Cheng S. A Short Course on the Lebesgue Integral and Measure Theory (web draft, 2004)(53s)_MCat_

A Short Course on the Lebesgue Integral and Measure Theory Steve Cheng August 5, 2004 Contents 1 Motivation for the Lebesgue integral 2 2 Basic measure theory 4 3 Measurable functions 8 4 Definition of the Lebesgue Integral 11 5 Convergence theorems 15 6 Some Results of Integration Theory 18 7 Lpspaces 23 8 Construction of Lebesgue Measure 28 9 Lebesgue Measure in Rn32 10 Riemann integrability implies Lebesgue integrability 34 11 Product measures and Fubini’s Theorem 36 12 Change of variables in Rn39 13 Vector-valued integrals 41 14 C∞ 0functions are dense in Lp(Rn) 42 15 Other examples of measures 47 16 Egorov’s Theorem 50 17 Exercises 51 1 18 Bibliography 52 Preface This article develops the basics of the Lebesgue integral and measure theory. In terms of content, it adds nothing new to any of the existing textbooks on the subject. But our approach here will be to avoid unduly abstractness and absolute generality, instead focusing on producing proofs of useful results as quickly as possible. Much of the material here comes from lecture notes from a short real analysis course I had taken, and the rest are well-known results whose proofs I had worked out myself with hints from various sources. I typed this up mainly for my own benefit, but I hope it will be interesting for anyone curious about the Lebesgue integral (or higher mathematics in general). I will be providing proofs of every theorem. If you are bored reading them, you are invited to do your own proofs. The bibliography outlines the background you need to understand this article. Copyright matters Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, with no Front-Cover Texts, and with no Back-Cover Texts. 1 Motivation for the Lebesgue integral If you have followed the rigorous definition of the Riemann integral in RorRn, you may be wondering why do we need to study yet another integral. After all, why should we even care to integrate nasty functions like: D(x) =/braceleftbigg1, x∈Q 0, x∈R\Q Rephrased in another way, D(x) is actually the indicator function1of the set S={x∈Q} ⊂R, and we want to find its “length”. Continuing to rephrase this question, sup- pose we are taking many real-valued measurements xof a particular physical phenomenon. What is the probability, say, that xis rational? If we assume 1For any set S, this is the function χSdefined by χS(x) = 1 ifx∈SandχS(x) = 0 if x /∈S. Math people call this the “characteristic function”, while probability people call it the “indicator function” instead. 2 the measurements are distributed normally with a mean of µand a standard deviation of σ, then this is given by: Pr[X∈Q] =/integraldisplay S1√ 2πσ2e−1 2(x−µ σ)2 dx So wild sets like Sare theoretically worth considering, and it does not work to use Riemann integral to evaluate the above probability. Another limitation to the Riemann integral is with limits. If a sequence of functionsfnis uniformly convergent (on a closed interval, or more generally a compact set A⊆Rn), then we can interchange limits for the Riemann integral: lim n→∞/integraldisplay Afn(x)dx=/integraldisplay Alim n→∞fn(x)dx, but the criterion of uniform convergence is often too restrictive, e.g. when integrating Fourier series. On the other hand, it can be proven with the Lebesgue integral that the interchange is valid under weaker conditions (e.g. the functions fnis bounded above somehow, and they converge pointwise ). As an added benefit, some sophisticated results concerning the Riemann integral, such as the Change of Variables Theorem in Rn, are more easily proven using the Lebesgue integral, with its arsenal of limit theorems. Finally, the Riemann integral does not deal with integration over “infinite bounds” very well. For example, the standard way to compute the probability integral/integraldisplay+∞ −∞e−1 2x2dx goes like this: /parenleftBig/integraldisplay+∞ −∞e−1 2x2dx/parenrightBig2 =/integraldisplay+∞ −∞e−1 2x2dx/integraldisplay+∞ −∞e−1 2y2dy =/integraldisplay+∞ −∞/integraldisplay+∞ −∞e−1 2(x2+y2)dxdy =/integraldisplay R2e−1 2(x2+y2)dxdy =/integraldisplay2π 0/integraldisplay∞ 0e−1 2r2rdrdθ (using polar coordinates) = 2π/bracketleftBig −e−1 2r2/bracketrightBigr=∞ r=0 = 2π. So /integraldisplay+∞ −∞e−1 2x2dx=√ 2π. The above computation seems easy, and although it can be justified using the Riemann integral alone, it is not entirely trivial, but it is with the Lebesgue 3 integral. (For example, why should/integraltext+∞ −∞/integraltext+∞ −∞be the same as/integraltext R2? Note that in the Riemann theory, the iterated integral and the area integral are proven to be equal only for bounded sets of integration.) You will probably be able to find other sorts of limitations with the Riemann integral. 2 Basic measure theory The setting of abstract integration is measure theory, which tells us what the areas or volumes of various sets are. Essentially we are given some function µof sets which returns the area or volume — formally called the measure — of the given set. i.e. We assume at the beginning that such a function µhas already been defined for us. The abstract approach of the Lebesgue integral has the obvious advantage that the theory can be applied to many other measures besides volume in Rn. We begin with the axioms of measure theory. Definition 2.1. LetXbe any non-empty set. A sigma algebra2of subsets of Xis a family Aof subsets of X, with the properties: 1.Ais non-empty. 2.IfE∈ A, thenX\E∈ A. 3.If{En}n∈Nis a sequence of sets in A, then their union is in A. That is, Ais closed under countable unions. The pair (X,A) is called a measurable space , and the sets in Aare called themeasurable sets . Notice that the axioms always imply that X∈ A. Also, by De Morgan’s laws,Ais closed under countable intersections as well as countable union. Needless to say, we cannot insist that Ais closed under arbitrary unions or intersections, as that would force A= 2XifAcontains all the singleton sets. That would be uninteresting. On the other hand, we want closure under countable set operations, rather than just finite ones, as we will want to take countable limits. Example 2.1.LetXbe any (non-empty) set. Then A= 2Xis a sigma algebra. Example 2.2.LetXbe any (non-empty) set. Then A={X,∅}is a sigma algebra. To get non-trivial sigma algebras to work with we need the following, a very unconstructive(!) construction: If we have a family of sigma algebras on X, then the intersection of all the sigma algebras from this family is also a sigma algebra on X. If all of the sigma 2I do not know why it has such a ridiculous name, other than the fact that it is often denoted by the Greek letter. 4 algebras from the family contains some fixed G ⊆ 2X, then the intersection of all the sigma algebras from the family, of course, is a sigma algebra containing G. Now if we are given G, and we take allthe sigma algebras on Xthat contain G, and intersect all of them, we get the smallest sigma algebra that contains G. Definition 2.2. The smallest sigma algebra containing any given G ⊆ 2X, as constructed above, is denoted /angbracketleftG/angbracketright, and is also called the sigma algebra generated byG. The following is an often-used sigma algebra. Definition 2.3. IfXis a topological space, we can construct the sigma algebra /angbracketleftT /angbracketright, where Tis the set of all open sets. This is called the Borel sigma algebra and is denoted B(X). When topological spaces are involved, we will always take the sigma algebra to be the Borel sigma algebra unless stated otherwise. B(X), being generated by the open sets, then contains all open sets, all closed sets, and countable unions and intersections of open sets and closed sets. It seems unlikely, however, that every set in B(X) is expressible as a countable union and/or intersection of open sets and closed sets, although it is tempting to think that. By the way, Theorem 9.3shows the Borel sigma algebra is generally not all of 2X. Sigma algebras are the domain on which measures are defined. Definition 2.4. Let (X,A) be a measurable space. A positive measure on this space is a function µ:A → [0,∞] such that 1.µ(∅) = 0 2.Countable additivity : For any sequence of mutually disjoint setsEn∈ A, µ/parenleftbigg∞/uniondisplay n=1En/parenrightbigg =∞/summationdisplay n=1µ(En). The set (X,A,µ) will be called a measure space . Whenever convenient we will abbreviate this expression, as in “let Xbe a measure space”, etc. Also, in this article, when we say “measure”, we will be dealing with positive measures only. (There are also theories about signed measures and complex measures.) Example 2.3.LetXbe an arbitrary set, and Abe a sigma algebra on X. Define µ:A → [0,∞] as µ(A) =/braceleftbigg|A|,ifAis a finite set ∞,ifAis an infinite set . This is called the counting measure . We will be able to model the infinite series/summationtext∞ n=1anin Lebesgue integration theory by using X=Nand the counting measure, since integrals are essentially sums of the integrand values weighted by areas or measures. 5 Example 2.4.X=Rn, andA=B(Rn). We can construct the Lebesgue measure λwhich assigns to the rectangle [ a1,b1]× ··· × [an,bn] inRnits expected n- dimensional volume ( b1−a1)···(bn−an). Of course this measure should also assign the correct volumes to the usual geometric figures, as well as for all the other sets in A. The existence of such a measure will be demonstrated later. Intuitively, defining the volume of the rectangle only should suffice to uniquely also determine the volume of the other sets, since the volume of every set can be approximated by the volume of many small rectangles. Indeed, we will later show this intuition to be true. In fact, you will see that most theorems using Lebesgue measure really depend only on the definition of the volume of the rectangle. Example 2.5.Any probability measure (as defined by the usual axioms of prob- ability) is actually a measure in our sense. For example, Pr[Z∈B] =µ(B) =/integraldisplay B1√ 2πe−1 2t2dt,3 where the integration, of course, is with respect to the Lebesgue measure on the real line. Other examples include the uniform distribution, the Poisson distribution, and so forth. Before we begin the prove more theorems, I must mention that we will be operating on the quantity ∞as if it were a number, even though you may have been told this is “wrong” by some teachers. It is true, of course, that certain algebraic properties of Rwould fail to hold with ∞included (i.e. R∪{∞,−∞} is not a field), but the crucial point in real analysis is that ∞obeys the usual ordering rules when used in inequalities. The rules we adopt are the following: a≤ ∞ ∞+∞=∞ a· ∞=∞(a/negationslash= 0) 0· ∞= 0 The first three rules are self-explanatory. The last rule may need explaining: when integrating functions, we often want to ignore “isolated” singularities, e.g. at zero for/integraltext1 0dx/√x. The point 0 is supposed to have “measure zero”, so even though the function is ∞there, the area contribution at that point should still be 0 = 0 · ∞. Hence the rule. At this point a warning should be issued: the additive cancellation rule will not work with∞. The danger should be sufficiently illustrated in the proofs of the following theorems. Theorem 2.1. The following are easy facts about measures: 1.It is finitely additive. 3I guess this is my favorite integral. It’s got all the important numbers in it — well, except fori. 6 2.Monotonicity: If E,F∈ A, andE⊆F, thenµ(E)≤µ(F). 3.IfE⊆Fhas finite measure ( µ(E)<∞), thenµ(F\E) =µ(F)−µ(E). 4.IfAorBhas finite measure, then µ(A∪B) =µ(A) +µ(B)−µ(A∩B). Proof. The first fact is obvious. For the second fact, we have µ(F) =µ((F\E)/unionmultiE) =µ(F\E) +µ(E), andµ(F\E)≥0. For the third fact, just subtract µ(E) from both sides. (The funny union symbol means that the union is disjoint.) Of course the fact that Ais a sigma algebra is used throughout to know that the new sets also belong toA.) For the fourth fact, we decompose each of A,B, andA∪Binto disjoint parts, to obtain the following: µ(A) =µ(A∩Bc) +µ(A∩B). µ(B) =µ(B∩Ac) +µ(B∩A). µ(A∪B) =µ(A∩Bc) +µ(Ac∩B) +µ(A∩B). Adding the first two equations and then substituting in the third one, µ(A) +µ(B) =µ(A∩Bc) +µ(A∩B) +µ(B∩Ac) +µ(B∩A) =µ(A∪B) +µ(A∩B). Since one of AorBhas finite measure, so does A∩B⊆A,B, by the second fact, so we may subtract µ(A∩B) from both sides. Of course if one of AorB has infinite measure, the resulting equation says nothing interesting. /square The preceding theorem, as well as the next ones, are quite intuitive and you should have no trouble remembering them. Theorem 2.2. Let(X,A,µ)be measure space, and let E1⊆E2⊆E3⊆ ··· be subsets in Awith union E. (The sets Enare said to increase to E, and henceforth we will write {En} /arrownortheastEfor this.) Then µ(E) =µ/parenleftbigg∞/uniondisplay n=0En/parenrightbigg = lim n→∞µ(En). Proof. The setsEkandEcan be written as the disjoint unions Ek=E1∪(E2\E1)∪(E3\E2)∪ ··· ∪ (Ek\Ek−1) E=E1∪(E2\E1)∪(E3\E2)∪ ···, (and setE0=∅), so that µ(E) =∞/summationdisplay k=1µ(Ek\Ek−1) = lim n→∞n/summationdisplay k=1µ(Ek\Ek−1) = lim n→∞µ(En)./square 7 Theorem 2.3. For anyEn∈ A, µ/parenleftbigg∞/uniondisplay n=1En/parenrightbigg ≤∞/summationdisplay n=1µ(En). Proof. µ/parenleftbigg∞/uniondisplay n=1En/parenrightbigg =µ/parenleftbigg∞/uniondisplay n=1En\(E1∪E2∪ ··· ∪En−1)/parenrightbigg =∞/summationdisplay n=1µ/parenleftbig En\(E1∪E2∪ ··· ∪En−1)/parenrightbig ≤∞/summationdisplay n=1µ(En)./square Theorem 2.4. Let{En} /arrowsoutheastE(that is,Enare decreasing and their intersection isE), andµ(E1)<∞. Then lim n→∞µ(En) =µ(E) =µ/parenleftbigg∞/intersectiondisplay n=1En/parenrightbigg . Proof. We have {E1\En} /arrownortheast(E1\E). So µ(E1\E) =µ/parenleftbigg∞/uniondisplay n=1(E1\En)/parenrightbigg = lim n→∞µ(E1\En), µ(E1)−µ(E) = lim n→∞[µ(E1)−µ(En)] =µ(E1)−lim n→∞µ(En), and cancel µ(E1) on both sides. /square 3 Measurable functions To do integration theory, we of course need functions to integrate. You should not expect that arbitrary functions can be integrated, but only the “measurable” ones. The following definition is not difficult to motivate. Definition 3.1. Let (X,A) and (Y,B) be measurable spaces. A map f:X→Y ismeasurable if for allB∈ B,the setf−1(B) = [f∈ B] is in A. Example 3.1.A constant map is always measurable, for f−1(B) is either ∅or X. Theorem 3.1. The composition of two measurable functions is measurable. Proof. Immediate from the definition. /square 8 Theorem 3.2. Let(X,A)and (Y,B)be measurable spaces, and suppose H generates the sigma algebra B:/angbracketleftH/angbracketright=B. A function f:X→Yis measurable if and only if for every V∈ H,f−1(V)is inA. Proof. The “only if” part is just the definition of measurability. For the “if” direction, define G={f−1(V) :V∈ H} , and also C={V∈ B:f−1(V)∈ /angbracketleftG/angbracketright} . It is easily checked that Cis a sigma algebra on Y, and it contains H, and hence it is actually equal to B. That is, for every V∈ B,f−1(V) is in/angbracketleftG/angbracketright ⊆ /angbracketleftA/angbracketright =A./square Corollary 3.3. All continuous functions (between topological spaces) are mea- surable. A comment about infinities again. There is a natural topology on [ −∞,+∞] and [0,∞] that make them look like closed intervals. Some denote [ −∞,+∞] by R(“the extended real numbers”). However, for convenience, I will just denote it as plain R. So keep in mind that when we prove our theorems, we have to make sure that they work (or do not work) when infinite quantities are introduced. Theorem 3.4. Let(X,A)be a measurable space. A map f:X→Ris mea- surable if and only if [f >c ] =f−1((c,+∞])∈ A for allc∈R. Proof. LetBbe the set of all open intervals ( a,b), along with {−∞},{+∞}. Let Hbe the set of all intervals ( c,+∞]. (a,b,c are finite.) Evidently Hgenerates B: (a,b) = [−∞,b)∩(a,+∞], [−∞,b) =∞/uniondisplay n=1[−∞,b−1 n] =∞/uniondisplay n=1R\(b−1 n,+∞]. {+∞}=∞/intersectiondisplay n=1(n,+∞]. {−∞} =R\∞/uniondisplay n=1(−n,+∞]. In turn, Bgenerates the Borel sigma algebra on R. Applying Theorem 3.2to the generator Hgives the result. (As B ⊆ /angbracketleftH/angbracketright andH ⊆ /angbracketleftB/angbracketright together mean /angbracketleftH/angbracketright=/angbracketleftB/angbracketright.) /square Remark 3.5.We can replace ( c,+∞], in the statement of the theorem, by [c,+∞], [−∞,c], etc. and there is no essential difference. The “countable union with 1 /n” trick used in the proof is widely applicable. It may be of interest to note that the Archimedean property of the real num- bers is being used here — the same proof will not work with non-Archimedean ordered fields. 9 Theorem 3.6. Letfnbe a sequence of measurable R-valued functions. Then the functions sup nfn,inf nfn,max nfn,min nfn,lim sup nfn,lim inf nfn (the limits are pointwise) are all measurable. Proof. Ifg(x) = supnfn(x), then [g>c ] =/uniontext n[fn>c], and we apply Theorem 3.4. Similarly, if g(x) = inf nfn(x), then [g < c ] =/uniontext n[fn< c]. The rest can be expressed as in terms of supremums and infimums (over a countable set), so they are measurable also. /square Not surprisingly, we will need to do arithmetic in integration theory, so we better know that Theorem 3.7. Iff,g:X→Rare measurable, then so are f+g,fg, andf/g. Proof. Consider the countable union [f+g<c ] =/uniondisplay r∈Q[f <c−r]∩[g<r ]. The set equality is justified as follows: Clearly f(x)<c−randg(x)<rtogether implyf(x) +g(x)< c. Conversely, if we set g(x) =t, thenf(x)< c−t, and we can increase tslightly to a rational number rsuch thatf(x)< c−r, and g(x)<t<r . This shows that f+gis measurable (by Theorem 3.4). Since [ −f <c ] = [f >−c], we see that −fis measurable. Therefore the functions f+(x) = max {+f(x),0}(positive part of f) f−(x) = max {−f(x),0}(negative part of f) are measurable (from Theorem 3.6and Example 3.1). Sincef=f+−f−,fis measurable if f+andf−are measurable separately also. Since fg= (f+−f−)(g+−g−) =f+g+−f+g−−f−g++f−g−, to prove that fgis measurable, it suffices to assume that fandgare both non-negative. Then just as with the sum, [fg<c ] =/uniondisplay r∈Q[f <c/r ]∩[g<r ]. Finally, for 1 /g, [1/g<c ] =/braceleftbigg [1/c<g,cg> 0]∪[1/c>g,cg> 0], c/negationslash= 0 [g<0], c = 0./square 10 Remark 3.8.You probably have already noticed there may be difficulty in defin- ing what the arithmetic operations mean when the operands are infinite (or when dividing by zero). The usual way to deal with these problems is to simply redefine the functions whenever they are infinite to be some fixed value. In particular, if the function gis obtained by changing the original measurable functionfon a measurable setAto be a constant c, we have: [g∈B] =/parenleftbig [g∈B]∩A/parenrightbig ∪/parenleftbig [g∈B]∩Ac/parenrightbig [g∈B]∩A=/braceleftbigg A, c ∈B ∅, c /∈B [g∈B]∩Ac= [f∈B]∩Ac, so the resultant function gis also measurable. Very conveniently, any sets like A= [f= +∞],[f= 0] are automatically measurable. Thus the gaps in the previous proof with respect to infinite values can be repaired with this device. As a final note, one intermediate result from the proof is quite useful and should be formally recognized: Theorem 3.9. AnR-valued function fis measurable if and only if f+andf− are measurable. Moreover, if fis measurable, so is |f|=f++f−. Remark 3.10.Of course the converse to the second statement is not true. You may construct a counterexample to convince yourself of this fact. 4 Definition of the Lebesgue Integral The idea behind Riemann integration is to try to measure the sums of area of the rectangles “below a graph” of a function and then take some sort of limit. The Lebesgue integral uses a similar approach: we perform integration on the “simple” functions first: Definition 4.1. A function is simple if its range is a finite set. AnR-valued simple function ϕalways has a representation ϕ=n/summationdisplay k=1akχEk, whereakare the distinct values of ϕ, andEk=ϕ−1({ak}). Conversely, any expression of the above form, where akneed not be distinct, and Ekis not neces- sarilyϕ−1({ak}), also defines a simple function. For the purposes of integration, however, we will require that Ekbe measurable, and that they partition X. It should be mentioned that χSis measurable if and only if Sis. 11 Definition 4.2. Let (X,µ) be a measure space. The Lebesgue integral, over X, of aR+-valued measurable simple function ϕis defined as /integraldisplay Xϕdµ =/integraldisplay Xn/summationdisplay k=1akχEkdµ=n/summationdisplay k=1akµ(Ek). (We restrict ϕto being non-negative for now, to avoid mixed + ∞,−∞ on the right-hand side.) Needless to say, the quantity on the right represents the sum of the areas below the graph of ϕ. It had better be the case that the value of the integral does not depend on the representation of ϕ. Ifϕ=/summationtext iaiχAi=/summationtext jbjχBj, whereAiandBj partitionX(soAi∩BjpartitionX), then /summationdisplay iaiµ(Ai) =/summationdisplay j/summationdisplay iaiµ(Ai∩Bj) =/summationdisplay j/summationdisplay ibjµ(Ai∩Bj) =/summationdisplay jbjµ(Bj). The second equality follows because the value of ϕisai=bjonAi∩Bj, so ai=bjwheneverAi∩Bj/negationslash=∅. So fortunately the integral is well-defined. Using the same algebraic manipulations just now, you can prove that if we have two simple functions ϕ≤ψ, then/integraltext Xϕdµ≤/integraltext Xψdµ (monotonicity of the integral). Theorem 4.1. The Lebesgue integral (for non-negative simple functions) is linear. Proof. Clearly/integraltext Xcϕdµ =c/integraltext Xϕdµ. And ifϕ=/summationtext iaiχAi,ψ=/summationtext jbjχBj, we have /integraldisplay Xϕdµ +/integraldisplay Xψdµ =/summationdisplay iaiµ(Ai) +/summationdisplay jbjµ(Bj) =/summationdisplay i/summationdisplay jaiµ(Ai∩Bj) +/summationdisplay j/summationdisplay ibjµ(Bj∩Ai) =/summationdisplay i/summationdisplay j(ai+bj)µ(Ai∩Bj) =/integraldisplay X(ϕ+ψ)dµ. /square Next, we integrate non-simple measurable functions like this: Definition 4.3. Letf:X→[0,+∞] be measurable. Consider te set Sfof all measurable simple functions 0 ≤ϕ≤f, and define the integral of foverXas /integraldisplay Xfdµ = sup ϕ∈Sf/integraldisplay Xϕdµ. 12 Intuitively, the simple functions in Sfare supposed to approximate fas close as we like, and we find the integral of fby computing the integrals of these approximations. But logically we need to know that these approximations really do exist. This is the essence of the following theorem. Theorem 4.2 (Approximation Theorem). Letf:X→[0,∞]be measur- able. Then there exists a sequence of non-negative functions {ϕn} /arrownortheastf, meaning ϕnare increasing pointwise and converging pointwise to f. Moreover, if fis bounded, it becomes possible for the ϕnto converge to funiformly. Proof. We prove the second statement first. Let Nbe any integer >supf, and set ϕn=N2n/summationdisplay k=1k−1 2nχEn,k, E n,k=f−1/parenleftbigg/bracketleftbiggk−1 2n,k 2n/parenrightbigg/parenrightbigg . We have 0 ≤f−ϕn<2−nuniformly. The detailed verification is left to the reader. The construction for the first statement is very similar. Set ϕn=n2n/summationdisplay k=1k−1 2nχEn,k+χFn, F n= [f≥n]. I leave it to you to check that 0 ≤f(x)−ϕn(x)<2−nwhenevern>f (x), and ϕn(x) =nwheneverf(x) =∞. /square In case you were worrying about whether this new definition of the integral agrees with the old one in the case of the non-negative simple functions, well, it does. Use monotonicity to prove this. Definition 4.4. Iffis not necessarily non-negative, we define /integraldisplay Xfdµ =/integraldisplay Xf+dµ−/integraldisplay Xf−dµ, provided that the two integrals on the right are not both ∞. Of course we will want to integrate over subsets of Xalso. This can be accomplished in two ways. Let Abe a measurable subset of X. Either we simply consider integrating over the measure space restricted to subsets of A, or we define /integraldisplay Afdµ =/integraldisplay XfχAdµ. Ifϕis non-negative simple, a simple working out of the two definitions of the integral over Ashows that they are equivalent. To prove this for the case of arbitrary measurable functions, we will need the tools of the next section. Let us note some other basic properties of our integral: 13 Letf,gbe non-negative. Since Scf=c·Sf={cϕ:ϕ∈Sf},0≤c<∞. we have (we freely omit the “ dµ” and/or the integration limit “ X” when they are implied by the context)/integraldisplay cf=c/integraldisplay f. This rule about constant multiplication also holds for fandcnot necessarily non-negative, as you can easily check, but proving linearity requires the tools of the next section. Moreover, if 0 ≤f≤g, thenSf⊆Sg, and therefore/integraltext f≤/integraltext g. In particular, ifA⊆B, thenfχA≤fχB, so /integraldisplay Af≤/integraldisplay Bf. Unsurprisingly/integraltext f≤/integraltext galso holds if f,gare not necessarily non-negative, and that is proven by considering the positive and negative parts of f,gseparately. Then since −|f| ≤f≤ |f|, we also obtain −/integraldisplay |f| ≤/integraldisplay f≤/integraldisplay |f|,i.e./vextendsingle/vextendsingle/vextendsingle/integraldisplay f/vextendsingle/vextendsingle/vextendsingle≤/integraldisplay |f|. (This last inequality is sometimes called the “generalized triangle inequality”, as integrals can be viewed as an advanced form of summing.) Next, we make one more definition related to integrals. Definition 4.5. A (µ-)measurable set is said to have ( µ-)measurable zero if µ(E) = 0. Typical examples of a measure-zero set are the singleton points in Rn, and lines and curves in Rn,n≥2. By countable additivity, any countable set in Rn has measure zero also. A particular property is said to hold almost everywhere if the set of points for which the property fails to hold is a set of measure zero. For example, “a function vanishes almost everywhere”. Clearly, if you integrate anything on a set of measure zero, you get zero. Assuming that linearity of the integral has been proved, we can demonstrate the following intuitive result. Theorem 4.3. A measurable function f:X→[0,∞]vanishes almost every- where if and only if/integraltext Xf= 0. Proof. LetA= [f= 0], andµ(Ac) = 0. Then /integraldisplay Xf=/integraldisplay Xf·(χA+χAc) =/integraldisplay XfχA+/integraldisplay XfχAc=/integraldisplay Af+/integraldisplay Acf= 0 + 0. 14 Conversely, if/integraltext Xf= 0, consider [ f >0] =/uniontext n[f >1 n]. We have µ[f >1 n] =/integraldisplay [f>1 n]1 =n/integraldisplay [f>1 n]1 n≤n/integraldisplay Xf= 0. Henceµ[f >0] = 0. /square 5 Convergence theorems The following theorems are another feature of the Lebesgue integral that make it so much better than the Riemann definition. Theorem 5.1 (Monotone Convergence Theorem). Let(X,µ)be a mea- sure space. Let fnbe non-negative measurable functions increasing pointwise to f. Then /integraldisplay Xfdµ =/integraldisplay X/parenleftBig lim n→∞fn/parenrightBig dµ= lim n→∞/integraldisplay Xfndµ. Proof.fis measurable because it is a limit of measurable functions. Since fnis an increasing sequence of functions bounded by f, their integrals is an increasing sequence of numbers bounded by/integraltext Xf; thus the following limit exists: lim n→∞/integraldisplay Xfndµ≤/integraldisplay Xfdµ. Next we show the inequality in the other direction. Take any 0 <t< 1. Given a fixed ϕ∈Sf, letAn= [fn−tϕ≥0]. TheAn are obviously increasing. If for a particular x∈X, we have ϕ(x) = 0, then x∈An, for alln. Otherwise, ϕ(x)>0, sof(x)≥ϕ(x)>tϕ (x), and there is going to be some n for whichfn(x)≥tϕ(x), i.e.x∈An. HenceX=/uniontext nAn. For allµ-measurable sets E, define a new measure on Xby ν(E) =/integraldisplay Etϕdµ. Then/integraldisplay Xtϕdµ =ν(X) =ν/parenleftBig/uniondisplay nAn/parenrightBig = lim n→∞ν(An) = lim n→∞/integraldisplay Antϕdµ ≤lim n→∞/integraldisplay Anfndµ, since onAnwe havetϕ≤fn ≤lim n→∞/integraldisplay Xfndµ. t/integraldisplay Xϕdµ≤lim n→∞/integraldisplay Xfndµ, and take limit t→1. /integraldisplay Xϕdµ≤lim n→∞/integraldisplay Xfndµ, and take sup over ϕ∈Sf. /square 15 Using this theorem, we are now in the position to prove linearity of the Lebesgue integral for non-simple functions. Given any two non-negative mea- surable functions f,g, by the approximation theorem (Theorem 4.2), we know that are non-negative simple functions {ϕn} /arrownortheastf, and {ψn} /arrownortheastg. Then {ϕn+ψn} /arrownortheastf+g, and so /integraldisplay f+g= lim n→∞/integraldisplay ϕn+ψn= lim n→∞/integraldisplay ϕn+/integraldisplay ψn=/integraldisplay f+/integraldisplay g. (The second equality follows because we already know the integral is linear for simple functions. For the first and third equality we apply the Monotone Convergence Theorem.) And iff,gnot necessarily non-negative, then /integraldisplay f+g=/integraldisplay (f+−f−) + (g+−g−) =/integraldisplay f++g+−(f−+g−) =/integraldisplay f++g+−/integraldisplay f−+g− =/integraldisplay f++/integraldisplay g+−(/integraldisplay f−+/integraldisplay g−) =/integraldisplay f+/integraldisplay g. (Only at the fourth equality we apply what we had just proved for non-negative functions. The third and last equality are just by the definition of the integral.) Here is another application of the Monotone Convergence Theorem. Theorem 5.2 (Beppo Levi). Letfn:X→[0,∞]be measurable. Then /integraldisplay∞/summationdisplay n=1fn=∞/summationdisplay n=1/integraldisplay fn. Proof. LetgN=/summationtextN n=1fn, andg=/summationtext∞ n=1fn. The Monotone Convergence Theorem applies to gN, and: /integraldisplay g=/integraldisplay lim N→∞gN= lim N→∞/integraldisplay gN= lim N→∞N/summationdisplay n=1/integraldisplay fn=∞/summationdisplay n=1/integraldisplay fn. /square There are many more applications like this. We postpone those for now, since you will probably be even more amazed by the next convergence theorem. We first need a lemma. Lemma 5.3 (Fatou’s Lemma). Letfn:X→[0,∞]be measurable. Then /integraldisplay lim inf n→∞fn≤lim inf n→∞/integraldisplay fn. 16 Proof. Setgn= inf k≥nfk, so thatgn≤fn, and{gn} /arrownortheastlim inf nfn. Then /integraldisplay lim inf nfn=/integraldisplay lim ngn= lim n/integraldisplay gn= lim inf n/integraldisplay gn≤lim inf n/integraldisplay fn./square Remark 5.4.By adding and subtracting a constant, the hypotheses may be weakened to allow functions that are bounded below by any fixed number, not just non-negative functions. (This lower bound condition cannot be dropped.) The same considerations apply to the Monotone Convergence Theorem. The following definition is used to formulate a crucial hypothesis of the theorem that is about to follow. Definition 5.1. A function f:X→Ris called integrable if it is measurable and/integraltext X|f|<∞. It is immediate that fis integrable if and only if f+andf−are both integrable. It is also helpful to know, that/integraltext |f|<∞must imply |f|<∞ almost everywhere. Theorem 5.5 (Dominated Convergence Theorem). Let(X,µ)be a mea- sure space. Let fn:X→Rbe a sequence of measurable functions converging pointwise to f. Moreover, suppose that there is an integrable functiongsuch that|fn| ≤g, for alln. Thenfnandfare also integrable, and lim n→∞/integraldisplay X|fn−f|dµ= 0. Proof. Obviouslyfnandfare integrable. Also, 2 g−|fn−f|is measurable and non-negative. By Fatou’s lemma, /integraldisplay lim inf n(2g− |fn−f|)≤lim inf n/integraldisplay (2g− |fn−f|). Sincefnconverges to f, the left-hand quantity is just/integraltext 2g. The right-hand quantity is: lim inf n/parenleftbigg/integraldisplay 2g−/integraldisplay |fn−f|/parenrightbigg =/integraldisplay 2g+ lim inf n/parenleftbigg −/integraldisplay |fn−f|/parenrightbigg =/integraldisplay 2g−lim sup n/integraldisplay |fn−f|. Since/integraltext 2gis finite, it may be cancelled from both sides. Then we obtain lim sup n/integraldisplay |fn−f| ≤0,i.e. lim n→∞/integraldisplay |fn−f|= 0. /square Remark 5.6.It obviously suffices to only require that fnconverge to fpointwise almost everywhere, or that |fn|is bounded above by galmost everywhere. (Of course, iffnonly converges to falmost everywhere, then the theorem would not automatically say that fis measurable.) 17 Remark 5.7.By the generalized triangle inequality, we conclude from the hy- potheses that also lim n→∞/integraldisplay Xfndµ=/integraldisplay Xfdµ, which is usually how this theorem is applied. Remark 5.8.The theorem also holds for continuous limits of functions, not just countable limits. That is, if we have a continuous sequence of functions, say ft, 0≤t<1, we can also say lim t→1/integraldisplay X|ft−f|dµ= 0 ; for given any sequence {an}convergent to 1, we can apply the theorem to fan. Since this can be done for any sequence convergent to 1, the above limit is established. 6 Some Results of Integration Theory This section contains some nice applications proven using the convergence the- orems from the last section. Theorem 6.1 (Generalization of Beppo Levi). LetXbe a measure space, andfn:X→Rbe measurable functions, with/integraltext/summationtext|fn|=/summationtext/integraltext |fn|<∞. Then ∞/summationdisplay n=1/integraldisplay fn=/integraldisplay∞/summationdisplay n=1fn. Proof. LetgN=/summationtextN n=1fn,g= lim supN→∞gN, andh=/summationtext∞ n=1|fn|. Then |gN| ≤h=|h|. Since/integraltext |h|<∞by hypothesis, we have |h|<∞almost everywhere, so/summationtext∞ n=1fnis absolutely convergent almost everywhere. That is, gNconverges pointwise to galmost everywhere. By the Dominated Convergence Theorem, lim N→∞/integraltext gN=/integraltext g, whence ∞/summationdisplay n=1/integraldisplay fn= lim N→∞N/summationdisplay n=1/integraldisplay fn= lim N→∞/integraldisplayN/summationdisplay n=1fn=/integraldisplay∞/summationdisplay n=1fn. /square Example 6.1.Here’s a perhaps unexpected application. Suppose we have a countable set of real numbers an,m,n,m∈N. Letµbe the counting measure on N. Then/integraltext m∈Nan,mdµ=/summationtext∞ m=1an,m. Moreover, Theorem 6.1says that we can sum either along nfirst ormfirst and get the same results (/summationtext∞ n=1/summationtext∞ m=1an,m=/summationtext∞ m=1/summationtext∞ n=1an,m) if the double sum is absolutely convergent. Of course, this fact can also be proven in an entirely elementary way. Theorem 6.2. Letg:X→[0,∞]be measurable in the measure space (X,A,µ). Let ν(E) =/integraldisplay Egdµ, E ∈ A. 18 Thenνis a measure on (X,A), and for any measurable function fonX, /integraldisplay Xfdν =/integraldisplay Xfgdµ, often written as dν=gdµ. Proof. We proveνis a measure. ν(∅) = 0 is trivial. For countable additivity, let{En}be measurable with union E, so thatχE=/summationtext∞ n=1χEn, and ν(E) =/integraldisplay Egdµ =/integraldisplay XgχEdµ =/integraldisplay X∞/summationdisplay n=1gχEndµ=∞/summationdisplay n=1/integraldisplay XgχEndµ=∞/summationdisplay n=1ν(En). Next, iff=χEfor someE∈ A, then /integraldisplay Xfdν =/integraldisplay XχEdν=ν(E) =/integraldisplay XχEgdµ =/integraldisplay Xfgdµ. By linearity, we see that/integraltext fdν =/integraltext fgdµ wheneverfis non-negative simple. For general non-negative f, we use a sequence of simple approximations {ϕn} /arrownortheast f, so{ϕng} /arrownortheastfg. Then by the monotone convergence, /integraldisplay Xfdν = lim n→∞/integraldisplay Xϕndν= lim n→∞/integraldisplay Xϕngdµ =/integraldisplay Xlim n→∞ϕngdµ =/integraldisplay Xfgdµ. Finally, for fnot necessarily non-negative, we apply the above to its positive and negative parts, and use linearity. /square The procedure of proving some fact about integrals by first reducing to the case of simple functions and non-negative functions is used quite often. (It will get quite monotonous if we had to detail the procedure every time we use it, so we won’t anymore if the circumstances permit.) Also, we should note that if fis only measurable but not integrable, then the integrals of f+orf−might be infinite. If both are infinite, the integral of f is not defined, although the equation of the theorem might still be interpreted as saying that the left-hand and right-hand sides are undefined at the same time. For this reason, and for the sake of the clarity of our exposition, we will not bother to modify the hypotheses of the theorem to state that fmust be integrable. Problem cases like this also occur for some of the other theorems we present, and there I will also not make too much of a fuss about these problems, trusting that you understand what happens when certain integrals are undefined. Theorem 6.3 (Change of variables). LetX,Y be measure spaces, and g:X→Y,f:Y→Rbe measurable. Then /integraldisplay X(f◦g)dµ=/integraldisplay Yfdν, whereν(B) =µ(g−1(B))is a measure defined for all measurable B⊆Y. 19 Proof. First suppose f=χB. LetA=g−1(B)⊆X. Thenf◦g=χA, and we have /integraldisplay Yfdν =/integraldisplay YχBdν=ν(B) =µ(g−1(B)) =µ(A) =/integraldisplay X(f◦g)dµ. Since both sides of the equation are linear in f, the equation holds whenever f is simple. Applying the “standard procedure” mentioned above, the equation is then proved for all measurable f. /square Remark 6.4.The change of variables theorem can also be applied “in reverse”. Suppose we want to compute/integraltext Yfdν, whereνis already given to us. Further assume that gis bijective and its inverse is measurable. Then we can define µ(A) =ν(g(A)), and it follows that/integraltext Yfdν =/integraltext X(f◦g)dµ. Our theorem (especially when stated in the reverse form) is clearly related to the usual “change of variables” theorem in calculus. If g:X→Yis a bijec- tion between open subsets of Rn, and both it and its inverse are continuously differentiable (i.e. g is a diffeomorphism ), andν=λis the Lebesgue measure inRn, then (as we shall prove rigorously in Lemma 12.1), µ(A) =λ(g(A)) =/integraldisplay A|det Dg|dλ. Appealing to Theorems 6.2and6.3, we obtain: Theorem 6.5 (Differential change of variables in Rn).Letg:X→Ybe a diffeomorphism of open sets in Rn. IfA⊆Xis measurable, and f:Y→R is measurable, then /integraldisplay g(A)fdλ =/integraldisplay A(f◦g)dµ=/integraldisplay A(f◦g)· |det Dg|dλ. The next two theorems are the Lebesgue versions of well-known results about the Riemann integral. Theorem 6.6 (First Fundamental Theorem of Calculus). LetI⊆Rbe an interval, and f:I→Rbe integrable (with Lebesgue measure in R). Then the function F(x) =/integraldisplayx af(t)dt is continuous. Furthermore, if fis continuous at x, thenF/prime(x) =f(x). Proof. To prove continuity, we compute: F(x+h)−F(x) =/integraldisplayx+h xf(t)dt=/integraldisplay If(t)·χ[x,x+h](t)dt. 20 (Naturally, when h<0,χ[x,x+h]should be interpreted as −χ[x+h,x]here.) Since |f·χ[x,x+h]| ≤ |f|, by the Dominated Convergence Theorem (along with remark 5.8), lim h→0F(x+h)−F(x) = lim h→0/integraldisplay If(t)·χ[x,x+h](t)dt =/integraldisplay Ilim h→0f(t)·χ[x,x+h](t)dt =/integraldisplay If(t)·χ{x}(t)dt= 0. The proof of differentiability is the same as for the Riemann integral: /vextendsingle/vextendsingle/vextendsingle/vextendsingleF(x+h)−F(x) h−f(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraltextx+h x(f(t)−f(x))dt h/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≤/integraltext [x,x+h]|f(t)−f(x)|dt |h| ≤supt∈[x,x+h]|f(t)−f(x)| · |h| |h|, which goes to zero as hdoes. /square The First Fundamental Theorem was easy, but the Second Fundamental Theorem (which states that/integraltextb af/prime=f(b)−f(a)) is not entirely trivial. The difficulty is that we should not assume as hypotheses that f/primeis continuous, or even that it is Lebesgue-integrable. It turns out that a theorem without such strong hypotheses is possible; we will not reproduce its proof here, but just settle for a weaker version: Theorem 6.7 (Second Fundamental Theorem of Calculus). Suppose f: [a,b]→Ris measurable and bounded above and below. If f=g/primefor some g, then/integraldisplayb af(x)dx=g(b)−g(a). Proof. We first note that/vextendsingle/vextendsingle/vextendsingleg(x+h)−g(x) h/vextendsingle/vextendsingle/vextendsinglecan be bounded by a constant using the Mean Value Theorem, and a constant is obviously integrable on a finite interval. 21 Then /integraldisplayb af(x)dx=/integraldisplayb alim h→0g(x+h)−g(x) hdx = lim h→0/integraldisplayb ag(x+h)−g(x) hdx = lim h→01 h/bracketleftBig/integraldisplayb+h a+hg(x)dx−/integraldisplayb ag(x)dx/bracketrightBig = lim h→01 h/bracketleftBig/integraldisplayb+h bg(x)dx−/integraldisplaya+h ag(x)dx/bracketrightBig =g(b)−g(a). (The last equality follows from the First Fundamental Theorem and that gmust be continuous at aandbif it is differentiable there.) /square Remark 6.8.Ifg/prime(x) exists, then it can also be computed as the countable limit limn→∞n(g(x+ 1/n)−g(x)), thus showing that g/primeis measurable. Thus we can drop the hypothesis that fis measurable in Theorem 6.7. Remark 6.9.You might have noticed that I cheated a bit in the proof, in as- suming the integral of a function is invariant under horizontal translations. But of course, this can be proven readily using the fact that Lebesgue measure is translation-invariant. The following theorems are often not found in calculus texts even though they are quite important for applications. Theorem 6.10 (Continous dependence on integral parameter). Let (X,µ)be a measure space, Tbe any metric space (e.g. Rn), andf:X×T→R, withf(·,t)being measurable for each t∈T. Consider the function F(t) =/integraldisplay x∈Xf(x,t). Then we have Fcontinuous at t0∈Tif the following conditions are met: 1.For eachx∈X,f(x,·)is continuous at t0∈I. 2.There is an integrable function gsuch that |f(x,t)| ≤g(x)for allt∈T. Proof. lim t→t0/integraldisplay x∈Xf(x,t) =/integraldisplay x∈Xlim t→t0f(x,t) =/integraldisplay x∈Xf(x,t0). /square Theorem 6.11 (Differentiation under the integral sign). Using the same notation as Theorem 6.10, withTbeing an open real interval, we have F/prime(t) =d dt/integraldisplay x∈Xf(x,t) =/integraldisplay x∈X∂ ∂tf(x,t) if the following conditions are satisfied: 22 1.For eachx∈X,∂ ∂tf(x,t)exists. 2.There is an integrable function gsuch that/vextendsingle/vextendsingle∂ ∂tf(x,t)/vextendsingle/vextendsingle≤g(x)for allt∈T. Proof. This theorem is often proven by using iterated integrals and switching the order of integration, but that method is theoretically troublesome because it requires more stringent hypotheses. It is easier, and better, to prove it directly from the definition of the derivative. The straightforward computation yields: lim h→0F(t+h)−F(t) h= lim h→0/integraldisplay x∈Xf(x,t+h)−f(x,t) h =/integraldisplay x∈Xlim h→0f(x,t+h)−f(x,t) h=/integraldisplay x∈X∂ ∂tf(x,t) (noting that/vextendsingle/vextendsingle/vextendsinglef(x,t+h)−f(x,t) h/vextendsingle/vextendsingle/vextendsingleis bounded by g(x)). /square Remark 6.12.It is easy to see that we may generalize Theorem 6.11 toTbeing any open set in Rn, taking partial derivatives. I won’t write it out in full because the notation is somewhat complicated. Example 6.2.Check that the function Γ( x) =/integraltext∞ 0e−ttx−1dt,x> 0 is continu- ous, and differentiable with the obvious formula for the derivative. 7 Lpspaces The contents in this section do not have applications in this article, but they are so well known that it would not do justice to omit them. Definition 7.1. LetXbe a measure space, and let p∈[1,∞). The space Lp consists of all measurable functions f:X→Rsuch that /integraldisplay |f|p<∞. Definition 7.2. For eachf∈Lp, define /bardblf/bardblp=/parenleftBig/integraldisplay |f|p/parenrightBig1/p . (Iff /∈Lp, this quantity is of course defined as ∞.) As suggested by the notation, /bardbl·/bardblpis a real norm on the vector space Lp, provided that we declare two functions to be equivalent if they differ only on a set of measure zero (so that /bardblf/bardblp= 0 if and only if f= 0 as equivalence classes). Only the verification of the triangle inequality presents any difficulties — this will be solved by the theorems below. Definition 7.3. Two numbers p,q∈(1,∞) are called conjugate exponents when1 p+1 q= 1. 23 Theorem 7.1 (H¨ older’s inequality). ForR-valued measurable functions f andg,/vextendsingle/vextendsingle/vextendsingle/integraldisplay fg/vextendsingle/vextendsingle/vextendsingle≤/integraldisplay |f||g| ≤ /bardblf/bardblp/bardblg/bardblq. Proof. The first inequality is trivial. For the second inequality, since it only involves absolute values, for the rest of the proof we may assume that f,gare non-negative. If/bardblf/bardblp= 0, then |f|p= 0 almost everywhere, and so f= 0 andfg= 0 almost everywhere too. Thus the inequality is valid in this case. (Similarly when /bardblg/bardblq= 0.) If/bardblf/bardblpor/bardblg/bardblqis infinite, the inequality is trivial. So we now assume these two quantities are both finite and non-zero. Define F=f//bardblf/bardblp,G=g//bardblg/bardblq, so that /bardblF/bardblp=/bardblG/bardblq= 1. We must then show that/integraltext FG≤1. To do this, we employ the fact that log is concave: 1 plogs+1 qlogt≤log/parenleftbiggs p+t q/parenrightbigg ,0≤s,t≤ ∞ or, s1 pt1 q≤s p+t q. Substitutes=Fp,t=Gq, and integrate both sides: /integraldisplay FG≤1 p/integraldisplay Fp+1 q/integraldisplay Gq=1 p/bardblF/bardblp p+1 q/bardblG/bardblq q=1 p+1 q= 1. /square You may have seen a special case of this theorem, for p=q= 2, as the Cauchy-Schwarz inequality. Theorem 7.2 (Minkowski’s inequality). ForR-valued measurable functions fandg, /bardblf+g/bardblp≤ /bardblf/bardblp+/bardblg/bardblp. Proof. The inequality is trivial when p= 1 or when /bardblf+g/bardblp= 0. Also, since /bardblf+g/bardblp≤ /bardbl|f|+|g|/bardblp, it again suffices to consider only the case when f,gare non-negative. Next, we employ the convexity of t/mapsto→tp,p>1, /parenleftbiggs+t 2/parenrightbiggp ≤sp+tp 2,0≤s,t≤ ∞. When we substitute s=f,t=g, we get (f+g)p≤2p−1(fp+gp). This inequality shows that if /bardblf+g/bardblpis infinite, then one of /bardblf/bardblpor/bardblg/bardblpis also infinite, so Minkowski’s inequality holds true in that case. We may now assume /bardblf+g/bardblpis finite. We write: /integraldisplay (f+g)p=/integraldisplay f(f+g)p−1+/integraldisplay g(f+g)p−1. 24 By H¨ older’s inequality, and noting that ( p−1)q=pfor conjugate exponents, /integraldisplay f(f+g)p−1≤ /bardblf/bardblp/vextenddouble/vextenddouble(f+g)p−1/vextenddouble/vextenddouble q=/bardblf/bardblp/parenleftBig/integraldisplay |f+g|p/parenrightBig1/q =/bardblf/bardblp/bardblf+g/bardblp/q p. A similar inequality holds for/integraltext g(f+g)p−1. Putting these together: /bardblf+g/bardblp p≤(/bardblf/bardblp+/bardblg/bardblp)/bardblf+g/bardblp/q p. Dividing by /bardblf+g/bardblp/q pyields the desired result. /square Definition 7.4. A measure space ( X,µ) has finite measure ifµ(X) is finite. Theorem 7.3. Let(X,µ)have finite measure. Then whenever 1≤r<p< ∞, Lp⊆Lr. Moreover, the inclusion map from LptoLris continuous. Proof. Iff∈Lp, apply the H¨ older inequality with conjugate exponentsp rand s=p p−r: /bardblf/bardblr r=/integraldisplay |f|r≤/parenleftBig/integraldisplay |f|r·p r/parenrightBigr/p/parenleftBig/integraldisplay 1s/parenrightBig1/s =/bardblf/bardblr pµ(X)1/s, and so /bardblf/bardblr≤ /bardblf/bardblpµ(X)1/rs=/bardblf/bardblpµ(X)1 r−1 p<∞. To show continuity of the inclusion map, replace fwithf−gabove where /bardblf−g/bardblp<ε. /square Example 7.1./integraltext1 0x−1 2dx= 2<∞, so automatically/integraltext1 0x−1 4dx < ∞. On the other hand, the condition that µ(X)<∞is indeed necessary:/integraltext∞ 1x−2dx<∞, but/integraltext∞ 1x−1dx=∞. Theorem 7.4. Letfn:X→Rbe measurable functions converging (almost everywhere) pointwise to f, and |fn| ≤gfor someg∈Lp. Thenf,fn∈Lp, andfnconverges to fin the Lpnorm, meaning: lim n→∞/parenleftBig/integraldisplay |fn−f|p/parenrightBig1/p = lim n→∞/bardblfn−f/bardblp= 0. Proof. |fn−f|pconverges to 0 and |fn−f|p≤(2g)p∈L1. Apply the Dominated Convergence Theorem on these functions. /square One wonders whether the converse is true: if fnconverges to fin the Lp norm, do the functions fnthemselves converge pointwise to f? The answer is no (the counterexamples are not difficult), but we do have the following. Theorem 7.5. Let{fn}be a Cauchy sequence in Lp. Then it has a subsequence converging pointwise almost everywhere. 25 Corollary 7.6. Iflim n→∞/bardblfn−f/bardblp= 0then there is a subsequence {fn(k)}con- verging tofpointwise almost everywhere. Corollary 7.6is used in the following result. Theorem 7.7. Lpis a complete metric space. (This means every Cauchy sequence in Lpconverges.) The proofs of these theorems are collected in the next section. It is also possible to define “ L∞”: Definition 7.5. LetXbe a measure space, and f:X→Rbe measurable. A numberM∈[0,∞] is an almost-everywhere upper bound for |f|if|f| ≤M almost everywhere. The infimum of all almost-everywhere upper bounds for |f| is denoted by /bardblf/bardbl∞. Definition 7.6. L∞is the set of all measurable functions fwith/bardblf/bardbl∞<∞. Its norm is given by /bardblf/bardbl∞. The use of the subscript “ ∞” is explained by the following theorem, whose proof is left to the reader: Theorem 7.8. Ifµ(X)<∞, then lim p→∞/bardblf/bardblp=/bardblf/bardbl∞. Remark 7.9.Observe that there is a similar thing for vectors /vector a= (a1,...,a n)∈ Rn: lim p→∞/bardbl/vector a/bardblp= lim p→∞/parenleftbig |a1|p+···+|an|p/parenrightbig1/p= max( |a1|,...,|an|) =/bardbl/vector a/bardbl∞. This remark may serve as a hint. If we (naturally) define the conjugate exponent of p= 1 to be q=∞, the H¨ older inequality remains valid, since |fg| ≤ |f|/bardblg/bardbl∞almost everywhere. Integrating both sides gives /bardblfg/bardbl1≤ /bardblf/bardbl1/bardblg/bardbl∞. Before closing, we mention two more results. From Theorem 7.4, it is clear that Theorem 7.10. Letf:X→R∈Lp(X),1≤p <∞. Then for any ε >0, there exist simple functions ϕ:X→Rsuch that /bardblϕ−f/bardblp=/parenleftBig/integraldisplay Rn|ϕ−f|pdλ/parenrightBig1/p <ε. (This is also true for p=∞.) This may be summarized by saying that the set of all simple functions is dense in Lp(in the topological sense). WhenX=Rnwith Lebesgue measure λ, we have another result. The set of infinitely differentiable functions on Rnwith compact support4, denoted C∞ 0, 4The support of a function ψ:X→Ris the closure of the set {x∈X:ψ(x)/negationslash= 0}. “ψhas compact support” means that the support of ψis compact (when X=Rn, same as closed and bounded). 26 is also dense in Lp(Rn,λ), 1≤p<∞. (This fact will be fully proven in Section 14.) These two facts are typically used as tools to prove other theorems about functionsf∈Lp. One first proves that a certain theorem holds for all simple ϕ(orϕ∈C∞ 0), and then prove that the same result holds for arbitrary f∈Lp by approximating such fbyϕ. An example of this procedure follows. Theorem 7.11 (Riemann-Lebesgue Lemma). For allf∈L1(R), lim |ω|→∞/integraldisplay Rf(x) sin(ωx)dx= 0. Proof. For convenience, assume ω>0. Suppose first that f=ψ∈C∞ 0. Sinceψhas compact support, integrating ψ(x) sin(ωx) over Ris the same as integrating over some compact interval [ a,b] containing the support of ψ. And since every function involved is infinitely differentiable, we may use integration by parts ( u=ψ(x),dv= sin(ωx)dx): /integraldisplayb aψ(x) sin(ωx)dx=−ψ(x)cos(ωx) ω/vextendsingle/vextendsingle/vextendsingle/vextendsingleb a+/integraldisplayb acos(ωx) ωψ/prime(x)dx. As|ψ|and|ψ/prime|are continuous on [ a,b], they are bounded by constants Mand M/primerespectively. We have, /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplayb aψ(x) sin(ωx)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 ω|ψ(b) cos(ωb)−ψ(a) cos(ωa)|+1 ω/integraldisplayb a|cos(ωx)ψ/prime(x)|dx ≤2M ω+(b−a)M/prime ω→0, ω→ ∞. Thus we have proven the result when f=ψ∈C∞ 0. Now suppose fis arbitrary. By the denseness of C∞ 0, for every ε >0 we can find ψ∈C∞ 0such that/bardblf−ψ/bardbl1<ε. Then /vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay Rf(x) sin(ωx)dx−/integraldisplay Rψ(x) sin(ωx)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤/integraldisplay R/vextendsingle/vextendsingle/vextendsingle/parenleftbig f(x)−ψ(x)/parenrightbig sin(ωx)/vextendsingle/vextendsingle/vextendsingledx ≤/integraldisplay R|f(x)−ψ(x)|dx <ε, or, /vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay Rf(x) sin(ωx)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle<ε+/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay Rψ(x) sin(ωx)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle. We take lim supω→∞of both sides: lim sup ω→∞/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay Rf(x) sin(ωx)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤ε+ 0. Butε>0 is arbitrary, so we must have: lim ω→∞/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay Rf(x) sin(ωx)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle= lim sup ω→∞/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay Rf(x) sin(ωx)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle= 0. /square 27 8 Construction of Lebesgue Measure We now come to actually construct Lebesgue measure, as promised. The idea is to extend an existing measure µwhich has been only partially defined, to an “outer measure” µ∗. The extension is remarkably simple and intuitive: µ∗(E) = inf A1,A2,...∈A E⊆S nAn/summationdisplay nµ(An),for allE⊆X. (One interesting point to note: the definition of µ∗“works” with pretty much any non-negative function µ, provided that µis defined on some set A ⊆ 2X with∅,X∈ A, andµ(∅) = 0. In fact, for the first few proofs, these are the only formal properties of µthat we need. Of course later we will need stronger conditions on µ, such as additivity.) Lemma 8.1. µ∗has the following properties: 1.µ∗(∅) = 0 . 2.It is monotone: µ∗(E)≤µ∗(F)whenE⊆F⊆X. 3.µ∗(A)≤µ(A)for allA∈ A. 4.µ∗is countably subadditive: if E1,E2,...⊆X, then µ∗(/uniontext nEn)≤/summationtext nµ∗(En). Proof. The first three properties are obvious. For the fourth, first observe that if any of the µ∗(En) is infinite, there is nothing to prove. Otherwise, letε >0. For each En, by the definition of µ∗, there are sets {An,m}m∈ A coveringEn, with/summationtext mµ(An,m)≤µ∗(En)+ε/2n. All of the An,mtogether cover/uniontext nEn, and/summationtext n,mµ∗(An,m)≤/summationtext nµ∗(En) +ε. Sinceεwas arbitrary, we have/summationtext n,mµ∗(An,m)≤/summationtext nµ∗(En). /square There is of course no guarantee that µ∗satisfies all the properties of a proper measure on 2X. (It often does not.) Instead we will claim that µ∗is a proper measure on the following subcollection of 2X: M={B∈2X|µ∗(B∩E) +µ∗(Bc∩E) =µ∗(E) for allE⊆X} ={B∈2X|µ∗(B∩E) +µ∗(Bc∩E)≤µ∗(E) for allE⊆X}. (The two subcollections are the same since by subadditivity we always have µ∗(B∩E) +µ∗(Bc∩E)≥µ∗(E).) Lemma 8.2. Mis a sigma algebra. Proof. It is immediate from the definition that Mis closed under taking com- plements, and that ∅ ∈ M . We first show Mis closed under finite intersection 28 (and hence under finite union). Let B,C∈ M . µ∗/parenleftbig (B∩C)∩E/parenrightbig +µ∗/parenleftbig (B∩C)c∩E/parenrightbig =µ∗(B∩C∩E) +µ∗/parenleftbig (Bc∩C∩E)∪(B∩Cc∩E)∪(Bc∩Cc∩E)/parenrightbig ≤µ∗(B∩C∩E) +µ∗(Bc∩C∩E) +µ∗(B∩Cc∩E) +µ∗(Bc∩Cc∩E) =µ∗(C∩E) +µ∗(Cc∩E),by definition of B∈ M =µ∗(E),by definition of C∈ M. ThusB∩C∈ M . We now have to show that if B1,B2,...∈ M ,/uniontext nBn∈ M . We may assume thatBnare disjoint, for otherwise we just consider B/prime n=Bn\(B1∪···∪Bn−1), which are in Mby the previous paragraph. We shall need to know that, for all N≥1, and allE⊆X, µ∗/parenleftbiggN/uniondisplay n=1Bn∩E/parenrightbigg =N/summationdisplay n=1µ∗(Bn∩E). The proof will be by induction on N. The statement is trivial for N= 1. For the induction step, let DN=/uniontextN n=1Bnwhich are increasing and are all in M. Then µ∗(DN+1∩E) =µ∗/parenleftbig DN∩(DN+1∩E)/parenrightbig +µ∗/parenleftbig Dc N∩(DN+1∩E)/parenrightbig =µ∗(DN∩E) +µ∗(BN+1∩E) =N+1/summationdisplay n=1µ∗(Bn∩E) (induction hypothesis). Using the fact just proven, we now have: µ∗(E) =µ∗(DN∩E) +µ∗(Dc N∩E) =N/summationdisplay n=1µ∗(Bn∩E) +µ∗(Dc N∩E) ≥N/summationdisplay n=1µ∗(Bn∩E) +µ∗/parenleftbigg/parenleftBig∞/uniondisplay n=1Bn/parenrightBigc ∩E/parenrightbigg (monotonicity). TakingN→ ∞ , we obtain: µ∗(E)≥∞/summationdisplay n=1µ∗(Bn∩E) +µ∗/parenleftbigg/parenleftBig∞/uniondisplay n=1Bn/parenrightBigc ∩E/parenrightbigg ≥µ∗/parenleftbigg/parenleftBig∞/uniondisplay n=1Bn/parenrightBig ∩E/parenrightbigg +µ∗/parenleftbigg/parenleftBig∞/uniondisplay n=1Bn/parenrightBigc ∩E/parenrightbigg (subadditivity). But this shows/uniontext∞ n=1Bn∈ M . /square 29 Lemma 8.3. IfB1,B2,...∈ M are disjoint, then µ∗(/uniontext nBn) =/summationtext nµ∗(Bn). Proof. The finite case µ∗(/uniontextN n=1Bn) =/summationtextN n=1µ∗(Bn) was proven in the previous lemma (set E=X). By monotonicity,/summationtextN n=1µ∗(Bn)≤µ∗(/uniontext∞ n=1Bn). Taking N→ ∞ gives/summationtext∞ n=1µ∗(Bn)≤µ∗(/uniontext∞ n=1Bn). Inequality in the other direction is implied by subadditivity of µ∗. /square We now know µ∗satisfies all the properties of a measure on M. In order for µ∗to be a sane extension of µtoM, we need to impose some conditions on µ. The ones that work from experience are: 1.Ashould be an algebra , meaning that it is non-empty, and closed under intersection and finite union (and intersection). 2.IfA1,...,A n∈ A are disjoint, then µ(/uniontext iAi) =/summationtext iµ(Ai). It follows that µis monotone and finitely subadditive. 3.Also, ifA1,A2,...∈ A are disjoint, and/uniontext iAihappens to be in A, the previous equation must also hold. (This is equivalent to requiring that µ(/uniontext iAi)≤/summationtext iµ(Ai).) Then we have the following important result. (By the way, the name of “Carath´ eodory Extension Process” is often used to refer to the constructions in this section.) Theorem 8.4. A ⊆ M , andµ∗(A) =µ(A)for allA∈ A. Thusµ∗is a measure extending of µonto the sigma algebra Mcontaining the algebra A. (Mmust also then contain the sigma algebra Bgenerated by A, although Mmay be larger than B.) Proof. FixA∈ A. For any E⊆Xandε > 0, by definition we can find A1,A2,...∈ A withE⊆/uniontext nAnand/summationtext nµ(An)≤µ∗(E) +ε. Then we have µ∗(A∩E) +µ∗(Ac∩E)≤µ∗/parenleftBig A∩/uniondisplay nAn/parenrightBig +µ∗/parenleftBig Ac∩/uniondisplay nAn/parenrightBig ≤/summationdisplay nµ∗(A∩An) +/summationdisplay nµ∗(Ac∩An) ≤/summationdisplay nµ(A∩An) +/summationdisplay nµ(Ac∩An) =/summationdisplay nµ(An) (finite additivity of µ) ≤µ∗(E) +ε. εbeing arbitrary, µ∗(A∩E) +µ∗(Ac∩E)≤µ∗(E), showing that A∈ M . 30 Now we show µ∗(A) =µ(A). Noteµ(A)≥µ∗(A) is always true. Con- sider anyA1,A2,...∈ A withA⊆/uniontext nAn. By countable subadditivity and monotonicity of µ, µ(A) =µ/parenleftBig/uniondisplay nA∩An/parenrightBig ≤/summationdisplay nµ(A∩An)≤/summationdisplay nµ(An). This implies µ(A)≤µ∗(A), directly from the definition of µ∗. /square Our final results for this section concern the uniqueness of this extension. Theorem 8.5. Assumeµ(X)<∞. LetBbe the sigma algebra generated from the algebra A. Ifνis another measure on B, which agrees with µ∗onA, then µ∗andνagree on Bas well. Proof. LetB∈ B. µ∗(B) = inf A1,A2,...∈A B⊆S nAn/summationdisplay nµ(An) = inf/summationdisplay nν(An)≥infν/parenleftBig/uniondisplay nAn/parenrightBig ≥infν(B). Soµ∗(B)≥ν(B). Similarly, we have µ∗(Bc)≥ν(Bc), so thatµ∗(X)−µ∗(B)≥ ν(X)−ν(B) =µ∗(X)−ν(B), orµ∗(B)≤ν(B). (In fact, this actually proves µ∗andνagree on Malso, ifνis defined on M.) /square But the hypothesis that µ(X)<∞is clearly too restrictive. The fix is easy: Definition 8.1. A measure space ( X,B,µ) issigma-finite , if there are mea- surable sets X1,X2,...⊆X, such that/uniontext nXn=Xandµ(Xn)<∞for all n. (Clearly, we may as well assume that the Xnare increasing in this definition.) Theorem 8.6. Theorem 8.5holds also in the case that (X,B,µ)is sigma-finite. Proof. Let{Xn} /arrownortheastX,µ(Xn)<∞as in the definition of sigma-finiteness. (Of course we also need to assume that the Xncan be chosen from A.) For each B∈ B, Theorem 8.5says thatµ∗(B∩Xn) =ν(B∩Xn). Taking limits as n→ ∞ givesµ∗(B) =ν(B). /square The restriction that Xbe sigma-finite is not too severe, since the usual spaces such as Rnaresigma-finite. Sigma-finiteness also comes back in the theorems of Section 11. Finally, the corollary below is just Theorem 8.6restated without reference to the outer measure: Corollary 8.7 (Uniqueness of measures). LetBbe the sigma algebra gen- erated by the algebra A. Then if two measures µandνagree on A, andXis sigma-finite (under either µorν), thenµ=νonB. 31 9 Lebesgue Measure in Rn In this section we rigorously construct the n-dimensional volume measure on Rn. As hinted before, the idea is to define the measure for rectangles and then use the extension process in Section 8. Unfortunately, there is some grunt work to do in order to verify that the hypotheses of those theorems are indeed satisfied. Our setting will be the collection Rof rectangles I1× ··· ×IninRn, where Ikis any open, half-open or closed, bounded or unbounded, interval in R. The following definition is merely an abstracted version of the formal facts we need about these rectangles. Definition 9.1. LetXbe any set. A semi-algebra is any R ⊆ 2Xwith the following properties: 1.The empty set is in R. 2.The intersection of any two sets in Ris also in R. 3.For any set in R, its complement is expressible as a finite disjoint union of other elements of R. It is easy to see, although tiresome to write down formally, that the collection of all rectangles is indeed a semi-algebra. But immediately from this, we can automatically construct an algebra A: Theorem 9.1. The set Aof all finite disjoint unions of elements of a semi- algebra Ris an algebra on X. Proof. We check the properties for an algebra: 1.The empty set is trivially in A. 2.IfA=/unionmultitext iRi, andB=/unionmultitext jSj, whereRiandSidenote a finite number of sets chosen from R, thenA∩B=/unionmultitext iRi∩/unionmultitext jSj=/unionmultitext i,jRi∩Sj∈ A. 3.IfA=/unionmultitext iRi, thenAc=/intersectiontext iRc i=/intersectiontext i/unionmultitext jSi,jfor someSi,j∈ R. The finite intersection belongs to Aby the previous step. So Ac∈ A. 4.Finally, given Ai=/unionmultitext jRi,j, for a finite number of i, letD0=∅, andDi= Di−1/unionmulti(Ai\Di−1)∈ A. Then/uniontext iAi=/uniontext iDi=/unionmultitext i(Di\Di−1)∈ A./square By the way, the sigma algebra generated by Awill contain the Borel sigma algebra: every open set U∈Rnobviously can be written as a union of open rectangles, and in fact we can use a countable union of open rectangles. For, given an arbitrary collection of rectangles covering U⊆Rn, there always exists acountable subcover5. The volume of A=I1×···×Inis naturally defined as λ(A) =λ(I1)·····λ(In), λ(Ik) being the length of the interval Ik, with the usual rules about multiplying zeroes and infinities together in force. 5If you are not aware of this theorem, you are invited to prove it yourself. 32 Now let’s suppose that the rectangle Ahas been partitioned into a disjoint smaller rectangles. Then the sum of the volumes of the smaller rectangles, as we have defined it, should equal the volume of A. This is true, of course, although it is again tedious to write down formally. Essentially, one draws a rectangular grid onAusing the boundaries of the smaller rectangles, and show that the sum of the volumes of each cell in the grid equals the volume of A, by applying the distributive property of multiplication over addition. In the even more general case, suppose A∈ Ais a disjoint union of rectangles, butAis not necessarily in the semi-algebra of rectangles. The volume of Ais defined as the sum of the volumes of the component rectangles. This is obvious — we mention it only to note that, although Amay certainly have different decompositions ( A=/uniontext iRi=/uniontext jSj), the volume sum is always the same. To see this, simply take the common refinement Ri∩Sj. Then/summationtext iλ(Ri) =/summationtext i/summationtext jλ(Ri∩Sj) by applying the result of the previous paragraph on each rectangleRi. But/summationtext jλ(Rj) equals this double sum also. It follows easily then, that λis finitely additive. Thus there is only one final thing left to show: if the disjoint union of A1,A2,...∈ A isC∈ A, thenλ(C) =/summationtext∞ i=1λ(Ai). From monotonicity and taking limits we always have λ(C)≥/summationtext∞ i=1λ(Ai). We showλ(C)≤/summationtext∞ i=1λ(Ai). Observe that this also ought to be true if C is contained in, but not necessarily equal to, the union of the Ai, andAineed not be disjoint at all. Henceforth these are our new hypotheses. Suppose first that Chappens to be compact, and Aiare all open. In other words, {Ai}form an open cover of the compact set C. So there is a finite subcoverA1,...,A n. By finite subadditivity, we have λ(C)≤/summationtextn i=1λ(Ai)≤/summationtext∞ i=1λ(Ai). Now continue to assume that Cis compact, but Aiare not open. But it is easy to make the Aiopen and still cover C, by slightly expanding each Ai. In particular, stipulate that the volume of each new Aigrows by at most ε/2i. (We can assume the Aiare plain rectangles, rather than disjoint unions of them, and that they are bounded, since Cis bounded.) Then we have λ(C)≤/summationtext∞ i=1λ(Ai) +ε, andε>0 is arbitrary. All that remains is the case that Cis not compact. If Cis bounded, so is its closure, and hence by the Heine-Borel theorem, Cis compact. Similarly take the closure of the Ai. But taking closures of elements of Adoes not change their volumes. Finally consider Cunbounded. But C∩[−N,N ]nis bounded for each N, so from the previous case we have λ(C∩[−N,N ]n)≤/summationtext∞ i=1λ(Ai). It is easily checked that the limit as N→ ∞ of the left side is exactly λ(C). Thus, using the theorems of Section 8, we can conclude: Theorem 9.2. Lebesgue measure in Rnexists, and it is uniquely determined, given our hypotheses. It is obvious from our constructions that Lebesgue measure is invariant under translations of sets. It is also invariant under other rigid motions (rotations, reflections); this will be a consequence of Lemma 12.1 and some linear algebra. 33 An interesting question to ask is whether there are any sets that are not Borel, or that cannot be assigned any volume. The following theorem gives a classic example (and should also serve to convince you why our strenuous efforts are necessary). Theorem 9.3 (Vitali). There exists a non-measurable set in [0,1]using Lebesgue measure. In other words, Lebesgue measure cannot be defined consistently for allsubsets of [0,1]. Proof. The key fact in this proof is translation-invariance. In particular, given any measurable H⊆[0,1], define its “shift with wrap-around”: H⊕x={h+x:h∈H, h +x≤1} ∪ {h+x−1 :h∈H, h +x>1}. Thenλ(H⊕x) =λ(H). Define two real numbers to be equivalent if their difference is rational. The interval [0,1] is partitioned by this equivalence relation. Compose a set H⊂ [0,1] consisting of exactly one element from each equivalence class, and also say 0/∈H. Then (0,1] equals the disjoint union of all H⊕r, forr∈[0,1)∩Q. Consequently, by countable additivity, 1 =λ((0,1]) =/summationdisplay r∈[0,1)∩Qλ(H⊕r) =/summationdisplay r∈[0,1)∩Qλ(H), a contradiction, because the sum on the right can only be 0 or ∞. HenceH cannot be measurable. /square A fact related to these matters is that Lebesgue measure is complete , mean- ing ifµ(A) = 0, then every B⊆Ais Lebesgue-measurable and µ(B) = 0. (This follows directly from the construction of the outer measure in the previous sec- tion.) On the other hand, one can show that the Lebesgue measure restricted to the Borel sets in Rnisnotcomplete. This means a slight complication in the theorems we prove about Lebesgue measure, but fortunately the extension process allows us to complete any (sigma-finite) measure if necessary. 10 Riemann integrability implies Lebesgue in- tegrability You have probably already suspected that any function that any Riemann- integrable function is also Lebesgue-integrable, and certainly with the same values for the two integrals. We shall prove this fact here. Let us first review the definition of the Riemann integral. LetA⊂Rmbe a (bounded) rectangle. Usually a bounded function f:A→ Ris said to be (proper-) Riemann-integrable if the supremum of its lower sums and the infimum of its upper sums are equal: sup P/braceleftBig/summationdisplay R∈Pµ(R)·inf x∈Rf(x)/bracerightBig = inf P/braceleftBig/summationdisplay R∈Pµ(R)·sup x∈Rf(x)/bracerightBig 34 (where Pdenotes a rectangular partition of A). We can rephrase the definition by considering not just lower sums and upper sums forf, but the integral of any simple function s≤fors≥f(simple with respect to a rectangular partition). Such simple functions are obviously both Riemann- and Lebesgue- integrable with the same values for the integral. It is also easily seen that for every such s≤fthere exists some lower sum for f(in the usual sense) such that the integral of sis less than or equal to that lower sum. Similarly for the upper simple functions and the upper sums. Therefore sup all simple s≤f/braceleftBig/integraldisplay As/bracerightBig = sup P/braceleftBig/summationdisplay R∈Pµ(R)·inf x∈Rf(x)/bracerightBig , inf all simple s≥f/braceleftBig/integraldisplay As/bracerightBig = inf P/braceleftBig/summationdisplay R∈Pµ(R)·sup x∈Rf(x)/bracerightBig , and we may equivalently define fto be Riemann-integrable if the supremum of the integrals of the lower simple functions is equal to the infimum of the integrals of the upper simple functions. It follows from the usual arguments, that if s1ands2are simple with s1≤ f≤s2, then/integraltext As1≤/integraltext As2, and thatfis Riemann-integrable if and only if there exists a sequence of lower simple functions ln≤f, and upper simple functions un≥fsuch that lim n→∞/integraldisplay Aln=/integraldisplay Af= lim n→∞/integraldisplay Aun. This more relaxed definition of Riemann integrability is easier to work with in the proof of the following theorem. Theorem 10.1. LetA⊂Rmbe a rectangle. If f:A→Ris proper Riemann- integrable, then it is also Lebesgue-integrable (with respect to Lebesgue measure) with the same value for the integral. Proof.fis Riemann-integrable, so choose a sequence of simple functions ln≤ f≤unwith lim n/integraltext Aln=/integraltext Af= lim n/integraltext Aun. Letgn(x) = max k≤nlk(x), so that gn(x) increase to g(x) = supngn(x). By our construction, we have ln≤gn≤g≤f≤un. Since the integrals of lnandunconverge onto each other, we know that g is Riemann-integrable. Riemann-integrating and applying limits to the above inequality, lim n→∞/integraldisplay Aln≤lim n→∞/integraldisplay Agn≤/integraldisplay Ag≤/integraldisplay Af≤lim n→∞/integraldisplay Aun. Thus the non-negative function f−ghas a Riemann integral of zero, and so f−g= 0 almost everywhere with respect to Lebesgue measure6. In turn,f−g must be measurable. (This follows from Lebesgue measure being complete.) 6This follows from a very famous theorem about Riemann integrability, whose proof you can find in [ Spivak2 ] or [ Munkres ]. 35 On the other hand, gis measurable, because gnandlnare, sofis measur- able. Since |f|is bounded and Ahas finite measure, fmust also be Lebesgue- integrable. Then the same inequality above with the Riemann integrals changed to Lebesgue integrals shows that the Lebesgue and Riemann integrals of fare the same. /square The following theorem concerns the absolutely convergent improper Rie- mann integral as defined in [ Munkres ]. Theorem 10.2. LetA⊆Rmbe open, and f:A→Rbe locally bounded on A and continuous almost everywhere on A. Iffis improper-Riemann-integrable (i.e./integraltext A|f|<∞), then its Lebesgue integral exists with the same value for the integral. Also/integraltext A|f|diverges simultaneously for the improper Riemann integral and the Lebesgue integral. Proof. Suppose first that f≥0. LetCnbe a sequence of compact Jordan- measurable subsets of Awhose union is AandCn⊂interiorCn+1. Then /integraldisplay Af= lim n→∞/integraldisplay Cnf. Note that/integraltext Cnfis valid as both a Riemann and Lebesgue integral, by Theo- rem10.1, and it can also be written as the Lebesgue integral/integraltext Af·χCnwhich converges monotonically, as n→ ∞ , to the Lebesgue integral/integraltext Af. This must of course be equal to the left side of the equation above, which is the improper Riemann integral. For general f, repeating the same reasoning for the non-negative functions |f|,f+,f−, in turn proves the theorem. /square 11 Product measures and Fubini’s Theorem Fubini’s theorem concerns integrals in “multiple dimensions” and their eval- uation using iterated integrals. The concept should be familiar from multi- dimensional calculus, so I won’t launch myself into an extended discussion here. But before we start writing down integral signs, we need to discuss the measurability of the sets involved in multiple integration. Definition 11.1. Let (X,A) and (Y,B) be two measurable spaces. A measur- able rectangle inX×Yis a set of the form A×B, whereA∈ A andB∈ B. The sigma algebra generated by all the measurable rectangles is denoted by A ⊗ B , and this will be the sigma algebra we use for X×Y. Theorem 11.1. LetEbe a measurable set from (X×Y,A ⊗ B ). Let Ex={y∈Y: (x,y)∈E}, x∈X. Ey={x∈X: (x,y)∈E}, y∈Y . ThenEy∈ A,Ex∈ B. 36 Proof. We prove the theorem for Ey; the proof for Exis the same. We consider the collection D={E∈ A ⊗ B :Ey∈ A} . SupposeE=A×Bis a measurable rectangle. Then Ey=Awheny∈ B, otherwiseEy=∅. In both cases Ey∈ A, soE∈ D. Dis a sigma algebra, because: 1.∅y={x∈X: (x,y)∈ ∅} =∅ ∈ A , so∅ ∈ D . 2.IfEn∈ D, then (/uniontextEn)y=/uniontextEn y∈ A. So/uniontextEn∈ D. 3.IfE∈ D, andF=Ec, thenFy={x∈X: (x,y)∈F}={x∈X: (x,y)/∈E}=X\Ey∈ A. SoEc∈ D. ThusDis a sigma algebra containing the measurable rectangles, i.e. Dis all of A ⊗ B . /square Theorem 11.2. Let(X×Y,A ⊗ B )and(Z,C)be measurable spaces. Iff:X×Y→Zis measurable, then the functions fy:X→Z,fx:Y→Z obtained by holding one variable fixed are also measurable. Proof. Again we consider only fy. LetIy:X→X×Ybe defined by Iy(x) = (x,y). Then given E∈ A ⊗ B , by Theorem 11.1.I−1 y(E) =Ey∈ A , soIyis a measurable function. But fy=f◦Iy. /square The next theorem on multiple integration requires the following technical tool, which comes equipped with a definition. Definition 11.2. A family Aof subsets of Xis amonotone class if it is closed under increasing unions and decreasing intersections. The intersection of any set of monotone classes is a monotone class. The smallest monotone class containing a given set Gis the intersection of all mono- tone classes containing G. This construction is analogous to the one for sigma algebras, and the result is also said to be the monotone class generated by G. Theorem 11.3 (Monotone Class Theorem). IfAbe an algebra on X, then the monotone class generated by Ais the same as the sigma algebra generated byA. Proof. Since a sigma algebra is a monotone class, the generated sigma algebra contains the generated monotone class M. So we only need to show Mis a sigma algebra. We first claim that Mis actually closed under complementation. Let M/prime= {S∈ M :X\S∈ M} ⊆ M . This is a monotone class, and it contains the algebra A. SoM=M/primeas desired. To prove that Mis closed under countable unions, we only need to prove that it is closed under finite unions, for it is already closed under countable increasing unions. First letA∈ A, andN(A) ={B∈ M :A∪B∈ M} ⊆ M . Again this is a monotone class containing the algebra A; thus N(A) =M. 37 Finally, let S∈ M , with the same definition of N(S). The last paragraph, rephrased, says that A ⊆ N (S). And N(S) is a monotone class containing A by the same arguments as the last paragraph. Thus N(S) =Mas desired. /square Theorem 11.4. Let(X,A,µ),(Y,B,ν)be sigma-finite measure spaces. IfE∈ A ⊗ B , then 1.ν(Ex)is a measurable function of x∈X. 2.µ(Ey)is a measurable function of y∈Y. Proof. We concentrate on µ(Ey). Let {Xm} /arrownortheastX, withµ(Xm)<∞. Fixm for now and let D={E∈ A ⊗ B :µ(Ey∩Xm) is a measurable function of y}. Dis equal to A ⊗ B , because: 1.IfE=A×Bis a measurable rectangle, then µ(Ey∩Xm) =µ(A∩Xm)· χB(y) which is a measurable function of y. IfEis a finite disjoint union of measurable rectangles En, thenµ(Ey∩ Xm) =/summationtext nµ(En y∩Xm) which is also measurable. The measurable rectangles form a semi-algebra (just like the rectangles in Rn). Therefore, applying Theorem 9.1,Dcontains the algebra of finite disjoint unions of measurable rectangles. 2.IfEnare increasing sets in D(not necessarily measurable rectangles), then µ((/uniontextEn)y∩Xm) =µ(/uniontextEn y∩Xm) = lim n→∞µ(En y∩Xm) is measurable, so/uniontextEn∈ D. Similarly, if Enare decreasing sets in D, then using limits we see that/intersectiontextEn∈ D. (Here it is crucial that En y∩Xmhave finite measure, for the limiting process to be valid.) 3.These arguments show that Dis a monotone class, and it contains the monotone class generated by the algebra of finite unions of measurable rectangles. By the Monotone Class Theorem, Dmust therefore be the same as the sigma algebra A ⊗ B . We now know that for each E∈ A⊗B ,µ(Ey∩Xm) is measurable, for every m. Taking limits as m→ ∞ , we conclude that µ(Ey) is also measurable. /square One thing has been deliberately left out of our discussion so far: the con- struction of the product measure µ⊗ν, which, as in the case of Rn, should assign a measure µ(A)ν(B) to the measurable rectangle A×B. The problem is that we cannot prove countable additivity of µ⊗νdefined this way. We take an indirect route instead, defining it by iterated integrals: Theorem 11.5. Let(X,A,µ),(Y,B,ν)be sigma-finite measure spaces. There exists a unique product measure µ⊗ν:A ⊗ B → [0,∞], with (µ⊗ν)(E) =/integraldisplay x∈X/integraldisplay y∈YχE(x,y)dν /bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright ν(Ex)dµ=/integraldisplay y∈Y/integraldisplay x∈XχE(x,y)dµ /bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright µ(Ey)dν. 38 Proof. Letλ1(E) denote the double integral on the left, and λ2(E) denote the one on the right. (These integrals exist by Theorem 11.4.) It is obvious that bothλ1andλ2are countably additive, so they are both measures on A ⊗ B . Moreover, if E=A×B, then just expanding the two integrals shows λ1(E) = µ(A)ν(B) =λ2(E). NowX×Yis sigma-finite if XandYare, so by uniqueness of measures7(Corollary 8.7),λ1=λ2on all of A ⊗ B . /square There’s not much work left for our final theorems: Theorem 11.6 (Fubini). Let(X,A,µ)and(Y,B,ν)be sigma-finite measure spaces. Iff:X×Y→Risµ⊗ν-integrable, then /integraldisplay X×Yfd(µ⊗ν) =/integraldisplay x∈X/bracketleftBig/integraldisplay y∈Yf(x,y)dν/bracketrightBig dµ=/integraldisplay y∈Y/bracketleftBig/integraldisplay x∈Xf(x,y)dµ/bracketrightBig dν. This equation also holds when f≥0(if it is merely measurable, not integrable). Proof. The casef=χEis just Theorem 11.5. Since all three integrals are additive, they are equal for non-negative simple f, and hence also for all other non-negative f, by approximation and monotone convergence. From linearity onf=f+−f−, we see that they are equal for any integrable f. (We will allow ∞ − ∞ to occur on a set of measure zero.) /square Theorem 11.7 (Tonelli). Let(X,A,µ)and(Y,B,ν)be sigma-finite measure spaces, and f:X×Y→Rbeµ⊗ν-measurable. Then fisµ⊗ν-integrable if and only if /integraldisplay x∈X/bracketleftBig/integraldisplay y∈Y|f(x,y)|dν/bracketrightBig dµ<∞ (or withXandYreversed). Consequently, if any one of these conditions hold, then it is valid to switch the order of integration when integrating f. Proof. Immediate from Fubini’s theorem applied to the function |f|. /square Remark 11.8.Note that it is possible that/integraltext y∈Y|f(x,y)|dν=∞on a set of measure zero in X, and still have integrability, and conversely. 12 Change of variables in Rn This section will be devoted to completing the proof of the differential change of variables formula, Theorem 6.5. As we noted in the remarks preceding that theorem, it suffices to prove the following. 7If you prefer, you can also prove this theorem using the Monotone Class Theorem instead. 39 Lemma 12.1. Letg:X→Ybe a diffeomorphism between open sets in Rn. Then for all measurable sets A⊆X, λ(g(A)) =/integraldisplay g(A)1 =/integraldisplay A|det Dg|. Proof. We first begin with two simple reductions. (I)It suffices to prove the lemma locally. That is, suppose there exists an open cover of X,{Uα}, so that the equation of the lemma holds for measurable Acontained inside one of the Uα. Then the equation actually holds for all measurable A⊆X. Proof. By taking a countable subcover, we may assume there are only countably many Ui. Define the disjoint measurable sets Ei=Ui\(U1∪ ··· ∪Ui−1), which cover X. Also define the two measures: µ(A) =λ(g(A)), ν(A) =/integraldisplay A|det Dg|. Now letA⊆Xbe any measurable set. We have A∩Ei⊆Ui, soµ(A∩ Ei) =ν(A∩Ei) by hypothesis. Therefore, µ(A) =µ/parenleftBig/uniondisplay iA∩Ei/parenrightBig =/summationdisplay iµ(A∩Ei) =/summationdisplay iν(A∩Ei) =ν(A). (II)Suppose the lemma holds for two diffeomorphisms gandh, and all mea- surable sets. Then it holds for the composition diffeomorphism g◦h(and all measurable sets). Proof. For any measurable A, /integraldisplay g(h(A))1 =/integraldisplay h(A)|det Dg|=/integraldisplay A|(det Dg)◦h|·|det Dh|=/integraldisplay A|det D(g◦h)|. The second equality follows from Theorem 6.5applied to the diffeomor- phismh, which is valid once we know λ(h(B)) =/integraltext B|det Dh|for all mea- surableB. We proceed to prove the lemma by induction, on the dimension n. Base case n= 1.CoverXby a countable set of bounded intervals IkinR. By Reduction I, it suffices to prove the lemma for measurable sets contained in each of the Ikindividually. By the uniqueness of measures (Corollary 8.7), it also suffices to show µ=νonly for the intervals [ a,b], (a,b), etc. But this is just the Fundamental Theorem of Calculus: /integraldisplay g([a,b])1 =|g(b)−g(a)|=|/integraldisplayb ag/prime|=/integraldisplayb a|g/prime|. 40 (For the last equality, remember that the g/primemust be either positive on all of [a,b] or negative on all of [ a,b]. If the interval is open or half-open, we may not be able to apply the Fundamental Theorem, but the preceding equation can still be obtained via a limiting procedure.) Induction step. Locally (i.e. on a sufficiently small open set around each pointx∈X),gcan always be factored8asg=hk◦ ··· ◦h2◦h1, where eachhiis a diffeomorphism and fixes one coordinate of Rn. By Reduction I, it suffices to consider this local case only. By Reduction II, it suffices to prove the lemma for each of the diffeomorphisms hi. So suppose gfixes one coordinate. For convenience in notation, assume g fixes the last coordinate: g(u,v) = (hv(u),v), foru∈Rn−1,v∈R, andhv are functions on (open subsets of) Rn−1, in fact diffeomorphisms. Clearly hvare one-to-one, and most importantly, det D hv(u) = det Dg(u,v)/negationslash= 0. Next, let a measurable set Abe given, and consider its projection V= {v∈R: (u,v)∈A}, and its cross-section Uv={u∈Rn−1: (u,v)∈A}. We now apply Fubini’s theorem and the induction hypothesis on hv: /integraldisplay g(A)1 =/integraldisplay v∈V/integraldisplay hv(Uv)1 =/integraldisplay v∈V/integraldisplay u∈Uv|det Dhv(u)| =/integraldisplay v∈V/integraldisplay u∈Uv|det Dg(u,v)|=/integraldisplay A|det Dg|. /square 13 Vector-valued integrals This section, in short, is a remark that everything we have done so far generalizes to vector-valued functions, the vectors being from real finite-dimensional spaces (orC). These often occur in applications. Givenf:X→Rnmeasurable, let {ek}denote the standard basis vectors inRn, and {fk}the components of fwith respect to this basis. Of course we define/integraldisplay f=n/summationdisplay k=1/parenleftBig/integraldisplay fk/parenrightBig ek, provided the integrals on the right exist. Generally, to say that/integraltext Xfexists, we do not allow any one of the components to be infinite, for this is usually not useful when n≥2. It is not hard to see that fis measurable if and only if each fkis. It follows immediately that this integral is linear, and hence, the definition is independent of the basis, as it ought to be. That is, if {ek}isanybasis of Rn, the sum in the definition does not change. 8The proof of this fact can be found in [ Munkres ], but it is not difficult to prove it yourself. Hint: Inverse Function Theorem. 41 Since by our convention that the components of the integral are not allowed to be infinite, it seems we do not need to define separately what it means for f to be integrable. But we define it, because we want to take note of some facts (admittedly they are not very interesting): fbeing integrable means/integraltext /bardblf/bardbl<∞. The norm can be arbitrary, for in Rn, every norm is equivalent: if /bardbl·/bardbl1and /bardbl·/bardbl2are any two given norms, then there always exist constants α,β > 0 such thatα/bardbl·/bardbl2≤ /bardbl·/bardbl 1≤β/bardbl·/bardbl2. Thus being integrable in one norm implies integra- bility in another norm. In particular, by using the norm /bardblx/bardblΣ=/summationtextn k=1|xk|, we see thatfis integrable if and only if its components fkare integrable. We want to show that /bardbl/integraltext f/bardbl ≤/integraltext /bardblf/bardbl; this basic inequality will enable us to make estimates without having to separate components. As a start, this is clearly true if fis a simple function, i.e. f=/summationtextm j=1ajχEjforaj∈Rn. It is also trivial if/integraltext /bardblf/bardbl=∞. To prove the inequality for the other f, we use the following easy lemma. Lemma 13.1. Letf:X→Rnbe integrable. There exists a sequence of simple functionsϕj:X→Rnconverging pointwise to f, with lim j→∞/integraldisplay /bardblϕj−f/bardbl= 0. Proof. By equivalence of norms, it suffices to prove this only for the norm /bardbl·/bardblΣ as defined above. For eachfk+, by the approximation theorem in R, there exists measurable simpleϕk+ jincreasing to fk+. Similarly for fk−. Letϕk j=ϕk+ j−ϕk− j. Using the Dominated Convergence Theorem applied to each component, lim j→∞/integraldisplay /bardblϕj−f/bardblΣ= lim j→∞/integraldisplayn/summationdisplay k=1|ϕk j−fk|= 0. We also note that by the Monotone Convergence Theorem, the ϕjsatisfy lim j→∞/integraltext ϕj=/integraltext f. /square We finish our demonstration of the generalized triangle inequality. Let ϕj as in the lemma. Then /vextenddouble/vextenddouble/vextenddouble/integraldisplay ϕj/vextenddouble/vextenddouble/vextenddouble≤/integraldisplay /bardblϕj/bardbl ≤/integraldisplay /bardblf/bardbl+/integraldisplay /bardblϕj−f/bardbl, so that lim j→∞/vextenddouble/vextenddouble/vextenddouble/integraldisplay ϕj/vextenddouble/vextenddouble/vextenddouble=/vextenddouble/vextenddouble/vextenddoublelim j→∞/integraldisplay ϕj/vextenddouble/vextenddouble/vextenddouble=/vextenddouble/vextenddouble/vextenddouble/integraldisplay f/vextenddouble/vextenddouble/vextenddouble≤/integraldisplay /bardblf/bardbl+ lim j→∞/integraldisplay /bardblϕj−f/bardbl=/integraldisplay /bardblf/bardbl. 14 C∞ 0functions are dense in Lp(Rn) This section is devoted to the result that the space of C∞ 0functions is dense in Lp(Rn), which was discussed at the end of Section 7. 42 Theorem 14.1. Letf:Rn→R∈Lp(Rn),1≤p <∞. Then for any ε >0, there exists ψ∈C∞ 0such that /bardblψ−f/bardblp=/parenleftBig/integraldisplay Rn|ψ−f|pdλ/parenrightBig1/p <ε. Our strategy for proving this theorem is straightforward. Since we already know that the simple functions ϕ=/summationtext iaiχEiare dense in Lp, we should try approximating χEibyC∞ 0functions. Since C∞ 0functions are non-zero on com- pact sets, it stands to reason that we should approximate the sets Eiby compact setsKi. If this can be done, then it suffices to construct the C∞ 0functions on the setsKi. Our constructions start with this last step. You might even have seen some of these constructions before. Lemma 14.2. LetAbe a compact rectangle in Rn. Then there exists φ∈C∞ 0 which is positive on the interior of Aand zero elsewhere. Proof. Consider the infinitely differentiable function f(x) =/braceleftbigg e−1/x2, x> 0 0, x ≤0. IfA= [0,1], thenφ(x) =f(x)·f(1−x) is the desired function of C∞ 0. (Draw pictures!) IfA= [a1,b1]× ··· × [an,bn], then we let φA(x) =φ/parenleftbiggx1−a1 b1−a1/parenrightbigg ···φ/parenleftbiggxn−an bn−an/parenrightbigg . /square Lemma 14.3. For anyδ >0, there exists an infinitely differentiable function h:R→[0,1]such thath(x) = 0 forx≤0andh(x) = 1 forx≥δ. Proof. Take the function φfrom Lemma 14.2 for the rectangle [0 ,δ], and let h(x) =/integraltextx −∞φ(t)dt/integraltext∞ −∞φ(t)dt. /square Theorem 14.4. LetUbe open, and K⊂Ucompact. Then there exists ψ∈ C∞ 0which is positive on Kand vanishes outside some other compact set L, K⊂L⊂U. Proof. For eachx∈U, letAx⊂Ube a bounded open rectangle containing x, whose closure Axlies inU. The {Ax}together form an open cover of K. Take a finite subcover {Axi}. Then the compact rectangles {Axi}also coverK. From Lemma 14.2, obtain functions ψi∈C∞ 0that are positive on Axiand vanish outside Axi. Letψ=/summationtext iψi∈C∞ 0.ψvanishes outside L=/uniontext iAxi, which is compact. /square 43 Corollary 14.5. In Theorem 14.4, it is even possible to require in addition that 0≤ψ(x)≤1for allx∈Rnandψ(x) = 1 forx∈K. Proof. Letψbe from Theorem 14.4. Sinceψis positive on the compact set K, it has a positive minimum δthere. Take the function hof Lemma 14.3 for this δ. The new candidate function is h◦ψ. /square As we have said, we must now approximate arbitrary Borel sets B∈ B(Rn) by compact sets. (We will also need approximation by open sets.) It turns out that this part of the proof is purely topological, and generalizes to other metric spacesXbesides Rn. Henceforth we consider the more general case. Letddenote the metric for the metric space X. Theorem 14.6. Let(X,B(X),µ)be a finite measure space, and let B∈ B(X). For everyε > 0, there exists a closed set Vand an open set Usuch that V⊆B⊆Uandµ(U\V)<ε. Proof. LetMbe the set of all B∈ B(X) for which the statement is true. We show that Mis a sigma algebra containing all the open sets in X. 1.LetB∈ M withVandUas above. Then Vcopen⊇Bc⊇Ucclosed, andµ(Vc)−µ(Uc)<ε. This shows Bc∈ M . 2.LetBn∈ M . ChooseVnandUnfor eachBnsuch thatµ(Un\Vn)<ε/2n. LetU=/uniontext nUnwhich is open, and V=/uniontext nVn, so thatV⊆/uniontext nBn⊆U. Of courseVis not necessarily closed, but WN=/uniontextN n=1Vnare, and these WNincrease to V. Henceµ(V\WN)→0 asN→ ∞ , meaning that for large enough N,µ(V\WN)<ε. Next, we have U\WN= (U\V)/unionmulti(V\WN) ⊆/uniondisplay n(Un\Vn)∪(V\WN), µ(U\WN) =µ(U\V) +µ(V\WN) ≤/summationdisplay n(Un\Vn) +µ(V\WN)<ε+ε. This shows that/uniontext nBn∈ M . 3.LetBbe open, and A=Bc. Also let d(x,A) = inf y∈Ad(x,y) be the distance from x∈XtoA. SetDn={x∈X:d(x,A)≥1/n}.Dnis closed, because d(·,A) is a continuous function, and [1 /n,∞] is closed. Clearlyd(x,A)≥1/n> 0 impliesx∈Ac=B, but since Ais closed, the converse is also true: for every x∈Ac=B,d(x,A)>0. Obviously the Dnare increasing, so we have just shown that they in fact increase to B. Henceµ(B\Dn)<εfor large enough n. ThusB∈ M . /square 44 The case that µis not a finite measure is taken care of, as you would expect, by taking limits like we did for sigma-finite measures in Section 8. But since compact and open sets are involved, we need stronger hypotheses: 1.There exists {Kn} /arrownortheastX, withKncompact and µ(Kn)<∞. 2.There exists {Xn} /arrownortheastX, withXnopen andµ(Xn)<∞. It is easily seen that these properties are satisfied by X=Rnand the Lebesgue measure λ, as well as many other “reasonable” measures µonB(Rn). We will discuss this more later. We assume henceforth that Xandµhave the properties just listed. Theorem 14.7. LetB∈ B(X)withµ(B)<∞. For every ε>0, there exists a compact set Vand an open set Usuch thatK⊆B⊆Uandµ(U\K)<ε. Proof. It suffices to show that µ(U\B)<εandµ(B\K)<εseparately. Existence of K.Since {B∩Kn} /arrownortheastB, there exists some nsuch thatµ(B)− µ(B∩Kn)<ε/2. For thisn, define the finite measure µKn(E) =µ(E∩Kn), forE∈ B(X). By Theorem 14.6, there are sets V⊆B⊆U,Vclosed, and µKn(B\V)≤ µKn(U\V)<ε/2. SinceXnis compact, it is closed. Then K=V∩Kn is also closed, and hence compact, because it is contained in the compact setKn. We have, µ(B\K) =µ(B)−µ(B∩Kn) +µ(B∩Kn)−µ(K) =µ(B)−µ(B∩Kn) +µKn(B)−µKn(V)<ε 2+ε 2. Existence of U.For everyn, define the finite measure µXn(E) =µ(E∩Xn), forE∈ B(X). By Theorem 14.6, there are sets Vn⊆B⊆Un,Unopen, andµXn(Un\B)≤µXn(Un\Vn)<ε/2n. LetU=/uniontext nUn∩Xn⊇B. We have, µ(U\B)≤/summationdisplay nµ(Un∩Xn\B) =/summationdisplay nµXn(Un\B)<ε. /square We return to the case of X=Rn. Theorem 14.8. LetB∈ B(Rn)withµ(B)<∞. For every ε>0, there exists ψ∈C∞ 0such that /bardblψ−χB/bardblp=/parenleftBig/integraldisplay Rn|ψ−χB|pdµ/parenrightBig1/p <ε. 45 Proof. By Theorem 14.7, there is compact Kand openU,K⊆B⊆U,µ(U\ K)< ε. From Corollary 14.5, there isψ∈C∞ 0such thatψ= 1 onK,ψ= 0 outsideU, and 0 ≤ψ≤1. Then /integraldisplay Rn|ψ−χB|p=/integraldisplay Rn\U0 +/integraldisplay U\Bψp+/integraldisplay B\K(1−ψ)p+/integraldisplay K0 ≤µ(U\B) +µ(B\K) =µ(U)−µ(B) +µ(B)−µ(K) <ε. /square Proof of Theorem 14.1.Letϕ=/summationtext iaiχEi,ai/negationslash= 0 be a simple function such that/bardblϕ−f/bardblp< ε/ 2. Letψi∈C∞ 0such that /bardblψi−χEi/bardblp< ε/ 2|ai|. (Note thatEimust have finite measure; otherwise ϕwould not be integrable.) Let ψ=/summationtext iaiψi. Then (Minkowski’s inequality), /bardblf−ψ/bardblp≤ /bardblf−ϕ/bardblp+/bardblϕ−ψ/bardblp ≤ /bardblf−ϕ/bardblp+/summationdisplay i|ai| · /bardblχEi−ψi/bardblp<ε. /square Actually, even the last part of theorem can be generalized to spaces other thanRn: instead of infinitely differentiable functions with compact support, we consider continuous functions, defined on the metric space X, with compact support. In this case, a topological argument must be found to replace Lemma 14.2. This is easy: Lemma 14.9. LetAbe any compact set in X. Then there exists a continuous functionφ:X→Rwhich is positive on the interior of Aand zero elsewhere. Proof. LetC=X\interiorA, soCis closed. Then φ(x) =d(x,C) works. (d(x,C) was defined in the proof of Theorem 14.6.) /square The proof of Theorem 14.4 goes through verbatim for metric spaces X, provided that Xislocally compact . This means: given any x∈Xand an open neighborhood Uofx, there exists another open neighborhood Vofx, such that Vis compact and V⊆U. Finally, we need to consider when properties (1) and (2) (in the remarks preceding Theorem 14.7) are satisfied. These properties are somewhat awkward to state, so we will introduce some new conditions instead. Definition 14.1. A measure µon a topological space Xislocally finite if for eachx∈X, there is an open neighborhood Uofxsuch thatµ(U)<∞. It is easily seen that when µis locally finite, then µ(K)<∞forevery compact set K. Definition 14.2. A topological space Xisstrongly sigma-compact if there exists a sequence of open sets Xnwith compact closure, and {Xn} /arrownortheastX. 46 IfXis strongly sigma-compact, and µis locally finite, then properties (1) and (2) are automatically satisfied. It is even true that strong sigma-compactness implies local compactness in a metric space. (The proof requires some topology and is left as an exercise.) Then we have the following theorem: Theorem 14.10. LetXbe a strongly sigma-compact metric space, and µbe any locally finite measure on B(X). Then the space of continuous functions with compact support is dense in Lp(X,B(X),µ),1≤p<∞. 15 Other examples of measures Since so far we have chiefly worked only in Rnwith Lebesgue measure, it should be of interest to give a few more useful examples of measures. k-dimensional volume of a k-dimensional manifold A manifold is a generalization of curves and surfaces to higher dimensions, and sometimes even to spaces other than Rn. But here we shall concentrate on differ- entiable manifolds inside Rn; the theory is elucidated in [ Spivak2 ] or [Munkres ]. Here we give a definition of the k-dimensional volume for k-dimensional mani- folds which does not require those dreaded “partitions of unity”. Suppose ak-dimensional manifold M⊆Rnis covered by a single coordinate chartα:U→M,U⊆Rkopen. Let D αdenote the n-by-kmatrix Dα=/bracketleftbiggdα dt1dα dt2...dα dtk/bracketrightbigg . (More precisely, each vectordα dtiis represented by a column vector in the standard basis of Rn. Actually our definition works using any orthonormal basis also.) Define, for any vectors v1,...,v k∈Rn(again represented in an orthnormal basis): V(v1,...,v k) =/radicalBig det/bracketleftbigv1v2... v k/bracketrightbigtr/bracketleftbigv1v2... v k/bracketrightbig =/radicalBig det [vi·vj]i,j=1,...,k, This is the k-dimensional volume of a k-dimensional parallelopiped spanned by the vectors v1,...,v kinRn. One easily shows that this volume is invariant under orthogonal transformations, and that it agrees with the usual k-dimensional volume (as defined by the Lebesgue measure) when the parallelopiped lies in the subspace Rk×0⊆Rn. Thek-dimensional volume of any E∈ B(M) is defined as: ν(E) =/integraldisplay α−1(E)V(Dα)dλ. (Sinceαis continuous, α−1(E)∈ B(Rk).) 47 The integrand, of course, is supposed to represent “infinitesimal” elements of surface area ( k-dimensional volume), or approximations of the surface area of E by polygons that are “close” to E. As indicated by the quotation marks, these assertions about “surface area” are completely non-rigorous, and we won’t be- labour to prove them, since the equation above isour definition of k-dimensional volume. But it should be pointed out that there are better theories of k- dimensional volume available, which are intrinsic to the sets being measured, instead of our computational theory. (I don’t know these other theories well enough though.) Back to our definitions. If Mis not covered by a single coordinate chart, but more than one, say αi:Ui→M,i= 1,2,..., then partition Mwith V1=α1(U1),Vi=αi(Ui)\Vi−1, and define ν(E) =/summationdisplay i/integraldisplay α−1 i(E∩Vi)V(Dα)dλ. It is left as an exercise to show that ν(E) is well-defined: it is independent of the coordinate charts αiused forM. Finally, the scalar integral of f:M→RoverMis simply /integraldisplay Mfdν. And the integral of a differential form ωon an oriented manifold Mis /integraldisplay p∈Mω/parenleftbig p;T(p)/parenrightbig dν, whereT(p) is an orthonormal frame of the tangent space of Matp, oriented according to the given orientation of M. (If you don’t know what I’m talking about, just ignore this definition — essentially it generalizes the line and surface integrals in calculus.) Again it is not hard to show that the formulae I have given are exactly equivalent to the classical ones for evaluating scalar integrals and integrals of differential forms, which are of course needed for actual computations. But there are several advantages to our new definitions. First is that they are elegant: they are mostly coordinate-free, and all the different integrals studied in calculus have been unified to the Lebesgue integral by employing different measures. In turn, this means that the nice properties and convergence theorems we have proven all carry over to integrals on manifolds. For example, everybody “knows” that on a sphere, any circular arc Chas “measure zero”, and so may be ignored when integrating over the sphere. To prove this rigorously using our definitions, we only have to remark that ν(C) = 0, sinceλ(α−1(C)) = 0 for a coordinate chart αfor the sphere. Stieltjes measure The definition of the Stieltjes measure is best motivated by probability theory. Suppose we have a random variable Zwith distribution µ, and the cumulative 48 distribution function F:R→[0,1] — by definition, they satisfy F(z) = Pr[Z≤ z] =µ/parenleftbig [−∞,z]/parenrightbig . It follows that Fis (non-strict) increasing, and µ((a,b]) = F(b)−F(a). The idea here is to try to reverse this procedure: given any increasing func- tionF, can we construct a measure µonB(R) that assigns, to any interval (a,b], a “length” of F(b)−F(a)? Actually we will need to impose some conditions on Ffirst. Since Fis in- creasing, it always has only a countable number of discontinuities, and these discontinuities must all be jump discontinuities. At these jumps, we will insist thatFis right-continuous, i.e. lim x/arrowsoutheastaF(x) =F(a). Otherwise, taking count- able limits may fail: for example, if at the point a,Fis left-continuous instead of right-continuous, then µ/parenleftbig (a,b]/parenrightbig =F(b)−F(a) /negationslash= lim x/arrowsoutheasta(F(b)−F(x)) = lim x/arrowsoutheastaµ/parenleftbig (a,x]/parenrightbig . (Of course, the preference of “right” over “left” comes from our convention that we used intervals ( a,b] that are open on the left and closed on the right.) We will also insist that F(x)<∞forx/negationslash=−∞,+∞, so that the subtraction F(b)−F(a) makes sense. However, it should be allowed that, say, F(−∞) =−∞ (andF(x) does not have to be in [0 ,1] either). This allows sigma-finite measures to be constructed. With the necessary conditions now stated, we can begin the construction of µ, which is not much different from the construction of the Lebesgue measure onR— not surprising, since F(x) =xis exactly the Lebesgue measure on R. First, it is easily checked, by drawing pictures of the intervals ( a,b], that they actually form a semi-algebra on R. (Just ignore the point −∞ for now.) The measure µon the generated algebra Ais defined in the obvious way, and it follows from the same arguments as in Section 9thatµis finitely additive. Countable additivity requires the typical approximation arguments. Suppose that we have Jn∈ A with infinite disjoint union I∈ A. By monotonicity we automatically have/summationtext∞ n=1µ(Jn)≤µ(I), so we only have to prove the other inequality. We can assume that Jnare simple intervals ( an,bn], instead of finite disjoint unions of intervals. Also assume, for now, that Iis the finite interval (a,b]. SinceFis right-continuous, for every ε>0, there exists δ>0 such that 0≤F(b)−F(a)<F(b)−F(a+δ) +ε. Also there exists δn>0 such that 0≤F(bn+δn)−F(an)<F(bn)−F(an) +ε 2n. There exists a finite set n1,...,n ksuch that (a+δ,b]⊆/uniontextk i=1(ani,bni+δni], 49 since the open sets ( an,bn+δ) cover the compact set [ a+δ,b]. Therefore, F(b)−F(a+δ)≤k/summationdisplay i=1F(bni+δni)−F(ani) ≤∞/summationdisplay n=1F(bn+δn)−F(an), F(b)−F(a)≤2ε+∞/summationdisplay n=1F(bn)−F(an). and we take ε→0. Finally, the proof for general I=I1/unionmulti ··· /unionmultiIk∈ A just follows from finite additivity and that the finite sum of limits equals the limit of the finite sum. Infinite intervals are handled in the same way as in Section 9. Thus using the theorems of Section 8,µcan thereby be extended to a measure onB(R). This is called the Stieltjes measure onR, and the integral /integraldisplay Rgdµ =/integraldisplay RgdF is the Stieltjes integral, and it generalizes the Riemann-Stieltjes integral that is sometimes studied in real analysis courses. (The Riemann-Stieltjes integral is defined by taking limits of Riemann-like sums/summationtext ig(ξi)·(F(xi)−F(xi−1)). Showing this limit exists requires some effort, however.) Of course, if Fis differentiable, the Stieltjes integral just reduces to /integraldisplay RgdF =/integraldisplay Rg·F/primedλ. Lastly, we should mention that if we admit signed measures , which are dif- ferences of two (positive) measures, then the condition that Fbe increasing can even be relaxed. We will not pursue that theory here though. 16 Egorov’s Theorem The following theorem does not really belong in a first course, but it is quite a surprising and interesting result, and I want to record its proof. Theorem 16.1 (Egorov). Let(X,µ)be a measure space of finite measure, and fn:X→Rbe a sequence of measurable functions convergent almost everywhere tof. Then given any ε>0, there exists a measurable subset A⊆Xsuch that µ(X\A)<εand the sequence fnconverges uniformly to fonA. Proof. First define Bn,m=∞/intersectiondisplay k=n/bracketleftBig |f−fk|<1 m/bracketrightBig . 50 Fixm. For mostx∈X,fn(x) converges to f(x), so there exists nsuch that |fk(x)−f(x)|<1/mfor allk≥n, sox∈Bn,m. Thus we see {Bn,m}n/arrownortheastX\C (Cis some set of measure zero). We construct the set Ainductively as follows. Set A0=X\C. For each m > 0, since {Am−1∩Bn,m}n/arrownortheastAm−1, we haveµ(Am−1\Bn,m)→0, so we can choose n(m) such that µ(Am−1\Bn(m),m)<ε 2m. Furthermore set Am=Am−1∩Bn(m),m. SinceAm/unionmulti(Am−1\Bn(m),m) =Am−1, we have µ(Am)>µ(Am−1)−ε 2m >µ(X)−ε 2−ε 4− ··· −ε 2m≥µ(X)−ε. The setsAmare decreasing, so letting A=∞/intersectiondisplay m=1Am=∞/intersectiondisplay m=1Bn(m),m, we haveµ(A)≥µ(X)−ε, orµ(X\A)≤ε. Finally, for x∈A,x∈Bn(m),mfor allm, showing that |f(x)−fk(x)|<1/mwheneverk≥n(m). This condition is uniform for all x∈A. /square 17 Exercises I have been suggested to provide some more exercises to this text. Here they are. 1.InRnwith Lebesgue measure, find an uncountable set of measure zero. 2.Show that if f:Rn→Ris continuous and equal to zero almost everywhere, thenfis in fact equal to zero everywhere. 3.Find a sequence of integrable functions fnsuch thatfn(x)→0 for every xbut/integraltext fn→ ∞ . 4.Letf∈L1(Rn), 1≤p<∞. Compute the limits lim h→0/integraldisplay |f(x+h)−f(x)|pdx, lim /bardblh/bardbl→∞/integraldisplay |f(x+h)−f(x)|pdx. 5.Letf∈L1(Rn). Show that lim /bardbly/bardbl→∞/integraldisplay Rnf(x)ei/angbracketleftx,y/angbracketrightdx= 0. 51 6.Letfn∈Lp,1<p< ∞be a sequence of functions converging to f∈Lp almost everywhere, and suppose there is a constant Msuch that /bardblfn/bardblp≤ Mfor alln. Then for each g∈Lq, /integraldisplay fg= lim n→∞/integraldisplay fng. Hint: Use Egovov’s Theorem and a density argument. Is this also true for p= 1? 7.Letfn∈Lp,1≤p <∞be a sequence of functions converging f∈Lep almost everywhere. Prove that fnconverges to fin theLepnorm if and only if /bardblfn/bardblp→ /bardblf/bardblp. 8.Letf∈Lp(R),g∈Lq(R), with 1 ≤p,q≤ ∞ . Show that the function F(x) =/integraltextx 0f(t)dtis defined and continuous for all x∈R, and that the functionh(x) = (|x|+ 1)−aF(x)g(x) is in L1(R), the constant abeing larger than 2 −1 p−1 q. 9.Letf∈L1(R). Show that the series ∞/summationdisplay n=11√nf(x√n) is convergent for almost all x∈R. 10.Letfbe in L1(R), andgbe a continuous periodic function with period 1. Show that lim n→∞/integraldisplay∞ −∞f(x)g(nx)dx=/integraldisplay∞ −∞f(x)dx/integraldisplay1 0g(y)dy. 18 Bibliography The following outlines the prerequesites for this article. (Although I’m not suggesting that you must first know everything here before you read this article; you could be learning these as you go along.) First, you need a respectable first-year calculus course, dealing with limits rigorously. The course I took used [ Spivak1 ], possibly the best math book ever. You probably should be at least somewhat familiar with multi-dimensional calculus, if only to have a motivation for the theorems we prove (e.g. Fu- bini’s Theorem, Change of Variables). I learned multi-dimensional calculus from [ Spivak2 ] and [ Munkres ]. As you’d expect, these are theoretical books, and not very practical, but we will need a few elementary results that these books prove. Point-set topology is also introduced in the study of multi-dimensional cal- culus. We will not need a deep understanding of that subject here, but just the 52 basic definitions and facts about open sets, closed sets, compact sets, continous maps between topological spaces, and metric spaces. I don’t have particular references for these, as it has become popular to learn topology with Moore’s method (as I have done), where you are given lists of theorems that you are supposed to prove alone. The last book, [ Rosenthal ], (not a prerequesite) is what I mostly referred to while writing up Section 8. It contains applications to probability of the abstract measure stuff we do here, and it is not overly abstract. I recommend it, and it’s cheap too. I don’t mention any of the standard real analysis or measure theory books here, since I don’t have them handy, and this text is supposed to supplant a fair portion of these books anyway. But surely you can find references elsewhere. References [Spivak1] Michael Spivak, Calculus (3rd ed.). Publish or Perish, 1994; ISBN 0-914098-89-6. [Spivak2] Michael Spivak, Calculus on Manifolds . Perseus, 1965; ISBN 0- 8053-9021-9. [Munkres] James R. Munkres, Analysis on Manifolds . Westview Press, 1991; ISBN 0-201-51035-9. [Rosenthal] Jeffrey S. Rosenthal, A First Look at Rigorous Probability Theory . World Scientific, 2000; ISBN 981-02-4303-0. 53