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Measures on Product σ-Algebras
5 Measures on Product -Algebras This chapter is devoted to the construction of measures on product σ -algebras. In Section 15 we construct the product of a finite number of σ -finite measures. The first part of Section 16 is the infinite version of Section 15, and in the second part of Section 16 we prove the Daniell-Kolmogorov extension theorem.
15
The Product of a Finite Number of Measures
In this section we construct the product of a finite number of σ -finite measures and prove versions of Fubini’s theorem. We first construct the product of two σ -finite measures. Theorem 15.1. A1 ⊗ A2 . Then
Let (X 1 , A1 ) and (X 2 , A2 ) be measurable spaces, and let A ∈
(i) A x1 ∈ A2 for any x1 ∈ X 1 and (ii) A x2 ∈ A1 for any x2 ∈ X 2 , where A xi is the section of A at xi , i = 1, 2. Proof. To prove (i), let A = {A1 × A2 : A1 ∈ A1 , A2 ∈ A2 } and M = {A ∈ A1 ⊗ A2 : A x1 ∈ A2 for any x1 ∈ X 1 }. For A1 × A2 ∈ A and x1 ∈ X 1 , in view of (1.2.30.b), we have A2 if x1 ∈ A1 , (A1 × A2 )x1 = / A1 . ∅ if x1 ∈ Thus A ⊂ M. If {An : n N } ⊂ M, then (∪n N An )x1 = ∪n N (An )x1 ∈ A2 for any x1 X 1 , and so ∪n N An ∈ M. If A ∈ M, then (Ac )x1 = (A x1 )c ∈ A2 for any x1 ∈ X 1 , and so Ac ∈ M. Therefore, M is a σ -algebra. Since A1 ⊗ A2 = σ (A) (7.17.a), we have A1 ⊗ A2 = M. This proves (i). The proof of (ii) is similar. Corollary 15.2. Let (X 1 , A1 ) and (X 2 , A2 ) be measurable spaces, and let f : X 1 × X 2 → R be an A1 ⊗ A2 -measurable function. Then (i) f x1 is A2 -measurable for any x1 ∈ X 1 Analysis and Probability. http://dx.doi.org/10.1016/B978-0-12-401665-1.00005-9 © 2013 Elsevier Inc. All rights reserved.
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and (ii) f x2 is A1 -measurable for any x2 ∈ X 2 , where f xi is the section of f at xi , i = 1, 2. Proof. If f = 1 A , where A ∈ A1 ⊗ A2 , then f xi = 1 A xi , i = 1, 2. Hence in this case (i) and (ii) follow at once from (15.1). The corollary is further proved by making use of the indicator function method. Theorem 15.3. Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be measure spaces, where μ1 and μ2 are σ -finite and let A ∈ A1 ⊗ A2 . Then the following assertions hold. (i) The function x1 → μ2 (A x1 ) from X 1 into R is A1 -measurable. (ii) The function x2 → μ1 (A x2 ) from X 2 into R is A2 -measurable. Proof. Since the proofs of (i) and (ii) are similar, we prove only (i). (I) Assume first that μ2 (X 2 ) < ∞. Let A be as in (15.1), and let D = {A ∈ A1 ⊗ A2 : assertion (i) holds}. For A1 × A2 ∈ A, we have μ2 ((A1 × A2 )x1 ) = μ2 (A2 )1 A1 (x1 ), x1 ∈ X 1 . Thus A ⊂ D. Evidently X 1 × X 2 ∈ D. If A, B ∈ D and A ⊃ B, then μ2 ((A − B)x1 ) = μ2 (A x1 − Bx1 ) = μ2 (A x1 ) − μ2 (Bx1 ), x1 ∈ X 1 , and so A − B ∈ D. If {An : n ∈ N } ⊂ D and Am ∩ An = ∅ for m = n, then μ2 ((∪n N An )x1 ) = μ2 (∪n N (An )x1 ) = n N μ2 ((An )x1 ), x1 X 1 , and so ∪n N An ∈ D. Therefore, D is a Dynkin system. Since A1 × A2 ∈ A and A 1 × A 2 ∈ A imply (A1 × A2 ) ∩ (A 1 × A 2 ) = (A1 ∩ A 1 ) × (A2 ∩ A 2 ) ∈ A, (7.36) shows that d(A) = σ (A) = A1 ⊗ A2 . Consequently, D = A1 ⊗ A2 . This proves (i). (II) Consider now the general case. Let {B j : j ∈ J } ⊂ A2 be a countable partition of X 2 such that μ2 (B j ) < ∞, j ∈ J . For A ∈ A1 ⊗ A2 , using (7.14),we have A ∩(X 1 × B j ) ∈ i X−11 ×B j (A1 ⊗A2 ) = A1 ⊗i B−1j (A2 ), j ∈ J . Thus, according to (I), the function x1 → μ2 ((A ∩ (X 1 × B j ))x1 ) is A1 -measurable for any j ∈ J . Therefore, the function x1 → μ2 (A x1 ) = μ2 (∪ j∈J (A∩(X 1 ×B j ))x1 ) = j∈J μ2 ((A∩(X 1 ×B j ))x1 ) is A1 -measurable. Corollary 15.4. Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be as in (15.3), and let f : X 1 × X 2 → [0, ∞] be an A1 ⊗A2 -measurable function. Then the following assertions hold. (i) The function x1 → X 2 f (x1 , x2 )dμ2 (x2 ) from X 1 into [0, ∞] is A1 -measurable. (ii) The function x2 → X 1 f (x1 , x2 )dμ1 (x1 ) from X 2 into [0, ∞] is A2 -measurable. Proof. If f = 1 A , where A ∈ A1 ⊗ A2 , then X 2 1 A (x1 , x2 )dμ2 (x2 )= X 2 1 A x1 (x2 ) dμ2 (x2 ) = μ2 (A x1 ), x1 ∈ X 1 , and X 1 1 A (x1 , x2 )dμ1 (x1 ) = X 1 1 A x2 (x1 )dμ1 (x1 ) = μ1 (A x2 ), x2 ∈ X 2 . Hence in this case (i) and (ii) follow immediately from (15.3). Therefore, if f is a simple function, then, using (11.6), (i) and (ii) obtain. Finally, in the X 1 ×X 2 of simple general case choose a nondecreasing sequence { f n : n ∈ N } ⊂ [0, ∞[ functions such that f n → f . Applying (11.22), we have X 2 f (x1 , x2 )dμ2 (x2 ) = limn X 2 f n (x1 , x2 )dμ2 (x2 ), x1 ∈ X 1 , and X 1 f (x1 , x2 )dμ1 (x1 ) = limn X 1 f n (x1 , x2 )dμ1 (x1 ), x2 ∈ X 2 . Hence, using (8.12.ii), assertions (i) and (ii) are proved.
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Theorem 15.5. Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be as in (15.3). For A ∈ A1 ⊗A2 , define μ(A) = X 1 μ2 (A x1 )dμ1 (x1 ) and μ (A) = X 2 μ1 (A x2 )dμ2 (x2 ). Then μ and μ are equal measures on A1 ⊗ A2 . Proof. Obviously, μ(∅) = 0. Let {An : n ∈ N } ⊂ A1 ⊗A2 be such that Am ∩ An = ∅ for m = n. Then, applying (11.23), we have μ(∪n N An ) = X 1 μ2 (∪n N (An )x1 )dμ1 (x1 ) = n∈N X 1 μ2 ((An )x1 )dμ1 (x 1 ) = n∈N μ(An ). Hence μ is a measure on in like manner. Let A be A1 ⊗ A2 . The fact that μ is a measure on A1 ⊗ A2 is proved as in (15.1). For A1 × A2 ∈ A, we obtain μ(A1 × A2 ) = X 1 μ2 ((A1 × A2 )x1 )dμ1 (x1 ) = X 1 μ2 (A2 )1 A1 dμ1 (x 1 ) = μ1 (A1 )μ2 (A2 ) = X 2 μ1 (A1 )1 A2 dμ2 (x 2 ) = X 2 μ1 ((A1 × A2 )x2 )dμ2 (x2 ) = μ (A1 × A2 ). Therefore, μ and μ coincide on A. Let B be the family of all finite unions of disjoint sets in A. Then B is an algebra of sets such that σ (B) = A1 ⊗ A2 , and μ(B) = μ (B) for any B ∈ B. Now let {Ai ) : i ∈ I } ⊂ A1 be a countable partition of X 1 such that μ1 (Ai ) < ∞, i ∈ I , and let {B j ) : j ∈ J } ⊂ A2 be a countable partition of X 2 such that μ2 (B j ) < ∞, j ∈ J . For i ∈ I and j ∈ J , we have Ai × B j ∈ A ⊂ B and μ(Ai × B j ) = μ (Ai × B j ) = μ1 (Ai )μ2 (B j ) < ∞. Consequently, μ and μ are σ -finite measures on B. Thus, using (10.9), it follows that μ = μ . Remark 15.6. The measure μ = μ in (15.5) is denoted by μ1 ⊗ μ2 , and it is called the product of μ1 and μ2 . μ1 ⊗ μ2 is the unique measure on A1 ⊗ A2 such that (μ1 ⊗ μ2 )(A1 × A2 ) = μ1 (A1 )μ2 (A2 ) for any A1 ∈ A1 and A2 ∈ A2 . The next theorem generalizes (15.5). Theorem 15.7. Let (X 1 , A1 ) and (X 2 , A2 ) be measurable spaces, let μ1 be a σ -finite measures on A1 , and let μ2 : X 1 × A2 → [0, ∞] be a function such that the section of μ2 at any x1 ∈ X 1 is a measure on A2 and the section of μ2 at any B ∈ A2 is A1 -measurable. Assume that there exists a countable partition {B j : j ∈ J } ⊂ A2 of X 2 such that μ2 (x1 , B j ) k j < ∞, x1 ∈ X 1 , for any j ∈ J . Then there exists a unique measure μ on A1 ⊗ A2 such that μ(A1 × A2 ) = A1 μ2 (x1 , A2 )dμ1 (x1 ) for any A1 ∈ A1 and A2 ∈ A2 . Proof. Let A ∈ A1 ⊗ A2 . Like in (15.3) it is proved that the function x1 → μ2 (x1 , A x1 ) from X 1 into R is A1 -measurable. Define μ(A) = X 1 μ2 (x1 , A x1 )dμ1 (x1 ). Like in (15.5) it is proved that μ is a measure on A1 ⊗ A2 . The details and the rest of the proof are left to the reader. In what follows we prove two versions of Fubini’s theorem. Theorem 15.8 (Fubini). Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be as in (15.3), and let f : X 1 × X 2 → [0, ∞] be an A1 ⊗ A2 -measurable function. Then (i) f d(μ1 ⊗ μ2 ) = f (x1 , x2 )dμ2 (x2 ) dμ1 (x1 ) X 1 ×X 2
X1
X
X2
X1
2
=
f (x1 , x2 )dμ1 (x1 ) dμ2 (x2 ).
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Proof. If f = 1 A , where A ∈ A1 ⊗ A2 , then X 2 ×X 2 1 A d(μ1 ⊗ μ2 ) = (μ1 ⊗ μ2 )(A) = X 1 μ2 (A x1 )dμ1 (x1 ) = X 1 ( X 2 1 A (x1 , x2 )dμ2 (x2 ))dμ1 (x1 ). Using (15.5), in like manner we get X 2 ×X 2 1 A d(μ1 ⊗ μ2 ) = X 2 ( X 1 1 A (x1 , x2 )dμ1 (x1 )) dμ2 (x2 ). Thus (i) holds in this case. Therefore, if f is a simple function, then, using (11.6), (i) obtains. Finally, in the general case choose a nondecreasing sequence { f n : n ∈ N } ⊂ [0, ∞[ X 1 ×X 2 of simple functions such that f n → f . Applying (11.22), we have
f d(μ1 ⊗ μ2 ) = lim
f n d(μ1 ⊗ μ2 )
n
X 1 ×X 2
X 1 ×X 2
f n (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
X1
f n (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
= lim n
X2
=
lim n
X1
= X1
Similarly we get (i) is proved.
X 2 ×X 2
X2
f (x1 , x2 )dμ2 (x2 ) dμ1 (x1 ).
X2
f d(μ1 ⊗ μ2 ) =
X2 ( X1
f (x1 , x2 )dμ1 (x1 ))dμ2 (x2 ), and so
Remark 15.9. The assumption that μ1 and μ2 are σ -finite in (15.8) is essential. To see this, let X 1 = X 2 = [0, 1], A1 = B([0, 1]), A2 = P([0, 1]), let μ1 be the Lebesgue on P([0, 1]), and let f = 1D , measure on B([0, 1]), let μ2 be the countingmeasure where D is the diagonal of [0, 1]2 . Then 1 = X 1 ( X 2 1D (x1 , x2 )dμ2 (x2 ))dμ1 (x1 ) = X 2 ( X 1 1D (x 1 , x 2 )dμ1 (x 1 ))dμ2 (x 2 ) = 0. Theorem 15.10 (Fubini). Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be as in (15.3), and let f : X 1 × X 2 → R be an A1 ⊗ A2 -measurable function such that one of the integrals X 1 ×X 2 | f | d(μ1 ⊗ μ2 ), X 1 ( X 2 | f (x1 , x2 )| dμ2 (x2 ))dμ1 (x1 ) and X 2 ( X 1 | f (x 1 , x 2 )|dμ1 (x 1 ))dμ2 (x 2 ) is finite. Then the following assertions hold. (i) There is A1 ∈ A1 such that μ1 (Ac1 ) = 0 and f x1 ∈ L1 (X 2 , A2 , μ2 ) for any x 1 ∈ A1 . (ii) There is A2 ∈ A2 such that μ2 (Ac2 ) = 0 and f x2 ∈ L1 (X 1 , A1 , μ1 ) for any x 2 ∈ A2 . (iii) The function ϕ1 (x1 ) = 1 A1 (x1 ) X 2 f (x1 , x2 )dμ2 (x2 ) ∈ L1 (X 1 , A1 , μ1 ). (iv) The function ϕ2 (x2 ) = 1 A2 (x2 ) X 1 f (x1 , x2 )dμ1 (x1 ) ∈ L1 (X 2 , A2 , μ2 ). (v) X 1 ×X 2 f d(μ1 ⊗ μ2 ) = X 1 ( X 2 f (x1 , x2 )dμ2 (x2 ))dμ1 (x1 ) = X 2 ( X 1 f (x1 , x2 )dμ1 (x1 ))dμ2 (x2 ).
Measures on Product σ -Algebras
Proof.
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In view of (15.8), we have | f | d(μ1 ⊗ μ2 ) = | f (x1 , x2 )| dμ2 (x2 ) dμ1 (x1 )
X 1 ×X 2
X1
X2
X2
X1
| f (x1 , x2 )| dμ1 (x1 ) dμ2 (x2 ) < ∞.
=
(1)
Thus, applying (11.15), there is A1 ∈ A1 such that μ1 (Ac1 ) = 0 and X 2 | f (x1 , x2 )| dμ2 (x2 ) < ∞ for any x1 ∈ A1 . Therefore, f x1 ∈ L1 (X 2 , A2 , μ2 ) for any x1 ∈ A1 , and so(i) is proved. Assertion (ii) is proved in like manner. Using (15.4), the functions x1 → X 2 f + (x1 , x2 )dμ2 (x2 ) and x1 → X 2 f − (x1 , x2 )dμ2 (x2 ) from X 1 into [0, ∞] are A1 -measurable, and so the function ϕ1 (x1 ) = 1 A1 (x1 ) X 2 f (x1 , x2 )dμ2 (x2 ) is A1 -measurable. As |ϕ1 (x1 )| X 2 | f (x1 , x2 )| dμ2 (x2 ), x1 ∈ X 1 , (1) shows that ϕ1 ∈ L1 (X 1 , A1 , μ1 ). Similarly it is shown that ϕ2 ∈ L1 (X 2 , A2 , μ2 ). Hence (iii) and (iv) are proved. Further, applying (15.8), we have f d(μ1 ⊗ μ2 ) = f + d(μ1 ⊗ μ2 ) − f − d(μ1 ⊗ μ2 ) X 1 ×X 2
X 1 ×X 2
= X1
X 1 ×X 2
f + (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
X2
− X1
= X1
−
f (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
= In like manner we get and so (v) is proved.
X 1 ×X 2
f + (x1 , x2 )dμ2 (x2 ) −
X2
X1
X2
X2
f − (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
f (x1 , x2 )dμ2 (x2 ) dμ1 (x1 ).
X2
f d(μ1 ⊗ μ2 ) =
X2 ( X1
f (x1 , x2 )dμ1 (x1 ))dμ2 (x2 ),
Theorem 15.11. Let (X 1 , A1 ), (X 2 , A2 ), μ1 , μ2 , and μ be as in (15.7), and let f : X 1 × X 2 → [0, ∞] be an A1 ⊗ A2 -measurable function. Then the following assertions hold. (i) The function x1 → X 2 f (x 1 , x 2 )μ2 (x 1 , d x 2 ) from X 1 into [0, ∞] is A1 -measurable.
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(ii) Proof.
X 1 ×X 2
f dμ =
X1 ( X2
f (x1 , x2 )μ2 (x1 , d x2 ))dμ1 (x1 ).
Exercise.
Theorem 15.12. Let (X 1 , A1 ), (X 2 , A2 ), μ1 , μ2 and μ be as in (15.7), and let f : X 1 ×X 2 → R be an A1 ⊗A2 -measurable function such that either X 1 ×X 2 | f | dμ < ∞ or X 1 ( X 2 | f (x1 , x2 )|μ2 (x1 , d x2 ))dμ1 (x1 ) < ∞. Then the following assertions hold. (i) There is A ∈ A1 such that μ1 (Ac ) = 0 and X 2 f (x1 , x2 )μ2 (x1 , d x2 ) ∈ R for any x1 ∈ A. (ii) The function ϕ1 (x1 ) = 1 A (x1 ) X 2 f (x1 , x2 )μ2 (x1 , d x2 ) ∈ L1 (X 1 , A1 , μ1 ). (iii) X 1 ×X 2 f dμ = X 1 ( X 2 f (x1 , x2 )μ2 (x1 , d x2 ))dμ1 (x1 ). Proof.
Exercise.
We now construct the product of a finite number of σ -finite measures. Theorem 15.13. For n 2, let (X i , Ai , μi ), 1 i n, be σ -finite measure spaces. n μ ] on A ⊗ · · · ⊗ A such Then there exists a unique measure μ1 ⊗ · · · ⊗ μn [⊗i=1 i 1 n that (i) (μ1 ⊗· · ·⊗μn )(A1 ×· · ·× An ) = μ1 (A1 ) · · · μn (An ),
A 1 ∈ A1 , . . . , A n ∈ An .
Proof. For n = 2 the result is true by (15.5) and (15.6). If the result is true for n, then, applying (15.5), there exists the measure (μ1 ⊗ · · · ⊗ μn ) ⊗ μn+1 on (A1 ⊗ · · · ⊗ An ) ⊗ An+1 such that ((μ1 ⊗ · · · ⊗ μn ) ⊗ μn+1 )((A1 × · · · × An ) × An+1 ) = (μ1 ⊗ · · · ⊗ μn )(A1 × · · · × An )μn+1 (An+1 ) = μ1 (A1 ) · · · μn (An )μn+1 (An+1 ) for Ai ∈ Ai , 1 i n +1. In accordance with (1.2.28) we identify the Cartesian products (X 1 × · · · × X n ) × X n+1 and X 1 × · · · × X n × X n+1 , and in view of (7.25) we identify the σ -algebras (A1 ⊗ · · · ⊗ An ) ⊗ An+1 and A1 ⊗ · · · ⊗ An ⊗ An+1 . Thus the existence of the measure μ1 ⊗ · · · ⊗ μn satisfying (i) is proved by induction. The uniqueness of this measure is proved as in (15.5), by making use of (10.9). The next theorem generalizes (15.13). Theorem 15.14. For n 2, let (X i , Ai , μi ), 1 i n, be measurable spaces, and let μ1 be a σ -finite measure on A1 . For 2 i n, let μi : X 1 × · · · × X i−1 × Ai → [0, ∞] be a function such that the section of μi at any (x1 , . . . , xi−1 ) ∈ X 1 ×· · ·× X i−1 is a measure on Ai and the section of μi at any A ∈ Ai is A1 ⊗· · ·⊗Ai−1 -measurable. For 2 i n, assume that there is a countable partition {Bi j : j ∈ Ji } ⊂ Ai of X i such that μi (x1 , . . . , xi−1 , Bi j ) ki j < ∞, (x1 , . . . , xi−1 ) ∈ X 1 × · · · × X i−1 , for any j ∈ Ji . Then there exists a unique measure μ on A1 ⊗ · · · ⊗ An such that μ(A1 ×· · ·×An ) = A1 μ1 (d x1 ) A2 μ2 (x1 , d x2 ) · · · An−1 μn−1 (x1 , . . . , xn−2 , d xn−1 ) An μn (x 1 , . . . , x n−1 , d x n ) for any A1 ∈ A1 , . . . , An ∈ An . Proof. Using (15.7) and (15.11), the theorem is proved by induction as in (15.13). The details are left to the reader. Exercise 15.15. (a) Find measurable spaces (X 1 , A1 ) and (X 2 , A2 ), and A ⊂ X 1 × X 2 such that A x1 ∈ A2 for any x1 ∈ X 1 , A x2 ∈ A1 for any x2 ∈ X 2 , but A ∈ / A1 ⊗ A2 .
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(b) Find measurable spaces (X 1 , A1 ) and (X 2 , A2 ), and f : X 1 × X 2 → R such that f x1 is A2 -measurable for any x1 ∈ X 1 , f x2 is A1 -measurable for any x2 ∈ X 2 , but f is not A1 ⊗ A2 -measurable. Exercise 15.16. Let (X 1 , A1 ) and (X 2 , A2 ) be measurable spaces, and let ∅ = A1 ⊂ X 1 and ∅ = A2 ⊂ X 2 . Show that A1 × A2 ∈ A1 ⊗ A2 if and only if A1 ∈ A1 and A2 ∈ A2 . Exercise 15.17.
Use (1.2.45.f) to give another proof of (15.2).
Exercise 15.18. Let λ be the Lebesgue measure on B(R). Prove that λ ⊗ λ is not a complete measure on B(R)λ ⊗ B(R)λ . [Use (10.17.d) and (15.16).] Exercise 15.19. Let (X, A, μ) be a σ -finite measure space, let f : X → [0, ∞] be an A-measurable function, and let V∗ f = {(x, y) ∈ X × [0, ∞] : y < f (x)} and V ∗ f = {(x, y) ∈ X × [0, ∞] : y f (x)}. Prove the following. (a) V∗ f ∈ A ⊗ B([0, ∞]), V ∗ f ∈ A ⊗ B([0, ∞]). (b) If λ is the Lebesgue measure on B([0, ∞]), then (μ ⊗ λ)(V∗ f ) = (μ ⊗ λ)(V ∗ f ) = X f dμ. Exercise 15.20. Let X 1 = X 2 = N , A1 = A2 = P(N ), and μ1 = μ2 = counting measure on P(N ). Define f : X 1 × X 2 → Rby f (i, i) = i, f (i, i +1) = −i, and f (i, j) = 0 if j = i or j = i + 1. Show that X 1 ( X 2 f (x1 , x2 )dμ2 (x2 ))dμ1 (x1 ) = 0 and X 2 ( X 1 f (x1 , x2 )dμ1 (x1 ))dμ2 (x2 ) = ∞. Exercise 15.21. Let f : R → R be a bijective function, let μ1 be the Lebesgue measure on B(R), let μ2 : R × B(R) → [0, ∞] be defined by μ2 (x, A) = ε f (x) (A) for any x ∈ R and A ∈ B(R), and let μ be as in (15.14). For (x1 , x2 ) ∈ R 2 , define F(x1 , x2 ) =
if f is strictly increasing x1 ∧ f −1 (x2 ) . (x1 − f −1 (x2 )) ∨ 0 if f is strictly decreasing
Prove the following. (a) F is continuous on R 2 . (b) ba F 0 whenever a, b ∈ R 2 are such that a b. (c) μ = μ F , where μ F is as in (3.4.24). (d) μ(Rc ) = 0, where R is the graph of f . Exercise 15.22. For n 2 and i = 1, . . . , n, let (X i , Ai ) be a measurable space, let μi and νi be σ -finite measures on Ai such that νi μi , and let f i = dνi /dμi . Put f (x1 , . . . , xn ) = f 1 (x1 ) · · · f n (xn ), (x1 , . . . , xn ) ∈ X 1 × · · · × X n . (a) Show that ν1 ⊗ · · · ⊗ νn μ1 ⊗ · · · ⊗ μn . (b) Use (11.28), (15.8) and (10.8) to prove that f = dν1 ⊗ · · · ⊗ νn /dμ1 ⊗ · · · ⊗ μn . Exercise 15.23. Let (X, A, μ) be a measure space, let f : X → [0, ∞] be an A-measurable function, and let a, p ∈ ]0, ∞[. ∞ (a) Use (15.19.a) and Fubini’s theorem (15.8) to show that X f dμ = 0 μ({ f > ∞ ∞ x})d x = 0 μ({ f x})d x, and that X ( f − a)+ dμ = a μ({ f > x})d x = ∞ a μ({ f x})d x.
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a (b) If μ(X ) < ∞, prove that { f a x})d x = p 0 x p−1 μ({ f x})d x, and that { f a} f p dμ + a p μ({ f > a}) = a p−1 a p−1 μ({ f > x})d x = p 0 x μ({ f x})d x. p 0 x Exercise 15.24. If f ∈ L p for some 0 < p < ∞, show that X | f | p dμ = ∞ ∞ p 0 x p−1 μ({| f | > x})d x = p 0 x p−1 μ({| f | x})d x, and infer that v p μ({| f | v v}) → 0 as v → ∞. [Hint. Use (15.23.a), then write μ({| f | v}) u x p−1 d x v p−1 ∞ p−1 μ({| f | x})d x u x μ({| f | x})d x, 0 < u < v < ∞.] u x Exercise 15.25. Assume that μ(X ) < ∞. Let L0 be the set of all measurable functions from X into R, and let ∼ be the equivalence relation on L0 defined as in 0 0 ˜ ˜ = X (1 ∧ | f − g|)dμ and d2 ( f˜, g) ˜ = (11.14.c). Put L = L /∼, and define d1 ( f , g) 0 ˜, g˜ ∈ L . Prove the following. | | f − g| /(1 + f − g|)dμ for f X (a) d1 and d2 are well-defined metrics for L 0 generating equal topologies. [Do not forget (6.88).] μ (b) For { f n : n ∈ N } ⊂ L 0 and f˜ ∈ L 0 , f n → f if and only if di ( f˜, f n ) → 0, i = 1, 2. (c) { f n : n ∈ N } is a Cauchy sequence in (L 0 , di ), i = 1, 2, if and only if { f n : n ∈ N } is Cauchy in μ-measure. f n : n ∈ N } is (d) (L 0 , d1 ) and (L 0 , d2 ) are complete metric spaces. [Hints. If { , choose a strictly increasing sequence {n : k 1} ⊂ N such that Cauchy relative to d 1 k f n − f n )dμ < 2−k , k 1, and so f n − f n )dμ < ∞. (1 ∧ (1 ∧ k+1 k k+1 k k1 X X Use Fubini’s theorem (15.8) to show that k1 f n k+1 − f n k < ∞ μ-a.e., and deduce that { f n k : k 1} is Cauchy μ-a.e. and converges μ-a.e. to some f ∈ L0 . Show that μ f n → f .] Exercise 15.26. Let (X, A, μ) be a measure space, let f : X → R be an A measurable function, and let t ∈ ]0, ∞[. Prove that X et f dμ = t R et x μ({ f > x})d x = t R et x μ({ f x})d x. [Use (15.8).]
16
The Product of Infinitely Many Measures
In this section we construct the product of infinitely many probabilities, and we prove the Daniell-Kolmogorov extension theorem. In (15.13) we constructed the product of a finite number of σ -finite measures. In what follows we construct the product of infinitely many probabilities. We begin with a lemma. Lemma 16.1. Let {(X i , Ai , Pi ) : i ∈ I } be a finite set of probability spaces, and let ∅ = J ⊂ I . Let PI be the product of the probabilities Pi , i ∈ I , and let PJ be the product of the probabilities Pi , i ∈ J . Then PI (π JI )−1 = PJ . Proof. In view of (15.13), PI is a probability on ⊗i∈I Ai , and PJ is a probability on ⊗i∈J Ai . Thus PI (π JI )−1 and PJ are probabilities on ⊗i∈J Ai . For i ∈ J , let Ai ∈ Ai .
Measures on Product σ -Algebras
Then (π JI )−1 (
Ai ) =
i∈J
(PI (π JI )−1 )
149
i∈I
Ai
= PI
Ai , where Ai =
i∈J
Ai
=
i∈I
Ai if i ∈ J . Therefore, / J X i if i ∈
Pi (Ai )
i∈I
=
i∈J
Pi (Ai ) = PJ
Ai .
i∈J
(1)
Since the family { i∈J Ai : Ai ∈ Ai for i ∈ J } generates ⊗i∈J Ai , (1) and (10.8) show that PI (π JI )−1 = PJ . Definition 16.2. Let {(X i , Ai , Pi ) : i ∈ I } be a nonempty set of probability spaces, and let J and A be as in (7.21). For each J ∈ J , let PJ denote the product of the probabilities P j , j ∈ J . For A ∈ A, choose J ∈ J such that A ∈ A J . Therefore, A = π J−1 (B) for some B ∈ ⊗ j∈J A j . Define Q(A) = PJ (B). Theorem 16.3. measure on A.
Notation is as in (16.2). Then Q is a well-defined finitely additive
Proof. Let us show that Q is well defined on A. For A ∈ A, let J1 ∈ J and J2 ∈ J be such that A ∈ A J1 and A ∈ A J2 . Therefore, A = π J−1 (B1 ) = π J−1 (B2 ) 1 2
(1)
for some B1 ∈ ⊗ j∈J1 A j and B2 ∈ ⊗ j∈J2 A j . Set J = J1 ∪ J2 . In view of (1.2.24), we have π J1 = π JJ1 ◦ π J and π J2 = π JJ2 ◦ π J . Hence, using (1), we get π J−1
π JJ1
−1
(B1 ) =
π J−1
π JJ2
−1
(B2 ) .
(2)
Since the function π J is surjective, (2) shows that −1 −1 π JJ1 (B1 ) = π JJ2 (B2 ).
(3)
By (16.1) and (3), we have PJ1 (B1 ) = (PJ (π JJ1 )−1 )(B1 ) = PJ ((π JJ1 )−1 (B1 )) = PJ ((π JJ2 )−1 (B2 )) = PJ2 (B2 ). Thus Q is well defined on A. Evidently, Q(∅) = 0. Now let A1 , A2 ∈ A be such that A1 ∩ A2 = ∅. Choose J1 ∈ J and J2 ∈ J such that A1 ∈ A J1 and A2 ∈ A J2 . Put J = J1 ∪ J2 . Then, in view of (7.20), we have A1 , A2 ∈ A J . Therefore, A1 = π J−1 (B1 ) and A2 = π J−1 (B2 ), where B1 , B2 ∈ ⊗ j∈J A j . We have ∅ = A1 ∩ A2 = π J−1 (B1 ∩ B2 ).
(4)
Since π J is surjective, (4) shows that B1 ∩ B2 = ∅. Since A1 ∪ A2 = π J−1 (B1 ∪ B2 ), we have Q(A1 ∪ A2 ) = PJ (B1 ∪ B2 ) = PJ1 (B1 ) + PJ2 (B2 ) = Q(A1 ) + Q(A2 ). Thus Q is finitely additive.
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Notation 16.4. Let A be as in (7.21), and let Q be as in (16.2). To emphasize the dependence of A and Q on the index set I we will write A I instead of A and Q I instead of Q. Theorem 16.5. Let {(X i , Ai , Pi ) : i ∈ I } be a set of probability spaces, let A ∈ A I , and let J be a nonempty finite subset of I. Then the following assertions hold:
(i) Ax ∈ A I −J for any x ∈ i∈J X i . Q I −J (Ax )d PJ (x), where PJ stands for the product of the (ii) Q I (A) =
i∈J
probabilities Pi , i ∈ J .
Xi
Proof. Let J be as in (7.21). There is K ∈ J such that A ∈ A K . Set L = J ∪K . Then,
I )−1 (B) = B × by (7.20), we have
A ∈ A L . Therefore, , where B ∈
A = (π L
i∈I −L X i
⊗i∈L Ai . Let x ∈ i∈J X i . Since i∈I X i = i∈J X i × i∈L−J X i × i∈I −L X i , we have Ax = (y, z) ∈ Xi × X i : (x, y, z) ∈ A = B × Xi
i∈L−J
i∈I −L
= (y, z) ∈
Xi ×
i∈L−J
= Bx ×
i∈I −L
X i : (x, y) ∈ B
i∈I −L
I −J −1 X i = (π L−J ) (Bx ).
(1)
i∈I −L
According to (15.1), we have Bx ∈ ⊗i∈L−J Ai . Consequently, (1) shows that Ax ∈ A I −J . Thus (i) holds. Moreover, we have Q I −J (Ax ) = PL−J (Bx ),
(2)
where PL−J is the product of the probabilities Pi , i ∈ L − J . Let PL denote the product of the probabilities Pi , i ∈ L. Then, using(2), we have Q I (A) = PL (B) = (PJ ⊗ PL−J )(B) = X i PL−J (Bx )d PJ (x) = X i Q I −J (Ax )d PJ (x). Hence i∈J i∈J (ii) holds. Theorem 16.6. Let {(X i , Ai , Pi ) : i ∈ I } be a denumerable set of probability spaces. Then there exists a unique probability P on ⊗i∈I Ai such that (i) Pπ J−1 = PJ for any nonempty finite set J ⊂ I , where PJ stands for the product of the probabilities Pj , j ∈ J . Proof. Without loss of generality, assume that I = N . For k ∈ N , denote Ik = {k +1, k +2, . . .}. Let A be as in (7.21), and let Q be as in (16.2). In view of (16.3), Q is a finitely additive measure on A. We next show that Q is a σ -additive measure on A. To do this, let {An : n ∈ N } ⊂ A be a nonincreasing sequence such that limn Q(An ) > 0. Then, applying (9.8), it will suffice to show that ∩n∈N An = ∅. Using (16.5), we have Q(An ) = Q I (An ) = X 1 Q I1 ((An )x1 )d P1 (x1 ). Thus the dominated convergence theo rem (11.25.ii) shows that 0 < limn Q(An ) = X 1 limn Q I1 ((An )x1 )d P1 (x1 ). Therefore,
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151
there is x1 ∈ X 1 such that limn Q I1 ((An )x1 ) > 0. If x1 ∈ X 1 , . . . , xk ∈ X k have been chosen so that limn Q Ik ((An )(x1 ,...,xk ) ) > 0, then, using (16.5) and (11.25.ii), we have Ik lim Q Ik+1 (An )(x1 ,...,xk ,xk+1 ) d Pk+1 (xk+1 ). 0 < lim Q (An )(x1 ,...,xk ) = n
X k+1 n
Consequently, there is xk+1 ∈ X k+1 such that limn Q Ik+1 ((An )(x1
,...,xk ,xk+1 ) ) > 0. Thus we have proved inductively that there is a point x = (xk )k∈N ∈ k∈N X k such that lim Q Ik ((An )(x1 ,...,xk ) ) > 0, k ∈ N . n
(1)
Equation (1) shows that (An )(x1 ,...,xk ) = ∅, k, n ∈ N .
(2)
For n ∈ N , select a nonempty finite set J ⊂ I such that An ∈ A J . Let k = sup J . −1 (B) = B × i>k X i , where B ∈ Then An ∈ A{1,...,k} . Therefore, An = π{1,...,k} A1 ⊗ · · · ⊗ Ak . Since
i>k X i if (x 1 , . . . , x k ) ∈ B , (An )(x1 ,...,xk ) = / B ∅ if (x1 , . . . , xk ) ∈ (2) shows that (x1 , . . . , xk ) ∈ B. Consequently, x ∈ An .
Thus x ∈ ∩n∈N An = ∅. Therefore, Q is a σ -additive measure on A. Obviously, Q( i∈I X i ) = 1. Using (7.21), Carathéodory’s extension theorem (10.9) shows that there exists a unique probability P on ⊗i∈I Ai such that P = Q on A. From this it follows at once that P is the only probability on ⊗i∈I Ai that satisfies (i). The probability P in (16.6) is called the product of the probabilities Pi , i ∈ I . Ai , Pi ) : i ∈ I } and P be as in (16.6). If Ai ∈ Ai for any Theorem 16.7. Let {(X i ,
i ∈ I , then P( i∈I Ai ) = i∈I Pi (Ai ). Proof. Without any loss
of generality, suppose that I = N . For n ∈ N , put × · · · × An × i>n X i . Then {Bn : n ∈ N } is a nonincreasing Bn = A1
sequence such that n∈N An = ∩n∈N Bn . Therefore, using (16.6), we have P( n∈N An ) = limn P(Bn ) = limn (Pπ{1,...,n}−1 )(A
1 × · · · × An ) = limn (P1 ⊗ · · · ⊗ Pn )(A1 × · · · × An ) = limn P1 (A1 ) · · · Pn (An ) = n∈N Pn (An ). The next theorem generalizes (16.6). Theorem 16.8 (Ionescu Tulcea). Let {(X n , An ) : n ∈ N } be a sequence of measurable spaces, and let P1 be a probability on A1 . For n 2, let Pn : X 1 × · · · × X n−1 × An → [0, 1] be a function such that the section of Pn at any (x1 , . . . , xn−1 ) ∈ X 1 × · · · × X n−1 is a probability on An and the section of Pn at any A ∈ An is A1 ⊗ · · · ⊗ An−1 -measurable. Then there exists a unique probability P on ⊗n∈N An −1 such that Pπ{1,...,n} = P (n) , n ∈ N , where P (n) is the probability on A1 ⊗ · · · ⊗ An constructed as in (15.14) starting from P1 , . . . , Pn . The proof of this theorem is similar to the proof of (16.6) and is left to the reader. In what follows we extend Theorem (16.6) to the case where I is an arbitrary infinite set.
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Analysis and Probability
Lemma 16.9. Let {(X i , Ai , Pi ) : i ∈ I } be a denumerable set of probability spaces, and let ∅ = J ⊂ I . Let PI be the product of the probabilities Pi , i ∈ I , and let PJ be the product of the probabilities P j , j ∈ J . Then (i) PI (π JI )−1 = PJ . Proof. PI (π JI )−1 and PJ are probabilities on ⊗ j∈J A j . Let K be a nonempty finite subset of J , and let PK be the product of the probabilities Pk , k ∈ K . Then, according to (16.6), we have PI (π JI )−1 (π KJ )−1 = PI (π KI )−1 = PK and PJ (π KJ )−1 = PK . Thus the uniqueness assertion in (16.6) shows that (i) holds. Definition 16.10. Let {(X i , Ai , Pi ) : i ∈ I } be a nonempty set of probability spaces, and let J ∗ be as in (7.22). For each J ∈ J ∗ , let PJ denote the product of the probabilities P j , j ∈ J . As proved in (7.22), ⊗i∈I Ai = ∪ J ∈J ∗ A J , where A J is as in (7.20). For A ∈ ⊗i∈I Ai , choose J ∈ J ∗ such that A ∈ A J . Therefore, A = π J−1 (B) for some B ∈ ⊗ j∈J A j . Define P(A) = PJ (B). Using (16.9), a reasoning similar to the one in the proof of (16.3) shows that P is well defined. Theorem 16.11. Notation is as in (16.10). Then P is the unique probability on ⊗i∈I Ai such that (i) Pπ J−1 = PJ for any nonempty finite set J ⊂ I , where PJ stands for the product of the probabilities Pj , j ∈ J .
Proof. Obviously, P(∅) = 0 and P( i∈I X i ) = 1. Let {An : n ∈ N } ⊂ ⊗i∈I Ai be such that Am ∩ An = ∅ for m = n. For n ∈ N , choose Jn ∈ J ∗ such that An ∈ A Jn . Set J = ∪n∈N Jn . Then J ∈ J ∗ , and An ∈ A J for any n ∈ N . Therefore, for n ∈ N , we have An = π J−1 (Bn ) for some Bn ∈ ⊗ j∈J A j . For m = n, we have −1 ∅ = Am ∩ An = π J−1 (Bm ∩ Bn ), and so Bm∩ Bn = ∅. Since ∪ n∈N An = π J (∪n∈N Bn ), we get P(∪n∈N An ) = PJ (∪n∈N Bn ) = n∈N PJ (Bn ) = n∈N P(An ). Thus P is a probability on ⊗i∈I Ai . Evidently, P satisfies (i). Using (7.21) and (10.9), it is easily seen that P is the only probability on ⊗i∈I Ai that satisfies (i). The probability P in (16.11) is called the product of the probabilities Pi , i ∈ I , and is denoted by ⊗i∈I Pi . If I = {n ∈ Z : n m} for some m ∈ Z , then we will write also ⊗nm Pn instead of ⊗i∈I Pi . One of the essential results of this chapter is the Daniell-Kolmogorov extension theorem which we will prove in (16.20). We begin with a definition. Definition 16.12. A family B ⊂ P(X ) is said to be compact if {Bn : n 1} ⊂ B n B = ∅ for any n 1 imply ∩ and ∩i=1 i i1 Bi = ∅. Example 16.13. (a) The family of all finite subsets of an arbitrary set X is compact. (b) Let X be a Hausdorff space, and let B be the family of all compact subsets of X . Then, according to (5.24) and (5.27), B is compact. Remark 16.14. (a) If B ⊂ P(X ) is compact, then B ∪ {X } is compact. (b) If B is compact and C ⊂ B, then C is compact.
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153
(c) For i ∈ I = ∅, let X i be a set, and let Bi
⊂ P(X i ) be a compact family. Then, in view of (2.44), the family ∪i∈I πi−1 (Bi ) ⊂ P( i∈I X i ) is compact. be the family of all denumerable Theorem 16.15. Let B be a compact family, and let B intersections of finite unions of sets in B. Then the following assertions hold. is compact. (i) B then ∩n1 An ∈ B. (ii) If {An : n 1} ⊂ B, then A ∪ B ∈ B. (iii) If A, B ∈ B, where C k = Ak ∪ · · · ∪ Proof. To prove (i), for k 1, let B k = ∩i1 Cik ∈ B, i i1 k n k Aili (k) , i 1, is such that ∩k=1 B = ∅ whatever n 1. We arrange the sets Cik =
i (k) k m Ai j , i, k 1, in a sequence {D m : m 1}, where D m = Am ∪lj=1 1 ∪ · · · ∪ Arm , m 1. Since ∅ = ∩nk=1 B k = ∩nk=1 ∩i1 Cik , n 1, it follows that ∩nm=1 D m = ∅ for any n 1. Consequently, if n 2, there is s1 with 1 s1 r1 such that A1s1 ∩ D 2 ∩ · · · ∩ D n = ∅. If s1 , . . . , sl have been chosen so that A1s1 ∩ · · · ∩ Alsl ∩ Dl+1 ∩ · · · ∩ D n = ∅ for any n l + 1, then there is sl+1 with 1 sl+1 rl+1 l+2 ∩ · · · ∩ D n = ∅, n l + 2. Thus it such that A1s1 ∩ · · · ∩ Alsl ∩ Al+1 sl+1 ∩ D follows inductively that ∩li=1 Aisi = ∅ for any l 1. As Aisi ∈ B, i 1, we have ∅ = ∩i1 Aisi ⊂ ∩m1 D m = ∩k1 B k . This completes the proof of (i). Assertion (ii) is obvious, and assertion (iii) follows from (ii).
Definition 16.16. Let A be an algebra of sets, let B ⊂ A be a compact family, and let μ be a finitely additive measure on A. We say that μ is regular [B-regular] if μ(A) = sup{μ(B) : A ⊃ B ∈ B} for any A ∈ A. Theorem 16.17. Let A be an algebra of subsets of X, let B ⊂ A be a compact family, and let μ be a finitely additive measure on A such that μ(X ) < ∞. If μ is B regular, then μ is σ -additive. Proof. Let {An : n ∈ N } ⊂ A be such that ∪n∈N An ∈ A and Am ∩ An = ∅ A ) assume for m = n. Evidently, μ(∪ n∈N n n∈N μ(An ). To get a contradiction that μ(∪n∈N An ) > n∈N μ(An ), and let 0 < ε μ(∪n∈N An ) − n∈N μ(An ). For n A ) ∈ A and n ∈ N , set Bn = ∪i>n Ai . Then Bn ⊃ Bn+1 , Bn = ∪i∈N Ai − (∪i=1 i μ(Bn ) = μ
i∈N
Ai
−
n i=1
μ(Ai ) μ
i∈N
Ai
−
μ(Ai ) ε, n ∈ N .
i∈N
(1) For n ∈ N , select Cn ∈ B such that Cn ⊂ Bn and μ(Bn ) − μ(Cn ) ε/2n+1 .
(2)
n C ) = μ(∩n B ) −μ(∩n B −(∩n C )) Then, using (1) and (2), we have μ(∩i=1 i i=1 i i=1 i i=1 i n n n C = ∅ μ(Bn )−μ(∪i=1 (Bi −Ci )) ε− i=1 ε/2i+1 > ε/2, n ∈ N . Therefore, ∩i=1 i for any n ∈ N . Since B is compact, we get ∩i∈N Ci = ∅. On the other hand, we have ∩n∈N Cn ⊂ ∩n∈N Bn = ∅. This contradiction shows that μ is σ -additive.
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Analysis and Probability
Theorem 16.18. Let (X, A, μ) be a finite measure space, and let B ⊂ A be such that {An : n ∈ N } ⊂ B implies ∩n∈N An ∈ B, and A, B ∈ B implies A ∪ B ∈ B. Let C = {A ∈ A : μ(A) = sup{μ(B) : A ⊃ B ∈ B}}. Then the family {A ∈ A : A ∈ C and Ac ∈ C} is a σ -algebra unless it is empty. Proof. Let {An : n ∈ N } ⊂ C. It suffices to show that ∪n∈N An ∈ C and ∩n∈N An ∈ C. Let α < μ(∪n∈N An ), and choose m ∈ N such that α < μ(∪m n=1 An ). For n = 1, . . . , m, n select Bn ∈ B such that Bn ⊂ An and μ(An ) − μ(Bn ) (μ(∪m n=1 An ) − α)/2 . Then we have m
m
m
m Bn = μ An − μ An − Bn μ n=1
μ μ
n=1 m
n=1 m
An −
n=1 m
n=1
μ(An − Bn )
n=1
An − μ
n=1
m
An − α
= α.
(1)
n=1
m m Since ∪m n=1 Bn ⊂ ∪n=1 An and ∪n=1 Bn ∈ B, (1) shows that ∪n∈N An ∈ C. Now let ε > 0. For n ∈ N , select Cn ∈ B such that Cn ⊂ An and μ(An ) − μ(Cn ) ε/2n . Then we have
Cn = μ An − μ An − Cn μ n∈N
n∈N
n∈N
μ μ
An −
n∈N
n∈N
μ(An − Cn )
n∈N
An − ε.
(2)
n∈N
Since ∩n∈N Cn ⊂ ∩n∈N An and ∩n∈N Cn ∈ B, (2) shows that ∩n∈N An ∈ C. Theorem 16.19. Let {(X i , Ai ) : i ∈ I } be a nonempty family of measurable spaces, and let μ be a finite measure on ⊗i∈I Ai such that μπi−1 is regular for any i ∈ I . Then μ is regular. Proof. For i ∈ I , let Bi ⊂ Ai be a compact family such that μπi−1 is B i -regular. In view of (16.14.a), we may and do assume that X i ∈ Bi . Let B = { i∈I Bi : Bi ∈ Bi , i ∈ I, and {i ∈ I : Bi = X i } is finite}. Obviously, B ⊂ ⊗i∈I Ai . be as in (16.15). Using (1.2.44), it follows at once that B is a compact family. Let B ⊂ ⊗i∈I Ai . We will show that μ is B-regular. Since ⊗i∈I Ai = Evidently, B σ (∪i∈I πi−1 (Ai )), and A ∈ ∪i∈I πi−1 (Ai ) implies Ac ∈ ∪i∈I πi−1 (Ai ), (16.18) for any shows that it suffices to prove that μ(A) = sup{μ(B) : A ⊃ B ∈ B} A ∈ ∪i∈I πi−1 (Ai ). Let A ∈ πi−1 (Ai ). Therefore, A = πi−1 (B) for some B ∈ Ai .
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For ε > 0, choose C ∈ Ai such that C ⊂ B and (μπi−1 )(C) > (μπi−1 )(B) − ε. Consequently (μπi−1 )(C) > (μπi−1 )(B) − ε = μ(A) − ε
(1)
(1) shows that μ is B-regular. Since πi−1 (C) ∈ B ⊂ B, Daniell-Kolmogorov extension theorem 16.20. Let {(X i , Ai ) : i ∈ I } be a nonempty family of measurable spaces, and let J be as in (7.21). For any J ∈ J , let μ J be a finite measure on ⊗ j∈J A j such that μ J (π KJ )−1 = μ K whenever ∅ = K ⊂ J . If μ{i} is regular for any i ∈ I , then there exists a unique measure μ on ⊗i∈I Ai such that (i) μπ J−1 = μ J for each J ∈ J . Proof. Let A be as in (7.21). For A ∈ A, choose J ∈ J such that A ∈ A J . Therefore, A = π J−1 (B) for some B ∈ ⊗ j∈J A j . Define μ(A) = μ J (B). As in (16.3), it is proved that μ is a well-defined finitely additive measure on A. For i ∈ I , let Bi ⊂ Ai be a compact family such that μ{i} is B i -regular. According to (16.14.a), we may and do assume that X i ∈ Bi . Let B = { i∈I Bi : Bi ∈ Bi , i ∈ I, and {i ∈ I : Bi = be as in (16.15). Let A = π −1 (B), where B ∈ ⊗ j∈J A j . In view X i } is finite}, and let B J J ⊂ ⊗ j∈J A j such that μ J is B J -regular. of (16.19), there exists a compact family B For ε > 0, select C ∈ B J such that C ⊂ B and μ J (C) > μ J (B) − ε. Therefore, μπ J−1 (C) > μπ J−1 (B) = μ(A) − ε.
(1)
Since π J−1 (C) ⊂ π J−1 (B), (1) shows that μ is regular with respect to the family J ) ⊂ A. As ∪ J ∈J π −1 (B J ) ⊂ B, (16.14.b) shows that ∪ J ∈J π −1 (B J ) is ∪ J ∈J π J−1 (B J J compact. Using (16.17), it follows that μ is σ -additive. Using (7.21), Carathéodory’s extension theorem (10.9) shows that μ has a unique extension to ⊗i∈I Ai . From this it follows at once that μ is the only measure on ⊗i∈I Ai that satisfies (i). Theorem (16.23) below shows that the Daniell-Kolmogorov extension theorem applies whenever each X i is a complete separable metric space and Ai = B(X i ); in particular, it applies whenever each X i is a countable set and Ai = P(X i ). We need the following intermediate results. Theorem 16.21. Let X be a metric space, let B be the family of all closed subsets of X and let μ be a finite measure on B(X ). Then μ(A) = sup{μ(B) : A ⊃ B ∈ B} for any A ∈ B(X ). Proof. Let A = B(X ). Notice that B fulfills the assumptions in (16.18). Further, let C be as in (16.18), and put D = {A ∈ A : A ∈ C and Ac ∈ C}. Clearly, it suffices to prove that C ⊃ B(X ). Let A ∈ B. Plainly, A ∈ C. Moreover, according to (6.85), there n B ∈ B, n ∈ N , is a sequence {Bn : n ∈ N } ⊂ B such that Ac = ∪n∈N Bn . Since ∪i=1 i n c c and μ(A ) = sup μ(∪i=1 Bi ) (9.9), it follows that A ∈ C. Thus A ∈ D, and so D ⊃ B. Hence, in view of (16.18), we see that D ⊃ B(X ). As C ⊃ D, we get C ⊃ B(X ).
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Theorem 16.22 (Ulam). Let (X, d) be a complete separable metric space, and let μ be a finite measure on B(X ). Then for each ε > 0, there exists a compact set C ⊂ X such that μ(X ) − μ(C) < ε. Proof. Since X is separable, there is a countable set A dense in X. The case in which A is finite is simpler and it is left to the reader. If A is denumerable, then put {A = xi : i 1}. For i, n 1, set Sin = S(xi , 1/n). For every n 1, as A is dense in X we get X = ∪i1 Sin , and so, by (9.9), there exists i n 1 such that i
n μ Sin > μ(X ) − ε/2n . (1) i=1 n Sin , n 1, and C = ∩n1 Cn . Obviously, C is a totally bounded Denote Cn = ∪ii=1 set. Since C is closed, and X is complete, (6.25.b) shows that (C, dC ) is a complete metric space. Hence, in view of (6.31), it follows that C is compact.Finally, on account of (1), we have μ(X ) − μ(C) = μ(∪n1 Cnc ) n1 μ(Cnc ) < n1 ε/2n = ε.
Theorem 16.23.
Let (X, d) and μ be as in (16.22). Then μ is regular.
Proof. Let C be the family of all compact subsets of X . Then, in view of (16.13.b), C is compact. We will show that μ is C-regular. Let A ∈ B(X ) and ε > 0. Applying (16.21), select a closed set B ⊂ X such that μ(A) − μ(B) < ε/2.
(1)
Then, by (6.25.b) and (6.70), (B, d B ) is a complete separable metric space. On the other hand, according to (7.14) and (7.48), we have B(B) = B ∩ B(X ) ⊂ B(X ). Hence (16.22) applies, and we can select a compact subset C of B such that μ(B) − μ(C) < ε/2.
(2)
By virtue of (4.31), C ∈ C. Moreover, combining (1) with (2), we get μ(A)−μ(C) < ε. Thus μ is C-regular. Exercise 16.24. For 1 < b ∈ N , let I = {0, . . . , b − 1}. For n ∈ N , set In = I and define a probability Pn on P(In ) by Pn ({i}) = 1/b, i ∈ In . Let λ be the Lebesgue measure on B([0, 1[), and let P be the product of the probabilities Pn , n ∈ N . Prove that λ(an )−1 n∈N = P, where the function (an )n∈N is as in (8.23). Exercise 16.25 (Kakutani). For n ∈ N , let (X n , An ) be a measurable space, let Pn and Q n be probabilities on An such that Pn Q n , and let f n = d Pn /d Q n . Let P and , and Q n , n ∈ N , respectively. Q denote the products of the probabilities
Pn , n ∈ N1/2 (a) Show that P Q whenever n∈N X n f n d Q n > 0. [Hints. For x =
and (x1 , x2 , . . .) ∈ n∈N X n , define gn (x) = ( f 1 (x1 ) · · · f n (xn ))1/2 . Use (11.29)
n ∈ N } is a Cauchy sequence in L 2 ( n∈N X n , (15.8) to prove that {gn :
⊗n∈N An , Q), and let g ∈ L2 ( n∈N X n , ⊗n∈N An , Q) be such that g − gn 2 → 0 −1 (13.16). For A ∈ π{1,...,m} (A ⊗ · · · ⊗ Am ), use (15.22) and (11.29) to show that 2 1 P(A) = limn A gn d Q = A g 2 d Q. Then apply (10.8).]
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1/2 (b) Prove that P ⊥ Q whenever n∈N X n f n d Q n = 0. [Hints. For ε > 0,
m
1/2 choose m ∈ N such that n=1 X n f n d Q n < ε. Put B = {x ∈ n∈N X n : gm (x) 1}. Show that Q(B) < ε and P(B c ) < ε, and apply (12.25).]
(c) Let ρ be as in (13.26). Use (13.5) and (12.23.d) to prove that ρ(P, Q) = n∈N ρ(Pn , Q n ). Exercise 16.26. For i ∈ I = ∅, let Ai be a σ -algebra, and let Pi and Q i be probabilities on Ai . Let P and Q stand for the products of the probabilities Pi , i ∈ I , and Q i , i ∈ I , respectively. Show that P Q implies that Pi Q i for any i ∈ I. Exercise 16.27.
If B ⊂ P(X ) is finite, then B is compact.
Exercise 16.28. Let Bi ⊂ P(X ), i = 1, 2, be compact families, and define B = {B1 ∪ B2 : B1 ∈ B1 and B2 ∈ B2 }. Is B compact? Exercise 16.29. Let A and B be as in (16.16), and let μ1 , . . . , μn be B-regular finitely additive measures on A. Show that μ1 + · · · + μn is regular. Exercise 16.30. Let (X, A, μ), (Y, B) and f be as in (9.26), let C ⊂ B be a compact family, and assume that f is surjective. Prove the following. (a) f −1 (C) is a compact family. (b) μ f −1 (B) is f −1 (C)-regular if and only if μf −1 is C-regular. Exercise 16.31. Let A be a σ -algebra, let μ be a σ -finite regular measure on A, and let ν be a finite measure on A such that ν μ. Prove that ν is regular. Exercise 16.32. Let X be a complete separable metric space, let μ be a finite measure on B(X ) such that μ({x}) = 0 for any x ∈ X , let B ∈ B(X ), and consider c ∈ [0, μ(B)]. Use (9.42.b) and (16.23) to prove that there exists a sequence {Cn : n ∈ N } of compact subsets of X such that Cn ⊂ B, n ∈ N , and μ(∪n∈N Cn ) = c. Exercise 16.33. Find a metric space X and a measure μ on B(X ) such that μ is not B-regular, where B stands for the family of all compact subsets of X . [Do not forget (9.55).] Exercise 16.34. Let (X, d) be a metric space, and let μ be a finite measure on B(X ). Show that μ(A) = inf{μ(U ) : A ⊂ U ∈ Td } for any A ∈ B(X ). Exercise 16.35. Let {(X i , Ai , Pi ) : i ∈ I } be a nonempty set of probability spaces, and let {Ja : a ∈ A} be a partition of I such that Ja = ∅, a ∈ A. Prove that ⊗i∈I Pi = ⊗a∈A (⊗ j∈Ja P j ). Exercise 16.36. For i ∈ I = ∅, let (X i , Ai , Pi ) be a probability space, let Bi ) be a measurable space, and let h i : X i → Yi be a measurable function. Define (Yi ,
h : i∈I X i → i∈I Yi by h((xi )i∈I ) = (h i (xi ))i∈I . Prove the following.
(a) For Bi ⊂ Yi , i ∈ I , h −1 ( i∈I Bi ) = i∈I h i−1 (Bi ). (b) h is (⊗i∈I Ai , ⊗i∈I Bi )-measurable. [Use (8.4) and (a).] (c) (⊗i∈I Pi )h −1 = ⊗i∈I Pi h i−1 . [Hints. For nonempty finite J ⊂ I , let π JX denote
the projection from i∈I X i onto j∈J X j , let π JY denote the projection from i∈I Yi
onto j∈J Y j , and let B j ∈ B j , j ∈ J . Then, by (a) and (16.11),
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Pi h −1 (π JY )−1 ⎝
i∈I
=
⎛
=⎝
⎞⎛
Pj ⎠ ⎝
⎞ ⎠ h −1 j (B j )
⎞
⎠ h −1 j (B j )
j∈J
Pj
i∈I
Bj⎠
j∈J
⎛ ⎞⎛ ⎞ ⎝ ⎠⎝ h −1 P j h −1 Bj⎠ j (B j ) = j
j∈J
=
⎞
j∈J
Pi (π JX )−1 ⎝
i∈I
⎛
j∈J
=
⎛
Pi h i−1
j∈J
⎞
⎛ Bj⎠ . (π JY )−1 ⎝
j∈J
j∈J
This means that ((⊗i∈I Pi )h −1 )(π JY )−1 = (⊗i∈I Pi h i−1 )(π JY )−1 . Apply next (16.11).] N } be a sequence Exercise 16.37. Let {(X j , d j ) : j ∈
of metric spaces. For n ∈ N , let π{1,...,n} denote the projection from j∈N X j onto nj=1 X j , let Cn be the family of
−1 all compact subsets of nj=1 X j , let Bn = π{1,...,n} (Cn ), and put B = ∪n∈N Bn . Show that the following assertions are equivalent: (i) B is compact; (ii) if {n k : k ∈ N } ⊂ N is strictly increasing and Bk ∈ Bn k , k ∈ N , with k B = ∅ for any k ∈ N , then ∩ ∩i=1 i i∈N Bi = ∅; n B = ∅ for any n ∈ N , then ∩ (iii) if Bn ∈ Bn , n ∈ N , with ∩i=1 i i∈N Bi = ∅. n B = ∅, n ∈ N . To prove that (ii) [Hints. Let {Bn : n ∈ N } ⊂ B be such that ∩i=1 i } < ∞, use (5.27) and (5.24). implies (i) in case sup{n ∈ N : Bm ∈ Bn for some m ∈ N
To show that (iii) implies (ii), assume that Bk = Ck × ( j>n k X j ), where Ck ∈ Cnk , k B = ∅ for any k ∈ N . Define B = B if n = n for some k, and and that ∩i=1 i k k n
{1,...,n } Bn = π{1,...,n}k (Ck ) × ( j>n X j ) if n k−1 < n < n k (with n 0 = 0) for some k. Then Bn ∈ Bn , n ∈ N , by (5.45), and Bn ⊃ Bn , n ∈ N . Apply (iii) to {Bn : n ∈ N }.] Exercise 16.38. that B is compact. [Hints. Let
Notation is as in (16.37). Prove n B = ∅ for n ∈ N . Let xn = Bn = Cn × ( j>n X j ), where Cn ∈ Cn , with ∩i=1 i n+ p
n+ p
n B , n ∈ N . Then xn+ p ∈ B , i.e. (x (x1n , x2n , . . .) ∈ ∩i=1 , . . . , xn ) ∈ Cn , for any i n 1 n ∈ N and p 0. Since x1n ∈ C1 , n ∈ N , and C1 is sequentially compact (6.19), choose {n 1k : k ∈ N } ⊂ N so that limk x1n 1k = x1 ∈ C1 . Since (x1n 1k , x2n 1k ) ∈ C2 , k ∈ N , and C2 is sequentially compact, choose {n 2k : k ∈ N } a subsequence of {n 1k : k ∈ N } such that limk (x1n 1k , x2n 1k ) = (x1 , x2 ) ∈ C2 . Continue this procedure and show that x = (x1 , x2 , . . .) = limk xn kk ∈ ∩i∈N Bi .]
Exercise 16.39. Let {(X j , d j ) : j ∈ N } be a sequence of complete separable metric spaces. For n ∈ N , let μn be a finite measure on B( nj=1 X j ) such that
Measures on Product σ -Algebras {1,...,n+1} −1 )
μn+1 (π{1,...,n}
159
= μn . Prove that there exists a unique measure μ on B( j∈N X j )
−1 such that μπ{1,...,n} = μn , n ∈ N . The following steps may be useful.
−1 (B( nj=1 X j )) is an algebra of sets. (a) The family A = ∪n∈N π{1,...,n}
−1 (b) For A ∈ A, choose n ∈ N so that A = π{1,...,n} (B) for some B ∈ B( nj=1 X j ), and put μ(A) = μn (B). Then μ is a well-defined finitely additive measure on A. (c) μ is σ -additive. [Hint. μ is B-regular, where B is as in (16.37) by (5.71.b), (6.98) and (16.23) applied to μn . Then use (16.17).]
(d) μ has a unique extension to B( j∈N X j ). [Use (10.9).]
Exercise 16.40. Let I = N or Z , and let X be a countable set with at least two elements. For each l 1 and x1 , . . . , xl ∈ X , suppose a nonnegative number . . . , xl ) is given such that: pl (x1 , (i) x∈X p1 (x) < ∞; (ii) x∈X pl+1 (x1 , . . . , xl , x) = pl (x1 , . . . , xl ); (iii) x∈X pl+1 (x, x1 , . . . , xl ) = pl (x1 , . . . , xl ). Let J denote the family of all finite subsets of I consisting of consecutive integers. If J ∈ J has l elements, define a measure PJ on P(X J ) by PJ ({(x1 , . . . , xl )}) = pl (x1 , . . . , xl ), (x1 , . . . , xl ) ∈ X J . Prove that there is a unique finite measure P on ⊗i∈I P(X i ), where X i = X for any i ∈ I , such that Pπ J−1 = PJ , J ∈ J . The next steps may be helpful. (a) The family A0 = ∪ J ∈J π J−1 (P(X J )) is an algebra of sets that generates ⊗i∈I P(X i ). (b) For A ∈ A0 , choose J ∈ J so that A = π J−1 (B) for some B ⊂ X J , and set P(A) = PJ (B). Then P is a well-defined finitely additive measure on A0 . [Use (i) and (ii).] (c) μ is σ -additive. [Hints. Let Bi be the family of all finite subsets of X i , where X i = X for any i ∈ I , and put B = ∪i∈I πi−1 (Bi ). Then B is compact by (16.13.a) and (16.14.c). Show that P is B-regular and use (16.17).] (d) P has a unique extension to ⊗i∈I P(X i ). [Use (10.9).] (e) P(X I ) = x∈X p1 (x).
15
The Product of a Finite Number of Measures
In this section we construct the product of a finite number of σ -finite measures and prove versions of Fubini’s theorem. We first construct the product of two σ -finite measures. Theorem 15.1. A1 ⊗ A2 . Then
Let (X 1 , A1 ) and (X 2 , A2 ) be measurable spaces, and let A ∈
(i) A x1 ∈ A2 for any x1 ∈ X 1 and (ii) A x2 ∈ A1 for any x2 ∈ X 2 , where A xi is the section of A at xi , i = 1, 2. Proof. To prove (i), let A = {A1 × A2 : A1 ∈ A1 , A2 ∈ A2 } and M = {A ∈ A1 ⊗ A2 : A x1 ∈ A2 for any x1 ∈ X 1 }. For A1 × A2 ∈ A and x1 ∈ X 1 , in view of (1.2.30.b), we have A2 if x1 ∈ A1 , (A1 × A2 )x1 = / A1 . ∅ if x1 ∈ Thus A ⊂ M. If {An : n N } ⊂ M, then (∪n N An )x1 = ∪n N (An )x1 ∈ A2 for any x1 X 1 , and so ∪n N An ∈ M. If A ∈ M, then (Ac )x1 = (A x1 )c ∈ A2 for any x1 ∈ X 1 , and so Ac ∈ M. Therefore, M is a σ -algebra. Since A1 ⊗ A2 = σ (A) (7.17.a), we have A1 ⊗ A2 = M. This proves (i). The proof of (ii) is similar. Corollary 15.2. Let (X 1 , A1 ) and (X 2 , A2 ) be measurable spaces, and let f : X 1 × X 2 → R be an A1 ⊗ A2 -measurable function. Then (i) f x1 is A2 -measurable for any x1 ∈ X 1 Analysis and Probability. http://dx.doi.org/10.1016/B978-0-12-401665-1.00005-9 © 2013 Elsevier Inc. All rights reserved.
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and (ii) f x2 is A1 -measurable for any x2 ∈ X 2 , where f xi is the section of f at xi , i = 1, 2. Proof. If f = 1 A , where A ∈ A1 ⊗ A2 , then f xi = 1 A xi , i = 1, 2. Hence in this case (i) and (ii) follow at once from (15.1). The corollary is further proved by making use of the indicator function method. Theorem 15.3. Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be measure spaces, where μ1 and μ2 are σ -finite and let A ∈ A1 ⊗ A2 . Then the following assertions hold. (i) The function x1 → μ2 (A x1 ) from X 1 into R is A1 -measurable. (ii) The function x2 → μ1 (A x2 ) from X 2 into R is A2 -measurable. Proof. Since the proofs of (i) and (ii) are similar, we prove only (i). (I) Assume first that μ2 (X 2 ) < ∞. Let A be as in (15.1), and let D = {A ∈ A1 ⊗ A2 : assertion (i) holds}. For A1 × A2 ∈ A, we have μ2 ((A1 × A2 )x1 ) = μ2 (A2 )1 A1 (x1 ), x1 ∈ X 1 . Thus A ⊂ D. Evidently X 1 × X 2 ∈ D. If A, B ∈ D and A ⊃ B, then μ2 ((A − B)x1 ) = μ2 (A x1 − Bx1 ) = μ2 (A x1 ) − μ2 (Bx1 ), x1 ∈ X 1 , and so A − B ∈ D. If {An : n ∈ N } ⊂ D and Am ∩ An = ∅ for m = n, then μ2 ((∪n N An )x1 ) = μ2 (∪n N (An )x1 ) = n N μ2 ((An )x1 ), x1 X 1 , and so ∪n N An ∈ D. Therefore, D is a Dynkin system. Since A1 × A2 ∈ A and A 1 × A 2 ∈ A imply (A1 × A2 ) ∩ (A 1 × A 2 ) = (A1 ∩ A 1 ) × (A2 ∩ A 2 ) ∈ A, (7.36) shows that d(A) = σ (A) = A1 ⊗ A2 . Consequently, D = A1 ⊗ A2 . This proves (i). (II) Consider now the general case. Let {B j : j ∈ J } ⊂ A2 be a countable partition of X 2 such that μ2 (B j ) < ∞, j ∈ J . For A ∈ A1 ⊗ A2 , using (7.14),we have A ∩(X 1 × B j ) ∈ i X−11 ×B j (A1 ⊗A2 ) = A1 ⊗i B−1j (A2 ), j ∈ J . Thus, according to (I), the function x1 → μ2 ((A ∩ (X 1 × B j ))x1 ) is A1 -measurable for any j ∈ J . Therefore, the function x1 → μ2 (A x1 ) = μ2 (∪ j∈J (A∩(X 1 ×B j ))x1 ) = j∈J μ2 ((A∩(X 1 ×B j ))x1 ) is A1 -measurable. Corollary 15.4. Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be as in (15.3), and let f : X 1 × X 2 → [0, ∞] be an A1 ⊗A2 -measurable function. Then the following assertions hold. (i) The function x1 → X 2 f (x1 , x2 )dμ2 (x2 ) from X 1 into [0, ∞] is A1 -measurable. (ii) The function x2 → X 1 f (x1 , x2 )dμ1 (x1 ) from X 2 into [0, ∞] is A2 -measurable. Proof. If f = 1 A , where A ∈ A1 ⊗ A2 , then X 2 1 A (x1 , x2 )dμ2 (x2 )= X 2 1 A x1 (x2 ) dμ2 (x2 ) = μ2 (A x1 ), x1 ∈ X 1 , and X 1 1 A (x1 , x2 )dμ1 (x1 ) = X 1 1 A x2 (x1 )dμ1 (x1 ) = μ1 (A x2 ), x2 ∈ X 2 . Hence in this case (i) and (ii) follow immediately from (15.3). Therefore, if f is a simple function, then, using (11.6), (i) and (ii) obtain. Finally, in the X 1 ×X 2 of simple general case choose a nondecreasing sequence { f n : n ∈ N } ⊂ [0, ∞[ functions such that f n → f . Applying (11.22), we have X 2 f (x1 , x2 )dμ2 (x2 ) = limn X 2 f n (x1 , x2 )dμ2 (x2 ), x1 ∈ X 1 , and X 1 f (x1 , x2 )dμ1 (x1 ) = limn X 1 f n (x1 , x2 )dμ1 (x1 ), x2 ∈ X 2 . Hence, using (8.12.ii), assertions (i) and (ii) are proved.
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143
Theorem 15.5. Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be as in (15.3). For A ∈ A1 ⊗A2 , define μ(A) = X 1 μ2 (A x1 )dμ1 (x1 ) and μ (A) = X 2 μ1 (A x2 )dμ2 (x2 ). Then μ and μ are equal measures on A1 ⊗ A2 . Proof. Obviously, μ(∅) = 0. Let {An : n ∈ N } ⊂ A1 ⊗A2 be such that Am ∩ An = ∅ for m = n. Then, applying (11.23), we have μ(∪n N An ) = X 1 μ2 (∪n N (An )x1 )dμ1 (x1 ) = n∈N X 1 μ2 ((An )x1 )dμ1 (x 1 ) = n∈N μ(An ). Hence μ is a measure on in like manner. Let A be A1 ⊗ A2 . The fact that μ is a measure on A1 ⊗ A2 is proved as in (15.1). For A1 × A2 ∈ A, we obtain μ(A1 × A2 ) = X 1 μ2 ((A1 × A2 )x1 )dμ1 (x1 ) = X 1 μ2 (A2 )1 A1 dμ1 (x 1 ) = μ1 (A1 )μ2 (A2 ) = X 2 μ1 (A1 )1 A2 dμ2 (x 2 ) = X 2 μ1 ((A1 × A2 )x2 )dμ2 (x2 ) = μ (A1 × A2 ). Therefore, μ and μ coincide on A. Let B be the family of all finite unions of disjoint sets in A. Then B is an algebra of sets such that σ (B) = A1 ⊗ A2 , and μ(B) = μ (B) for any B ∈ B. Now let {Ai ) : i ∈ I } ⊂ A1 be a countable partition of X 1 such that μ1 (Ai ) < ∞, i ∈ I , and let {B j ) : j ∈ J } ⊂ A2 be a countable partition of X 2 such that μ2 (B j ) < ∞, j ∈ J . For i ∈ I and j ∈ J , we have Ai × B j ∈ A ⊂ B and μ(Ai × B j ) = μ (Ai × B j ) = μ1 (Ai )μ2 (B j ) < ∞. Consequently, μ and μ are σ -finite measures on B. Thus, using (10.9), it follows that μ = μ . Remark 15.6. The measure μ = μ in (15.5) is denoted by μ1 ⊗ μ2 , and it is called the product of μ1 and μ2 . μ1 ⊗ μ2 is the unique measure on A1 ⊗ A2 such that (μ1 ⊗ μ2 )(A1 × A2 ) = μ1 (A1 )μ2 (A2 ) for any A1 ∈ A1 and A2 ∈ A2 . The next theorem generalizes (15.5). Theorem 15.7. Let (X 1 , A1 ) and (X 2 , A2 ) be measurable spaces, let μ1 be a σ -finite measures on A1 , and let μ2 : X 1 × A2 → [0, ∞] be a function such that the section of μ2 at any x1 ∈ X 1 is a measure on A2 and the section of μ2 at any B ∈ A2 is A1 -measurable. Assume that there exists a countable partition {B j : j ∈ J } ⊂ A2 of X 2 such that μ2 (x1 , B j ) k j < ∞, x1 ∈ X 1 , for any j ∈ J . Then there exists a unique measure μ on A1 ⊗ A2 such that μ(A1 × A2 ) = A1 μ2 (x1 , A2 )dμ1 (x1 ) for any A1 ∈ A1 and A2 ∈ A2 . Proof. Let A ∈ A1 ⊗ A2 . Like in (15.3) it is proved that the function x1 → μ2 (x1 , A x1 ) from X 1 into R is A1 -measurable. Define μ(A) = X 1 μ2 (x1 , A x1 )dμ1 (x1 ). Like in (15.5) it is proved that μ is a measure on A1 ⊗ A2 . The details and the rest of the proof are left to the reader. In what follows we prove two versions of Fubini’s theorem. Theorem 15.8 (Fubini). Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be as in (15.3), and let f : X 1 × X 2 → [0, ∞] be an A1 ⊗ A2 -measurable function. Then (i) f d(μ1 ⊗ μ2 ) = f (x1 , x2 )dμ2 (x2 ) dμ1 (x1 ) X 1 ×X 2
X1
X
X2
X1
2
=
f (x1 , x2 )dμ1 (x1 ) dμ2 (x2 ).
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Proof. If f = 1 A , where A ∈ A1 ⊗ A2 , then X 2 ×X 2 1 A d(μ1 ⊗ μ2 ) = (μ1 ⊗ μ2 )(A) = X 1 μ2 (A x1 )dμ1 (x1 ) = X 1 ( X 2 1 A (x1 , x2 )dμ2 (x2 ))dμ1 (x1 ). Using (15.5), in like manner we get X 2 ×X 2 1 A d(μ1 ⊗ μ2 ) = X 2 ( X 1 1 A (x1 , x2 )dμ1 (x1 )) dμ2 (x2 ). Thus (i) holds in this case. Therefore, if f is a simple function, then, using (11.6), (i) obtains. Finally, in the general case choose a nondecreasing sequence { f n : n ∈ N } ⊂ [0, ∞[ X 1 ×X 2 of simple functions such that f n → f . Applying (11.22), we have
f d(μ1 ⊗ μ2 ) = lim
f n d(μ1 ⊗ μ2 )
n
X 1 ×X 2
X 1 ×X 2
f n (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
X1
f n (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
= lim n
X2
=
lim n
X1
= X1
Similarly we get (i) is proved.
X 2 ×X 2
X2
f (x1 , x2 )dμ2 (x2 ) dμ1 (x1 ).
X2
f d(μ1 ⊗ μ2 ) =
X2 ( X1
f (x1 , x2 )dμ1 (x1 ))dμ2 (x2 ), and so
Remark 15.9. The assumption that μ1 and μ2 are σ -finite in (15.8) is essential. To see this, let X 1 = X 2 = [0, 1], A1 = B([0, 1]), A2 = P([0, 1]), let μ1 be the Lebesgue on P([0, 1]), and let f = 1D , measure on B([0, 1]), let μ2 be the countingmeasure where D is the diagonal of [0, 1]2 . Then 1 = X 1 ( X 2 1D (x1 , x2 )dμ2 (x2 ))dμ1 (x1 ) = X 2 ( X 1 1D (x 1 , x 2 )dμ1 (x 1 ))dμ2 (x 2 ) = 0. Theorem 15.10 (Fubini). Let (X 1 , A1 , μ1 ) and (X 2 , A2 , μ2 ) be as in (15.3), and let f : X 1 × X 2 → R be an A1 ⊗ A2 -measurable function such that one of the integrals X 1 ×X 2 | f | d(μ1 ⊗ μ2 ), X 1 ( X 2 | f (x1 , x2 )| dμ2 (x2 ))dμ1 (x1 ) and X 2 ( X 1 | f (x 1 , x 2 )|dμ1 (x 1 ))dμ2 (x 2 ) is finite. Then the following assertions hold. (i) There is A1 ∈ A1 such that μ1 (Ac1 ) = 0 and f x1 ∈ L1 (X 2 , A2 , μ2 ) for any x 1 ∈ A1 . (ii) There is A2 ∈ A2 such that μ2 (Ac2 ) = 0 and f x2 ∈ L1 (X 1 , A1 , μ1 ) for any x 2 ∈ A2 . (iii) The function ϕ1 (x1 ) = 1 A1 (x1 ) X 2 f (x1 , x2 )dμ2 (x2 ) ∈ L1 (X 1 , A1 , μ1 ). (iv) The function ϕ2 (x2 ) = 1 A2 (x2 ) X 1 f (x1 , x2 )dμ1 (x1 ) ∈ L1 (X 2 , A2 , μ2 ). (v) X 1 ×X 2 f d(μ1 ⊗ μ2 ) = X 1 ( X 2 f (x1 , x2 )dμ2 (x2 ))dμ1 (x1 ) = X 2 ( X 1 f (x1 , x2 )dμ1 (x1 ))dμ2 (x2 ).
Measures on Product σ -Algebras
Proof.
145
In view of (15.8), we have | f | d(μ1 ⊗ μ2 ) = | f (x1 , x2 )| dμ2 (x2 ) dμ1 (x1 )
X 1 ×X 2
X1
X2
X2
X1
| f (x1 , x2 )| dμ1 (x1 ) dμ2 (x2 ) < ∞.
=
(1)
Thus, applying (11.15), there is A1 ∈ A1 such that μ1 (Ac1 ) = 0 and X 2 | f (x1 , x2 )| dμ2 (x2 ) < ∞ for any x1 ∈ A1 . Therefore, f x1 ∈ L1 (X 2 , A2 , μ2 ) for any x1 ∈ A1 , and so(i) is proved. Assertion (ii) is proved in like manner. Using (15.4), the functions x1 → X 2 f + (x1 , x2 )dμ2 (x2 ) and x1 → X 2 f − (x1 , x2 )dμ2 (x2 ) from X 1 into [0, ∞] are A1 -measurable, and so the function ϕ1 (x1 ) = 1 A1 (x1 ) X 2 f (x1 , x2 )dμ2 (x2 ) is A1 -measurable. As |ϕ1 (x1 )| X 2 | f (x1 , x2 )| dμ2 (x2 ), x1 ∈ X 1 , (1) shows that ϕ1 ∈ L1 (X 1 , A1 , μ1 ). Similarly it is shown that ϕ2 ∈ L1 (X 2 , A2 , μ2 ). Hence (iii) and (iv) are proved. Further, applying (15.8), we have f d(μ1 ⊗ μ2 ) = f + d(μ1 ⊗ μ2 ) − f − d(μ1 ⊗ μ2 ) X 1 ×X 2
X 1 ×X 2
= X1
X 1 ×X 2
f + (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
X2
− X1
= X1
−
f (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
= In like manner we get and so (v) is proved.
X 1 ×X 2
f + (x1 , x2 )dμ2 (x2 ) −
X2
X1
X2
X2
f − (x1 , x2 )dμ2 (x2 ) dμ1 (x1 )
f (x1 , x2 )dμ2 (x2 ) dμ1 (x1 ).
X2
f d(μ1 ⊗ μ2 ) =
X2 ( X1
f (x1 , x2 )dμ1 (x1 ))dμ2 (x2 ),
Theorem 15.11. Let (X 1 , A1 ), (X 2 , A2 ), μ1 , μ2 , and μ be as in (15.7), and let f : X 1 × X 2 → [0, ∞] be an A1 ⊗ A2 -measurable function. Then the following assertions hold. (i) The function x1 → X 2 f (x 1 , x 2 )μ2 (x 1 , d x 2 ) from X 1 into [0, ∞] is A1 -measurable.
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Analysis and Probability
(ii) Proof.
X 1 ×X 2
f dμ =
X1 ( X2
f (x1 , x2 )μ2 (x1 , d x2 ))dμ1 (x1 ).
Exercise.
Theorem 15.12. Let (X 1 , A1 ), (X 2 , A2 ), μ1 , μ2 and μ be as in (15.7), and let f : X 1 ×X 2 → R be an A1 ⊗A2 -measurable function such that either X 1 ×X 2 | f | dμ < ∞ or X 1 ( X 2 | f (x1 , x2 )|μ2 (x1 , d x2 ))dμ1 (x1 ) < ∞. Then the following assertions hold. (i) There is A ∈ A1 such that μ1 (Ac ) = 0 and X 2 f (x1 , x2 )μ2 (x1 , d x2 ) ∈ R for any x1 ∈ A. (ii) The function ϕ1 (x1 ) = 1 A (x1 ) X 2 f (x1 , x2 )μ2 (x1 , d x2 ) ∈ L1 (X 1 , A1 , μ1 ). (iii) X 1 ×X 2 f dμ = X 1 ( X 2 f (x1 , x2 )μ2 (x1 , d x2 ))dμ1 (x1 ). Proof.
Exercise.
We now construct the product of a finite number of σ -finite measures. Theorem 15.13. For n 2, let (X i , Ai , μi ), 1 i n, be σ -finite measure spaces. n μ ] on A ⊗ · · · ⊗ A such Then there exists a unique measure μ1 ⊗ · · · ⊗ μn [⊗i=1 i 1 n that (i) (μ1 ⊗· · ·⊗μn )(A1 ×· · ·× An ) = μ1 (A1 ) · · · μn (An ),
A 1 ∈ A1 , . . . , A n ∈ An .
Proof. For n = 2 the result is true by (15.5) and (15.6). If the result is true for n, then, applying (15.5), there exists the measure (μ1 ⊗ · · · ⊗ μn ) ⊗ μn+1 on (A1 ⊗ · · · ⊗ An ) ⊗ An+1 such that ((μ1 ⊗ · · · ⊗ μn ) ⊗ μn+1 )((A1 × · · · × An ) × An+1 ) = (μ1 ⊗ · · · ⊗ μn )(A1 × · · · × An )μn+1 (An+1 ) = μ1 (A1 ) · · · μn (An )μn+1 (An+1 ) for Ai ∈ Ai , 1 i n +1. In accordance with (1.2.28) we identify the Cartesian products (X 1 × · · · × X n ) × X n+1 and X 1 × · · · × X n × X n+1 , and in view of (7.25) we identify the σ -algebras (A1 ⊗ · · · ⊗ An ) ⊗ An+1 and A1 ⊗ · · · ⊗ An ⊗ An+1 . Thus the existence of the measure μ1 ⊗ · · · ⊗ μn satisfying (i) is proved by induction. The uniqueness of this measure is proved as in (15.5), by making use of (10.9). The next theorem generalizes (15.13). Theorem 15.14. For n 2, let (X i , Ai , μi ), 1 i n, be measurable spaces, and let μ1 be a σ -finite measure on A1 . For 2 i n, let μi : X 1 × · · · × X i−1 × Ai → [0, ∞] be a function such that the section of μi at any (x1 , . . . , xi−1 ) ∈ X 1 ×· · ·× X i−1 is a measure on Ai and the section of μi at any A ∈ Ai is A1 ⊗· · ·⊗Ai−1 -measurable. For 2 i n, assume that there is a countable partition {Bi j : j ∈ Ji } ⊂ Ai of X i such that μi (x1 , . . . , xi−1 , Bi j ) ki j < ∞, (x1 , . . . , xi−1 ) ∈ X 1 × · · · × X i−1 , for any j ∈ Ji . Then there exists a unique measure μ on A1 ⊗ · · · ⊗ An such that μ(A1 ×· · ·×An ) = A1 μ1 (d x1 ) A2 μ2 (x1 , d x2 ) · · · An−1 μn−1 (x1 , . . . , xn−2 , d xn−1 ) An μn (x 1 , . . . , x n−1 , d x n ) for any A1 ∈ A1 , . . . , An ∈ An . Proof. Using (15.7) and (15.11), the theorem is proved by induction as in (15.13). The details are left to the reader. Exercise 15.15. (a) Find measurable spaces (X 1 , A1 ) and (X 2 , A2 ), and A ⊂ X 1 × X 2 such that A x1 ∈ A2 for any x1 ∈ X 1 , A x2 ∈ A1 for any x2 ∈ X 2 , but A ∈ / A1 ⊗ A2 .
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147
(b) Find measurable spaces (X 1 , A1 ) and (X 2 , A2 ), and f : X 1 × X 2 → R such that f x1 is A2 -measurable for any x1 ∈ X 1 , f x2 is A1 -measurable for any x2 ∈ X 2 , but f is not A1 ⊗ A2 -measurable. Exercise 15.16. Let (X 1 , A1 ) and (X 2 , A2 ) be measurable spaces, and let ∅ = A1 ⊂ X 1 and ∅ = A2 ⊂ X 2 . Show that A1 × A2 ∈ A1 ⊗ A2 if and only if A1 ∈ A1 and A2 ∈ A2 . Exercise 15.17.
Use (1.2.45.f) to give another proof of (15.2).
Exercise 15.18. Let λ be the Lebesgue measure on B(R). Prove that λ ⊗ λ is not a complete measure on B(R)λ ⊗ B(R)λ . [Use (10.17.d) and (15.16).] Exercise 15.19. Let (X, A, μ) be a σ -finite measure space, let f : X → [0, ∞] be an A-measurable function, and let V∗ f = {(x, y) ∈ X × [0, ∞] : y < f (x)} and V ∗ f = {(x, y) ∈ X × [0, ∞] : y f (x)}. Prove the following. (a) V∗ f ∈ A ⊗ B([0, ∞]), V ∗ f ∈ A ⊗ B([0, ∞]). (b) If λ is the Lebesgue measure on B([0, ∞]), then (μ ⊗ λ)(V∗ f ) = (μ ⊗ λ)(V ∗ f ) = X f dμ. Exercise 15.20. Let X 1 = X 2 = N , A1 = A2 = P(N ), and μ1 = μ2 = counting measure on P(N ). Define f : X 1 × X 2 → Rby f (i, i) = i, f (i, i +1) = −i, and f (i, j) = 0 if j = i or j = i + 1. Show that X 1 ( X 2 f (x1 , x2 )dμ2 (x2 ))dμ1 (x1 ) = 0 and X 2 ( X 1 f (x1 , x2 )dμ1 (x1 ))dμ2 (x2 ) = ∞. Exercise 15.21. Let f : R → R be a bijective function, let μ1 be the Lebesgue measure on B(R), let μ2 : R × B(R) → [0, ∞] be defined by μ2 (x, A) = ε f (x) (A) for any x ∈ R and A ∈ B(R), and let μ be as in (15.14). For (x1 , x2 ) ∈ R 2 , define F(x1 , x2 ) =
if f is strictly increasing x1 ∧ f −1 (x2 ) . (x1 − f −1 (x2 )) ∨ 0 if f is strictly decreasing
Prove the following. (a) F is continuous on R 2 . (b) ba F 0 whenever a, b ∈ R 2 are such that a b. (c) μ = μ F , where μ F is as in (3.4.24). (d) μ(Rc ) = 0, where R is the graph of f . Exercise 15.22. For n 2 and i = 1, . . . , n, let (X i , Ai ) be a measurable space, let μi and νi be σ -finite measures on Ai such that νi μi , and let f i = dνi /dμi . Put f (x1 , . . . , xn ) = f 1 (x1 ) · · · f n (xn ), (x1 , . . . , xn ) ∈ X 1 × · · · × X n . (a) Show that ν1 ⊗ · · · ⊗ νn μ1 ⊗ · · · ⊗ μn . (b) Use (11.28), (15.8) and (10.8) to prove that f = dν1 ⊗ · · · ⊗ νn /dμ1 ⊗ · · · ⊗ μn . Exercise 15.23. Let (X, A, μ) be a measure space, let f : X → [0, ∞] be an A-measurable function, and let a, p ∈ ]0, ∞[. ∞ (a) Use (15.19.a) and Fubini’s theorem (15.8) to show that X f dμ = 0 μ({ f > ∞ ∞ x})d x = 0 μ({ f x})d x, and that X ( f − a)+ dμ = a μ({ f > x})d x = ∞ a μ({ f x})d x.
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a (b) If μ(X ) < ∞, prove that { f
16
The Product of Infinitely Many Measures
In this section we construct the product of infinitely many probabilities, and we prove the Daniell-Kolmogorov extension theorem. In (15.13) we constructed the product of a finite number of σ -finite measures. In what follows we construct the product of infinitely many probabilities. We begin with a lemma. Lemma 16.1. Let {(X i , Ai , Pi ) : i ∈ I } be a finite set of probability spaces, and let ∅ = J ⊂ I . Let PI be the product of the probabilities Pi , i ∈ I , and let PJ be the product of the probabilities Pi , i ∈ J . Then PI (π JI )−1 = PJ . Proof. In view of (15.13), PI is a probability on ⊗i∈I Ai , and PJ is a probability on ⊗i∈J Ai . Thus PI (π JI )−1 and PJ are probabilities on ⊗i∈J Ai . For i ∈ J , let Ai ∈ Ai .
Measures on Product σ -Algebras
Then (π JI )−1 (
Ai ) =
i∈J
(PI (π JI )−1 )
149
i∈I
Ai
= PI
Ai , where Ai =
i∈J
Ai
=
i∈I
Ai if i ∈ J . Therefore, / J X i if i ∈
Pi (Ai )
i∈I
=
i∈J
Pi (Ai ) = PJ
Ai .
i∈J
(1)
Since the family { i∈J Ai : Ai ∈ Ai for i ∈ J } generates ⊗i∈J Ai , (1) and (10.8) show that PI (π JI )−1 = PJ . Definition 16.2. Let {(X i , Ai , Pi ) : i ∈ I } be a nonempty set of probability spaces, and let J and A be as in (7.21). For each J ∈ J , let PJ denote the product of the probabilities P j , j ∈ J . For A ∈ A, choose J ∈ J such that A ∈ A J . Therefore, A = π J−1 (B) for some B ∈ ⊗ j∈J A j . Define Q(A) = PJ (B). Theorem 16.3. measure on A.
Notation is as in (16.2). Then Q is a well-defined finitely additive
Proof. Let us show that Q is well defined on A. For A ∈ A, let J1 ∈ J and J2 ∈ J be such that A ∈ A J1 and A ∈ A J2 . Therefore, A = π J−1 (B1 ) = π J−1 (B2 ) 1 2
(1)
for some B1 ∈ ⊗ j∈J1 A j and B2 ∈ ⊗ j∈J2 A j . Set J = J1 ∪ J2 . In view of (1.2.24), we have π J1 = π JJ1 ◦ π J and π J2 = π JJ2 ◦ π J . Hence, using (1), we get π J−1
π JJ1
−1
(B1 ) =
π J−1
π JJ2
−1
(B2 ) .
(2)
Since the function π J is surjective, (2) shows that −1 −1 π JJ1 (B1 ) = π JJ2 (B2 ).
(3)
By (16.1) and (3), we have PJ1 (B1 ) = (PJ (π JJ1 )−1 )(B1 ) = PJ ((π JJ1 )−1 (B1 )) = PJ ((π JJ2 )−1 (B2 )) = PJ2 (B2 ). Thus Q is well defined on A. Evidently, Q(∅) = 0. Now let A1 , A2 ∈ A be such that A1 ∩ A2 = ∅. Choose J1 ∈ J and J2 ∈ J such that A1 ∈ A J1 and A2 ∈ A J2 . Put J = J1 ∪ J2 . Then, in view of (7.20), we have A1 , A2 ∈ A J . Therefore, A1 = π J−1 (B1 ) and A2 = π J−1 (B2 ), where B1 , B2 ∈ ⊗ j∈J A j . We have ∅ = A1 ∩ A2 = π J−1 (B1 ∩ B2 ).
(4)
Since π J is surjective, (4) shows that B1 ∩ B2 = ∅. Since A1 ∪ A2 = π J−1 (B1 ∪ B2 ), we have Q(A1 ∪ A2 ) = PJ (B1 ∪ B2 ) = PJ1 (B1 ) + PJ2 (B2 ) = Q(A1 ) + Q(A2 ). Thus Q is finitely additive.
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Analysis and Probability
Notation 16.4. Let A be as in (7.21), and let Q be as in (16.2). To emphasize the dependence of A and Q on the index set I we will write A I instead of A and Q I instead of Q. Theorem 16.5. Let {(X i , Ai , Pi ) : i ∈ I } be a set of probability spaces, let A ∈ A I , and let J be a nonempty finite subset of I. Then the following assertions hold:
(i) Ax ∈ A I −J for any x ∈ i∈J X i . Q I −J (Ax )d PJ (x), where PJ stands for the product of the (ii) Q I (A) =
i∈J
probabilities Pi , i ∈ J .
Xi
Proof. Let J be as in (7.21). There is K ∈ J such that A ∈ A K . Set L = J ∪K . Then,
I )−1 (B) = B × by (7.20), we have
A ∈ A L . Therefore, , where B ∈
A = (π L
i∈I −L X i
⊗i∈L Ai . Let x ∈ i∈J X i . Since i∈I X i = i∈J X i × i∈L−J X i × i∈I −L X i , we have Ax = (y, z) ∈ Xi × X i : (x, y, z) ∈ A = B × Xi
i∈L−J
i∈I −L
= (y, z) ∈
Xi ×
i∈L−J
= Bx ×
i∈I −L
X i : (x, y) ∈ B
i∈I −L
I −J −1 X i = (π L−J ) (Bx ).
(1)
i∈I −L
According to (15.1), we have Bx ∈ ⊗i∈L−J Ai . Consequently, (1) shows that Ax ∈ A I −J . Thus (i) holds. Moreover, we have Q I −J (Ax ) = PL−J (Bx ),
(2)
where PL−J is the product of the probabilities Pi , i ∈ L − J . Let PL denote the product of the probabilities Pi , i ∈ L. Then, using(2), we have Q I (A) = PL (B) = (PJ ⊗ PL−J )(B) = X i PL−J (Bx )d PJ (x) = X i Q I −J (Ax )d PJ (x). Hence i∈J i∈J (ii) holds. Theorem 16.6. Let {(X i , Ai , Pi ) : i ∈ I } be a denumerable set of probability spaces. Then there exists a unique probability P on ⊗i∈I Ai such that (i) Pπ J−1 = PJ for any nonempty finite set J ⊂ I , where PJ stands for the product of the probabilities Pj , j ∈ J . Proof. Without loss of generality, assume that I = N . For k ∈ N , denote Ik = {k +1, k +2, . . .}. Let A be as in (7.21), and let Q be as in (16.2). In view of (16.3), Q is a finitely additive measure on A. We next show that Q is a σ -additive measure on A. To do this, let {An : n ∈ N } ⊂ A be a nonincreasing sequence such that limn Q(An ) > 0. Then, applying (9.8), it will suffice to show that ∩n∈N An = ∅. Using (16.5), we have Q(An ) = Q I (An ) = X 1 Q I1 ((An )x1 )d P1 (x1 ). Thus the dominated convergence theo rem (11.25.ii) shows that 0 < limn Q(An ) = X 1 limn Q I1 ((An )x1 )d P1 (x1 ). Therefore,
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151
there is x1 ∈ X 1 such that limn Q I1 ((An )x1 ) > 0. If x1 ∈ X 1 , . . . , xk ∈ X k have been chosen so that limn Q Ik ((An )(x1 ,...,xk ) ) > 0, then, using (16.5) and (11.25.ii), we have Ik lim Q Ik+1 (An )(x1 ,...,xk ,xk+1 ) d Pk+1 (xk+1 ). 0 < lim Q (An )(x1 ,...,xk ) = n
X k+1 n
Consequently, there is xk+1 ∈ X k+1 such that limn Q Ik+1 ((An )(x1
,...,xk ,xk+1 ) ) > 0. Thus we have proved inductively that there is a point x = (xk )k∈N ∈ k∈N X k such that lim Q Ik ((An )(x1 ,...,xk ) ) > 0, k ∈ N . n
(1)
Equation (1) shows that (An )(x1 ,...,xk ) = ∅, k, n ∈ N .
(2)
For n ∈ N , select a nonempty finite set J ⊂ I such that An ∈ A J . Let k = sup J . −1 (B) = B × i>k X i , where B ∈ Then An ∈ A{1,...,k} . Therefore, An = π{1,...,k} A1 ⊗ · · · ⊗ Ak . Since
i>k X i if (x 1 , . . . , x k ) ∈ B , (An )(x1 ,...,xk ) = / B ∅ if (x1 , . . . , xk ) ∈ (2) shows that (x1 , . . . , xk ) ∈ B. Consequently, x ∈ An .
Thus x ∈ ∩n∈N An = ∅. Therefore, Q is a σ -additive measure on A. Obviously, Q( i∈I X i ) = 1. Using (7.21), Carathéodory’s extension theorem (10.9) shows that there exists a unique probability P on ⊗i∈I Ai such that P = Q on A. From this it follows at once that P is the only probability on ⊗i∈I Ai that satisfies (i). The probability P in (16.6) is called the product of the probabilities Pi , i ∈ I . Ai , Pi ) : i ∈ I } and P be as in (16.6). If Ai ∈ Ai for any Theorem 16.7. Let {(X i ,
i ∈ I , then P( i∈I Ai ) = i∈I Pi (Ai ). Proof. Without any loss
of generality, suppose that I = N . For n ∈ N , put × · · · × An × i>n X i . Then {Bn : n ∈ N } is a nonincreasing Bn = A1
sequence such that n∈N An = ∩n∈N Bn . Therefore, using (16.6), we have P( n∈N An ) = limn P(Bn ) = limn (Pπ{1,...,n}−1 )(A
1 × · · · × An ) = limn (P1 ⊗ · · · ⊗ Pn )(A1 × · · · × An ) = limn P1 (A1 ) · · · Pn (An ) = n∈N Pn (An ). The next theorem generalizes (16.6). Theorem 16.8 (Ionescu Tulcea). Let {(X n , An ) : n ∈ N } be a sequence of measurable spaces, and let P1 be a probability on A1 . For n 2, let Pn : X 1 × · · · × X n−1 × An → [0, 1] be a function such that the section of Pn at any (x1 , . . . , xn−1 ) ∈ X 1 × · · · × X n−1 is a probability on An and the section of Pn at any A ∈ An is A1 ⊗ · · · ⊗ An−1 -measurable. Then there exists a unique probability P on ⊗n∈N An −1 such that Pπ{1,...,n} = P (n) , n ∈ N , where P (n) is the probability on A1 ⊗ · · · ⊗ An constructed as in (15.14) starting from P1 , . . . , Pn . The proof of this theorem is similar to the proof of (16.6) and is left to the reader. In what follows we extend Theorem (16.6) to the case where I is an arbitrary infinite set.
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Lemma 16.9. Let {(X i , Ai , Pi ) : i ∈ I } be a denumerable set of probability spaces, and let ∅ = J ⊂ I . Let PI be the product of the probabilities Pi , i ∈ I , and let PJ be the product of the probabilities P j , j ∈ J . Then (i) PI (π JI )−1 = PJ . Proof. PI (π JI )−1 and PJ are probabilities on ⊗ j∈J A j . Let K be a nonempty finite subset of J , and let PK be the product of the probabilities Pk , k ∈ K . Then, according to (16.6), we have PI (π JI )−1 (π KJ )−1 = PI (π KI )−1 = PK and PJ (π KJ )−1 = PK . Thus the uniqueness assertion in (16.6) shows that (i) holds. Definition 16.10. Let {(X i , Ai , Pi ) : i ∈ I } be a nonempty set of probability spaces, and let J ∗ be as in (7.22). For each J ∈ J ∗ , let PJ denote the product of the probabilities P j , j ∈ J . As proved in (7.22), ⊗i∈I Ai = ∪ J ∈J ∗ A J , where A J is as in (7.20). For A ∈ ⊗i∈I Ai , choose J ∈ J ∗ such that A ∈ A J . Therefore, A = π J−1 (B) for some B ∈ ⊗ j∈J A j . Define P(A) = PJ (B). Using (16.9), a reasoning similar to the one in the proof of (16.3) shows that P is well defined. Theorem 16.11. Notation is as in (16.10). Then P is the unique probability on ⊗i∈I Ai such that (i) Pπ J−1 = PJ for any nonempty finite set J ⊂ I , where PJ stands for the product of the probabilities Pj , j ∈ J .
Proof. Obviously, P(∅) = 0 and P( i∈I X i ) = 1. Let {An : n ∈ N } ⊂ ⊗i∈I Ai be such that Am ∩ An = ∅ for m = n. For n ∈ N , choose Jn ∈ J ∗ such that An ∈ A Jn . Set J = ∪n∈N Jn . Then J ∈ J ∗ , and An ∈ A J for any n ∈ N . Therefore, for n ∈ N , we have An = π J−1 (Bn ) for some Bn ∈ ⊗ j∈J A j . For m = n, we have −1 ∅ = Am ∩ An = π J−1 (Bm ∩ Bn ), and so Bm∩ Bn = ∅. Since ∪ n∈N An = π J (∪n∈N Bn ), we get P(∪n∈N An ) = PJ (∪n∈N Bn ) = n∈N PJ (Bn ) = n∈N P(An ). Thus P is a probability on ⊗i∈I Ai . Evidently, P satisfies (i). Using (7.21) and (10.9), it is easily seen that P is the only probability on ⊗i∈I Ai that satisfies (i). The probability P in (16.11) is called the product of the probabilities Pi , i ∈ I , and is denoted by ⊗i∈I Pi . If I = {n ∈ Z : n m} for some m ∈ Z , then we will write also ⊗nm Pn instead of ⊗i∈I Pi . One of the essential results of this chapter is the Daniell-Kolmogorov extension theorem which we will prove in (16.20). We begin with a definition. Definition 16.12. A family B ⊂ P(X ) is said to be compact if {Bn : n 1} ⊂ B n B = ∅ for any n 1 imply ∩ and ∩i=1 i i1 Bi = ∅. Example 16.13. (a) The family of all finite subsets of an arbitrary set X is compact. (b) Let X be a Hausdorff space, and let B be the family of all compact subsets of X . Then, according to (5.24) and (5.27), B is compact. Remark 16.14. (a) If B ⊂ P(X ) is compact, then B ∪ {X } is compact. (b) If B is compact and C ⊂ B, then C is compact.
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(c) For i ∈ I = ∅, let X i be a set, and let Bi
⊂ P(X i ) be a compact family. Then, in view of (2.44), the family ∪i∈I πi−1 (Bi ) ⊂ P( i∈I X i ) is compact. be the family of all denumerable Theorem 16.15. Let B be a compact family, and let B intersections of finite unions of sets in B. Then the following assertions hold. is compact. (i) B then ∩n1 An ∈ B. (ii) If {An : n 1} ⊂ B, then A ∪ B ∈ B. (iii) If A, B ∈ B, where C k = Ak ∪ · · · ∪ Proof. To prove (i), for k 1, let B k = ∩i1 Cik ∈ B, i i1 k n k Aili (k) , i 1, is such that ∩k=1 B = ∅ whatever n 1. We arrange the sets Cik =
i (k) k m Ai j , i, k 1, in a sequence {D m : m 1}, where D m = Am ∪lj=1 1 ∪ · · · ∪ Arm , m 1. Since ∅ = ∩nk=1 B k = ∩nk=1 ∩i1 Cik , n 1, it follows that ∩nm=1 D m = ∅ for any n 1. Consequently, if n 2, there is s1 with 1 s1 r1 such that A1s1 ∩ D 2 ∩ · · · ∩ D n = ∅. If s1 , . . . , sl have been chosen so that A1s1 ∩ · · · ∩ Alsl ∩ Dl+1 ∩ · · · ∩ D n = ∅ for any n l + 1, then there is sl+1 with 1 sl+1 rl+1 l+2 ∩ · · · ∩ D n = ∅, n l + 2. Thus it such that A1s1 ∩ · · · ∩ Alsl ∩ Al+1 sl+1 ∩ D follows inductively that ∩li=1 Aisi = ∅ for any l 1. As Aisi ∈ B, i 1, we have ∅ = ∩i1 Aisi ⊂ ∩m1 D m = ∩k1 B k . This completes the proof of (i). Assertion (ii) is obvious, and assertion (iii) follows from (ii).
Definition 16.16. Let A be an algebra of sets, let B ⊂ A be a compact family, and let μ be a finitely additive measure on A. We say that μ is regular [B-regular] if μ(A) = sup{μ(B) : A ⊃ B ∈ B} for any A ∈ A. Theorem 16.17. Let A be an algebra of subsets of X, let B ⊂ A be a compact family, and let μ be a finitely additive measure on A such that μ(X ) < ∞. If μ is B regular, then μ is σ -additive. Proof. Let {An : n ∈ N } ⊂ A be such that ∪n∈N An ∈ A and Am ∩ An = ∅ A ) assume for m = n. Evidently, μ(∪ n∈N n n∈N μ(An ). To get a contradiction that μ(∪n∈N An ) > n∈N μ(An ), and let 0 < ε μ(∪n∈N An ) − n∈N μ(An ). For n A ) ∈ A and n ∈ N , set Bn = ∪i>n Ai . Then Bn ⊃ Bn+1 , Bn = ∪i∈N Ai − (∪i=1 i μ(Bn ) = μ
i∈N
Ai
−
n i=1
μ(Ai ) μ
i∈N
Ai
−
μ(Ai ) ε, n ∈ N .
i∈N
(1) For n ∈ N , select Cn ∈ B such that Cn ⊂ Bn and μ(Bn ) − μ(Cn ) ε/2n+1 .
(2)
n C ) = μ(∩n B ) −μ(∩n B −(∩n C )) Then, using (1) and (2), we have μ(∩i=1 i i=1 i i=1 i i=1 i n n n C = ∅ μ(Bn )−μ(∪i=1 (Bi −Ci )) ε− i=1 ε/2i+1 > ε/2, n ∈ N . Therefore, ∩i=1 i for any n ∈ N . Since B is compact, we get ∩i∈N Ci = ∅. On the other hand, we have ∩n∈N Cn ⊂ ∩n∈N Bn = ∅. This contradiction shows that μ is σ -additive.
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Theorem 16.18. Let (X, A, μ) be a finite measure space, and let B ⊂ A be such that {An : n ∈ N } ⊂ B implies ∩n∈N An ∈ B, and A, B ∈ B implies A ∪ B ∈ B. Let C = {A ∈ A : μ(A) = sup{μ(B) : A ⊃ B ∈ B}}. Then the family {A ∈ A : A ∈ C and Ac ∈ C} is a σ -algebra unless it is empty. Proof. Let {An : n ∈ N } ⊂ C. It suffices to show that ∪n∈N An ∈ C and ∩n∈N An ∈ C. Let α < μ(∪n∈N An ), and choose m ∈ N such that α < μ(∪m n=1 An ). For n = 1, . . . , m, n select Bn ∈ B such that Bn ⊂ An and μ(An ) − μ(Bn ) (μ(∪m n=1 An ) − α)/2 . Then we have m
m
m
m Bn = μ An − μ An − Bn μ n=1
μ μ
n=1 m
n=1 m
An −
n=1 m
n=1
μ(An − Bn )
n=1
An − μ
n=1
m
An − α
= α.
(1)
n=1
m m Since ∪m n=1 Bn ⊂ ∪n=1 An and ∪n=1 Bn ∈ B, (1) shows that ∪n∈N An ∈ C. Now let ε > 0. For n ∈ N , select Cn ∈ B such that Cn ⊂ An and μ(An ) − μ(Cn ) ε/2n . Then we have
Cn = μ An − μ An − Cn μ n∈N
n∈N
n∈N
μ μ
An −
n∈N
n∈N
μ(An − Cn )
n∈N
An − ε.
(2)
n∈N
Since ∩n∈N Cn ⊂ ∩n∈N An and ∩n∈N Cn ∈ B, (2) shows that ∩n∈N An ∈ C. Theorem 16.19. Let {(X i , Ai ) : i ∈ I } be a nonempty family of measurable spaces, and let μ be a finite measure on ⊗i∈I Ai such that μπi−1 is regular for any i ∈ I . Then μ is regular. Proof. For i ∈ I , let Bi ⊂ Ai be a compact family such that μπi−1 is B i -regular. In view of (16.14.a), we may and do assume that X i ∈ Bi . Let B = { i∈I Bi : Bi ∈ Bi , i ∈ I, and {i ∈ I : Bi = X i } is finite}. Obviously, B ⊂ ⊗i∈I Ai . be as in (16.15). Using (1.2.44), it follows at once that B is a compact family. Let B ⊂ ⊗i∈I Ai . We will show that μ is B-regular. Since ⊗i∈I Ai = Evidently, B σ (∪i∈I πi−1 (Ai )), and A ∈ ∪i∈I πi−1 (Ai ) implies Ac ∈ ∪i∈I πi−1 (Ai ), (16.18) for any shows that it suffices to prove that μ(A) = sup{μ(B) : A ⊃ B ∈ B} A ∈ ∪i∈I πi−1 (Ai ). Let A ∈ πi−1 (Ai ). Therefore, A = πi−1 (B) for some B ∈ Ai .
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For ε > 0, choose C ∈ Ai such that C ⊂ B and (μπi−1 )(C) > (μπi−1 )(B) − ε. Consequently (μπi−1 )(C) > (μπi−1 )(B) − ε = μ(A) − ε
(1)
(1) shows that μ is B-regular. Since πi−1 (C) ∈ B ⊂ B, Daniell-Kolmogorov extension theorem 16.20. Let {(X i , Ai ) : i ∈ I } be a nonempty family of measurable spaces, and let J be as in (7.21). For any J ∈ J , let μ J be a finite measure on ⊗ j∈J A j such that μ J (π KJ )−1 = μ K whenever ∅ = K ⊂ J . If μ{i} is regular for any i ∈ I , then there exists a unique measure μ on ⊗i∈I Ai such that (i) μπ J−1 = μ J for each J ∈ J . Proof. Let A be as in (7.21). For A ∈ A, choose J ∈ J such that A ∈ A J . Therefore, A = π J−1 (B) for some B ∈ ⊗ j∈J A j . Define μ(A) = μ J (B). As in (16.3), it is proved that μ is a well-defined finitely additive measure on A. For i ∈ I , let Bi ⊂ Ai be a compact family such that μ{i} is B i -regular. According to (16.14.a), we may and do assume that X i ∈ Bi . Let B = { i∈I Bi : Bi ∈ Bi , i ∈ I, and {i ∈ I : Bi = be as in (16.15). Let A = π −1 (B), where B ∈ ⊗ j∈J A j . In view X i } is finite}, and let B J J ⊂ ⊗ j∈J A j such that μ J is B J -regular. of (16.19), there exists a compact family B For ε > 0, select C ∈ B J such that C ⊂ B and μ J (C) > μ J (B) − ε. Therefore, μπ J−1 (C) > μπ J−1 (B) = μ(A) − ε.
(1)
Since π J−1 (C) ⊂ π J−1 (B), (1) shows that μ is regular with respect to the family J ) ⊂ A. As ∪ J ∈J π −1 (B J ) ⊂ B, (16.14.b) shows that ∪ J ∈J π −1 (B J ) is ∪ J ∈J π J−1 (B J J compact. Using (16.17), it follows that μ is σ -additive. Using (7.21), Carathéodory’s extension theorem (10.9) shows that μ has a unique extension to ⊗i∈I Ai . From this it follows at once that μ is the only measure on ⊗i∈I Ai that satisfies (i). Theorem (16.23) below shows that the Daniell-Kolmogorov extension theorem applies whenever each X i is a complete separable metric space and Ai = B(X i ); in particular, it applies whenever each X i is a countable set and Ai = P(X i ). We need the following intermediate results. Theorem 16.21. Let X be a metric space, let B be the family of all closed subsets of X and let μ be a finite measure on B(X ). Then μ(A) = sup{μ(B) : A ⊃ B ∈ B} for any A ∈ B(X ). Proof. Let A = B(X ). Notice that B fulfills the assumptions in (16.18). Further, let C be as in (16.18), and put D = {A ∈ A : A ∈ C and Ac ∈ C}. Clearly, it suffices to prove that C ⊃ B(X ). Let A ∈ B. Plainly, A ∈ C. Moreover, according to (6.85), there n B ∈ B, n ∈ N , is a sequence {Bn : n ∈ N } ⊂ B such that Ac = ∪n∈N Bn . Since ∪i=1 i n c c and μ(A ) = sup μ(∪i=1 Bi ) (9.9), it follows that A ∈ C. Thus A ∈ D, and so D ⊃ B. Hence, in view of (16.18), we see that D ⊃ B(X ). As C ⊃ D, we get C ⊃ B(X ).
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Theorem 16.22 (Ulam). Let (X, d) be a complete separable metric space, and let μ be a finite measure on B(X ). Then for each ε > 0, there exists a compact set C ⊂ X such that μ(X ) − μ(C) < ε. Proof. Since X is separable, there is a countable set A dense in X. The case in which A is finite is simpler and it is left to the reader. If A is denumerable, then put {A = xi : i 1}. For i, n 1, set Sin = S(xi , 1/n). For every n 1, as A is dense in X we get X = ∪i1 Sin , and so, by (9.9), there exists i n 1 such that i
n μ Sin > μ(X ) − ε/2n . (1) i=1 n Sin , n 1, and C = ∩n1 Cn . Obviously, C is a totally bounded Denote Cn = ∪ii=1 set. Since C is closed, and X is complete, (6.25.b) shows that (C, dC ) is a complete metric space. Hence, in view of (6.31), it follows that C is compact.Finally, on account of (1), we have μ(X ) − μ(C) = μ(∪n1 Cnc ) n1 μ(Cnc ) < n1 ε/2n = ε.
Theorem 16.23.
Let (X, d) and μ be as in (16.22). Then μ is regular.
Proof. Let C be the family of all compact subsets of X . Then, in view of (16.13.b), C is compact. We will show that μ is C-regular. Let A ∈ B(X ) and ε > 0. Applying (16.21), select a closed set B ⊂ X such that μ(A) − μ(B) < ε/2.
(1)
Then, by (6.25.b) and (6.70), (B, d B ) is a complete separable metric space. On the other hand, according to (7.14) and (7.48), we have B(B) = B ∩ B(X ) ⊂ B(X ). Hence (16.22) applies, and we can select a compact subset C of B such that μ(B) − μ(C) < ε/2.
(2)
By virtue of (4.31), C ∈ C. Moreover, combining (1) with (2), we get μ(A)−μ(C) < ε. Thus μ is C-regular. Exercise 16.24. For 1 < b ∈ N , let I = {0, . . . , b − 1}. For n ∈ N , set In = I and define a probability Pn on P(In ) by Pn ({i}) = 1/b, i ∈ In . Let λ be the Lebesgue measure on B([0, 1[), and let P be the product of the probabilities Pn , n ∈ N . Prove that λ(an )−1 n∈N = P, where the function (an )n∈N is as in (8.23). Exercise 16.25 (Kakutani). For n ∈ N , let (X n , An ) be a measurable space, let Pn and Q n be probabilities on An such that Pn Q n , and let f n = d Pn /d Q n . Let P and , and Q n , n ∈ N , respectively. Q denote the products of the probabilities
Pn , n ∈ N1/2 (a) Show that P Q whenever n∈N X n f n d Q n > 0. [Hints. For x =
and (x1 , x2 , . . .) ∈ n∈N X n , define gn (x) = ( f 1 (x1 ) · · · f n (xn ))1/2 . Use (11.29)
n ∈ N } is a Cauchy sequence in L 2 ( n∈N X n , (15.8) to prove that {gn :
⊗n∈N An , Q), and let g ∈ L2 ( n∈N X n , ⊗n∈N An , Q) be such that g − gn 2 → 0 −1 (13.16). For A ∈ π{1,...,m} (A ⊗ · · · ⊗ Am ), use (15.22) and (11.29) to show that 2 1 P(A) = limn A gn d Q = A g 2 d Q. Then apply (10.8).]
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1/2 (b) Prove that P ⊥ Q whenever n∈N X n f n d Q n = 0. [Hints. For ε > 0,
m
1/2 choose m ∈ N such that n=1 X n f n d Q n < ε. Put B = {x ∈ n∈N X n : gm (x) 1}. Show that Q(B) < ε and P(B c ) < ε, and apply (12.25).]
(c) Let ρ be as in (13.26). Use (13.5) and (12.23.d) to prove that ρ(P, Q) = n∈N ρ(Pn , Q n ). Exercise 16.26. For i ∈ I = ∅, let Ai be a σ -algebra, and let Pi and Q i be probabilities on Ai . Let P and Q stand for the products of the probabilities Pi , i ∈ I , and Q i , i ∈ I , respectively. Show that P Q implies that Pi Q i for any i ∈ I. Exercise 16.27.
If B ⊂ P(X ) is finite, then B is compact.
Exercise 16.28. Let Bi ⊂ P(X ), i = 1, 2, be compact families, and define B = {B1 ∪ B2 : B1 ∈ B1 and B2 ∈ B2 }. Is B compact? Exercise 16.29. Let A and B be as in (16.16), and let μ1 , . . . , μn be B-regular finitely additive measures on A. Show that μ1 + · · · + μn is regular. Exercise 16.30. Let (X, A, μ), (Y, B) and f be as in (9.26), let C ⊂ B be a compact family, and assume that f is surjective. Prove the following. (a) f −1 (C) is a compact family. (b) μ f −1 (B) is f −1 (C)-regular if and only if μf −1 is C-regular. Exercise 16.31. Let A be a σ -algebra, let μ be a σ -finite regular measure on A, and let ν be a finite measure on A such that ν μ. Prove that ν is regular. Exercise 16.32. Let X be a complete separable metric space, let μ be a finite measure on B(X ) such that μ({x}) = 0 for any x ∈ X , let B ∈ B(X ), and consider c ∈ [0, μ(B)]. Use (9.42.b) and (16.23) to prove that there exists a sequence {Cn : n ∈ N } of compact subsets of X such that Cn ⊂ B, n ∈ N , and μ(∪n∈N Cn ) = c. Exercise 16.33. Find a metric space X and a measure μ on B(X ) such that μ is not B-regular, where B stands for the family of all compact subsets of X . [Do not forget (9.55).] Exercise 16.34. Let (X, d) be a metric space, and let μ be a finite measure on B(X ). Show that μ(A) = inf{μ(U ) : A ⊂ U ∈ Td } for any A ∈ B(X ). Exercise 16.35. Let {(X i , Ai , Pi ) : i ∈ I } be a nonempty set of probability spaces, and let {Ja : a ∈ A} be a partition of I such that Ja = ∅, a ∈ A. Prove that ⊗i∈I Pi = ⊗a∈A (⊗ j∈Ja P j ). Exercise 16.36. For i ∈ I = ∅, let (X i , Ai , Pi ) be a probability space, let Bi ) be a measurable space, and let h i : X i → Yi be a measurable function. Define (Yi ,
h : i∈I X i → i∈I Yi by h((xi )i∈I ) = (h i (xi ))i∈I . Prove the following.
(a) For Bi ⊂ Yi , i ∈ I , h −1 ( i∈I Bi ) = i∈I h i−1 (Bi ). (b) h is (⊗i∈I Ai , ⊗i∈I Bi )-measurable. [Use (8.4) and (a).] (c) (⊗i∈I Pi )h −1 = ⊗i∈I Pi h i−1 . [Hints. For nonempty finite J ⊂ I , let π JX denote
the projection from i∈I X i onto j∈J X j , let π JY denote the projection from i∈I Yi
onto j∈J Y j , and let B j ∈ B j , j ∈ J . Then, by (a) and (16.11),
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Analysis and Probability
Pi h −1 (π JY )−1 ⎝
i∈I
=
⎛
=⎝
⎞⎛
Pj ⎠ ⎝
⎞ ⎠ h −1 j (B j )
⎞
⎠ h −1 j (B j )
j∈J
Pj
i∈I
Bj⎠
j∈J
⎛ ⎞⎛ ⎞ ⎝ ⎠⎝ h −1 P j h −1 Bj⎠ j (B j ) = j
j∈J
=
⎞
j∈J
Pi (π JX )−1 ⎝
i∈I
⎛
j∈J
=
⎛
Pi h i−1
j∈J
⎞
⎛ Bj⎠ . (π JY )−1 ⎝
j∈J
j∈J
This means that ((⊗i∈I Pi )h −1 )(π JY )−1 = (⊗i∈I Pi h i−1 )(π JY )−1 . Apply next (16.11).] N } be a sequence Exercise 16.37. Let {(X j , d j ) : j ∈
of metric spaces. For n ∈ N , let π{1,...,n} denote the projection from j∈N X j onto nj=1 X j , let Cn be the family of
−1 all compact subsets of nj=1 X j , let Bn = π{1,...,n} (Cn ), and put B = ∪n∈N Bn . Show that the following assertions are equivalent: (i) B is compact; (ii) if {n k : k ∈ N } ⊂ N is strictly increasing and Bk ∈ Bn k , k ∈ N , with k B = ∅ for any k ∈ N , then ∩ ∩i=1 i i∈N Bi = ∅; n B = ∅ for any n ∈ N , then ∩ (iii) if Bn ∈ Bn , n ∈ N , with ∩i=1 i i∈N Bi = ∅. n B = ∅, n ∈ N . To prove that (ii) [Hints. Let {Bn : n ∈ N } ⊂ B be such that ∩i=1 i } < ∞, use (5.27) and (5.24). implies (i) in case sup{n ∈ N : Bm ∈ Bn for some m ∈ N
To show that (iii) implies (ii), assume that Bk = Ck × ( j>n k X j ), where Ck ∈ Cnk , k B = ∅ for any k ∈ N . Define B = B if n = n for some k, and and that ∩i=1 i k k n
{1,...,n } Bn = π{1,...,n}k (Ck ) × ( j>n X j ) if n k−1 < n < n k (with n 0 = 0) for some k. Then Bn ∈ Bn , n ∈ N , by (5.45), and Bn ⊃ Bn , n ∈ N . Apply (iii) to {Bn : n ∈ N }.] Exercise 16.38. that B is compact. [Hints. Let
Notation is as in (16.37). Prove n B = ∅ for n ∈ N . Let xn = Bn = Cn × ( j>n X j ), where Cn ∈ Cn , with ∩i=1 i n+ p
n+ p
n B , n ∈ N . Then xn+ p ∈ B , i.e. (x (x1n , x2n , . . .) ∈ ∩i=1 , . . . , xn ) ∈ Cn , for any i n 1 n ∈ N and p 0. Since x1n ∈ C1 , n ∈ N , and C1 is sequentially compact (6.19), choose {n 1k : k ∈ N } ⊂ N so that limk x1n 1k = x1 ∈ C1 . Since (x1n 1k , x2n 1k ) ∈ C2 , k ∈ N , and C2 is sequentially compact, choose {n 2k : k ∈ N } a subsequence of {n 1k : k ∈ N } such that limk (x1n 1k , x2n 1k ) = (x1 , x2 ) ∈ C2 . Continue this procedure and show that x = (x1 , x2 , . . .) = limk xn kk ∈ ∩i∈N Bi .]
Exercise 16.39. Let {(X j , d j ) : j ∈ N } be a sequence of complete separable metric spaces. For n ∈ N , let μn be a finite measure on B( nj=1 X j ) such that
Measures on Product σ -Algebras {1,...,n+1} −1 )
μn+1 (π{1,...,n}
159
= μn . Prove that there exists a unique measure μ on B( j∈N X j )
−1 such that μπ{1,...,n} = μn , n ∈ N . The following steps may be useful.
−1 (B( nj=1 X j )) is an algebra of sets. (a) The family A = ∪n∈N π{1,...,n}
−1 (b) For A ∈ A, choose n ∈ N so that A = π{1,...,n} (B) for some B ∈ B( nj=1 X j ), and put μ(A) = μn (B). Then μ is a well-defined finitely additive measure on A. (c) μ is σ -additive. [Hint. μ is B-regular, where B is as in (16.37) by (5.71.b), (6.98) and (16.23) applied to μn . Then use (16.17).]
(d) μ has a unique extension to B( j∈N X j ). [Use (10.9).]
Exercise 16.40. Let I = N or Z , and let X be a countable set with at least two elements. For each l 1 and x1 , . . . , xl ∈ X , suppose a nonnegative number . . . , xl ) is given such that: pl (x1 , (i) x∈X p1 (x) < ∞; (ii) x∈X pl+1 (x1 , . . . , xl , x) = pl (x1 , . . . , xl ); (iii) x∈X pl+1 (x, x1 , . . . , xl ) = pl (x1 , . . . , xl ). Let J denote the family of all finite subsets of I consisting of consecutive integers. If J ∈ J has l elements, define a measure PJ on P(X J ) by PJ ({(x1 , . . . , xl )}) = pl (x1 , . . . , xl ), (x1 , . . . , xl ) ∈ X J . Prove that there is a unique finite measure P on ⊗i∈I P(X i ), where X i = X for any i ∈ I , such that Pπ J−1 = PJ , J ∈ J . The next steps may be helpful. (a) The family A0 = ∪ J ∈J π J−1 (P(X J )) is an algebra of sets that generates ⊗i∈I P(X i ). (b) For A ∈ A0 , choose J ∈ J so that A = π J−1 (B) for some B ⊂ X J , and set P(A) = PJ (B). Then P is a well-defined finitely additive measure on A0 . [Use (i) and (ii).] (c) μ is σ -additive. [Hints. Let Bi be the family of all finite subsets of X i , where X i = X for any i ∈ I , and put B = ∪i∈I πi−1 (Bi ). Then B is compact by (16.13.a) and (16.14.c). Show that P is B-regular and use (16.17).] (d) P has a unique extension to ⊗i∈I P(X i ). [Use (10.9).] (e) P(X I ) = x∈X p1 (x).