Extension of Measure

Extension of Measure

1 Extension of measure Sayan Mukherjee Dynkin’s π − λ theorem We will soon need to define probability measures on infinite and possible uncountable sets, like the power set of the naturals. This is hard. It is easier to define the measure on a much smaller collection of the power set and state consistency conditions that allow us to extend the measure to the the power set. This is what Dynkin’s π − λ theorem was designed to do. Definition 0.0.1 (π-system) Given a set Ω a π system is a collection of subsets P that are closed under finite intersections. 1) P is non-empty; 2) A \ B 2 P whenever A; B 2 P. Definition 0.0.2 (λ-system) Given a set Ω a λ system is a collection of subsets L that contains Ω and is closed under complementation and disjoint countable unions. c S1 1) Ω 2 L ; 2) A 2 L ) A 2 L ; 3) An 2 L ; n ≥ 1 with Ai \ Aj = ; 8i 6= j ) n=1 An 2 L : Definition 0.0.3 (σ-algebra) Let F be a collection of subsets of Ω. F is called a field (algebra) if Ω 2 F and F is closed under complementation and countable unions, c Sn S1 1) Ω 2 F; 2) A 2 F ) A 2 F; 3) A1; :::; An 2 F ) j=1 Aj 2 F; 4) A1; :::; An; ::: 2 F ) j=1 Aj 2 F. A σ-algebra is a π system. A σ-algebra is a λ system but a λ system need not be a σ algebra, a λ system is a weaker system. The difference is between unions and disjoint unions. Example 0.0.1 Ω = fa; b; c; dg and L = fΩ; ;; fa; bg; fb; cg; fc; dg; fa; dg; fb; dg; fa; cgg L is closed under disjoint unions but not unions. However: Lemma 0.0.1 A class that is both a π system and a λ system is a σ-algebra. Proof. We need to show that the class is closed under countable union. Consider A1;A2; :::; An 2 L note that these Ai are not disjoint. Sn−1 Tn−1 c We disjointify them, B1 = A1 and Bn = An n i=1 Ai = An i=1 Ai :Bn 2 L by π system and complement property of λ system. S1 S1 i=1 Ai = i=1 Bi but the fBig are disjoint so 1 1 [ [ Ai = Bi 2 L : i=1 i=1 We will start constructing probability measures on infinite sets and we will need to push the idea of extension of measure. In this the following theorem is central. Theorem 0.0.1 (Dynkin) 1) If P is a π system and L is a λ system, then P ⊂ L implies σ(P) ⊂ L . b) If P is a π system σ(P) = L (P): Proof of a). Let L0 be the λ system generated by P , the intersection of all λ systems containing P. This is a λ system and contains P and is contained in every λ system that contains P. So P ⊂ L0 ⊂ L . If L0 is also a π system, it is a λ system and so is a σ algebra. By minimality of σ(P) it follows σ(P) ⊂ L0 so P ⊂ σ(P) ⊂ L0 ⊂ L . So we need to show that L0 is a π system. Set A 2 σ(P) and define GA = fB 2 σ(P): A \ B 2 L (P)g: 2 1) We show if A 2 L (P) then GA is a λ system. i) Ω 2 GA since A \ Ω = A 2 L (P). c ii) Suppose B 2 GA. B \A = An(A\B) but B 2 GA means A\B 2 L (P) and since A 2 L (P) it holds A n (A \ B) = Bc \ A 2 L (P) by complementarity. Since Bc \ A 2 L (P) it holds c that B 2 GA. iii) fBjg are mutually disjoint and Bj 2 GA then 1 1 \ [ [ A ( Bj) = (A \ Bj) 2 L (P): j=1 j=1 2) We show if A 2 P then L (P) 2 GA. Since A 2 P ⊂ L (P) from 1) GA is a λ system, For B 2 P we have A \ B 2 P since A 2 P and P is a π system. So if B 2 P then A \ B 2 P ⊂ L (P) implies B 2 GA so P ⊂ GA and since GA is a λ system L (P) ⊂ GA. 3) We rephrase 2) to state if A 2 P and B 2 L (P) then A\B 2 L (P) (if we can extend A to L (P) we are done). 4) We show that if A 2 L (P) then L (P) ⊂ GA. If B 2 P and A 2 L (P) then A \ B 2 L (P) then from 3) it holds A \ B 2 L (P). So when A 2 L (P) it holds P 2 GA so from 1) GA is a π system and L (P) 2 GA. 5) If A 2 L (P) then for any B 2 L (P), B 2 GA then A \ B 2 L (P) . The following lemma is a result of Dynkin’s theorem and highlights uniqueness. Lemma 0.0.2 If Pi and P2 are two probability measures on σ(P) where P is a π system. If P1 and P2 agree on P, then they agree on σ(P): Proof L = fA 2 P : P1(A) = P2(A)g; is a λ system and P ⊂ L so σ(P) ⊂ L : We will need to define a probability measure, µ, for all sets in a σ algebra. It will be convenient and almost necessary to define a (pre)-measure, µ0, on a smaller collection F0 that extends to µ on σ(F0) and µ is unique, µ1(F ) = µ2(F ); 8F 2 F0 implies µ1(F ) = µ2(F ); 8F 2 σ(F0). Restatement of Dynkin’s theorem in the above context is that the smallest σ algebra generated from a π system is the λ system generated from that π system. Theorem 0.0.2 Let F0 be a π system then λ(F0) = σ(F0). We now define the term extension and develop this idea of extension of measure. The main working example we will use in the development is. Ω = R; S = f(a; b]: −∞ ≤ a ≤ b ≤ 1g; P((a; b]) = F (b) − F (a): Remember, that the Borel sets B(R) = σ(S): Goal: we have a probability measure on S we want to extend it to B(R). Extension: If G1 and G2 are collections of subsets of Ω with G1 ⊂ G2 and P1 : G1 ! [0; 1]; P2 : G2 ! [0; 1]: P2 is an extension of P1 if P2jG1 = P1; P2(A1) = P1(Ai) 8Ai 2 G1: 3 Few terms: A measure is P additive if for disjoint sets A1; :::; An 2 G n n n n X [ X X Ai ≡ Ai 2 G ) P( Ai) = P(Ai): i=1 i=1 i=1 i=1 A measure is σ additive if for disjoint sets A1; :::; An; ::: 2 G 1 1 1 X X X Ai 2 G ) P( Ai) = P(Ai): i=1 i=1 i=1 Definition 0.0.4 (Semialgebra) A class S of subsets of Ω is a semialgebra if 1) ;; Ω 2 S 2) S is a π system c Pn 3) If A 2 S there exists disjoint sets C1; :::; Cn with each Ci 2 S such that A = i=1 Ci. Example of Semialgebra: Intervals Ω = R; S = f(a; b]: −∞ ≤ a ≤ b ≤ 1 ;g: c If I1;I2 2 S then I1 \ I2 2 S and I is a union of disjoint intervals. The idea/plan: 1. Start with a semialgebra S 2. Show a unique extension to an algebra A(S) generated by S 3. Show a unique extension to σ(A(S)) = σ(S). If S is a semialgebra of Ω then X A(S) = f Si : I is finite ; fSig is disjoint ;Si 2 Sg; i2I a system of finite collections of mutually disjoint sets. An example of extension of measure for a countable set. X 1 Ω = f!1; ::::; !n; :::g 8i ! pi pi ≥ 0; −i = 1 p1 = 1: F = P(Ω) the power set of Ω. We define for all A 2 F the following extension of measure X P(A) = pi: !i2A For this to be a proper measure the follow need to hold (they do) 1. P(A) ≥ 0 for all A 2 F P1 2. P(Ω) = i=1 pi = 1 3. If fAjg with j ≥ 1 are disjoint then 1 X [ X X X X P( Aj) = pi = pi = P(Aj): j=1 wi2[j Aj j !i2Aj j Is the power set P(Ω) countable ? What was the semialgebra above ? What would we do if Ω = R ? As we consider the following theorems and proofs or proof sketches a good example to keep in mind is the case where Ω = R and S == f(a; b]: −∞ ≤ a < b ≤ 1 ;g: 4 Theorem 0.0.3 (First extension) S is a semialgebra of Ω and P : S! [0; 1] is σ-additive on S and P(Ω) = 1. There is a unique extension P0 of P to A(S) defined by 0 X P (Si) = P(Si); i P0(Ω) = 1 and P0 is σ additive on A(S).

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