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Measure and Integration Appendix A Measure and Integration A.1 Measures and the Carathéodory Extension Let S be a nonempty set. A class F of subsets of S is a field,oranalgebra if (i) ∅∈F, S ∈ F, (ii) A ∈ F implies Ac ∈ F, (iii) A, B ∈ F implies A ∪ B ∈ F. Note that (ii) and (iii) imply that F is closed under finite unions and finite intersections. If (iii) ∈ F ( = , ,...) ∪∞ ∈ F F is replaced by (iii) : An n 1 2 implies n=1 An , then is said to be a σ-field,oraσ-algebra. Note that (iii) implies (iii), and that a σ-field is closed under countable intersections. A function μ : F→[0, ∞] is said to be a measure on a field F if μ(∅) = 0 (∪∞ ) = ∞ ( ) ∈ F and μ n=1 An n=1 μ An for every sequence of pairwise disjoint sets An ( = , ,...) ∪∞ ∈ F n 1 2 such that n=1 An . Note that this property, known as countable additivity, implies finite additivity (by letting An =∅for n ≥ m for some m, say). A measure μ on a field F is σ-finite if there exists a sequence An ∈ F (n = 1, 2,...) ∪∞ = ( )<∞ such that n=1 An S and μ An for every n. F ∈ F ( ≥ ) ⊂∪ , ∈ F If μ is a measure on a field , and An n 1 , A n An A , ( ) ≤ ∞ ( ) = , = c ∩ then μ A n=1 μ An (subadditivity). To see this write B1 A1 Bn A1 ···∩ c ∩ ( ≥ ) ( ≥ ) ∪∞ =∪∞ An−1 An n 2 . ThenBn n 1 are disjoint, n=1 An n=1 Bn, so that ( ) = ( ∩ (∪∞ )) = ∞ ( ∩ ) ≤ ∞ ( ) ⊂ μ A μ A n=1 Bn n=1 μ A Bn n=1 μ An (since Bn An for all n). Let μ be a measure on a σ-field F on S. Then F is said to be μ-complete if all subsets of μ-null sets in F belong to F : N ∈ F, μ(N) = 0, B ⊂ N implies B ∈ F. In this case the measure μ is also said to be complete. Given any measure μ on a σ-field F, it is simple to check that the class of subsets F = {C = A ∪ B : A ∈ F, B ⊂ N for some N ∈ F such that μ(N) = 0} is a μ-complete σ-field, μ(A ∪ B) := μ(A)(A ∈ F, B ⊂ N, μ(N) = 0) is well defined, and μ is a measure on F extending μ. This extension of μ is called the completion of μ. © Springer International Publishing AG 2016 225 R. Bhattacharya and E.C. Waymire, A Basic Course in Probability Theory, Universitext, DOI 10.1007/978-3-319-47974-3 226 Appendix A: Measure and Integration Wenow derive one of the most basic results in measure theory, due to Carathéodory, which provides an extension of a measure μ on a field A to a measure on the σ-field F = σ(A), the smallest σ-field containing A.First,ontheset2S of all subsets of S, call a set function μ∗ : 2S →[0, ∞] an outer measure on S if ∗ ∗ ∗ (1) μ (∅) = 0, (2) (monotonicity) A ⊂ B implies μ (A) ≤ μ (B), and (3) (subad- ∗(∪∞ ) ≤ ∞ ∗( ) ( = , ,...) ditivity) μ n=1 An n=1 μ An for every sequence An n 1 2 . Proposition 1.1 Let A be a class of subsets such that ∅∈A, S ∈ A, and let μ : A →[0, ∞] be a function such that μ(∅) = 0. For every set A ⊂ S, define ∗ μ (A) = inf μ(An) : An ∈ A∀nA ⊂∪n An . (1.1) n Then μ∗ is an outer measure on S. Proof (1) Since ∅⊂∅∈A, μ∗(∅) = 0. (2) Let A ⊂ B. Then every countable collection {An : n = 1, 2,...}⊂A that covers B (i.e., B ⊂∪n An) also covers A. ∗ ∗ ∗ Hence μ (A) ≤ μ (B).(3)LetAn ⊂ S (n = 1, 2,...), and A =∪n An.Ifμ (An) = ∗ ∗ ∞ for some n, then by (2), μ (A) =∞. Assume now that μ (An)<∞∀n.Fix > { : = , , ···} ⊂ A ε 0 arbitrarily. For each nthere exists a sequence An,k k 1 2 ⊂∪ ( )< ∗( ) + / n ( = , ,...) such that An k An,k and k μ An,k μ An ε 2 n 1 2 . Then ⊂∪ ∪ ∗( ) ≤ ( )< ∗( ) + A n k An,k , and therefore μ A n,k μ An,k n μ An ε. The technically simplest, but rather unintuitive, proof of Carathéodory’s theorem given below is based on the following notion. Let μ∗ be an outer measure on S.Aset A ⊂ S is said to be μ∗-measurable if the following “balance conditions” are met: μ∗(E) = μ∗(E ∩ A) + μ∗(E ∩ Ac) ∀ E ⊂ S. (1.2) Theorem 1.2 (Carathéodory Extension Theorem) (a) Let μ∗ be an outer measure on S. The class M of all μ∗-measurable sets is a σ-field, and the restriction of μ∗ to M is a complete measure. (b) Let μ∗ be defined by (1.1), where A is a field and μ is a measure on F. Then σ(A) ⊂ M and μ∗ = μ on A. (c) If a measure μ on a field A is σ-finite, then it has a unique extension to a measure on σ(A), this extension being given by μ∗ in (1.1) restricted to σ(A). Proof (a) To show that M is a field, first note that A =∅trivially satisfies (1.2) and that if A satisfies (1.2), so does Ac. Now, in view of the subadditivity property of μ∗, (1.2) is equivalent to the inequality μ∗(E) ≥ μ∗(E ∩ A) + μ∗(E ∩ Ac) ∀ E ⊂ S. (1.3) To prove that M is closed under finite intersections, let A, B ∈ M. Then ∀ E ⊂ S, Appendix A: Measure and Integration 227 μ∗(E) = μ∗(E ∩ B) + μ∗(E ∩ Bc) (since B ∈ M) = μ∗(E ∩ B ∩ A) + μ∗(E ∩ B ∩ Ac) + μ∗(E ∩ Bc ∩ A) +μ∗(E ∩ Bc ∩ Ac) (since A ∈ M) ≥ μ∗(E ∩ (B ∩ A)) + μ∗(E ∩ (B ∩ A)c). For the last inequality, use (B ∩ A)c = Bc ∪ Ac = (Bc ∩ A) ∪ (Bc ∩ Ac) ∪ (B ∩ Ac), and subadditivity of μ∗. By the criterion (1.3), B ∩ A ∈ M. Thus M is a field. ∗ Next, we show that M is a σ-field and μ is countably additive on M.LetBn ∈ M ( = , ,...) M =∪m n 1 2 be a pairwise disjoint sequence in , and write Cm n=1 Bn (m ≥ 1). We will first show, by induction on m, that m ∗ ∗ μ (E ∩ Cm) = μ (E ∩ Bn) ∀ E ⊂ S. (1.4) n=1 This is true for m = 1, since C1 = B1. Suppose (1.4) holds for some m. Since Bm+1 ∈ M, one has for all E ⊂ S, ∗( ∩ ) = ∗(( ∩ ) ∩ ) + ∗(( ∩ ) ∩ c ) μ E Cm+1 μ E Cm+1 Bm+1 μ E Cm+1 Bm+1 ∗ ∗ = μ (E ∩ Bm+1) + μ (E ∩ Cm) m ∗ ∗ = μ (E ∩ Bm+1) + μ (E ∩ Bm), n=1 using the induction hypothesis for the last equality. Thus (1.4) holds for m + 1in =∪∞ place of m, and the induction is complete. Next, writing A n=1 Bn one has, for all E ⊂ S, ∗( ) = ∗( ∩ ) + ∗( ∩ c ) ∈ M μ E μ E Cm μ E Cm (since Cm ) m m = ∗( ∩ ) + ∗( ∩ c ) ≥ ∗( ∩ ) + ∗( ∩ c), μ E Bn μ E Cm μ E Bn μ E A n=1 n=1 c ⊃ c →∞ since Cm A . Letting m , one gets ∞ ∗ ∗ ∗ c ∗ ∗ c μ (E) ≥ μ (E ∩ Bn) + μ (E ∩ A ) ≥ μ (E ∩ A) + μ (E ∩ A ), (1.5) n=1 using the subadditivity property for the last inequality. This shows that A ≡ ∪∞ ∈ M M { : = ,...} n=1 Bn , i.e., is closed under countable disjoint unions. If An n 1 M ≡∪∞ =∪∞ is an arbitrary sequence in , one may express A n=1 An as A n=1 Bn, where = = c ∩ = c ∩···∩ c ∩ ( > ) B1 A1, B2 A1 A2, Bn A1 An−1 An n 2 , are pairwise disjoint sets in M. Hence A ∈ M, proving that M is a σ-field. To prove countable additivity ∗ of μ on M,letBn (n ≥ 1) be a pairwise disjoint sequence in M as before, and take 228 Appendix A: Measure and Integration = ≡∪∞ ∗(∪∞ ) ≥ ∞ ∗( ) E A n=1 Bn in the first inequality in (1.5) to get μ n=1 Bn n=1 μ Bn . ∗ ∗(∪∞ ) = ∞ ∗( ) By the subadditive property of μ , it follows that μ n=1 Bn n=1 μ Bn . We have proved that μ∗ is a measure on the σ-field M. Finally, if A ⊂ N ∈ M, μ∗(N) = 0, then μ∗(E ∩ A) ≤ μ∗(A) ≤ μ∗(N) = 0, and μ∗(E ∩ Ac) ≤ μ∗(E),so that (1.3) holds, proving A ∈ M. Hence M is μ∗-complete. (b) Consider now the case in which A is a field, μ is a measure on A, and μ∗ is the outer measure (1.1). To prove A ⊂ M,letA ∈ A.FixE ⊂ S and > ∈ A ( = , ,...) ⊂∪∞ ε 0 arbitrarily. There exists An n 1 2 such that E n=1 An and ∗( ) ≥ ∞ ( ) − μ E n=1 μ An ε.Also, ∞ ∞ ∗ ∗ μ (E ∩ A) ≤ μ A ∩ An ≤ μ(A ∩ An), n=1 n=1 ∞ ∞ ∗ c ∗ c c μ (E ∩ A ) ≤ μ A ∩ An ≤ μ(A ∩ An), n=1 n=1 ∞ ∗ ∗ c c μ (E ∩ A) + μ (E ∩ A ) ≤ μ(A ∩ An) + μ(A ∩ An) n=1 ∞ ∗ = μ(An) ≤ μ (E) + ε. n=1 Hence (1.3) holds, proving that A ∈ M. To prove μ = μ∗ on A,letA ∈ A.By ∗( ) ≤ ( ) = =∅ ≥ definition (1.1), μ A μ A (letting A1 A and An for n 2, be a cover ( ) ≤ ∞ ( ) ∈ A ( ≥ ) of A). On the other hand, μ A n=1 μ An for every sequence An n 1 ⊂∪∞ ∗( ) ≥ ( ) A such that A n=1 An, so that μ A μ A (by subadditivity of μ on ).
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