUEM Ux 1 Proposition 97 Every open cover of a compact metric space possesses a subordinate partition of unity Proof Let u be the given cover It has a Lebesgue number 6 gt 0 Theorem 44 By compactness there is a finite cover of X by balls of radius 62 Let 51113N be the centers of these balls Let 1 i 1N be the function 52 dW i dz lt 52 was O d13 13 2 62 For each 139 there is Ui e M such that pi is supported in U For each U G Ll define U 12111 in TIt is possible that Uz Uj even if 2 7E j The denominator is gt O everywhere because of the covering property and clearly the gbU form a partition of unity D Proof of Theorem 94 We shall prove that a bounded eduicon tinuous set J g CX is totally bounded Then Proposition 46 will complete the proof Let 6 gt 0 be given By definition of equicontinuity for each 1 E X there is a ball Ux B136x such that if x E Ux and f E 7quot then f13 f13 lt 62 Pick a finite subcover U1Un and a subordinate partition of unity 1 n For a1an 6 CC write 0177anx Ziaing This is a continuous function Observe that lt 62 fOI all 33 ii if azgt bi lt 62 for all i then alruyanx oblmbnx lt 62 for all 13 Here is the proof of item i write lfa fx1fxnltxl lynx fxi ixl lt f fz z The proof of ii is similar Now suppose that J is bounded say by M Choose a finite subset A of the ball of radius M in C such that every member of that ball lies within 62 of a member of A Then by i and ii above the functions palman a e A form a finite set and every f E 7quot is within 6 of a member of this set Thus 7quot is totally bounded D Exercise 91 Prove the converse of the Arzela Ascoli theorem a compact subset of CX is closed bounded and eduicontinu ous Remark 98 Why should we be interested in compact subsets of function spaces Compactness is an element of proving the existence of a minimizer like the shortest curve between two given points or the curve of given perimeter enclosing the greatest area Cf Fermat s principle and other ideas of applied mathematics Lecture 10 Normed Vector Spaces Definition 101 Let V be a vector space real or complex A norm on V is a function V gt R denoted v l gt v such that a v 2 O for all 2 and v 0 iff v O b gtw gtv where A is a scalar real or complex C llv v ll lt llvll llv ll Clearly dvv 2 HQ v is then a metric Because of b this metric has an affine structure not present in a general metric Example 102 Let x 51113n be a vector in R or C The expressions X1 lx1llxnl Xl2 lx1l2 lxnl2 llxlloo max ml lxnl gt12 all define norms Exercise 101 Let a and b be positive real numbers and let p gt 1 Show that inft1pap 1 01 19191 t 6 01 a b Hence or otherwise show that the expression llxllp lx1lp lxnlp1p is also a norm on R or C Example 103 If V CX we may define llfll suplfxl a e X This is a norm and it gives rise to the usual metric on CX Since it is complete theorem 52 it is a Banach space as defined below Definition 104 If a normed vector space is complete in its metric it is called a Banach space Proposition 105 A linear mapping T V gt W between normed vector spaces is continuous if and only if there is a constant k such that llTvll lt kllvll one then says that T is bounded Proof If T is bounded then Tu Tv lt ku v so T is contin uous If T is continuous then there is some 6 gt 0 such that u lt 6 implies Tu lt 1 Take 1 16 D Definition 106 The best constant k in the above proposition that is the quantity SUIDTv i llvll lt 1 is called the norm of T and denoted T Exercise 102 Show that with the above norm the collection 304 W of bounded linear maps from V to W is a normed vector space and that it is a Banach space if W is a Banach space Exercise 103 Show that the norm of linear maps is submulti plicative under composition ie SoT g ST The most interesting examples like CX are infinite dimen sional vector spaces In fact we shall prove that any two norms on a finite dimensional vector space are equivalent in the sense that each is bounded by a constant multiple of the other We need some tools to do this Definition 107 A linear functional on a vector space V over a field is is a linear map V gt k In what follows we ll consider complex vector spaces it is trivial to replace C by R Proposition 108 A linear functional on a normed vector space is continuous if and only if its Kernel is closed Proof Let V be a normed vector space and gl V gt C a linear functional If gl is continuous then Kergl gb 1O is the inverse image of a closed set under a continuous map hence closed Suppose that Kergb W is closed Assume wog that gr 72 0 Then there is u e V such that gbu 1 Clearly u e W and since W is closed there is 6 gt 0 such that Bu6 does not meet W By linearity Bgtu gt6 does not meet W for any scalar A Now for any 2 E V v gbvu e W Therefore v 2 gbv6 It follows that for all 2 vl lt 6 1v which implies that gr is continuous D Theorem 109 Any finite dimensional normed vector space is complete and any two norms on such a space are equivalent Proof Induction on the dimension the 1 dimensional case being obvious Let V be n dimensional and suppose that the n 1 dimensional case is proved Pick a basis 22 vn for V Then there are linear functionals 1 n the dual basis such that v 1vvl I39 I39 nvvn for all 1 E V The kernels of the gr are n 1 dimensional hence complete by the inductive hypothesis hence closed By proposition 108 the gin are all continuous Hence the mapping CD 1 n V gt C is a continuous linear isomorphism and its inverse A1gtn l gt A1221 Anon is continuous also triangle in equality It follows that the given norm on V is equivalent to the norm obtained by transporting that on C via CD Completeness is preserved under equivalence of norms so V is complete D Lecture 11 Differentiation The basic idea of differentiation is that of best linear approx imation Normed spaces provide a systematic way to express this Notation 111 Let f be a function defined on some ball B0 6 in a normed space V and having values in another normed space W We shall write fh oh to mean that the limit limhgt0h 17h exists and equals zero Similarly for ex pressions like fh gh 0h Exercise 111 Suppose that T V gt W is a linear map Show that Th oh if and only if T 0 Now let f V gt W be a continuous need not be linear map between normed vector spaces In fact we need only suppose f is defined on some open subset U of V Definition 112 With above notation let 1 e V We say that f is differentiable at 1 if there is a bounded linear map T V gt W such that fxh fx Th0llhll By the exercise T is unique if it exists It is called the derivative of f at 1 and written Df13 Example 113 If V W IR then every linear map V gt W is multiplication by a scalar ie we identify V W 2 IR Under this identification our definition of the derivative corresponds to the usual one from Calculus I Example 114 If V W and W IRquot then Q3VW is the space of n x m matrices The matrix entries of the derivative of f are the partial derivatives of the components of f as defined in Calculus III Remark The existence of the partial derivatives does not by itself imply differentiability in the sense of our def inition above This just shows that our definition is right and partials are wrong Exercise 112 Show that the derivative of a linear map is always equal to the map itself Proposition 115 Mean value theorem Suppose that the func tion f is differentiable throughout a ball B130 r g V and that Dfa g k for a a E B130 7 Then fx fxo lt kllx woll for a a E B130 r Proof Let 6 gt 0 By definition of the derivative each y E B130 7 has a neighborhood Uy g B130r such that fy fy lt k 6lly yll Vy e Uy Consider the line segment 130513 g B130r The Uy form an open cover of this compact metric space so there is a Lebesgue number 6 gt 0 Theorem 44 Now subdivide the line segment into finitely many subihtervals 51305131 131332 and so on with xN 1 for some N and mi xi1 lt 6 Using the triangle inequality N l f f0 lt Z fz fz 1ll i0 N l lt 0 6 2 llxixi1k 0 i0 Since 6 was arbitrary the proof is complete D Remark 116 From the proof one sees that the theorem is true if we replace the ball by any region that is star shaped about 0 It is easy to see that the derivative of fg is ij Dg Another familiar calculus result that generalizes easily is the chain rule Proposition 117 Chain rule Let VWX be normed vector spaces and et a e V Let f V gt W be differentiable at 1 and et 9 W gt X be differentiable at y f13 Then 9 o f is differentiable at 1 and D9 O fx D9y O Dfx Note that f need only be defined near 13 and 9 need only be defined near y Proof Let kh f13 h f13 Df13 h oh Note that there is a constant A such that kh lt Ah for small h Now write 9 o fa h g o fx 9y 1601 9y D9y Mil 0llkhll DQQJ Dfx h 0lth giving the result D Remark 118 Note that the derivative of f Df is itself a function having values in a normed space namely the space 304 W Thus we can differentiate it and define second and higher deriva tives if we require Exercise 113 Open ended Formulate precisely the advanced calculus result that the mixed derivatives are symmetric that is 82f8x8y 82f8y8x and prove it under suitable hypotheses for maps between normed vector spaces Lecture 12 The inverse function theorem Let VV be normed vector spaces The space 304 V is then a normed vector space also Say T e V V is invertible if it has an inverse which is a bounded linear map from V to V If V V are Banach spaces we shall see later that this is just the same as saying that T is bijective this will follow from the Closed Graph Theorem but in general the boundedness of the inverse is an extra condition Proposition 121 Let VV be Banach spaces Then the in vertibes form an open subset of V V and the inverse oper ation T l gt T L is continuous on this subset from 304 V to Q3V V Proof Without loss of generality V V Look first at a nbhd of the invertible operator I Suppose S lt 1 For y e V consider the map oyiV gtV va Sv Since S lt 1 this map is a strict contraction so it has a unique fixed point 13 Theorem 54 This fixed point is an 1 such that I S13 y By the triangle inequality 1 llyll 2 1 lSlllxl SO llwll lt 1 llSll llyll and the map y l gt 13 is bounded We have shown that if S lt 1 1 8 is invertible Moreover Hy I S1yll31 S1y lt S1 S1y which shows that the map S l gt IS 3L is continuous at S O We have shown that the identity is an interior point of the set of invertibles and that the inverse is continuous there However left multiplication by a fixed invertible is a homeomorphism from the set of invertibles to itself so the same results apply to any point of the set of invertibles D Exercise 121 Let V be a normed space and let W be the normed space 230 V Show that the mapping 139 T l gt T L is differentiable where defined on W and that its derivative is DiT H T 1HT 1 Definition 122 A map f from an open subset of a normed space V to a normed space W is continuously differentiable or 01 if it is differentiable everywhere and the map 13 l gt Df13 is continuous The inverse function theorem says that under suitable conditions if Dfa is invertible then f is locally invertible near 13 Theorem 123 Let f be as above defined near a E V and suppose that f is continuously differentiable and that Df13 6 304 W is invertible Suppose moreover that VW are Banach spaces Then there is an open set U containing 1 such that f is a bijection of U onto fU which is an open set containing f13 and such that its inverse g fU gt U is also continuously differentiable Proof It is an application of Banach s fixed point theorem 54 We will construct the inverse function by looking at the fixed points of a suitable map Let A Df131 For y e W define a map gby V gt V by yz z A 34 fz A fixed point of gby is a solution to fz y We have magZ I A o Dfz A o Dfx Dfz Since Df is continuous there is r gt 0 such that if z e U B13 r then D yzll lt Fix zO e U and let yo fz0 Choose 6 gt 0 such that the closed ball Fz06 is contained in U Choose 6 A 16 If z e Pz06 and y e By0 6 then we may compute ll yz 20 llcbyCZ yozoll lt ll yz yzoll ll yzo y0zoll lt Ilz zoll IIAy 90 lt 6 6 6 using the mean value theorem 115 We conclude that for these y the map gby is a contraction of the complete metric space Pz06 and thus has a unique fixed point there We have shown that each yo 6 fU has an e neighborhood contained in fU thus fU is open The contraction property of the gby ensures that each y e fU has only one inverse image in U thus f is a bijection of U onto fU Now to show that the inverse function g f l fU gt U is differentiable at y f13 write gy h gy uh say The calculations of the previous paragraph show that uh lt 2Ah for h small Then h fgy h y fgy uh y Dfa uh oh Apply A Dfx1 to get uh A h This gives differentiability of g with Dgf13 Df131 Con tinuous differentiability now follows from the continuity of the inverse operation on the space of bounded linear maps Propo sition 121 D Lecture 13 Measure Spaces Rudin chapter 1 to 121 This is the central part of the course We develop an abstract theory of integration which is complete in a sense analogous to the completeness of R Let X be a set The notation 73X refers to the set of all subsets of X sometimes called the power set Definition 131 A subset A g 73X that is a collection of subsets of X is called an algebra of subsets ofX if it is nonempty and closed under the formation of complements pairwise unions and pairwise intersections if A A E A then the sets XA AUA AnA must also all belong to A Note that an algebra must contain Q and X Definition 132 An algebra A of subsets of X is a a agebra if in addition it is closed under the formation of countable unions and intersections if A1A2 E A then the sets 00 00 U A717 H An n21 n21 both belong to A Remark 133 In both cases it is sufficient to require closure for unions the corresponding fact for intersections follows on taking complements Definition 134 A positive measure on a a algebra A is a function 12 A gt Ooo such that a zQl O and b 1 is countaby additive if A1A2 are disjoint members of A then n1 1 8 An i lAn n21 A measure is finite if pX lt 00 and its is a finite if X UAn with An 6 A and uAn lt 00 We will use the term measure space for a set X equipped with a a algebra A and a measure 1 on it We call members of A the measurable sets for the measure space This is an abstraction of the idea of measuring the size of a set One can for example use this as an abstract basis for probability theory With this in mind define a probability measure as one for which 1X 1 Example 135 If X is any set and A 73X fix a point x0 6 X 1 if 5130 e A and define 1A I 0 If 130 it A Example 136 Let X N and A 73X Let an be any sequence of positive reals Define 1A ZnEA an Less trivial examples will be given later In particular there is a measure A called Lebesgue measure on a a algebra of subsets ofR such that every interval a b is measurable and its measure iS b a Remark 137 The intersection of any collection of algebras or a algebras is itself an algebra or a algebra Thus for any 8 g 73X we may define the a algebra generated by S to be the intersection of all the a algebras containing 8 aka the smallest a algebra containing 8 Definition 138If X is a metric space the a algebra of Borel subsets of X is the a algebra generated by the open subsets Exercise 131 Show that the a algebra of Borel subsets of R is generated by the intervals aoo with a E Q Proposition 139 Let X1 be a measure space Then i If AB are measurable and disjoint then 1A U B 1A 1B ii If AB are measurable and A g B then 1A g 1B iii If A1A2 are measurable then 1 U An g ZzAn iv IfA1A2 are measurable and A1 g A2 g then 1 U An lim 1An Proof These are trivial i Write AUB AUBUQUQU and use countable additvity ii Write AUB AU BA a disjoint union and use i iii Write UAn UBn where Bn 2 An AL U LJ An1 a disjoint union and use countable additivity and ii iv Proceed as in iii but now 1Bn 1An 1An1 D Exercise 132 Let An be a monotone decreasing sequence of measurable sets A1 2 A2 2 and suppose that 1A1 is finite Show that then 1 Anim1An Give an example to show that the finiteness hypothesis is necessary Exercise 133 Let An be a sequence of measurable subsets of X The limit superior lim sup An is the set of those 1 E X that belong to infinitely many of the An the limit inferior limiann is the set of those 1 e X that belong to all but finitely many of the An a Show that lim iann and limsupAn are measurable b Show that zIim iann g im infzAn c Show that zIimsupAn 2 im sup 1An provided that UAn has finite measure See exercise 62 for the definitions of Iiminf and Iim sup for sequences of real numbers Definition 1310 Let X Au be a measure space A function f X gt R is measurable if for every open subset U of R f 1U is measurable Notice that 1 plays no role here Compare Theorem 27 this shows of course that every continuous function is Borel mea surable It makes no difference to the definition if we require f 1U measurable for every Bore set U exercisewhy Example 1311 Let A e A The characteristic function of A is the function 1 ifxEA 33 o ifxg A Clearly this is a measurable function Proposition 1312 Iffn is a sequence of measurable functions then each of the functions inffn sulDfn liminffn limsupfn is measurable also In particular iflim fn exists it is a measurable function Proof Let g sup fn for example Then g1aoo Ufn1aoo is measurable By Exercise 131 that is enough to make 9 mea surable Similarly for inf fn and the last two follow since lim sup fn infsupfn k ngtk and similarly for lim inf D Lecture 14 Integration Rudin chapter 1 to 130 Proposition 141 The collection of measurable functions on a measure space X is a vector space closed under the formation of linear combinations Proof Easy to see that if f is measurable so is cf c constant Suppose f and g are measurable and let U be open in R Let h 2 g X gt R2 Then f g1U h1V where V xy 1 y e U is an open subset of R2 It suffices then to show that V is a countable union of rectangles of the form a b x ed Enumerate the rational points in V as aimyn For each n let 6n gt 0 be the distance dxnynR2 V and let Rn be the rectangle xn 6nxn6n Xyn 6nyn6n By construction Rn g V Moreover if xy e V with dxyR2V e gt 0 then there is a rational point mmyn with d13y xnyn lt 6 Then 6n gt 6 and so 13y e Rn Thus V URn as required D Definition 142 The vector space of measurable functions on a measure space X spanned by the characteristic functions of measurable sets with finite measure is called the space of simple functions Remark 143 This definition is really only adequate in a a finite measure space if there exist sets of infinite measure that can t be approximated by sets of finite measure one needs to consider their characteristic functions to be simple functions too then one has to take various precautions to avoid oo 00 problems Since we are not going to be using any non a finite measures we won t address this issue The same simple function may be represented in different ways as a linear combination of characteristic functions The following lemma is therefore important Lemma 144 Let X1 be a measure space and let m n f Z aiXAi Z ijBj i1 j1 be two representations of the same simple function f Then i1 91 Proof Consider first a special case assume that the sets B partition X this means that they are disjoint and their union is X and refine the sets A this means that each A is a union of B s say Ai UjEJiBj Then by considering 1 6 Bi one sees that bi Z ijeji a1 Now we have i i jEJi j tjeJi j using the additivity of the measure The general case follows from this since given any two finite families of subsets one can find a third family which partitions X and refines both of them D Definition 145 The common value of the sum appearing in the lemma is called the integral of the simple function f and denoted fd1 Sometimes we mayjust write f if the measure is clear from context Proposition 146 The integral is a positive linear functional on the vector space of simple functions Positive means that if f 2 0 then fde 2 0 Proof Exercise D Definition 147 Let X1 be a measure space and let f be a nonnegative measurable function on X The integral of f written fd1 is the least upper bound of the set of integrals of simple functions that are g f If this integral is finite we say that f is integrable Exercise 141 Consistency Verify that a nonnegative simple function is integrable and that the two senses of integral agree for such functions Exercise 142 Verify that if fg are nonnegative measurable functions with f g 9 then fde lt fgdu Verify also that fcfdl 2 cfde for a positive constant c It is also true that the integral is additive ff gd1 fde fgdu but that is not so easy to check right now We will prove it as a consequence of the monotone convergence theorem The following is a preliminary version of that result which is the central theorem of measure theory Proposition 148 Let fn be a sequence of integrable functions on a measure space X1 such that a For each 1 e X the sequence of real numbers fn13 is mono tone increasing and convergent say to f13 b The integrals ffndz are bounded independent of n Then f is an integrable function and fde lim ffndv Proof Clearly fde 2 limffndu To prove the converse fix 6 gt O and let 9 be a nonnegative simple function with g g f Write g ZaiXAi where the AZ partition X Notice that for all 1 6 Ai we have fn13 gt f13 2 a1 Consequently the sets Ban 1 13 6 Ai fna gt 1 dart increase to Ari as n gt 00 Now the simple function Zil ea XBin is less than or equal to fn so fnd1 2 1 ZaiVB in By Proposition 139iv zBm gt 1Ai as n gt 00 Taking limits as n gt 00 therefore gives limfndz 2 1 egd1 Since the integral of f is the supremum of the integrals of simple functions 9 lt f it follows that limfndz 2 1 efd1 and letting e gt O completes the proof D From the monotone convergence theorem we have in particular that fde limfgndz where gn is any monotone sequence of nonnegative simple functions converging to f It is easy to see that such sequences always exist exercise Now we can prove Lemma 149 The integral is additive ff f d1 ff dz ff d1 for any two nonnegative integrable functions f f Proof If 9 9 are monotone sequences of simple functions ap proximating f f then ggg is a monotone sequence of simple functions approximating f f Use the additivity of the integral for simple functions D Exercise 143 Verify that the integral of the characteristic func tion of any measurable A g X is equal to the measure of A With our definitions you will need to assume that the measure is a fihite Lecture 15 Integrable functions and the convergence theorems Rudin chapter 1 Definition 151 A real valued function on a measure space X1 is integrable if it is the difference of two nonnegative in tegrable functions The integral of an integrable function f is defined to be fg fh where f g h gh nonnegative integrable The additivity of the integral for nonnegative integrable functions shows that this is well defined iff g h g h then gI h hI g so fgfh fhf9 Exercise 151 Show that the integrable functions form a vector space and that the integral is a linear functional on this vector space Exercise 152 Give a sensible definition of the integral for com plex valued functions Lemma 152 Let f be a measurable function on a measure space X1 The following are equivalent a f is integrable b f is integrable C f g g for some nonnegative integrable function 9 Proof a implies c If f is integrable write f f f where f f are nonnegative integrable then take 9 f f c implies b Every simple function g f has integral lt fg so the supremum of the integrals of such simple functions is finite and f is integrable b implies a Write f f f where f13 maxfx0 and f13 minfx0 and note that ff are nonneg ative measurable and bounded in absolute value by f hence integrable D In measure theory one often encounters phenomena that happen for all 1 not belonging to some set of measure zero For instance if a sequence of functions fn13 converges for all 1 not in a set of measure zero one says that it converges aImost everywhere or for almost all x Theorem 153 Monotone convergence Let fn be a mono tone increasing sequence of real valued integrable functions on a measure space X1 Ifffndz is bounded above then fn13 converges almost everywhere to a limit f1339 the function f is integrable and fd1iTrLTifndl Proof We can use proposition 148 there is no loss of generality in assuming the fn are nonnegative once we have established that fn13 converges equivalently is bounded for almost all 13 Suppose ffn lt M Then the measure of the set Bk as Iimrnoz 2 k is bounded by Mk Consequently B k Bk is a set of measure zero and fn13 is bounded for all 13 g B D Exercise 153 Show that in for any integrable function f ltlfldv Exercise 154 Let fn be a sequence of nonnegative integrable functions on a measure space Show that if 00 Z fnd1 lt oo n21 then fn13 gt O for almost all 13 Exercise 155 Ifa nonnegative integrable function f has fde 0 show that f13 O for almost all 13 Apply the MCT to the sequence nf Lemma 154 Fatou If fn is a sequence of nonnegative in tegrable functions on a measure space X1 then lim inffndz g lim inffndu Proof Let gn inffnfn1 Then gn is measurable bounded above by fn and below by 0 so it is integrable The sequence gn is monotone increasing and its limit 9 is liminffn The monotone convergence theorem gives gd1 IiWgndu g limninffndu Since fgn lt ffn D Strict inequality can occur Theorem 155 Dominated convergence theorem Let X1 be a measure space and let fn be a sequence of integrable functions on X converging almost everywhere to a function f If there is an integrable function 9 called a dominating function such that fn g g for a n then fnd1 gtfd1 ash gt00 Easy examples exercise show that without some extra hy pothesis fn gt f ae does not imply ffn gt ff Proof Lemma 152 shows that f is integrable Applying Fatou s lemma 154 to the sequence 9 fn gives 9 new lt Iimninfg floaty and so fdz g limninffndu A similar argument using the sequence 9 fn instead gives fdz 2 Iimrsupfndz and together these inequalities imply the result D Remark 156 Applying the theorem to fn f gives the stronger result that ffn fdu gt O Lecture 16 Functions and Spaces Rudin chapter 1 Definition 161 Let X1 be a measure space The space L1X1 is the space of equivalence classes of integrable func tions on X modulo the equivalence relation equality almost everywhere Warning In many measure spaces singleton sets 13 have mea sure zero In such a space the notion the value of f e L1 at the point x is not well defined Proposition 162 The norm f ffdu makes L1X1 into a normed vector space Proof Mostly obvious Use Exercise 155 to show that if dfg 0 then f 9 almost everywhere D We are going to prove that L1 is a Banach space First a useful lemma Lemma 163 Let V be a normed vector space Then V is a Banach space iff Whenever on is a sequence in V with on lt 2 quot the sum 00 Z w n21 converges in V Proof If on g 2 quot then the partial sums of 222m form a Cauchy sequence hence they converge if V is complete Suppose that V satisfies the stated condition For any Cauchy sequence am in V one can find a subsequence xnk such that xnk xnk1 lt 2quot Take 22k xnk xnk Then 222k converges and so xnk is a convergent sequence But a Cauchy sequence with a convergent subsequence is itself convergent D Theorem 164 For any measure space X y the normed space L1X1 is complete Proof Let fn be a sequence in L1 with fn g 2 quot Then 2 lfnd1 is finite and hence by the monotone convergence the orem 153 Zfnx converges for almost all 1 to an integrable function gx It follows that anm converges almost everywhere say to f13 Let gNa Zgzlf x then gNa gt f13 for almost all 13 and lgNl g 9 By the dominated convergence theorem 155 and remark following f is integrable and f gN fd1 gt O which is to say that an converges to f in L1X1 Lemma 163 completes the proof D Remark 165 A sequence fn convergent in L1 need not have fn13 convergent for any particular 13 classic example using Lebesgue measure is the sequence of characteristic functions of the intervals 01 07 51 07 5 71 7m However from the above proofs one can see that a sequence con vergent in L1 must at least have a subsequence that converges almost everywhere We will prove this again below Definition 166 Let fn be a sequence of measurable functions on a measure space X We say fn converges in measure to the measurable function f if for every 6 gt O 1331 fna fa gt6 gt O as n gtoo Remark 167 In probability theory this is called convergence in probability One thinks of the fn as corresponding to random variables Xn Then the assertion is that the probability that Xn X gt 6 tends to zero as n gt 00 Proposition 168 A sequence of functions that converges in L1 also converges in measure Proof If f fn fd1 an then W 1 fnx fx gt 5 lt 5 1an which gives the result D Lemma 169 Let Nk be a sequence of measurable sets with ZzNk lt 00 Then N lim sup Nk is a null set measure zero This result is known especially by probabilists as the Bore Canteli lemma It is a set version of Exercise 154 Proof By definition N nAn where An UkgtnNk We have zAn g ZkgtnuNk which tends to zero as n gt 00 since it is the tail of a convergent series Since N g An for all n 1N g zAn and consequently 1N lt6 1An O D Proposition 1610 A sequence of functions converging in mea sure has a subsequence that converges almost everywhere PrOOf SUDDOSG that fn gt f in measure Pass to a subsequence fnk with the property that 1331 fnkx fx gt 2k lt 2 Let Nk be the set appearing in the display above then 21Nk lt 00 and thus N limsuka is a null set by the Borel Cantelli lemma If 13 g N then 1 belongs to only finitely many of the sets Nk and it follows that fnk13 gt f13 D It follows from these results that as stated above every se quence converging in L1 has a subsequence converging ae Proposition 1611 In a finite measure space in particular in a probability space a sequence of functions that converges almost everywhere also converges in measure Proof Given 6 gt 0 consider the sets An 13 f13 gt 6 Bk 2 U An ngtk The sets Bk form a monotone decreasing sequence If fn13 gt f13 then 1 g Bk for some k thus Bk is a null set Because X has finite measure 1Bk gt 0 it follows that zAk lt zBk tends to 0 also D Exercise 161 Give counterexamples to other possible implica tions among notions of convergence That is give examples to show that neither convergence in measure nor convergence ae necessarily imply convergence in L1 and that convergence ae does not imply convergence in measure in case 1X 00 Exercise 162 Show that if fn gt f in measure and fn g g for some integrable 9 then fn gt f in L1 Lecture 17 Constructing Measures I Before proving any more abstract theorems let s begin the con struction of some more interesting examples of measures the most important one is Lebesgue measure on R The two stage process starts with a premeasure an elementary idea of the size of some sets It extends this to an outer measure defined on all sets and then restricts the outer measure to a a algebra of measurable sets to obtain the desired measure Additional regularity conditions ensure that the constructed measure coin cides with the original elementary notion of size when both are de ned Definition 171 Let X be a set An outer measure on X is a function u 73X gt 000 such that 3 MW O b monotonicity if A g B then uA g uB C countable subadditivity if A UAn then uA g ZuAn Definition 172 Carathe odory Let u be an outer measure on X A subset A g X is u measurabe if for all B Q X MB uB 0 A uB A ie A splits B evenly according to u measure Exercise 171 Show that any set of outer measure 0 is mea surable Exercise 172 Let X be an uncountable set and for A g X 39f A 39s co ntable define uA O U Show that u IS an 1 if A IS uncountable outer measure Which subsets of X are u measurable and what is their measure Lemma 173 Suppose uquot is an outer measure on X and A1 g A2 g is a monotone sequence of measurable sets with union A Then uA limuAn More generally ifB is any set measurable or not then uA F B limuAn F B Proof Write A m B UAj1Aj B Countable subadditivity gives WM 0 B lt ZMAj1Aj B 339 Since A7 is measurable uAj1 m B uAj1Ajm B uAj m B The sum in the display above therefore collapses to imuAj B and it follows that uA B g imuAn B On the other hand uA B 2 uAnt for all n by monotonicity The result follows E Theorem 174 Let u be an outer measure on X Then the measurable subsets ofX form a a agebra and the restriction of u to this a agebra is a measure Proof Let M be the collection of measurable sets First we ll show that M is Closed under finite unions Let A1A2 E M and B be a test set We have MB MB 0 A1 nB A1 MB A1 nB A1 0 A2 nB A1 U A2 MB 0 A1 U 142 MB 0 A1 MB A1 0 A2 using three different test sets in the definition of measurability for A1 A2 and A1 respectively These give uB uB A1LJ A2 uB A1 U A2 so A1 U A2 is measurable This argument can be visualized in a table split B into four disjoint pieces as follows A2 NOt A2 A1 W X Not A1 Y Z We want to prove that the decomposition B WXY U Z is even and this follows from the evenness of the decompositions WXUYZ YUZ and WXUY Now deal with countable unions Let Ari be a sequence in M with o UglAj and let oj ugzlAi Each oj belongs to M Let B be a test set For any j MB MB Cj MB Cj 2 MB Cj MB C Apply Lemma 173 to the sequence B Cj to see that limjgt00 uB OJ 2 uB m C and so MB 2 MB 0 C MB C By subadditivity we have the opposite inequality MB lt MB 0 C MB C so the two sides are equal and C is measurable We have now proved M is a a algebra That u is a measure on M follows from Lemma 173 D Definition 175 A measure space is said to be complete if any subset of a set of measure zero is measurable and has measure zero Proposition 176 The measure spaces produced from outer measures by Carathe odory s construction are complete Proof Exercise D Exercise 173 Let X 812 be a a finite measure space Show that uA infzB A g BB 6 8 defines an outer measure on X and that measure produced from u by Carathe odory s construction agrees with 1 on B Deduce that X can be extended to a complete measure space Where might outer measures come from Proposition 177 Let X be a set and let X be any family of subsets of X Let 739 be a function X gt 000 With 79 2 0 We may call this a premeasure Then the function u on 73X defined by 00 uA inf Z 7Xn Xn E X and A g Ej Xn n21 n21 is an outer measure on X We take the infimum of an empty set to be oo Proof Only countable subadditivity presents any problems Let Am be a sequence in 73X Let 6 gt O For each m there is a sequence an in X with ZnTan lt uAm 2 me Organiz ing these an into a single sequence gives a countable collection covering A Um Am and MA lt Z Z 7an m1n1 lt Z MAm2m gt lte 2 Mann Since 6 is arbitrary we get A lt ZMAm as required D Lecture 18 Lebesgue Measure Now we can define Lebesgue measure Let X R and let 739 be the premeasure which associates to each interval a b its length 19 61 The extension restriction process of Proposition 177 and Theorem 174 produces a measure on a a algebra of subsets of R which we call Lebesgue measure A on R We can similarly define Lebesgue measure on R2 or Rquot starting with the area or volume of open rectangles or cuboids We ll get back to this Proposition 181 Any countable subset ofR is Lebesgue mea surable and has measure 0 Proof A one point set can be included in an interval of arbitrarily small length so has outer measure zero Countable subadditivity now tells us that any countable set has outer measure zero It is measurable by Exercise 171 D Lemma 182 Suppose that a closed interval a b g R is covered by finitely many open intervals ab i i 1 n Then b alt ilt iai i1 This expresses an intuitively obvious property of lengths the point is that we have to prove it without making use of our intuition about length and area Proof Induction on n The case n 1 is obvious In the general case there is one of the intervals 04155 that contains b assume WLOG that it is 04m n so that on lt b lt n Then the intervals a1 1ozn1 n1 cover mom and so n l Zlt iaz gtan agt b a Bn Oln i1 which completes the induction D Proposition 183 An interval in R open half open or closed is Lebesgue measurable and its measure is equal to its length Proof Since one point sets have measure 0 there is no differ ence between the various kinds of intervals Consider I a b Let B Q R Given 6 gt 0 there is a sequence Jn of open intervals such that B Q UJn and 220 lt AB e 2 denotes length For each n the sets Jn ooa Jn m a b and Jn boo are open intervals possibly empty whose length adds up to the length of Jn The sets of the first and third types cover B a b those of the second type cover B m a b hence B 0 a b AB 6 19 lt AB 6 Since a b has measure zero we can replace the open interval by the closed interval here then let 6 gt O to get B 0 a7 19 AB 6 19 lt AB The opposite inequality follows from subadditivity so we ve shown that a b is measurable Now to prove that Aa b b a Clearly Aa b lt b a On the other hand suppose a b has a countable cover u by open intervals By compactness we may replace u by a finite subcover by intervals 1 ably i 1n Then from Lemma 182 220 2bi ai gt b a Thus Aba 2 b a and the proof is complete D Since intervals are Lebesgue measurable any open set is Lebesgue measurable and consequently Lebesgue measure is a Bore mea sure on R Proposition 184 Let A be any subset ofR measurable or not Then a For any 6 gt 0 there is an open set U 2 A with MU lt xA I e b There is a Bore set G 2 A with MG AA Proof a follows from the definition of outer measure and the fact that the Lebesgue measure of a countable union of intervals is at most the sum of their lengths for b let Un be constructed as in a with e 171 and take G Un D A set such as G above which is a countable intersection of open sets is traditionally called a G5 set The dual notion a countable union of closed sets is called a F0 set Exercise 181 Show that for a measurable subset A ofR having finite measure and any 6 gt 0 there is a closed set F g A with MA lt AF e The measurability requirement cannot be dropped this is tricky to see Exercise 182 Show that a subset A g a b is Lebesgue mea surable if and only if its outer measure AA is equal to its inner measure AA 2 b a gta b A This connects the defi nition of Lebesgue measure to the classical exhaustion method of Eudoxus and Archimedes IfX is a subset of R we define X I a to be 13 I a a E X This set is called a transate of X Proposition 185 Lebesgue measure is translation invariant MX a MX for all measurable X and all a e R Moreover the only translation invariant Borel measure on R that assigns measure 1 to the unit interval is Lebesgue measure Proof Lebesgue measure is translation invariant because it is constructed via Carath odory s extension restriction process from a translation invariant premeasure If u is a translation invariant Borel measure that gives mass 1 to the unit interval then by subdividing the interval appropriately we see that u gives mass 0 to a 1 point set and that ua b 9 0 if the length 9 a is rational The same statement is then true for all intervals by approximation Thus u agrees with Lebesgue measure on all intervals and hence on all Borel sets D Example 186 We have seen that all countable sets have Lebesgue measure zero However there exist uncountable sets of measure zero The standard example is the Cantor set 00 C 13 6 01 113 2 Z 2 where each an 6 01 n21 3n which we have already encountered in Example 36 Distinct sequences an yield distinct real numbers 1 so 0 is uncountable On the other hand the complement 01C is a union of disjoint open intervals 1 2 7 8 and so on of total measure gas gas 00 Z 2n 13 n 1 n1 Thus 0 has measure zero Lecture 19 Lebesgue Integration If a b g R then the collection of Lebesgue measurable subsets of a b forms a a algebra and Lebesgue measure is a finite measure on it Integrability for f on a b is the same as integrability for fXa7b on Write b b Afxdgtx orjust Afadx for the integral If a gt b we define f fxdx to be fb f13d13 This makes the addition law from bcfxdx Cfada true in all cases Proposition 191 Every continuous function a b gt R is inte grable Moreover iff is continuous the function F013 fydy is differentiable and F 13 f13 Proof Continuous functions are Borel and therefore measur able Since a continuous function on ab must be bounded Lemma 152 shows that it will be integrable We have Fa h F13 fghfydy and thus Fa h xh h fx g my mndy Given any 6 gt 0 there is 6 gt 0 such that x yl lt 6 implies f13 fy lt e and thus if h lt 6 1 xh Wm M fxdy lt e by Exercise 153 D The latter part of the proof is the usual argument for the funda mental theorem of calculus We shall later be very interested to see how far the hypothesis that f is continuous can be weakened Proposition 191 means that Lebesgue integrals can be com puted by the familiar antidifferentiation technique However the monotone and dominated convergence theorems are now also available Example 192 Show that 1 313 33de 11 22 33 Solution Write x x e x39ogx z zow The terms in the series are positive continuous functions on 01 so the partial sums form a monotone sequence converging to the func tion x x By the MCT 153 1 0 1 1 quot IO n 33 510de MM 0 n20 O n The integral on the right can be evaluated by repeated integra tion by parts and it is 71 1 quot1 Example 193 Let F13y be continuous on R x R Suppose there is g e L1Rgt such that F13y lt gx for all y Then the function y H Fazydx is continuous To see this note that if yn gt y then Fayn gt Fay pointwise in 13 and Fayn g gx Apply the DCT 155 and Exercise 21 Lebesgue integrals and improper integrals should be carefully distinguished Proposition 194 Let f Ooo gt R be a continuous function Then a Iff 2 0 then f is Lebesgue integrable if and only if the limit I R INT 0 f13d13 R gtoo is finite and in that case the value of the limit is equal to the Lebesgue integral b In general f is Lebesgue integrable ifand only iff is which may be tested for as in a Proof a Apply the MCT to the functions fXMR b This is contained in Lemma 152 D One can give a similar discussion at oo or at an isolated finite point of discontinuity Exercise 191 Let f13 Sigx Use the alternating series test from calculus to show that limRgt00 f fxdx exists later we shall see that the value of this limit is 7r2 Nevertheless show that f is not Lebesgue integrable The above exercise is an integral version of the series Z 21 1quotn which is convergent but not absolutely convergent Exercise 192 Differentiating under the integral sign Sup pose that the function F13y is defined on R x R is integrable as a function of 13 for each fixed y and is differentiable as a function of y for each fixed 13 suppose also that its partial derivative Fy satisfies lexy lt 933 for some integrable function g and all y Show that then the function Go Frxydx is differentiable and that G y nyxydx Same method as Example 193 together with the mean value theorem Exercise 193 Normal numbers Define functions fn on 01 as follows 1 if 2 is even Mm 1 if pm is odd Show that 1 m n O m 75 n and deduce that if gm 2 f1 fnn then fggxdx 1n Conclude using Exercise 154 that gk2a gt O for almost all 1 as k gt 00 and hence that gna gt O for almost all 1 as n gt 00 This conclusion is usually expressed by saying that almost all 1 6 01 are normal in base 2 ie have on average as many O s as 1 s in their binary expansion Can you show that almost all 1 are normal in every base simultaneously No explicit example of such an 13 is known 01 fnxfmxdx Lecture 20 Set Theoretic Questions There are some natural questions we can ask about Lebesgue measure a Are there measurable subsets of R that are not Borel sets b Is every subset of R measurable The answers involve ideas from set theory They will require that we assume the axiom of Choice which gives rise to the theory of ordinal numbers Proposition 201 The collection of Bore subsets ofR has car dinality c 2N0 Proof We need to give a construction of the Borel sets Let J quotO denote the family of open intervals Define families fa of subsets of R parameterized by ordinals or using transfinite induction as follows o if or is a successor ordinal then Fa is the collection of all subsets of R that can be formed from the members of J quot by taking countable unions countable intersections and complements o if or is a limit ordinal then fa is the union of the J quot for all lt or All the sets in every fa are Borel Let S2 be the first uncountable ordinal Then J quotQ is a a algebra this follows from the fact that the least upper bound of a countable collection of countable ordinals is countable Consequently J quotQ is exactly the a algebra of Borel sets Now J quotO has cardinality c and by transfinite induction one sees that fa has cardinality c for each countable ordinal 1 Finally then J quotQ has cardinality ltN1cltccc D On the other hand we have Proposition 202 The collection of Lebesgue measurable sub sets ofR has cardinality 2C Proof The Cantor set 0 Example 186 has cardinality c and measure zero Every subset of C is therefore measurable and has measure zero and there are 2C such subsets D From Propositions 202 and 201 we get the answer to question a There exist many Lebesgue measurable sets that are not Borel Now we turn to question b There exist nonmeasurable subsets of R Example 203 Vitali Let N be the equivalence relation on 01 defined by 1 N y if and only ifx y e Q Let X g 01 be a set containing exactly one representative from each equivalence class This uses the axiom of choice Let qn be an enumeration of the rationals in 11 and let Xn 2 gm X The sets Xn are disjoint and their union A contains 01 and is contained in 1 2 IfX were measurable then each Xn would be measurable too and have the same measure translation invariance so A would have measure either 0 or 00 which is impossible Vitali s example depends on the translation invariance of Lebesgue measure But this is not in fact essential to the construction it is simply a matter of cardinality Proposition 204 Uam Let X O 2 where S2 is the first uncountable ordinal There is no nonzero a finite measure it on all subsets ofX that gives measure 0 to each singleton subset If we assume the continuum hypothesis there is a bijection be tween X and R It follows immediately that Lebesgue measure is not defined on all subsets of R ie nonmeasurable sets exist Proof For or e X let Pa B lt a be the set of predecessors Each Pa is countable choose an injection la Pa gt N For B E X n E N set B57 2 Oz B lt 04 MW 2 n Clearly for distinct one has B yn B317 2 Q Since u is a finite this means that there are only countably many ordinals B E X for which any MB yn gt 0 Let 7 e X be greater than all of these ordinals Put B UnBWL we have MB 2 0 Every ordinal greater than 7 is in B Hence is the union of a countable set and a set of measure zero so has measure zero D Remark 205 We have assumed the axiom of choice throughout It is known that there are models for set theory Without the axiom of choice in which all subsets of R are measurable Lecture 21 The Riesz Representation Theorem Rudin chapter 2 Let X be a compact metric space Given a finite Borel measure 1 on X we can integrate any continuous function However this process can be reversed given an integral on continuous functions on X we can construct a measure that represents it Definition 211 An outer measure 12 defined on subsets of a metric space is a metric outer measure if 1A U B 1A 1B whenever A and B are separated that is infda b a E A b e B gt O Exercise 211 Show that Lebesgue outer measure is metric Lemma 212 Let 1 be a finite metric outer measure on subsets of X Suppose that A g X is written as an increasing union A U311 An and suppose that the difference sets Bn 2 An1 An have the property that Bn and Bm are separated Whenever n m 2 2 Then 1A limngtoo1An Proof Monotonicity gives 1A 2 Iimngt00 11An To prove the opposite inequality write A 2 AN U UgozN Bn By subadditivity WA lt WAN Z Man nN The result will follow provided ZzBn converges If it does not then either ZkuB2k or 2k 1B2k1 diverges say the first Since the sets B2k are all separated 12 U521 B2kgt diverges to 00 as K gt 00 but this sequence is bounded by 1X assumed finite so this is a contradiction D Definition 213 Let A CX gt R be a linear functional is positive if whenever f13 2 O for all 13 then f 2 0 Lemma 214 A positive linear functional is continuous wrt the norm of CX Proof Let a 2 0 be 1 1 is the constant function on X Then for any f E CX f1 lt f lt llfll1 and so by positivity f lt af as required D Theorem 215 Let be a positive linear functional on CX Then there is a Bore measure 1 on X such that f fdz for a f E CX Proof Let X be the family of open subsets of X Define a mapping 71X R by 7U suDf1f E CX O lt f lt 1 Supportf Q U For brevity we shall write f lt U iff and U are related as above Using 739 as a premeasure carry out Carath odory s construction 177 and 174 to construct an outer measure 1 and then a measure 1 defined on a a algebra A We claim that A con tains the Borel sets and that 1 integrates continuous functions correctly The proof has several steps Step 1 739 is countaby subadditive on open sets Let U U311 Un and let f lt U Since the support of f is compact it is contained say in U1 U U UN There is a partition of unity 1 N defined on the support of f and subordinate to the cover U1UN Then onf lt Un so gbnf lt 7Un and therefore 00 N f Z nf lt 2 TM n21 n21 Since this holds for all f lt U 7U lt 27Un as required It follows easily from Step 1 that 7U 1U for any open set U and more generally that 1A infTU U 2 AU open Step 2 If U and V are disjoint open sets then 7U U V 7U TV If f lt U and g lt V then fIg lt UUV it follows that f g lt 7U U V and since this holds for all fg 7U 7V lt 7U U V The opposite inequality follows from monotonicity We deduce that the outer measure 12 is metric For if A B are separated then there exist disjoint open U V with U 2 A and V 2 B If W is open and contains AUB then W U and W V are open and contain A and B Thus 1A 1B 7W U I 7 W V 7W UUV lt Passing to the limit we get 1A 1B g 1ALJB and the opposite inequality follows from subadditivity Step 3 All closed subsets of X and therefore all Bore subsets are z measurabe Let K be closed and let Fn 13 d13K 2 171 the Fn form an increasing sequence of sets each of which is separated from K whose union is XK and whose successive differences satisfy the separation condition of Lemma 212 Let B be a test set as in Definition 172 Then 1B 2 1B D K 1B Fn for each n monotonicity and separation Apply Lemma 212 to the sets B m Fn whose union is B K We get NB 2 uB n K uB K as required At this point we have constructed a Borel measure 1 that agrees with 739 on all open sets Step 4 Show that f fde for all f e CX Fix 6 gt O and let En f1ne 71 1 The E s are disjoint Borel sets covering X and all but finitely many are empty wlog say EOEN1 are nonempty By definition of 12 there are open sets Un 2 En with 1Un lt zEn 1N 22 and f lt n 2e on Un Choose a partition of unity on subordinate to the cover Um Observe that on lt Un and thus Mon lt 1Un 7Un We compute N l f lt Z nf n20 N l N 1 1 lt goo 2mm lt goo 2gt6 WEN m using the positivity of the functional On the other hand we have the inequality f 2 212720 neXEn Thus N l fd1 2 Z nezEn n20 by the positivity of integration Combining the two previous displays we get f lt W 2ezX e Let 6 gt O to obtain f lt de Applying the argument to f instead gives the opposite inequal ity so f fde and the proof is complete D Exercise 212 The measures 1 constructed by the Riesz rep resentation theorem have various regularity properties Verify that for every z measurabe set E and every 6 gt 0 there exist an open set U and a closed set K with K g E Q U and 1UK lt e Deduce that every z measurabe set differs from a Borel set by a set of z measure zero Prove also that the continuous functions form a dense subspace of the normed vector space L1X1 Exercise 213 Show that we have uniqueness in the Riesz repre sentation theorem two Borel measures that determine the same linear functional are the same Suggestion Let A be the class of sets on which the two measures agree Show that A is a a algebra and contains the open sets Lecture 22 Lebesgue Stieltjes Measures Rudin chapter 2 Remark 221 Instead of defining Lebesgue measure from scratch we could have used the Riesz representation theorem starting with some other theory of integration for continuous functions eg Cauchy s define integration as a positive linear functional and then define Lebesgue measure as the measure corresponding to this functional by the Riesz representation theorem That is what is done in Rudin s book Let I a b be an interval in R A partition of I is a finite subdivision a x0 lt x1 lt lt am 2 b and the mesh of the partition is the maximum value of xi 13 1 for z 1n Proposition 222 With notation as above let f I gt R be a continuous function and let 9 I gt R be a monotone increasing function Then the expression i has 9 gown i1 tends to a limit Agf as the mesh of the partition 51013n tends to zero Moreover fixing g the mapping Ag CI gt R is a positive linear functional According to the proposition and the Riesz representation theo rem then there is a measure dg on a b such that Mr fdg This is called the Stietjes measure with distribution function 9 For example if gx 13 we recover Lebesgue measure exer cise O a lt O 1 a gt 0 What is the corresponding Stieltjes measure Does the value of g at O affect your answer Exercise 221Take I 11 and let gx Proof of Proposition 222 Fix 9 and leave it out of the notation For a partition 73 define 121 mm i Milgm gwol A720 i milgxi gx 1l i1 where Mi and mi are respectively the supremum and infimum of f on x 113 Clearly the sum that the proposition talks about is sandwiched between these we ll show they both converge to the same thing The continuous function f is uniformly continuous Exercise 43 Thus for any 6 gt 0 there is 6 gt 0 such that if 73 has mesh smaller than 6 then M m lt e for each 139 Hence Mr A720 lt e Z 9er gem Ke i1 where K gb 9a Now notice that if a partition Q refines a partition 73 meaning that every breakpoint of 73 is a breakpoint of Q then Ago 2 A730 and K90 g K730 Since any two covers 73 73 of mesh g 6 have a common refinement we deduce that A7gtf Apf lt Ke and the same for of course Thus if 737 is a sequence of partitions of mesh tending to O the corresponding s form a Cauchy sequence which converges to the desired limit Linearity and positivity of the limit functional are obvious D Different functions 9 may determine the same measure compare Exercise 221 But the correspondence is almost unique Exercise 222 With dg a Stieltjes measure as above show that dgc d 9d 9c for every open interval cd g a b Here gd limxdga and gc limxcga the limits exist because 9 is monotone Find corresponding formulae for closed and half open intervals Hence show that if g is continuous from the right or from the left then dg determines g uniquely up to an additive constant Exercise 223 Conversely suppose that 1 is a finite Borel mea sure on a b such that 1a 0 Define gx 1a 13 with 9a O Show that 1 dg Thus every Borel measure on a b arises from the Stieltjes construction Exercise 224Show that if g is continuous then dgE AgE for any Borel subset E of a b The construction is linear in g as well as in f we have dgl92 dgl dgg This suggests defining a linear functional no longer positive Ah A91 A92 whenever h gl 92 is the difference of two monotonic functions But how to recognize such an h Definition 223 Let f a b gt R be a function not necessar ily continuous The variation Vf ab of f on a b is the supremum of Z fz fz 1 i1 taken over all partitions 73 0 xn of a b If Vf ab lt 00 we say that f is of bounded variation Obviously a monotonic function is of bounded variation The sum telescopes Moreover by the triangle inequality W i g cm b lt Vf cm b Vg a7b Thus the BV functions form a vector space Proposition 224 A function is of bounded variation if and only if it is a difference of monotonic increasing functions Proof Let f be of BV on a b Consider the total variation function defined by Tx Vf ax The BV property of f shows that T is finite Obviously T is increasing on a b On the other hand if y gt 1 we also have Ty T012 2 y fx because among the partitions defining Ty are those whose last two breakpoints are xy Rearranging the above inequality T f is an increasing function and therefore f T T f is a difference of two such functions D This discussion shows that we could define dg as a signed mea sure whenever g is of bounded variation In the next lectures we will look at signed measures in general Remark 225 Notice that dg need not have the same null sets as Lebesgue measure This is clear from Exercise 221 for ex ample where O is not a null set for dg In fact one can find such examples even when 9 is continuous such as the Cantor function 37 The relevant notion here is absolute continuity to be discussed next Exercise 225 Let f 01 gt R be a function One says that f is absolutely continuous if for any 6 gt 0 there is 6 gt O with the following property given any finite or countable collection akbk of disjoint closed subintervals of 01 if E lak bkl lt 5 the Z fak fbk lt 6 k k Show that every absolutely continuous function is continuous and of bounded variation Show that the Cantor function 37 is continuous and of bounded variation but not absolutely contin uous Lecture 23 Signed measures Rudin chapter 6 Let A be a a algebra of subsets of X Definition 231 A signed measure on A is a countably additive function 1 A gt R such that 1 0 Here countable additivity means that if A UAn with the An disjoint then 1A 21An with the series absolutely conver gent Compare Definition 134 Our definition requires signed measures to be finite everywhere one can allow infinite values but then extra care is needed to avoid oo oo difficulties Let 1 be a signed measure on A Define functions 11 A gt 000 by 1A sup1B B e A B c A 1A sup 1B B E A B Q A These are called the upper and lower variations of 1 Theorem 232 Jordan With notation as above 1 and 1 are finite positive measures on A and 1 1 1 Proof First show that 1i are countany additive Suppose A UAn is a disjoint union Then whenever Bn g An B UBn g A and the Bn are disjoint Given 6 gt 0 choose Bn with zBn 2 11An 62 Then thmn lt 221037 e lt 1B e lt WA e and since 6 is arbitrary Efrain g 1A The opposite in equality is even easier so we have equality and 1 is a measure It is easy to modify the argument in the case 1A 00 In fact this can t happen but we don t know that yet Now show that 1 is finite Suppose not and put A A0 with 1A0 oo Inductively define sequences AnBn E A with the following properties b VBn 9 71 C An is chosen to be one of i Bn or ii An1Bn in such a way that 1An 2 00 One of these two sets must have infinite 1 measure because of the additivity of 1 There are now two possibilities either choice cii is made infinitely often or choice ci is made for all sufficiently large n In the first case infinitely many of the Bn are disjoint so their union has infinite 1 measure a contradiction In the sec ond case we have BN 2 BN1 2 BN2 2 for some N Put B BNk Then by Exercise 132 whose proof remains valid for signed measures 1B limk 1BNk 2 00 again a con tradiction Finally we must prove that 1 VT 1 For any B Q A we have 1B 1A 1A B and therefore zA WA lt 24B lt M u A Taking suprema over all B we get WA lt M mm from the second inequality and u A 2 24A WA from the first inequality Together these give the desired con clusion 1A 1A 1A D Thus a signed measure is the difference of two positive measures just as a BV function is the difference of two increasing functions It was proved by Hahn that this decomposition of a signed mea sure in fact comes from a decomposition of the underlying space Theorem 233 Hahn Let 1 be a signed measure on a a agebra A of subsets of a set X and let 1 1 1 be its Jordan decomposition Then there are disjoint subsets RN 6 A with X P U N and 1A 14A m P 1A 1A m N Proof For each n choose Pn such that 1Pn 2 11X 2 quot Then it is easy to see that VPn lt2n7 VXPnlt2n Put Qk Ungtk Pm and put P I k We have 1Qk g Zngtk 2 quot 22 Since P g Qk for each k 1P 0 We have XQk g XPn for all n 2 k so 1XQk g 2 for n 2 k and therefore 1XQk 0 Since N XP UXQk it follows that 1N O The theorem follows E Definition 234 Two positive measures 11 and 12 on the same a algebra are mutually singular if one can decompose X A1UA2 with 12A1 11A2 0 One may assume A1A2 disjoint Thus the positive and negative parts of the Jordan decomposi tion of a signed measure are mutually singular Definition 235If 1 is a signed measure with Jordan decom position 1 1 1 the absolute value of 1 is the positive measure 1 1 1 By definition a null set for 1 means a null set for M We say that signed measures are mutually singular if their abso lute values are mutually singular Exercise 231 Let 11 12 and u be finite measures on the same a algebra Suppose that 1L and u are mutually singular and that 12 and u are mutually singular Prove that then 1112 and u are mutually singular Lecture 24 Absolute continuity Rudin Chapter 6 Let A be a a algebra of subsets of X and let u be a measure on A Definition 241 A signed measure 1 on A is absolutely con tinuous with respect to u if MA 0 implies 1A O for all A e A Example 242 If f e L1XL then the mapping EHEfdu is a signed measure absolutely continuous with respect to u The next proposition shows why absolute continuity is indeed a form of continuity Proposition 243 1 is absolutely continuous with respect to u iff for each 6 gt 0 there is 6 gt 0 such that if ME lt 6 then lt 6 Proof Clearly the condition implies absolute continuity Con versely suppose the condition fails Then for some 6 gt 0 there is a sequence En E A with uEn lt 2 quot and 1En gt e Let Fm U oszn and let F anm We have Harm lt 2m1 so uF 0 On the other hand 1Fm gt e and since Fm is a monotone decreasing sequence of sets of finite z measure remember that by our definition signed measures are finite it follows that 1F 2 e gt 0 We conclude that absolute continuity fails too D For Stieltjes measures on R absolute continuity for measures is related to absolute continuity of functions in the sense of Exercise 225 Proposition 244 Letg be continuous and nondecreasing Then 9 is absolutely continuous if and only if the associated Stietjes measure dg is absolutely continuous with respect to Lebesgue measure Proof Recall that dgE AgE Exercise 224 The result now follows from Proposition 243 D It was proved by Lebesgue that every absolutely continuous func tion F on an interval a b is an indefinite integral ie Fx 2 fffydy for some f e L1a b This is a special case of the following general theorem Theorem 245 LetXAu be a finite measure space and let 1 be a finite signed measure on X A Then a Ifz is absolutely continuous with respect to u then there is a unique f e L1Xu such that zA fAfdu for all A e A b In general 1 can be expressed as 11 I 12 where 11 is absolutely continuous with respect to u so that part a applies to it and 12 and u are mutually singular Definition 234 This decomposition is unique Part a is called the Radon Nikodym theorem part b is the Lebesgue decomposition theorem We are deferring the proof of this result until Lecture 27 Then we shall give a proof due to von Neumann which depends on some simple facts about Hilbert spaces We shall also discuss the generalization of the Radon Nikodym theorem to a finite measure spaces The function f in a is called the Radon Nikodym derivative of 1 wrt u and written f dudu Exercise 241 Let A be the a algebra of Lebesgue measurable subsets of 01 let u be the counting measure on A and let A be Lebesgue measure Show that A is absolutely continuous with respect to u Does the Radon Nikodym theorem apply If it does find the R N derivative dAdu if it does not say what goes wrong Exercise 242 Show that if 1 is both absolutely continuous and mutually singular with respect to u then it is zero This gives the uniqueness in b above Exercise 243 Show that every signed measure is absolutely continuous with respect to its own absolute value What can you say about the Radon Nikodym derivative in this case How does it relate to the Hahn decomposition Here is an example of the application of the Radon Nikodym theorem Proposition 246 Let Xu be a finite or a finite measure space Every continuous linear map T from the Banach space L1Xu to R is of the form fHfgdu Where g X gt R is a measurable function with g13 g T for all 13 Proof Assume the measure is finite exercise generalize to the a finite case For each measurable set E let XE E L1Xu be its characteristic function and consider the map This is a signed measure absolutely continuous with respect to u Reason for countable additivity if E is the disjoint union of E1E2 and Fn E1 U U En then the characteristic functions of En converge to the characteristic function of E in L1Xu by the MCT Thus by the Radon Nikodym theorem there is a function g e L1Xu such that TXE ngdu for all E We shall show that g is bounded For n 12 let En 13 933 gt Then mm E gem 2 T 1nultEngt T 1nllgtltEnll This contradicts the definition of T unless uEn 0 Taking the union over n and arguing similarly for the lower bound M331 9x gt T 0 Thus we can modify 9 on a null set to make it bounded by T We have proved the identity Tf f9 du for every characteristic function f and therefore by linearity for every simple function Now for any integrable f there is a sequence fn of simple functions converging to it in L1 norm Then Tfn gt Tf and on the other hand ffngdu gt ffgdu now that we know that g is bounded It follows that the identity holds for all f 6 L1 as asserted D Lecture 25 Inner Products and L2Xu Rudin chapter 4 Let V be a real or complex vector space An inner product on V is a scalar valued function on V x V written gt which has the properties that ugt1v1 A2222 2 A1uv1gt A2uv2gt that uvgt 2W and that uugt is a nonnegative real number zero only when u 0 These properties are modeled on those of the dot product in Euclidean space Lemma 251 Let V be an inner product space that is a vector space equipped with an inner product Then the expression v vvgt12 defines a norm on V and we have the Cauchy Schwarz inequality ltuvgt lt llullllvll with equality if and only if uv are linearly dependent Proof We show Cauchy Schwarz first Take uv independent Then the expression llAu lWH2 lAl2llull2 mom 22gt lnl2llvll2 is always strictly positive Thus the eigenvalues of the hermitian matrix Hull2 um vugt llvll2 are strictly positive That means that its determinant is strictly positive as well which is the Cauchy Schwarz inequality If uv are linearly dependent it is easy to see that equality holds in Cauchy Schwarz In that case the matrix has at least one zero eigenvalue Now from Cauchy Schwarz u vll2 llull2 llvll2 2m 11gt lt llull2 llvll2 2llullllvll The right side is u v2 and taking square roots gives the triangle inequality Compare Proposition 18 D An inner product space is called a Hilbert space if it is complete in its norm Thus a Hilbert space is a particular kind of Banach space Example 252 The space 62 of sequences a 2 an of complex numbers such that Zan2 lt 00 is a Hilbert space the inner product is a b Za nbn The completeness of this space is a very special case of the next proposition Example 253 Let Xu be a measure space The space L2Xu consists of all functions f X gt C such that f2 is integrable For f e L2Xu we define f by f2 fxl2dux It is easy to see that this is the norm corresponding to an inner product ltfggt mgltxgt duos one needs to check that the expression under the integral sign belongs to L1 which follows from the elementary inequality fxgx lt f vl2 9x2 Proposition 254 The inner product space L2Xu is a Hilbert space Proof According to Lemma 163 it suffices to show that if fn E L2 with fn lt 2 quot then the series an converges in L2 Let gN 2172721 fn be the partial sum of the absolute values Then N 912v Z ltlfml lfngt lt Z 2ltmngt lt 00 771 m7n By the monotone convergence theorem then 912V converges almost everywhere to an integrable function which we write 92 Then gN converges ae to g 6 L2 and consequently hN 212721 fn converges ae say to a function h We must show that hN converges to h in the norm of L2 By the triangle inequality hN g gN g g ae and letting N gt 00 we also get h g 9 Thus hN h2 lt 4g2 is dominated by an integrable function so llhn hll2 was hxl2dux a o by the DCT D Remark 255 If X N with counting measure we obtain the space 62 described above In fact for any set S equipped with counting measure let us use 628 to denote the L2 space asso ciated to S with counting measure That is 228 is the space of maps 5 l gt as from S to C such that 2865 ch2 lt 00 The sum makes sense for any 8 as the supremum of the sums over finite subsets of X Exercise 251 Show that the simple functions form a dense subspace of L2Xu for any a finite measure space Xu Exercise 252If u is a finite Borel measure on the compact space X show that the continuous functions are dense in L2X u Definition 256 A subset C of a vector space is called convex if for all uv E C and t 6 01 the vector tu I 1 tv also belongs to C Proposition 257 A closed convex subset of a Hilbert space contains a unique element of smallest norm Proof Let C be a closed convex set and let G inf13 113 6 0 Notice the parallelogram law Ila yll2 Ila yll2 2ll vll2 2y2 We deduce that if xy e 0 both have norm 6 then 13 I y lt c unless 1 y But a I y e C by convexity so has norm 2 c Thus a point of minimum norm if it exists is unique To show existence let am e C be a sequence with am gt c Applying the parallelogram law llxn mm lt 2llxnll2 2llxmll2 4c2 since an 33m 6 C The right hand side tends to zero as nm gt 00 so am is a Cauchy sequence and hence converges to some 13 which has norm c Since 0 is closed a E C D Lecture 26 The Geometry of Hilbert Space Proposition 257 has many important implications Theorem 261 Let V be a closed subspace of a Hilbert space H Then H Veavi where Vi 13 e H 31513 OVy e V To put it another way for any subspace V there is a bounded linear projection operator P H gt H projection means P2 P whose range is V and whose kernel is Vi Because of this interpretation call the result the projection theorem Proof Let z e H By Proposition 257 applied to V z there is a unique vector y e V that minimizes the distance from V to z I claim that 1 z y belongs to Vi to see this let 2 E V and consider IIZ y Av2 llw Mll2 ll vll2 2 mltwvgt A2llvll2 This must attain its minimum at A O which implies that 12513 O This is true for all 2 E V so a 6 Vi D Exercise 261In the Hilbert space L2 11 with respect to Lebesgue measure show that the set V of functions f that satisfy f13 f 13 ae is a closed subspace and find its orthogonal complement Exercise 262 Let S be any subset of a Hilbert space Show that SA is the closed linear span of S that is the smallest closed subspace that contains 8 Let H be a Hilbert space A continuous linear functional on H is a continuous that is bounded linear map H gt C Some examples are easy to obtain for any 1 e H the Cauchy Schwarz inequality shows that the map oxinltxwgt is a continuous linear functional of norm x Proposition 262 Every continuous linear functional on a Hilbert space is of the form ox for some uniquely determined 13 Confusihgly this result is also sometimes called the Riesz rep resentation theorem Proof Let gi be a nonzero continuous linear functional and K its kernel which is a closed subspace of H By simple algebra KiHKc so Ki is one dimensional Pick a unit vector u e Ki and define a Then ltx Augt MKWHUIF Wu so gi agrees with ox on Ki It also agrees with ox on K since both vanish there so gi ox on K ea Ki By the projection theorem that is all of H D Lemma 263 If T is a bounded linear operator on a Hilbert space H then T SUDltxTygt i llxll llyll lt 1 Proof By Cauchy Schwarz SUDIltxzgt I llxll lt 1 llzll Thus SUDIltxTygt I llxll llyll lt 1 SUDTy I llyll lt 1 llTll as required D Proposition 264 For every bounded linear operator T on a Hilbert space H there is a bounded linear operator T on H such that ltTxygt xITw for all xy e H T is called the adjoint ofT Proof Fix 13 The mapping y H x7Tygt is a continuous linear functional on H so it is of the form gbz for some z depending on 13 of course Define T13 to be z Then the equation Txygt 13Tygt is satisfied Linearity of T follows from the antilinearity of the inner product in the first variable for instance ltTx 23 ygt lt33 x 7Tygt lt33 Tygt 33 Tygt T53 Tx ygt Boundedness follows from Lemma 263 which also shows that llTll llTll 5 Exercise 263 Show that TT T2 for any bounded oper ator T on a Hilbert space Exercise 264 Show that a bounded operator U on a Hilbert space H gives an isometry from H onto itself if and only if UU I UU in this case we say U is unitary What is the corresponding condition for an operator V to give an isom etry of H into but not necessarily onto itself Exercise 265 Let Xu be a measure space and let f be a bounded measurable function X gt C Show that the multiplica tion operator Mf defined by Mf9 2 f9 is bounded on L2Xu What is the adjoint of Mf Lecture 27 Proof of the Radon Nikodym Theorem Exercise 271 Let g be a real valued function in L1X1 If ng 2 O for every measurable set A then 9 2 0 almost every where Let us recall the statement of the Radon Nikodym theorem Theorem 271 Let X Au be a finite measure space and let 1 be a finite signed measure on X A Then a Ifz is absolutely continuous With respect to u then there is a unique f e L1Xu such that IA fAfdu for a A e A b In general 1 can be expressed as 11 I 12 Where 11 is absolutely continuous With respect to u so that part a applies to it and 12 and u are mutually singular Definition 234 This decomposition is unique We have already proved the uniqueness statement in b Exer cise 242 Proof Without loss of generality assume that 1 is also a positive measure Then w u I I is a positive measure also Let H be the Hilbert space L2Xu Consider the linear mapping A f l gt ffdu We have lfl ltlfldv ltificw ltllfllll1ll where the norms on the right are in H The norms on the right are in the Hilbert space H and H1 2 MX 2 is finite because X has finite measure Thus is a continuous linear functional on H By Proposition 262 there is g e H representing A that is fdv fgdw fngfgd1 Taking f XA for any set A E A gives 0 lt 49cm lt MA Since this is true for all A O g g lt 1 almost everywhere by Exercise 271 and let us modify 9 on a set of measure zero so that this is true everywhere We can rewrite the fundamental identity above as L l QMVRWM Let B1 be the set of points where g lt 1 and let B2 its comple ment be the set of points where g 1 Put WM 1A Bi 139 12 Then uB2 0 take f to be the characteristic func tion of B2 to see this and thus 12 is singular wrt M Now let E be a subset of B1 Taking f 1 I I gnXE gives 1 g 1dv 91gquotdu E E Let n gt 00 and apply the MCT to get 9 E d V E1gu which is the desired result with h gl g integrable by the MCT D Remark 272 The Radon Nikodym theorem remains valid if u is only a finite rather than finite To prove it write X as a countable disjoint union of measurable subsets Xn of finite mea sure apply the previous version of theRadon Nikodym theorem to each Xn separately form the sum of the corresponding Radon Nikodym representations of 1an to get a Radon Nikodym repre sentation of 1 Exercise 272 Formulate and prove a version of the Radon Nikodym theorem that is valid when both 11 and 1 are a finite If 1 is absolutely continuous with respect to u then one some times writes 1 ltlt 11 In this case write dudu for the function defined by the Radon Nikodym theorem such that dz 1E E 1111 for all measurable E Proposition 273 In the above situation let f be a u integrabe function Then d fdu f Vdu dn Proof Take 1 to be a positive measure If f is a simple function the result follows from the definition of dVdu For monotone limits of simple functions it follows from the Monotone Conver gence theorem and L1X1 is spanned by such limits D In particular we can consider the case f XEdAdzx where A ltlt 1 is a third measure and E is any measurable set This gives and thus we get the chain rule for Radon Nikodym derivatives dgt dAdz E E almost everywhere Lecture 28 Orthonormal Bases and Fourier Series Let H be a Hilbert space Lemma 281 Pythagoras theorem Ifu v e H are orthogonal then llu vll2 llull2 llvll2 Proof Expand the norm in terms of the inner product D Definition 282 A subset S of H is orthonormal if each 2 e S has norm 1 and 22111 0 for distinct 22111 e S Proposition 283 Let S be an orthonormal set in a Hilbert space H Then for every 13 e H X lt8 vgt2 lt llxll2 86539 with equality if and only ifa belongs to the closed linear span of S that is the smallest closed subspace containing 8 The inequality is called Bessel s inequality Proof For a finite subset 51 sn of 8 let y 1sjxgtsj Using orthonormality one sees that 15513 2 Spy for z 1 n Thus y is orthogonal to 1 y and so 2 2 2 llxll y IIxyll Pythagorasl In particular n 2 2 2 llyll Z lt5ja33gt lt llxll j1 which is Bessel s inequality for a finite set The infinite case follows by passing to the limit If we have equality then given any 6 gt O we can find a finite sum y as above with 13 y lt e this implies that 13 is a limit of linear combinations of members of 8 so is in the closed linear span of 8 Conversely suppose that 1 is in the linear span of S ie is a finite linear combination of members of S Then 1 y for some y as above so Bessel s inequality is an equality for such 13 Both sides of the inequality are continuous as functions of 13 so this equality extends to the closure D It follows from Bessel s inequality that if S is an orthonormal set and 1 e H then the Fourier coefficients 5513 considered as depending on s e 8 form an element of 209 In particular only countany many of them are nonzero If 51 is any enumeration of the countably many 5 for which the corresponding Fourier coefficient is nonzero then the corresponding Fourier series 00 Z lt82 xgts i1 is convergent its tails satisfy n n ll 2 lt8i7xgtsill2 Z lltsixgtl2 e o as mane oo im im so its partial sums form a Cauchy sequence It follows from the discussion in the proof of Bessel s inequality that the sum of this series is the orthogonal projection of 1 onto the closed linear span of 8 Proposition 284 The following conditions on an orthonormal set S are equivalent 1 The closed linear span ofS is H 2 The only vector orthogonal to each member ofS is the zero vector 3 The Fourier series based on S of any 1 e H converges to 13 itself 4 The map which sends 1 e H to its Fourier coefficients sets up an isometric isomorphism from H to 4428 5 For every 13 e H we have equality in Bessel s inequality This result is called Parseval s theorem Proof 1 and 2 are equivalent by Exercise 262 The discus sion above shows they re both equivalent to 3 which obviously implies 5 and 5 implies 1 by the condition for equality in Propo sition 283 Finally 4 and 5 arejust two ways of saying the same thing D An orthonormal set satisfying these conditions is called complete Proposition 285 Every Hilbert space admits a complete or thonormal set Proof Apply Zorn s Lemma to the collection of all orthonormal sets partially ordered by inclusion We obtain a maximal or thonormal set S This must be complete because if not there is a unit vector orthogonal to each member of it this unit vec tor could be added to to S to obtain a larger orthonormal set contradicting maximality D Exercise 281 Prove that a Hilbert space is separable if and only if it admits a countable complete orthonormal set We need some examples of complete orthonormal sets Proposition 286 The functions ena 27f12emx n E Z form a complete orthonormal set in L2 7r7r Proof It is easy to check that these functions form an orthonor mal set By the Stone Weierstrass theorem their linear span is dense in C 7r7r with the sup norm topology Now suppose that f e L2 7r7r is orthogonal to all the en then it s orthog onal to every continuous function Thus the bounded linear functional on CX 7T 9H fgdA 7T is the zero functional By uniqueness in the Riesz representation theorem Exercise 213 this functional is given by integration with respect to some unique Borel measure which is fdA Thus fdA is the zero measure which is to say f 0 almost everywhere This proves completeness in the form 2 above D Thus as a corollary to the general theory we get the Riesz Fischer theorem Proposition 287 Let f be an L2 function on 7r7r in par ticular it might be a continuous function Then the Fourier series 00 7t Z cnemx on faemxda noo 27139 7t converges in L2 to f In fact the Fourier series converges almost everywhere under these conditions This is a deep theorem Carleson 1966 Exercise 282 Part of the complication of the above proof arises from the fact that we have not yet shown that the continuous functions are dense in L2 Prove this for any finite Borel mea sure on a compact metric space Lecture 29 Pointwise results about Fourier Series In classical Analysis there was greater interest in the pointwise convergence of Fourier series than in their uniform convergence Proposition 291 Riemann Lebesgue lemma Letf E L1 7r7r Then the sequence of Fourier coefficients on 1 7r faemxda 27139 7139 tends to zero as n gt 00 Proof The statement is obviously true if f is a trigonometric polynomial a finite linear combination of the functions emit By the Stone Weierstrass theorem the trigonometric polynomials are dense in C 7r7r with respect to the uniform norm and C 7r7r is dense in L1 7r7r with respect to the L1 norm The statement is preserved under L1 limits so it is true for all f 6 L1 D Remark 292 It follows of course that the real forms of the Fourier coefficients fa COSnxdx sinnxdx also tend to zero as n gt 00 To see whether the Fourier series of f converges at a point write a formula for its partial sums N rms 1 7r 2 cne g 7T Dive ymymy n N where 1 DW ethSInN t nN sint is the Dirichlet Kernel ZU i A f AVAV1 UVVZVK Proposition 293 Suppose that f is continuous on 7r7r and that for some point 1 e 7r7r the function fy fx y x y gt is Lebesgue integrable Then the Fourier series off converges at 13 to f13 In particular the Fourier series converges iff is differentiable at 13 Proof Write N 7139 fx 2 enem2 DNx yfaz fydy n N 7r Plugging in the trigonometric formula for the Dirichlet kernel we get 1 7r 1 2 smaN 5X3 ygltydy 7T 7139 where g is the integrable function fx fy 9W The result therefore follows from the Riemann Lebesgue lemma D Exercise 291 Show that the Fourier series of the function f13 x13 converges to zero at the origin despite the fact that it is not differentiable there Exercise 292 Suppose that f is a 02 function on some subin terval ab of 7r7r Show that the Fourier series of f con verges to f uniformly on compact subsets of a b We now work towards showing that some extra condition be sides continuity is necessary for the convergence of Fourier se ries there is a continuous function whose Fourier series does not converge at some point The proof is a Baire category argument For a continuous function f let SNf denote the sum of the first N terms of its Fourier series evaluated at 0 That is twf Jmmenw 7T 7139 Lemma 294 The L1 norm of DN satisfies wwmmmsm Nsm Proof Since siny g y for y E 7r7r 2 7r 1 dy 2 N7r dt DN1 2 O ism N gty O smt7 2 N 1 kn 4 1 gt 2 Sintdt 2 gtoo szl k7r k 17r 7r kzlk Notice that one needs to consider all the peaks of DN to prove this it is not enough simply to consider the central summit D Lemma 295 There is no constant C such that ISNf lt Cllfll for all continuous functions f and all N Proof For fixed N one can find a sequence fk of continuous functions with values in 11 that tend pointwise to the sign of DN By the dominated convergence theorem the corresponding SNfk 2W1ffkyDNydy tend to 277 1HDNH1 AS we have seen this is not bounded in N D Proposition 296 There exists a continuous function f for which the sequence SNfN172W is unbounded and in particular does not converge Proof Suppose not Let F be the set of functions f e C 7r7r such that SNf lt 1 for all N Clearly F is a closed subset of the complete metric space C 7r7r By hypothesis the sequence SNf is bounded for every function f That is to say the union U311 TF 2 C 7r7r By the Baire category theorem one and hence all of the sets rF must have nonempty interior Since F F g 2F we see that in fact F must contain some ball BOe Therefore we would have ISNW lt e lllfll for all N But this contradicts lemma 295 D Exercise 293 The Cesaro means of the Fourier series for f are the averages of the partial sums Slf Snf Show that for any continuous function f the Cesaro means converge uniformly to f Construct a kernel called the Fejer Kernel Fn such that iFncr yfydy represents the n th Cesaro mean of the Fourier series for f Show that Fn is a positive function Lecture 30 Coverings and Maximal Inequalities Rudin chapter 8 Many delicate theorems in analysis depend on the properties of various maximal functions In general a maximal function associated to a function f is a supremum of averages of f over various different scales A maximal theorem estimates the size of a maximal function of f in terms of the size of f itself Notice that because of the supremum the maximal operator which takes f to its maximal function is non linear Here is a basic example Definition 301 Let f E L1R The Hardy Littlewood maximal function Mf is defined by 1 Mfx sup altxltbb a b a lfyldy Exercise 301 Show that the Hardy Littlewood maximal func tion is measurable Hint it suffices to consider intervals with rational endpoints Show that it is usually not integrable Definition 302 We say that a measurable function f on R or on any measure space is of weak L1 type if there is a constant A such that Mm i fx gt 7 lt Ai for all r gt O Exercise 302 Show that the functions of weak L1 type form a vector space If f is integrable then the estimate M331 fx gt 7 lt llflllr shows that f is of weak L1 type The function 13 l gt 151 shows that the converse is false Theorem 303 Hardy Littlewood maximal theorem Iff e L1R then the Hardy Littewood maximal function Mf is of weak L1 type In fact Mm i Mfx gt 7 lt 5f17 Proof Fix 7 and let E be the set 13 Mf13 gt 7 By definition if 1 e E then there is some interval Ix away containing 1 such that bac lfaldx 2 bx ax max 036 Notice that all the points of Ix now belong to E If leIxQ is a sequence of disjoint intervals of this sort whose union is U then we can sum to obtain f1 2 U lfalda 2 MW Thus the proof will be finished if we can show that there are disjoint intervals Ix whose union occupies a substantial fraction at least one fifth of E To reach this conclusion we use a Vitai covering argument which depends only on the facts that 1 e Ix and that the lengths of the Ix are bounded above think why this is the case For each interval Ix let Jx be the interval with the same center and five times the length Inductively define a sequence 131332 of points of E and a sequence E1E2 of subsets of E by setting E0 E and following the procedure 0 x1 is a point of E141 such that the length M1332 is greater than supgtIx 1 6 E141 and 39 E Ei 1zquot If Ek Q for some is the procedure terminates Because LUZ contains a point outside 39 for any j lt i and is less than twice as long as Ixj the intervals Ix and Ix are disjoint We have Summing over i we see that it suffices to prove that the sz cover E If the induction terminated after finitely many steps this is clear Otherwise note that XML lt r1f lt 00 Thus the lengths M1332 tend to zero But if 1 e E does not belong to Uin then M133 lt 2gtIxz for each i so Aux O which is a contradiction D We can abstract the last part of the proof to obtain the following Vitai covering lemma Lemma 304 Let I be a collection of nonclegeneratei intervals in R whose union contains a measurable set E Then one iThat is not reduced to a single point can find a finite or countable sequence I Z 1727m of pairwise disjoint intervals in I such that f M 2 lME i1 5 Exercise 303 Let E be a subset of R A collection I of closed nondegenerate intervals is called a Vitai cover of E if for every 1 e E and every 6 gt 0 there is an interval I e I of length less than 6 and containing 13 Prove that from any Vitali cover one can select a finite or countable disjoint subfamily that covers all of E except possibly for a set of Lebesgue measure zero Apply the previous result inductively Exercise 304 Prove that an arbitrary even uncountable union of nondegenerate closed intervals is Lebesgue measurable Use the previous exercise Lecture 31 The Lebesgue Differentiation Theorem Rudin Chapter 8 Here is a typical application of the Hardy Littlewood maximal theorem Proposition 311 Let f e L1R Then for almost all a the limit 1 gm 1 my randy exists and equals zero In this notation the limit is taken over all nondegenerate intervals I containing 13 and we mean that the expression on the right can be made arbitrarily small by taking the length MI small enough Proof Fix 6 gt O and let E5 513 limsup jfy fa dy 2 6 1316 It will suffice to show that for each 6 E5 is a null set Now fix 6 gt 0 Since the continuous functions are dense in L1 Exercise 212 we may write f g h with 9 continuous and h lt 6 Write now 1 m I my fxldy lt 1 1 mA gy gady ha j hyldy As I shrinks to 13 the first term tends to zero because 9 is continuous The second term is constant and the third term is bounded by the maximal function Mha We conclude that if 1 e E5 then either h13 gt 62 or Mha gt 62 Thus 26 106 A E g 6 6 6 using the Hardy Littlewood maximal theorem for the second es timate But 6 is arbitrary so AE5 O as asserted D Exercise 311 Let E be a measurable subset of R A point p is a point of density of E if the limit MI 0 E lim Iap AI exists and equals 1 Show that if E has positive measure then almost every point of E is a point of density of E The basic result of differentiation theory is Proposition 312 Let f e L1R Then the function F013 fydy is differentiable almost everywhere and its derivative is equal almost everywhere to f Proof It follows immediately from Proposition 311 In fact this theorem implies that for almost all 13 xh hum my faldy o In particular w m2 3 Wow render h h 16 tends to zero as the positive number h tends to zero for almost all 13 There is a similar statement for negative h and thus F is differentiable ae with derivative f D Corollary 313 If F is absolutely continuous then it is differ entiable almost everywhere and it is equal to the integral of its derivative Proof If F is absolutely continuous then F de where dF is a Lebesgue Stieltjes measure absolutely continuous with respect to Lebesgue measure By the Radon Nikodym theorem there is a function f integrable on each finite interval such that Fb Fa f fxdx Apply the previous result to see that F is differentiable ae with derivative f D The two previous results together constitute the generalization of the fundamental theorem of calculus to Lebesgue integrals The example of the Cantor function shows that a function that is differentiable almost everywhere with derivative in L1 nev ertheless need not be the integral of its derivative the missing condition here is absolute continuity The Stieltjes measure as sociated to the Cantor function is nevertheless covered by the following result Proposition 314 Let u be a finite positive Bore measure on an interval ab Let N g a b be a u nul set Then I limsupL O 1316 AI for A amost all a E N Proof Fix 6 gt O and let MU E xeNlimsup gt6 5 13x MI It suffices to show that E has Lebesgue measure 0 for each 6 Fix 6 gt 0 By regularity for u there is an open set U containing N with uU lt e For each 1 e E5 there is an open interval Ix containing 1 and contained in U such that Max 2 6M5 Let W UIx which is an open subset of U By the Vitali covering lemma 304 there is a countable disjoint subfamily 1 such that Exam 2 gtW Thus 5 5 W 5 AE5 lt MW lt 52mm lt 3341 lt ES k 36 Since 6 is arbitrary we see that E has measure zero as asserted D We say that a function g of bounded variation is singular if its associated Stieltjes measure dg is singular with respect to Lebesgue measure Proposition 315 A singular function is differentiable almost everywhere with derivative 0 Proof Assume g is monotonic If g is singular and defined on a b say then a b ALIB where A has zero Lebesgue measure and B has zero dg measure The previous proposition shows that 923 h 9x h for Lebesgue almost all a E B D Iimsup 0 h gtO This gives us Corollary 316 A function of bounded variation is differentiable almost everywhere Proof By the Radon Nikodym theorem and the Lebesgue de composition theorem a function of bounded variation can be written as the sum of an absolutely continuous function and a singular function Our previous results show that functions of both these sorts are differentiable almost everywhere D Lecture 32 Product Measures Rudin Chapter 7 Let X and Y be sets and let A and B be a algebras of subsets of X and Y respectively A set of the form A x B Q X x Y with A E A and B E B is called a measurable rectangle Definition 321 In the above situation let A x 8 denote the a algebra of subsets of X x Y generated by the measurable rect angles If E is a subset of X x Y we define its cross sections Exyixy E Y Eyxixy E X Lemma 322 IfE belongs to A x B then every Ex belongs to B and every E3 belongs to A Proof The collection of sets E satisfying the conclusion of the lemma is a a algebra and contains all measurable rectangles B Let f be a function on X x Y We can define cross sections for it fxy fxy fyx fxy Proposition 323 The cross sections of a measurable function are measurable Remark 324 The converse is false one can find nonmeasurable functions say on R2 all of whose cross sections are measurable Proof Any measurable function is the pointwise limit of simple functions so it suffices to prove the result for simple functions A simple function in turn is a linear combination of characteristic functions so it suffices to prove the result for a characteristic function But now it follows from the preceding lemma D We shall need a technical result A collection M of subsets of some set is called a monotone class if it is closed under the for mation of countable increasing unions and countable decreasing intersections For instance every a algebra is a monotone class Lemma 325 Monotone class lemmaIfa monotone class contains an algebra A it also contains the a agebra generated by A Recall that an algebra is closed under complements and finite unions Proof Let M be the smallest monotone class containing A ex ercise why does this exist For any set E let E be the class of all sets F for which E F EF and FE are in M Clearly E is a monotone class Suppose that E belongs to A Then A g E and therefore since E is a monotone class M g E Hence for every F E M F E E but this is the same as to say that E e F because of the symmetric nature of the definition of the operation Since E was arbitrary A g F for every F E M and then M g F since F is a monotone class We have shown that M is closed under intersections and complements so it is an algebra But it is easy to see that a monotone class which is also an algebra is in fact a a algebra D Proposition 326 In the situation above let u and 1 be a finite measures on X A and Y B respectively Then for any E E A x B the functions 13 gt 1Ex and y i gt ME are measurable and we have X 1Exdiix Amy dIy Moreover considered as a function of the set E the common value of these two integrals is a measure on X x Y We call the measure so defined the product measure and write it u x 1 Proof For simplicity we ll assume that the measures are finite The extension to the a finite case is routine Let D be the class of all sets that are finite disjoint unions of measurable rectangles It is easy to see that the equality of the theorem holds for every measurable rectangle and hence for every E e D Moreover D is an algebra of sets This follows because an intersection of measurable rectangles is a measurable rectangle and the complement of a measurable rectangle XxYltAxB XA xYlLJle YB is a disjoint union of two measurable rectangles Let C be the class of all sets E that satisfy the equality of the theorem We see that C 2 D Moreover C is a monotone class if E is a monotone increasing sequence of sets in C with union E then 1 E d lim E d X an MOI Xu M Ma n gtoo um yquundzy YMEyd1y n gtoo by the Monotone Convergence Theorem Thus by the mono tone class lemma C contains the a algebra generated by D But this is A x B so we are done The fact that the common value of the two integrals defines a measure is a consequence of the MCT as in the identity displayed just above D Exercise 321 Prove associativity for the product measure con struction if XAu z 1 2 3 are finite or a finite measure spacesthen 1 X 2 X 3 1 X 2 X 3 As a result it makes unambiguous sense to speak of the product of any finite number of measure spaces Theorem 327 Tonelli s theorem Suppose that u x 1 is a product measure as above and let f be a nonnegative measurable function on X x Y Then the extended real vaued functions mag mwwox w meW are measurable and Aig mwwunmm Af nawwogwg Xxy fdm gtlt 1 equality of extended real numbers Proof If f is a characteristic function this is exactly Proposi tion 326 The theorem is true for characteristic functions hence for simple functions and hence for all nonnegative measurable functions by approximating with simple functions and using the Monotone Convergence Theorem D Theorem 328 Fubini s Theorem Suppose that u x 1 is a product measure as above and let f be a measurable function on X x Y Then the extended real valued functions a H Y fxyldvy y H X fxyldnx are measurable and if either iterated integral X Y Ifltxygtldulty dnx Y X Wampum My is finite then f is integrable In that case f5 belongs to L1Y1 for almost all 13 fy belongs to L1Xu for almost ally and X fxydvygt duct Y ltXfxyduxgt lly Xxy fdm gtlt 1 equality of real numbers Proof Apply the preceding result to f and f separately D In practical calculations using Fubini s theorem what this means is that one can exchange the order of integration in integrat ing a measurable function f13y of two variables provided that either iterated integral of the absolute value f13y is fi nite Even if the iterated integrals themselves ff f13ydxdy and fffxydydx exist they may not be equal unless fff13y is finite See Exercise 332 in the next lecture for a specific exam ple Exercise 322 Assume the continuum hypothesis for this exer cise According to the continuum hypothesis one can well order 01 in such a way that each element has only countably many predecessors Let lt be such a well ordering and define a func tion f13y on 01 x 01 to be the characteristic function of the set xy 1 lt y Show that each x section and each y section of f is bounded and Lebesgue measurable but that nevertheless fxydAagtdMy 72 frxydAydAx The point here is that f is nonmeasurabe as a function of two variables Lecture 33 Applications of Fubini s Theorem The most important application of the theory of product mea sures is to Lebesgue measure It is natural to think that Lebesgue measure on R should just be the product ofn copies of Lebesgue measure on R Unfortunately this has a defect it is not com plete Example 331 Consider a set of the form A N x 0 where N is a non Lebesgue measurable subset of 01 Then A has a non measurable section so A itself is nonmeasurable for the product measure lemma 322 On the other hand A is contained in the subset 01 x O which is measurable and has measure zero Thus the product measure is not complete Definition 175 For this reason we define Lebesgue measure on R to be the completion of the product measure A x x A That is one extends the product measure u to a complete measure using Caratheodory s construction Exercise 173 first defining an outer measure for a set A as the infimum of the u measure of all u measurable sets containing A It is easy to see that Borel sets in R are measurable for the prod uct measure and therefore for Lebesgue measure Conversely one has Lemma 332 For every subset A ofRquot and every 6 gt 0 there is an open set U containing A with AA 2 MU 6 Consequently there is a 05 set G containing A with MG AA Proof By definition WA infwB A c B taken over all sets B measurable for the product measure M It suffices therefore to show that every such B is contained in an open U with uB 2 MU e If B is a measurable box this follows from the regularity of 1 dimensional Lebesgue measure Proposition 184 The class B of all B for which the desired result is true is a monotone class and it contains the ring of all finite unions of measurable boxes Thus by the monotone class lemma 8 contains the a algebra generated by the measurable boxes which comprises all sets measurable for the product mea sure The second statement follows from the first just as in the 1 dimensional case Proposition 184 D Corollary 333 If A g Rquot has Lebesgue measure zero then almost all cross sections ofA are Lebesgue measurable and have Lebesgue measure zero Proof A is contained in a Borel set B having measure zero for the product measure The cross sections of B are measurable and by Fubini s theorem almost all of them have measure zero Each cross section Ax is contained in the corresponding cross section Bag and if Bx has measure zero then Ax is Lebesgue measurable and has measure zero by completeness D The example at the beginning of the lecture shows that almost all cannot be strengthened to all here Exercise 331 Long Show that the following procedures yield an identical definition of Lebesgue measure on R 0 Define a functional 739 on the class of rectangular parallelepipeds in the obvious way use it as a premeasure and extend by Caratheodory s procedure 0 The same but start with the class of measurable rectangles 0 Define a linear functional on the class of compactly sup ported continuous functions by iterated integration fI gtfx1xndx1dxn and apply the Riesz representation theorem The completion process in the definition of Lebesgue measure has the following effect if f13y is a Lebesgue measurable function say on R2 for simplicity we can only assert that fx is measurable for almost all 13 and that f is measurable for almost all y Reason see this for characteristic functions first using the preceding results and then prove it for all functions by approximation Except for this change Fubini s and Tonelli s theorems hold as before Exercise 332 Show that 1 1 x2y2 7T 1 1 x2y2 7T dd dd 2 OO x2y22 my 4 O 05132 I y22 ya 4 What went wrong here Why did Fubini s theorem not work Let g 6 L1R Consider the product function If H ftgx t for some constant 13 It is measurable but it might not be inte grable the product of two integrable functions is not integrable in general example However we can use Fubini s theorem to prove Proposition 334 In the above situation the function it l gt ftga t is integrable for almost all 13 The convolution f we rouge wait is an integrable function and f glll g f1g1 Proof Let h13t ftg13 t The iterated integral Ihltx7tgtldx dt In lgl llfllllgll is finite so h e L1R2 Then by Fubini s theorem t H MN is integrable for almost all 1 so that f g is defined almost everywhere and f91 lhltxtgtdt dx lt lhxtldt dx f191 as required D Exercise 333 Prove that the operation of convolution is com mutative and associative Proposition 335 Let g 6 L1 with 9 continuous and bounded Then f g is continuous Proof Let M supg Let xn be a sequence of real numbers converging to 13 Then f gm rouge udt a rouge udt f go by the dominated convergence theorem with dominating func tion Mf This implies that f g is continuous by exercise 21 D Lecture 34 Change of variable in multiple integrals We begin to investigate the effect of various transformations on Lebesgue measure Proposition 341 Let T R gt R be a linear transformation Then for every Lebesgue measurable set E the set TE is also Lebesgue measurable and ATE detTgtE Proof If the result holds for linear transformations T1 and T2 it also holds for their composite T1T2 Now by the theory of ele mentary row operations every linear transformation can be writ ten as a composite of elementary operations multiplying a coor dinate by a scalar interchanging two coordinates or adding one coordinate to another one For each of these the result follows from Fubini s theorem together with the translation invariance and scaling properties of one dimensional Lebesgue measure For example check this for elementary operations of the third type SUCh as 5131 xn i gt 2131132132 xn If f131 xn iS the characteristic function of E then f131 13213213n is the characteristic function of TE According to Fubini s theorem then we need to check that fa1xndx1dxnf 131 5132 I3nd 131 dCBn This is true because for any positive measurable function g 91dx1 9x1 x2d1 by the translation invariance of Lebesgue measure D Now consider more general coordinate transformations Let U U be open subsets of Rquot A diffeomorphism from U to U is a continuously differentiable bijection f U gt U whose inverse is also continuously differentiable Definition 342 Let f U gt U be a diffeomorphism The Jacobian of f is the real valued function Jf on U defined by Jf13 det Df13 where Df13 is the derivative of f at 13 which is a linear transformation R gt Rquot We are going to prove Theorem 343 Let f U gt U be a diffeomorphism Then map U iJfltxgtidegt More generally for any 9 e L1U UgltydAy UgfxiJfxdAx Example 344 For n 1 functions of a single variable and taking f to be increasing for simplicity of notation the theorem says 1 b 4a gltydy gfxf meta which is the usual formula for integration by substitution Re call this can be proved for continuous g by showing that both sides have the same derivative with respect to b then extended to all g e L1 by the usual approximation Example 345 Polar coordinates For an integrable function 2 g on R oo 00 27139 00 gxydxdy 2 gr cosQrsin9rdrd9 00 OO O 0 Exercise 341 Let I fEOOO ew2dx By writing I2 ex292dxdy in polar coordinates show that I Proof of Theorem 343 If the theorem is true for two diffeo morphisms f1 and f2 it is also true for their composite As with the special case of linear transformations Proposition 341 this allows us to prove the result in special cases and build up the general case by composition A key role is played by the Inverse Function Theorem 123 Notice also that it is sufficient to prove the theorem localy that is to show that every point 1 e U has a neighborhood Ux such that the theorem holds for all functions 9 supported in Ux This is because any nonnegative function 9 can be broken up using a partition of unity into a countable sum of nonnegative functions gj each of which is supported in one of the sets an By the monotone convergence theorem the truth of the theorem for each gj implies its truth for g Suppose that f is a diffeomorphism of the special form 21113n i gt hx1xnx2xngt where h Rquot gt R is a smooth function We will call such a diffeomorphism triangular The Jacobian of f is then just the partial derivative ail8131 According to Fubini s theorem then we need to check that Ajlgx1xndx1dxn Ughag1xnxnI8h8x1dx1dxn This follows from the formula for integration by substitution Example 344 applied in the innermost iterated integral Thus Theorem 343 holds for triangular diffeomorphisms To complete the proof it suffices to show that any diffeomor phism is locally a composite of triangular diffeomorphisms and linear transformations Localy means that every x E U has an open neighborhood on which this is true This follows from the inverse function theorem We may assume that we are looking locally near the point 0 and that f0 0 By composing with the inverse of the linear transformation DfO we may further assume that DfO I the identity transformation Write f13113n y1yn where each yii Rquot gt R is a smooth function of n variables and BM8370 67 By the inverse function theorem the smooth maps Chi x17quot397xn H y17quot397yi7xi17quot 7xn are all local diffeomorphisms near 13 The diffeomorphisms 191 2O 117 3O 21W are all triangular and their composite is f Thus Theorem 343 holds for f as required D Lecture 35 Fourier Transforms Rudin Chapter 9 Let f be an integrable function on R With little extra effort we could consider functions on Rn The Fourier transform of f is the function foe 27 e r mfoodx By the DCT f is continuous though it might not be integrable The Fourier transform is a key tool in the study of partial differ ential equations This is because it converts differentiation into multiplication as expressed below Proposition 351 Suppose that f is differentiable and that both f and g f are integrable Then the Fourier transforms off and g are related by t it t Proof Assume first that g is continuous and compactly sup ported in this case integrate by parts The general case of the result follows from an approximation argument D Proposition 352 Suppose that f and g xf are integrable then f is differentiable and ddt t z t Proof Differentiate under the integral sign using Exercise 192 D From these results we can compute a useful Fourier transform Lemma 353 Let f13 5322 Then its Fourier transform is foe et22 Proof The function f satisfies the differential equation dfdaaf 0 Taking the Fourier transform and using the two previous propo sitions f satisfies the differential equation itf z dfdt O which is the same It has general solution Ce tQQ and the constant C may be computed by evaluating f0 1 using Ex ercise 341 D Exercise 351 Show that the Fourier transform converts con volution into multiplication the Fourier transform f g is equal to a constant times fg Definition 354 A sequence gn in L1 is called an approximate identity if 9n 2 O fgn 1 and for every 6 gt 0 one has xgt6gnxdx gt o aSn gtOO The Fejer kernel Exercise 293 is an example of an approximate identity Another example is the sequence of functions gna rt212e WQ2 Gaussian bumps We will use these below Proposition 355 Let gn be an approximate identity and sup pose that f is continuous and bounded Then f gn converges to f pointWise as n gt 00 Proof Fix 1 and let 6 gt 0 be given There is 6 gt 0 such that ft f13 lt e if x t lt 6 Then there is N such that flugt5 gnudu lt e for n gt N NOW write r gum M ft fxgnx ndt Thus If gm fxl lt ft rltxgtgnltx wait HM ft rltxgtgnltx wait HM W aws tdt e I 2esupf for n gt N It follows that f gna gt f13 as required D Exercise 352 Let gn be an approximate identity Show that for every f 6 L1 fgtlltgn gt f in L1 as n gt oo Hint First prove it for continuous functions of compact support The key result about Fourier transforms is the Fourier inversion formula which states that f can be recovered from f by 1 ix m2 E et ftdt Just as with the summation of Fourier series there are various ways of interpreting what this means and various overlapping conditions under which it is true We ll prove Theorem 356 The Fourier inversion formula above is true in the naive sense if both f and f are integrable and continuous and tend to zero at infinity Some of the hypotheses are redundant For instance the inte grability off automatically implies that f is continuous and tends to zero at infinity this is the Riemann Lebesgue lemma But the statement above is memorably symmetrical Proof We want to evaluate 1 ix gm E et ftdt Because f is integrable we can use the dominated convergence theorem to write 933 2 nigtmOO27reimet22nftdt Expand the definition of the Fourier transform in the integral appearing on the right hand side here to get eitx y t22nfydydt The integrand belongs to L1R2 why so we may apply Fu bini s theorem to rearrange it as 1 gt 2 eztx y t 2ndtgt fydy 27r From lemma 353 the inner integral the one in parentheses is equal to x27tnenxy22 so the whole integral is n 12 2 lt2 gt e x 3 2fydy 7T But the sequence of functions u l gt n27T12 nu22 is an approx imate identity so this tends to f13 as n gt 00 This completes the proof D
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