Facts: Let V and S be nonempty sets. 1) If C is a collection of subsets of V then C CO(C) with equality if and only if C

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Facts: Let V and S be nonempty sets. 1) If C is a collection of subsets of V then C CO(C) with equality if and only if C

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Facts Let V And S Be Nonempty Sets 1 If C Is A Collection Of Subsets Of V Then C Co C With Equality If And Only If C 1
Facts Let V And S Be Nonempty Sets 1 If C Is A Collection Of Subsets Of V Then C Co C With Equality If And Only If C 1 (121.49 KiB) Viewed 36 times
Facts: Let V and S be nonempty sets. 1) If C is a collection of subsets of V then C CO(C) with equality if and only if C is a sigma algebra on V. 2) If C and D are collections of subsets of V that satisfy C C D, then o(C) CO(D). 3) Let X: S +R be a function. For all sets A CR define the inverse image of A with respect to the function X, denoted x-1(A), as the following set: x-1(A) = {WES : X(w) E A} It can be shown that if A C R then X-1(A)° = X-1(AC); If {A;}}, is a sequence of subsets of R then x-1(UFLA;) = VX-1(Ai)

Let (S, F, P) be a probability triplet: S is the sample space; F is a sigma algebra on S; P: F + R is a probability measure. Define C as the set of all intervals in R of the type (-0, x] for some x E R. The set o(C) is called the Borel sigma algebra on R. A subset of R is said to be a Borel measurable set if it is in o(C). A function X : S + R is said to be a Borel measurable function if: x-1(A) E F VAEC (1) We want to show that if X:S + R is a Borel measurable function then x-1(A) E F VAE O(C) (2) In particular, (2) implies that if A € 0(C) then {X E A} is an event (and hence has a well defined probability P[X E A]). That is, a function X : S + R with the property that {X < x} is an event for all x E R must also have the property that {X E A} is an event for all Borel measurable subsets A CR. For the tasks below, assume that X:S + R satisfies (1). a) Define H as the following set of subsets of R: H = {A CR: X-1(A) € F} Show that C CH. It immediately follows (by Fact 2) that o(C) Co(H). b) Argue that H is a sigma algebra on R. It immediately follows (by Fact 1) that o(H) = H. c) Argue that o(C) CH. Explain why this yields our desired conclusion (2).
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