REAL ANALYSIS I Spring 2016 Product Measures

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1 REAL ANALSIS I Spring 216 Product Measures We assume that (, M, µ), (, N, ν) are σ- finite measure spaces. We want to provide the Cartesian product with a measure space structure in which all sets of the form A B with A M, B N are measurable and have measure µ(a)ν(b). We then want to see how you integrate with respect to this product measure. Finally, we discuss the case of more than two factors. We begin with some notation. We will denote (in this section) by E the family of all cartesian products of sets of M by sets of N. That is, E = {A B : A M, B N }. The elements of E are will be called measurable rectangles. Lemma.1 E is a semi-ring in containing. Proof. Everything is immediate, one just has to remember what has to be proved. The following rather immediate formulas and facts do the job. E. (A B) (C D) = (A C) (B D). ( )\(A B) = ((\A) ) (A ( \B)). Because E is a semi-ring, we know that finite unions of elements of E constitute a ring; since is contained in E it is an algebra. Every element of this algebra can be written as a union of a pairwise disjoint finite number of elements of E. Let us call this algebra A; n A = { E i : E 1,..., E n E} i=1 n = { (A i B i ) : A 1,..., A n M, B 1,..., B n N } i=1 It is time to begin defining the product measure. It is denoted by µ ν. We begin defining it on E and then extend it step by step to a σ-algebra of measurable sets. We will keep on using the same symbol for the (presumed or putative) measure as we extend it, but one should remember that at each extension, one has to verify that the new definition still assigns the same measure to the sets on which the measure had been defined so far. Our first definition is µ ν(a B) = µ(a)ν(b) if A M, B N, with the proviso that if one of µ(a), µ(b) is, the other one, then µ ν(a B) =. This defines µ ν : E [, ]. We now need to prove that we have a pre-measure on E; proving the next lemma could be the hardest step, even if the result is intuitively obvious. Lemma.2 Let {A n } be a sequence of sets in M, {B n } be a sequence of sets in N ; assume (A n B n ) (A m B m ) = if n m and (A n B n ) = A B 1

2 for some A M, B M. Then that is, µ ν(a B) = µ ν(a n B n ); µ(a)ν(b) = µ(a n )ν(b n ). Proof. Let y B (This is a statement that normally has to be preceded with an argument explaining why B is not empty; in our case, if B or A- -is empty one can see that the lemma reduces to = + +, so we can assume B not empty). We now single out all rectangles A n B n such that y B n, which is done by singling out the indices n for which this happens. We define P (y) = {n N : y B n }. The set P (y) is, of course, not empty (once we assume B ). We now claim: 1. If n, m P (y), n m, then A n A m =. 2. A = A n. n P (y) In fact, let x A n A m, n, m P (y). Then (x, y) A n B n and (x, y) A m B m, hence n = m. This takes care of 1. Let x A. Then (x, y) A B and hence there exists n N with (x, y) A n B n ; i.e., x A n, y B n. By definition of P (y) we have n P (y). This takes care of 2; the claim is established. By the claim, we have µ(a) = µ(a n ). n P (y) Now n P (y) if and only if y B n, if and only if χ Bn (y) = 1; we canthus write the expression we found for µ(a) in the form µ(a) = µ(a n )χ Bn (y). Finally, multiplying by χ B (y) = 1 we get µ(a)χ B (y) = µ(a n )χ Bn (y). This last equality is true not only for y B but also, quite trivially, for y \B (where all terms are ). In other words, it holds for all y and we can integrate with respect to ν to get (by Beppo-Levi s or Lebesgue s monotone convergence theorem), i.e., µ(a) χ B dν = µ(a)ν(b) = µ(a n ) χ Bn dν; µ(a n )ν(b n ) dν. 2

3 To complete the proof that µ ν is a premeasure on E it suffices to see that it is countably monotone on E; i.e, if R, R 1, R 2,... are in E and R R n, then (1) µ ν(r) µ ν(r n ). But this involves some technicalities. It would be nice to be able to say now: We may assume that {R n } is a disjoint sequence of measurable rectangles. Possibly one can say this, but I think however messy the technicalities to come are, actually proving that one can assume one can replace the measurable rectangles {R n } by a disjoint sequence of such rectangles without increasing the sum on the right hand side of (1) seems to require even more contortions. I could be wrong. I will now repeat part of the proof of the extension of measure theorem I had in the notes on extension of measures. We begin extending µ ν to the algebra A. If E A we can write E = N (A n B n ) where A n M, B n N for 1 n N, and (A n B n ) (A k B k ) = if n k. Define (2) µ ν(e) = N µ(a n )ν(b n ) We have to see, of course, that this is a valid definition; that it does not depend on the choid=ce of representation for E. So assume we have a second representation M E = (C m D m ), m=1 where C m M, D m N for 1 m M, and (C m D m ) (C j D j ) = if m j. We can then form a third such representation by just intersecting sets from one family with sets from the other. We have hence A n B n = A n B n M m=1 M (A n B n ) (C m D m ) = m=1 C m D m, M (A n C m ) (B n D m ). The sets {(A n C m ) (B n D m )} M m=1 are pairwise disjoint and we can apply Lemma.2 (if necessary add empty sets to this family so as to make a sequence of pairwise disjoint sets out of it) to get µ(a n B n ) = m=1 M µ(a n C m )ν(b n D m ). m=1 Switching the roles of the families, we get similarly µ(c m D m ) = N µ(a n C m )ν(b n D m ). 3

4 It follows that N µ(a n B n ) = = N M µ(a n C m )ν(b n D m ) m=1 M N µ(a n C m )ν(b n D m ) = m=1 m=1 M µ(c m )ν(d m ). This shows that µ ν(e) is well defined on A. Clearly it coincides with the previous definition if E E. It is now easy to prove that µ ν is σ-additive in A; the only difficulty is writing out the obvious in detail. We recall (see the notes on extension of measures) that in an algebra a finitely additive function is σ- additive if and only if it is countably monotone. Anyway, here goes σ-additivity. Let s call it a Lemma. Lemma.3 µ ν is σ-additive in A. Proof. Let E, E 1, E 2,... A, E = E n, E n E m = if n m. Then µ ν(e) = µ ν(e n ). So assume E, E 1, E 2,... as stated. We can write E = K A k B k k=1 where A 1,... A K M, B 1,..., B K N, (A k B k ) (A k B k ) = if k k. Similarly, for each n N we can write E n = J n j=1 A n,j B n,j where A n,1,... A n,jn M, B n,1,..., B n,jn N, (A n,j B n,j ) (A n,j B n,j ) = if j j. We now have, since A k B k = (A k B k ) E, A k B k = J n j=1 for k = 1, 2,..., K. The family of sets (A k A n,j ) (B k B n,j ), {(A k A n,j ) (B k B n,j )} n N,1 j Jn consists of pairwise disjoint sets, so that by Lemma.2 we get µ(a k )ν(b k ) = J n j=1 µ(a k A n,j )ν(b k B n,j ), for k = 1,..., K. On the other hand, we also have for n N, 1 j J n, A n,j B n,j = (A n,j B n,j ) E, hence A n,j B n,j = K (A k A n,j ) (B k B n,j ). k=1 4

5 Applying again Lemma.2 (any two sets in the union of the right hand side being disjoint), we get Putting it all together: µ(a n,j )ν(b n,j ) = K µ(a k A n,j )ν(b k B n,j ). k=1 µ ν(e) = = = K K µ(a k )ν(b k ) = J n k=1 k=1 j=1 J n j=1 k=1 J n K µ(a k A n,j )ν(b k B n,j ) µ(a n,j )ν(b n,j ) = j=1 µ(a k A n,j )ν(b k B n,j ) µ ν(e n ). Thanks to our extension theory, we are basically done. We will denote by M N the σ-algebra generated by E in (which is also the σ-algebra generated by A).. We recall that we are assuming, that µ, ν are σ-finite; then µ ν is clearly σ-finite on A. By Lemma.2 and the Caratheodory Extension Theorem we have Theorem.4 Let (, µ, ν), (, N, ν) be σ-finite measure spaces. There exists a unique measure µ ν defined on the σ-algebra M N generated by the family of measurable rectangles A B, A M, B N such that if A M, B N. µ ν(a B) = µ(a)ν(b) We will denote by M N the σ-algebra one obtains when completing the measure space (, M N, µ ν); that is, for example, E M N if and only if E = F G where F M N, G N M N, µ ν(n) =. We will say that a subset E of is measurable iff E M N. Definition 1 Let E. If x, we define E x by E x = {y : (x, y) E}. Similarly, if y, we define E y by Lemma.5 Let E M N. Then E y = {x : (x, y) E}. 1. E x N for all x, the map x ν(e x ) : [, ] is measurable and ν(e x ) dµ(x) = µ ν(e). 2. E y M for all y, the map y µ(e y ) : [, ] is measurable and µ(e y ) dν(y) = µ ν(e). 5

6 We are here once more having to prove something for all sets in a σ-algebra generated by some family of sets. In this case the σ-algebra is M N = σ(e). We know very little about the sets in this σ-algebra, so the only way to proceed is what should be by now a familiar one. We let Σ (to give it a name) be the class of all sets satisfying the property. We show E Σ. Next we try to show Σ is a σ-algebra. It turns out that the easiest way of doing this is by introducing a new class of sets, a class known as a monotone class. Hoping this whole subject will not be too boring, here is the definition. We postpone the proof of Lemma.5 for a while. Definition 2 Let be a set. A family Q of subsets of is a monotone class (in ) iff whenever {E n } is a monotone sequence of sets in Q, then lim n E n Q. (A sequence of sets {E n } is monotone if either E 1 E 2 or E 1 E 2.) Recall that if E 1 E 2, then lim n E n = E n; if E 1 E 2, then lim n E n = E n. It should be clear and trivial that σ-algebras are monotone classes. Another definition we need is: Definition 3 Let C be a family of subsets of. The monotone class generated by C is defined by mon(c) = {Q : Qis a monotone class and C Q}. In other words mon(c) is characterized by the following properties 1. mon(c) is a monotone class. 2. C mon(c). 3. If Q is a monotone class and C Q then mon(c) Q. Lemma.6 Let A be an algebra in. Then mon(a) = σ(a). Proof. Let Q = mon(a) and let Σ = σ(a). Since σ-algebras are monotone classes, it is clear that Q Σ. To get the converse inclusion we begin showing that if A Q, then A c Q and A, B Q implies A B Q. (As it turns out, that s really all that needs to be proved.) Showing that something is in this generated monotone class causes difficulties similar to proving things are in σ-algebras; similar techniques are needed. We begin showing E Q implies E c in Q. For this we introduce a new family of sets, say D = {E Q : E c Q}. Because A Q and A is an algebra, it is clear that A D. We also see that D is a monotone class; in fact let {E n } be a monotone sequence in D. It is easily verified that in this case {En} c is also monotone and ( ) c lim E n = lim n n Ec n Q. It follows that D is a monotone class containing A, hence Q D, hence D = Q. This means that if E Q, then E D, hence E c Q. To prove that A, B Q implies A B Q we define more families of sets. First of all, for an arbitrary subset of A Q set D A = {B Q : A B Q}. 6

7 It is easy to see that D A is always a monotone class (maybe consisting of the single set ). This is simply due to the fact that Q is a monotone class and if {E n } is a monotone sequence, then so is {A E n } with lim n A E n = A lim n E n. Assume now A A. Then B A implies A B A Q so that B D A. It follows that A D A ; hence also Q D A, hence Q = D A. We proved that D A = Q for all A A. Let now F = {A Q : D A = Q}. We just proved F A; let us see it is a monotone class. For this, assume {A n } is a monotone sequence of sets in F; then A n Q and D An = Q for all n. If B Q, the fact that D An = Q implies A n B Q; since {A n B} is monotone we conclude that lim n (A n B) = lim n A B Q. Since B Q is arbitrary, we just proved D limn A n = Q; i.e., lim n A n F. Thus F is a monotone class containing A, hence F = Q. It follows that if A, B Q, then D A = Q, hence B D A, hence A B Q. All that remains to be seen is that if A n Q for all n N, then n A n Q. By induction, we have m B m = A n Q for all m N; since {B m } is monotone and lim m B m = n A n, we proved n A n Q. We are ready to prove Lemma.5. Proof. Actual starting point of the proof. For simplicity, we only prove Property 1; the proof of 2. is identical. We will assume first that µ() <, ν( ) <. Let Σ consist of all sets E M N that satisfy Property 1; i.e., sets E such that E x N for all x, the map x ν(e x ) : [, ] is measurable and ν(e x ) dµ(x) = µ ν(e). Step 1.E Σ. Let E = A B, A ]sm, B N. From the immediately verified formulas { B, x A, (A B) x =, x / A. we see that if E = A B E, then E x N for all x, E y M for all y. Moreover, ν(e x ) = ν(b)χ A (x) for all x. The map x ν(e x ) is the map ν(b)χ A, clearly measurable since A is measurable and ν(e x ) dµ(x) = ν(b) χ A dµ = ν(b)µ(a) = µ ν(e). It follows that E Σ. In particular Σ. Step 2. We next show that Σ is closed under finite disjoint unions. By induction, it suffices to show that E, F Σ, E F =, implies E F Σ. So let E, F be disjoint elements of E. If x, we see that (E F ) x = E x F x, so (E F ) x N for all x. Moreover, E x F x = for all x so that ν((e F ) x ) = ν(e x ) + ν(f x ) from which measurability of the map x ν((e F ) x ) and ν((e F ) x ) dµ(x) = ν(e x ) dµ(x) + ν(f x ) dµ(x) = µ ν(e) + µ ν(f ) = µ ν(e F ) 7

8 follows. This proves that E F Σ. Step 3. A Σ. Since E Σ by Step 1, and every element of A can be written as a finite union of pairwise disjoint elements of E, this is an immediate consequence of Step 2. Step 4. Σ is closed under limits of increasing sequences of sets. Assume E n Σ for n N, E 1 E 2 E 3. We want to see that E = n E n Σ. Since E x = n (E n) x it follows that E x N for all x. From (E 1 ) x (E 2 ) x we get ν(e x ) = lim ν((e n) x ) n for all x, hence the function x ν((e) x, being the pointwise limit of the sequence of measurable functions {x ν((e n ) x )}, is measurable. Moreover, by Lebesgue s monotone convergence theorem, ν(e x ) dµ(x) = lim n ν((e n ) x ) dµ(x) = lim n µ ν(e n) = µ ν(e). This completes the proof of E Σ, so that Σ is closed under the limit of increasing sequences of sets. Step 5. Σ is closed under countable disjoint unions: Let E n Σ for n N and assume E n E m = if n m. Then E = n E n Σ. To see this, set F N = N E n for N N. By Step 2, F N Σ for all N N. In addition, {F N } is an increasing family of sets converging to n E n so the result follows FROM Step 4. Step 6. Σ is closed under limits of decreasing sequences of sets. Let E n Σ for each n N, E 1 E 2, let E = n E n. We need to see that E Σ. In fact, since E x = n (E n) x it follows that E x N for all x. We have, moreover, (E 1 ) x (E 2 ) x and because we are assuming ν is a finite measure, we can affirm that ν(e x ) = lim n ν((e n ) x ), hence x ν(e x ) is measurable on. Since ν((e n ) x ) ν( ) < for all n N, x, and x ν( ) is in L 1 (µ) (constant functions being integrable if the space has finite measure), the dominated convergence theorem now gives ν(e x ) dµ(x) = lim ν((e n ) x ) dµ(x) = lim µ ν(e n). n n Because we are assuming that µ ν is a finite measure, we have that lim n µ ν(e n ) = µ ν(e) and we proved that µ ν(e) = ν(e x ) dµ(x). It follows that E Σ. This concludes the proof that Σ is a monotone class, under the assumption that µ() <, ν( ) <. Thus Σ mon(a) = σ(a) = M N, hence Σ = M N, concluding the proof in the finite measure case. If all we know is that the spaces are σ-finite, we can write = n n, where n M, n N, µ( n ) <, ν( n ) < for all n N, and ( n n ) ( m m ) = if n m. If E M N, let E n = E ( n n ). Then E n satisfies all the statements of the Lemma and, as is fairly (or even very) easy to see, so does E. 8

9 Definition 4 If f : S, S some set, then for every x we define f x : S by f x (y) = f(x, y). For y we define f y : S by F y (x) = f(x, y). Theorem.7 (Tonelli) Let f : [, ] be measurable with respect to the σ-algebra M N. Then 1. f x is measurable with respect to N for every x and f y is measurable with respect to M for every y. 2. The maps x f x dν and y f y dµ are measurable with respect to M and N, respectively and ( ) ( ) f x dν dµ(x) = f dµ ν = f y dµ dν(y). Proof. We use a standard approach; to prove something holds for non-negative measurable functions, we first prove it for characteristic functions of measurable sets, then extend the result to non-negative, measurable simple functions by linearity, finally use the fact that non-negative measurable functions are limits of increasing sequences of measurable simple functions to establish the result for non-negative measurable functions. This procedure does not always work; in this case it does. So here we go. Let T denote the set of all M N measurable functions f : [, ] for which the conclusions of the theorem are true. Step 1. Assume f = χ E, E M N. Since (χ E ) x = χ Ex, (χ E ) y = χ E y, the theorem reduces to Lemma.5 in this case. Thus χ E T for all E M N. Step 2. If f, g T and a, b [, ), then af + bg T. Because (af + bg) x = af x + bg x, (af + bg) y = af y + bg y for all x, y, this is clear. Step 3. If {f n } is a sequence of functions in T and f 1 f 2 f 3, then f = lim n f n T. Once more, this is immediate. All we need to observe is that we will have (f 1 ) x (f 2 ) x, lim n (f n) x = f x to conclude that f x : [, ] is measurable for every x. Integrating we get (f 1 ) x dν (f 2 ) x dν, lim (f n ) x dν = f x dν, n the last equality being a consequence of Lebesgue s Monotone Convergence Theorem. The last equality also proves that x f x dν is measurable on and, because it is a limit of increasing, non- negative measurable functions, ( ) ( ) f x dν dµ(x) = lim (f n ) x dν dµ(x) = lim f n dµ ν n n = f dµ ν, 9

10 the last equality coming from yet another application of Lebesgue s Monotone Convergence Theorem. One deals similarly with f y. We are done; by Step 1 and 2, all non-negative measurable simple functions satisfy the conclusion of the theorem, thus by Step 3 (and the fact that nonnegative measurable functions are the limits of an increasing sequence of simple functions) all non-negative measurable functions satisfy the conclusions of the theorem. Tonelli s theorem deals only with non-negative functions. However, since every measurable function can be decomposed into combinations of non-negative ones, we get at once Theorem.8 Let f : : C be measurable with respect to M N. Then f x : C is measurable for every x, f y : C is measurable for every y. It can be of interest to remember occasionally that every theorem about integration also says something about summations; one simply assumes that the measure (or some of the measures) are counting measures. For example, if, are sets, if a x,y R, a x,y for all (x, y), if we consider, as measure spaces with counting measures (µ equal counting measure on, ν on and all subsets measurable), we get a x,y = = ( ) a x,y. x y y x (x,y) If (, M, µ) is an arbitrary measure space and for the second measure space we take (N, P(N),counting meas.), then it is easy to verify that a function F : N C is measurable if and only if the map x F (x, n) : C is measurable. Actually, that s not a too bad exercise, so let s make it into one. Exercise 1 Prove the assertion that was just made; i.e., prove that F as defined is measurable if and only if F (, n) : C is measurable for all n N. Tonelli s Theorem now implies: If F : N [, ] and F (x, ) is measurable for each n N, then F (x, n) dµ(x) = F (x, n) dµ(x). I hope it is clear that this is precisely Beppo Levi s Theorem. Of course, we would never have been able to obtain Tonelli without the aid of Beppo Levi (or its equivalent twin, Lebesgue s Monotone Theorem). Our next Theorem is Fubini s Theorem. Fubini and Tonelli go so hand in hand that they are frequently stated as one, and then called the Fubini- Tonelli Theorem. Theorem.9 (Fubini) Let (, M, µ), (, N, ν) be σ-finite measure spaces and let f L 1 (, M N, µ ν). Then 1. f x L 1 (ν) for a.e. x and the a.e. defined map x f x dν is in L 1 (µ). 2. f y L 1 (µ) for a.e. y and the a.e. defined map y f y dµ is in L 1 (ν). 1

11 3. ( ) ( ) f x dν dµ(x) = f dµ ν = f y dµ dν(y). Proof. Most of the serious work has already been done. As is usual, when working with complex valued functions, one breaks them up into real and imaginary parts, then into positive and negative parts. But a bit of care must be exercised. Our hypothesis is that f : C is measurable with respect to the σ-algebra M N and f dµ ν <. By Theorem.8, f x is measurable for every x. By Tonelli s Theorem, f x dν dµ(x) = f dµ ν so that x f x dν L 1 (µ). But a function in L 1 (µ) must be a.e. finite hence f x dν < a.e., implying f x L 1 (ν) a.e. x. Similarly we see that f y L 1 (µ) for a.e. y. Now write f = u + iv, with u, v real valued. Tonelli s theorem implies that the functions x u + x dν, x u x dν, x v x + dν, x vx dν, are measurable from to[, ). They are all bounded by x f x dν, so they are a.e. finite valued and the equality ( (3) f x dν = u + x dν u x dν + i ) v x + dν vx dν makes sense a.e. and shows that x f x dν is measurable and, in fact in L 1 (µ) since each one of the four non- negative functions of the right hand side of (3) are in L 1 (µ) by Tonelli s Theorem. Integrating both sides of (3) with respect to µ gives (once more due to Tonelli) f x dν dµ = = = u + x dν dµ u + dµ ν f dµ ν. ( u x dν dµ + i ( u dµ ν + i 11 v x + dν dµ v + dµ ν ) vx dν dµ ) v dµ ν

12 One deals with f y in a similar fashion. Before continuing, and seeing the version of Tonelli-Fubini for the completed measure space (, M N, µ ν), we should do some exercises and applications. For integrals involving Lebesgue measure it is traditional to use the Riemann notation, and we ll follow this tradition. Application. Here is a nice application, involving a well known integral. We will compute sin x x dx. Actually, x sin x/x is not in L 1 (m): Exercise 2 Prove that So what we are going to compute is sin x x R lim R dx =. sin x x There are many ways of getting this limit, one of them being by residues. In the one we will use we observe that 1 x = dx. e xy dy for all x >. This allows us to write R sin x R ( ) x dx = e xy dy sin x dx. All functions involved are continuous, so measurability is no problem. We would like to change the order of integration. By Tonelli, that is always legal if the integrand is ; unfortunately it isn t. So we need Fubini, which tells us that the interchange is legal if the integrand is integrable with respect to the product measure. To show it is integrable, we have to show the absolute value has finite integral. But once we take the absolute value, Tonelli kicks in allowing us to integrate in any order we wish. So we begin doing (,R) (, ) e xy sin x dx dy = = R R R ( e xy sin x ) dy dx ( sin x dx = R <, ) R e xy dy dx = sinx x, dx where we used the fact that sin x /x 1 for all x >. These calculations, justified by Tonelli, show that the integrand is in L 1 of the product measure. Thus, by Fubini, R ( ( ) R R sin x x dx = ) e xy dy sin x dx = e xy sin x dx We might remember that e xy sin x dx = y 2 e xy (cos x + y sin x) + C 12 dy.

13 so that We thus get R R e xy sin x dx = y 2 ( 1 e Ry (cos R + y sin R) ). sin x x dx = 1 ( 1 e Ry 1 + y 2 (cos R + y sin R) ) dy. Taking lim R, the integrand on the right hand side converges pointwise to 1/1 + y 2 ; it is also always bounded by 1/1 + y 2, so that by the Dominated Convergence Theorem we get R sin x lim R x dx = 1 dy = arctan y 1 + y2 = π 2. This result that we just derived is usually written in the form sin x x dx = π 2. Exercise 3 Let f L 1 (µ), g L 1 (ν). We define the function f g : C by f g(x, y) = f(x)g(y). Show that fotimesg L 1 (µ ν) and ( ) ( ) f g dµ ν = f dµ g dν. The connection with Lebesgue measure is important enough to deserve special mention. We will denote by B n the σ-algebra B(R n ) of Borel subsets of R n ; m n denotes Lebesgue measure in R n. We consider the two measure spaces (R n, B n, m n ) and (R k, M k, m k ), and we identify R n R k in the obvious way: If x = (x 1,..., x n ) R n, y = (y 1,..., y n ) R k we identify (x, y) with the element ξ = (ξ 1,..., ξ n+k ) given by In other words We then have ξ j = x j, j = 1,..., n; ξ j = y j n, j = n + 1,..., n + k. ((x 1,..., x n ), (y 1,..., y k )) = (x 1,..., x n, y 1,..., y k ). Theorem.1 With the identification described above, (R n R k, B n B k, m n m k ) = (R n+k, B n+k, m n+k ). Proof. The main thing that needs to be proved is that B n+k = B n B k. To see, B n+k B n B k it suffices to prove (by Lemma.5) that if E B n+k, then E x B k for all x R n (and declare that the fact that E y R n for all y R k is done the same way, or is just a matter of switching the roles of x and y, or is the same thing). To achieve this, we fix x R n and then look at all subsets E of R n+k such that E x B k. These sets constitute a σ algebra containing all open subsets of R n+k ; it follows that this σ-algebra contains B n+k. To see that B n+k B n B k requires a more multi-step approach. One way of doing it is as follows. Step 1. A R k B n+k for all A B n. This is proved in the standard way; the family of all subsets of A of R n having this property is easily seen to be a σ-algebra containing all open sets. Similarly, R n B B n+k for all B B k. 13

14 Step 2. For A R n, define M(A) = {B B k : A B B n+k }. For a general set A, all one can be sure that M(A) contains is the empty set. But, we claim: If A is open in R n, then M(A) is a σ-algebra in containing all open subsets of R k, hence containing B k. In fact, assume A is open in R n. Then A R k is open in R n+k, hence a Borel set, hence R k M(A). Let B M(A). Then A B B n+k and the equation A B c = [(A B) c ] (A R k ). Now A R k is open, hence a Borel set, in R n+k, (A B) c is the complement of a Borel set, so also a Borel set. It follows that A B c B n+k, hence B M(A). Assume now B l M(A) for all l N. Then A B l = (A B l ) l=1 shows that l=1 B l M(A). Thus M(A) is a σ-algebra in R k. Since A is open, A B is open, hence a Borel set, in R n+k for every open subset B of R k ; thus M(A) contains all open subsets of R k. The claim is established. Step 3. Let N = {A B n :, M(A) B k }. We claim that N is a σ-algebra in R n containing all open sets. In fact, N contains all open sets by Step 2. In particular, R n, N. Assume A N. To see that A c N, we have to prove: If B B k, then A c B B n+k. This time we use A c B = (A B) c (R n B); A B is a Borel set because B B k M(A), hence (A B) c is a Borel set; R n B is a Borel set by Step 1. It follows that A c B is a Borel set. Thus A c N if A N. Finally, if M(A l ) B k for all l N, let B B k. Then A l B B n+k for every l N, hence ( ) ( ) A l B = A l B B n+k. l=1 l=1 It follows that l=1 A l N. This shows that N is a σ-algebra in R n and establishes the claim. By Step 3, B n N. Thus A B n implies A N which implies A B B n+k for all B B k Thus B n+k contains all measurable rectangles, hence also the σ-algebra B n B k generated by them. Having proved B n+k = B n B k, consider the set E n+k consisting of all sets of the form n+k j=1 (a j, b j ]; a j, b j R, a j b j for j = 1,..., n + j. We see that m n m k and m n+k coincide on E n+k. By the uniqueness of Lebesgue measure, we proved m n m k = m n+k. Let us return to the theory. One problem with this theory so far is posed by the null sets. Suppose, for example, that there is A M, A, and µ(a) =, and assume there is B such that B / N. Then A B / M N. In fact, let x A; then (A B) x = B / N, and our assertion follows from Lemma.5. On the other hand A B A M N and µ ν(a ) = µ(a)ν( ) =. It follows that if (, M, µ) contains non-empty null sets (as happens for Lebesgue measure), and contains non-measurable sets (as happens in the Lebesgue case), then (, M N, µ ν) is not a complete measure space. Since l=1 14

15 working in non-complete measure spaces tends to be somewhat of an annoyance, one usually works in the completion (, M N, µ ν). This means that we must have the analogue of the Fubini-Tonelli theorems in this case. But first, let us do a warm-up exercise, something we could have done long ago. We formulate it as a lemma, so we can refer to it later on. Lemma.11 Let (, M, mu) be a measure space and let (, M, µ) be its completion. Let f : C. Show that f is measurable with respect to M if and only if there exists g : C such that g is measurable with respect to M and g = f a.e. (to be precise:{g f} is a null set in M). Exercise 4 Prove Lemma.11 Hints. One direction should be trivial: If f = g a.e. and g is M-measurable, then f is M-measurable. For the converse direction, one can reduce this to the case f : [, ]. Then one does the familiar division into cases; case 1, f = χ E, E M. Case 2, f is simple M-measurable. Case 3, f is the limit of an increasing sequence of simple Mmeasurable functions. Back to our product spaces. From now on, and until further notice, if E, we will say E is measurable if and only if it is an element of M N. If E M N then E is, of course, measurable in this sense. But if we want to emphasize that it is in the smaller σ-algebra, we ll call it M N -measurable. As usual, a subset of (of ) will be called measurable if and only if it is in M (in N ). We go directly to the main results. Theorem.12 Let f : C be measurable. 1. f x : C is measurable for a.e. x and f y : C is measurable for almost every y. 2. (Tonelli) If f is real valued, non-negative, then the almost everywhere defined functions x f x dν : [, ], y f y dµ : [, ], are measurable and ( ) f x dν dµ(x) = f dµ ν = ( ) f y dµ dν(y). 3. If f L 1 (µ ν), then f x L 1 (ν) for a.e. x, f y L 1 (µ) for a.e. y, and the a.e. defined maps x f x dν : [, ], y f y dµ : [, ], are in L 1 (µ), L 1 (ν), respectively; moreover ( ) f x dν dµ(x) = f dµ ν = ( ) f y dµ dν(y). 15

16 Proof. By Lemma.11 there is a M N -measurable function g such that g = f a.e. This means that the set on which g and f differ is contained in a null set of M N ; i.e., there exists E M N such that µ ν(e) = and f(x, y) = g(x, y) if (x, y) / E. Now, by Lemma.5, = µ ν(e) = ν(e x ) dµ(x) so that ν(e x ) = a.e. Since y / E x implies (x, y) / E, we see that f x (y) = g x (y) for all y / E x so that whenever E x is a null set, we have f x = g x a.e.[ν]. This proves that f x = g x a.e. with respect to ν, for a.e. x. Similarly, f y = g y a.e. with respect to µ, for a.e. y. Since the statements are true for g instead of f, it is now easy to see that they also hold for f. Many of the results that led to the original, non-complete, version of Fubini- Tonelli are actually particular cases of Fubini or Tonelli. We did not need their analogues to prove them first to get the complete space version of Fubini- Tonelli and we could derive them now as corollaries. Or as exercises. Exercise 5 Let E be a measurable subset of. Then E x is a measurable subset of for a.e. x, E y is a measurable subset of for a.e. y. Moreover, the (almost everywhere defined) maps are measurable and x ν(e x ) : [, ], y µ(e y ) : [, ] ν(e x ) dµ(x) = µ ν(e) = Hint: Apply Tonelli s theorem to χ E. µ(e y ) dν(y). The case of several factors. The case of a finite number ( 1, M 1, µ 1 ),..., ( r, M r, µ r ) of σ-finite measure spaces can be treated similarly. One can begin with the elementary family {A 1 A r : A i M i, i = 1,..., r}, defining µ 1 µ r (A 1 A r ) = µ 1 (A 1 ) µ r (A r ), and then proceeding in an obvious fashion. Or, one can just take the products two at a time. In all events one gets a complete measure space r r ( i, M i, r i=1µ i ) i=1 i=1 in which all sets of the form r i=1 A i with A i M i for i = 1,..., r are measurable and r r r i=1µ i ( A i ) = µ(a i ). i=1 One gets a lots of Fubini-Tonelli theorems but basically they all say that the order of integration doesn t matter. A different situation comes up when one has an infinite number of factors. Finding a suitable notion of a product measure plays an important role in several areas of mathematics, most especially in Probability Theory and applications of Probability Theory to Functional Analysis. We just give a very brief description of how one proceeds. i=1 16

17 Assume ( λ, M λ, µ λ ) is a measure space for every λ in some index set Λ. There is very little hope to define a measure ν on some σ-algebra of = λ Λ λ such that ( ) ν A λ = µ λ (A λ ). λ Λ λ Λ In the first place infinite products, especially over uncountable index sets, are not as easy to define as infinite sums. So one has to explain first what is meant by λ Λ c λ when c λ [, ] for all λ Λ. There are, however, standard ways of doing this; for example here is one. Let c λ [, ] for λ Λ. Let c [, ]. We write c λ = c λ Λ if and only if for every neighborhood V of c in [, ] there exists a finite subset F of Λ such that c λ V λ F for all finite subsets F of Λ such that F F. We recall that a neighborhood of in [, ] is any set V [, ] such that there is a R with (a, ] V. A neighborhood of c (, ) is any set V such that there exists ϵ > with (c ϵ, c + ϵ) V ; a neighborhood of is any set containing an interval [, ϵ) for some ϵ >. One thing we observe with this definition is that if all c λ < and a single c λ =, then the product of all is. This is perhaps as it should be, but this makes a very special entry. For example, consider the case in which Λ = N and we have c 2 = c 3 = = 2. Then {, if c1 >, c n =, if c 1 =. n N Opposite to what happens with sums, having zero factors is sort of bad; having too many small factors is also sort of bad. Products of small factors results in even smaller objects; the more factors, the smaller. That is why one does not say a product converges to, but that it diverges to if the limit as defined above is. The possibility of having some c λ s equal to, others equal to also has to be addressed. One can, of course, define once more =. But the problems keep on mounting. While there exists a well developed theory of infinite products, which plays a big role in several areas of mathematics, keeping on this track will not lead us to a coherent and useful product measure in the product space. Just to give one example of a serious problem in the countable case (and any countable problem tends to become an unsurmountable monster of a problem when transported to the uncountable case), assume that Λ = N and ( λ, M λ, µ λ ) = (R, L, m) for la = 1, 2,.... We take A λ = [, 1/λ]. Then, for any n N, if F is a finite subset of N which contains {1, 2,..., n}, we have m(a λ ) 1 n! λ F so that λ N m(a λ) diverges to. Should n [, 1 j ] j=1 17

18 be a null set in R N? Notice also that if we take A λ = [, 1] for all λ N, we get the product set to have measure 1; if we take A λ = [, 1 + ϵ], with ϵ >, it has infinite measure. In view of these and other problems, one follows a different tack. One assumes first that all the spaces ( λ, M λ, µ λ ) are probability spaces; i.e., one assumes µ λ ( λ ) = 1 for each λ Λ. One now defines the family E of measurable rectangles, frequently called now elementary measurable cylinders by: a set A is in E iff A = λ Λ A λ where A λ M λ for every λ Λ and A l a = λ for all but a finite number of λ s in La. One can then define ν(a) = λ Λ µ λ (A λ ) and since all but a finite number of the factors on the left hand side of the equation defining ν(a) are equal to 1, the product is really a finite product. All problems have been removed. It is fairly easy to see that E is a semi-ring, so that what needs to be done next is obvious. One has to first extend ν to the algebra of finite unions of elements of E; see that it is well defined on that algebra and that it is a measure. Then Caratheodory s extension theory takes over. The main technical difficult (and it is not major) is the analogue of Lemma.2. A cylinder set is a subset A of which in all but a finite number of directions coincides with the whole space. A formal definition is that there exists a finite subset F of Λ, say F = {λ 1,..., λ n } and a subset B n j=1 λ j such that x = (x λ ) λ Λ A if and only if (x λ1,..., x λn ) B. (B is like the base of the cylinder.) Somewhat incorrectly, we could write A = B λ/ F λ to indicate this. It should be clear that the product measure ν (assuming it constructed) coincides with µ λ1 µ λn on the family of all cylinder sets A = B λ/ F λ, F = {λ 1,..., λ n }, B M λ1 M λn. A very important example of this construction is the case in which λ = R, M λ = L for each λ Λ and µ λ = γ is Gaussian measure for each λ Λ, defined by 2 γ(a) = e x2 /2 dx π A for Lebesgue subsets A of R. 18