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REAL ANALYSIS I Spring 2016 Product Measures REAL ANALYSIS I Spring 2016 Product Measures We assume that (X; M; µ), (Y; N ; ν) are σ- finite measure spaces. We want to provide the Cartesian product X ×Y with a measure space structure in which all sets of the form A×B with A 2 M, B 2 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 = fA × B : A 2 M;B 2 N g: The elements of E are will be called measurable rectangles. Lemma 0.1 E is a semi-ring in X × Y containing X × Y . Proof. Everything is immediate, one just has to remember what has to be proved. The following rather immediate formulas and facts do the job. • X × Y 2 E. • (A × B) \ (C × D) = (A \ C) × (B \ D). • (X × Y )n(A × B) = ((XnA) × Y ) [ (A × (Y nB)). Because E is a semi-ring, we know that finite unions of elements of E constitute a ring; since X × Y 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 = f Ei : E1;:::;En 2 Eg i=1 [n = f (Ai × Bi): A1;:::;An 2 M;B1;:::;Bn 2 N g 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 2 M;B 2 N ; with the proviso that if one of µ(A), µ(B) is 0, the other one 1, then µ ⊗ ν(A × B) = 0. This defines µ ⊗ ν : E! [0; 1]. 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 0.2 Let fAng be a sequence of sets in M, fBng be a sequence of sets in N ; assume (An × Bn) \ (Am × Bm) = ; if n =6 m and [1 (An × Bn) = A × B n=1 1 for some A 2 M, B 2 M. Then X1 µ ⊗ ν(A × B) = µ ⊗ ν(An × Bn); n=1 that is, X1 µ(A)ν(B) = µ(An)ν(Bn): n=1 Proof. Let y 2 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 0 = 0 + 0 + ··· , so we can assume B not empty). We now single out all \rectangles" An × Bn such that y 2 Bn, which is done by singling out the indices n for which this happens. We define P (y) = fn 2 N : y 2 Bng: The set P (y) is, of course, not empty (once we assume B =6 ;). We now claim: 1. If n; m 2 P (y), n =6 m, then An \ Am = ;. 2. [ A = An: n2P (y) In fact, let x 2 An \ Am, n; m 2 P (y). Then (x; y) 2 An × Bn and (x; y) 2 Am × Bm, hence n = m. This takes care of 1. Let x 2 A. Then (x; y) 2 A × B and hence there exists n 2 N with (x; y) 2 An × Bn; i.e., x 2 An, y 2 Bn. By definition of P (y) we have n 2 P (y). This takes care of 2; the claim is established. By the claim, we have X µ(A) = µ(An): n2P (y) 2 2 Now n P (y) if and only if y Bn, if and only if χBn (y) = 1; we canthus write the expression we found for µ(A) in the form X1 µ(A) = µ(An)χBn (y): n=1 Finally, multiplying by χB(y) = 1 we get X1 µ(A)χB(y) = µ(An)χBn (y): n=1 This last equality is true not only for y 2 B but also, quite trivially, for y 2 Y nB (where all terms are 0). In other words, it holds for all y 2 Y and we can integrate with respect to ν to get (by Beppo-Levi's or Lebesgue's monotone convergence theorem), Z X1 Z µ(A) χB dν = µ(An) χBn dν; Y n=1 Y i.e., X1 µ(A)ν(B) = µ(An)ν(Bn) dν: n=1 2 To complete the proof that µ ⊕ ν is a premeasure on E it suffices toS see that E E ⊂ 1 it is countably monotone on ; i.e, if R; R1;R2;::: are in and R n=1 Rn, then X1 (1) µ ⊕ ν(R) ≤ µ ⊕ ν(Rn): n=1 But this involves some technicalities. It would be nice to be able to say now: f g1 We may assume that Rn n=1 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 fRng 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 2 A we can write [N E = (An × Bn) n=1 where An 2 M, Bn 2 N for 1 ≤ n ≤ N, and (An × Bn) \ (Ak × Bk) = ; if n =6 k. Define XN (2) µ ⊗ ν(E) = µ(An)ν(Bn) n=1 We have to see, of course, that this is a valid definition; that it does not de- pend on the choid=ce of representation for E. So assume we have a second representation [M E = (Cm × Dm); m=1 where Cm 2 M, Dm 2 N for 1 ≤ m ≤ M, and (Cm × Dm) \ (Cj × Dj) = ; if m =6 j. We can then form a third such representation by just intersecting sets from one family with sets from the other. We have [M An × Bn ⊂ Cm × Dm; m=1 hence [M [M An × Bn = (An × Bn) \ (Cm × Dm) = (An \ Cm) × (Bn \ Dm): m=1 m=1 f \ × \ gM The sets (An Cm) (Bn Dm) m=1 are pairwise disjoint and we can apply Lemma 0.2 (if necessary add empty sets to this family so as to make a sequence of pairwise disjoint sets out of it) to get XM µ(An × Bn) = µ(An \ Cm)ν(Bn \ Dm): m=1 Switching the roles of the families, we get similarly XN µ(Cm × Dm) = µ(An \ Cm)ν(Bn \ Dm): n=1 3 It follows that XN XN XM µ(An × Bn) = µ(An \ Cm)ν(Bn \ Dm) n=1 n=1 m=1 XM XN XM = µ(An \ Cm)ν(Bn \ Dm) = µ(Cm)ν(Dm): m=1 n=1 m=1 This shows that µ ⊗ ν(E) is well defined on A. Clearly it coincides with the previous definition if E 2 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 0.3 µ ⊗ ν is σ-additive in A. S 2 A 1 \ ; 6 Proof. Let E; E1;E2;::: , E = n=1 En, En Em = if n = m. Then X1 µ ⊗ ν(E) = µ ⊗ ν(En): n=1 So assume E; E1;E2;::: as stated. We can write [K E = Ak × Bk k=1 0 where A1;:::AK 2 M, B1;:::;BK 2 N ,(Ak × Bk) \ (Ak0 × Bk0 ) = ; if k =6 k . Similarly, for each n 2 N we can write [Jn En = An;j × Bn;j j=1 2 M 2 N × \ 0 × 0 where An;1;:::An;Jn , Bn;1;:::;Bn;Jn ,(An;j Bn;j) (An;j Bn;j ) = 0 ; if j =6 j . We now have, since Ak × Bk = (Ak × Bk) \ E, [1 [Jn Ak × Bk = (Ak \ An;j) × (Bk \ Bn;j); n=1 j=1 for k = 1; 2;:::;K. The family of sets f(A \ A ) × (B \ B )g k n;j k n;j n2N;1≤j≤Jn consists of pairwise disjoint sets, so that by Lemma 0.2 we get X1 XJn µ(Ak)ν(Bk) = µ(Ak \ An;j)ν(Bk \ Bn;j); n=1 j=1 for k = 1;:::;K. On the other hand, we also have for n 2 N, 1 ≤ j ≤ Jn, An;j × Bn;j = (An;j × Bn;j) \ E, hence [K An;j × Bn;j = (Ak \ An;j) × (Bk \ Bn;j): k=1 4 Applying again Lemma 0.2 (any two sets in the union of the right hand side being disjoint), we get XK µ(An;j)ν(Bn;j) = µ(Ak \ An;j)ν(Bk \ Bn;j): k=1 Putting it all together: XK XK X1 XJn µ ⊗ ν(E) = µ(Ak)ν(Bk) = µ(Ak \ An;j)ν(Bk \ Bn;j) k=1 k=1 n=1 j=1 X1 XJn XK = µ(Ak \ An;j)ν(Bk \ Bn;j) n=1 j=1 k=1 X1 XJn X1 = µ(An;j)ν(Bn;j) = µ ⊗ ν(En): n=1 j=1 n=1 Thanks to our extension theory, we are basically done.
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