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I have been reading Billingsleys book where I came across this theorem and proof. I am having difficulty understanding the theorem/proof. I feel there is a better, more complete way to prove it. Does anyone know how?

Theorem: If P and Q are probability measures on S so that PF=QF for every closed set F, then PA=QA for every A in S.

Proof: The collection of sets where they agree is seen to be a sigma algebra, so containing all the closed sets is enough to get equality everywhere.

Remark: This theorem shows us that the probability measure is determined entirely by its action on open and closed sets.

MathLover123
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  • I am having difficulty understanding the theorem/proof. Is the statement of the theorem clear to you? Why do you think that the proof is incomplete? – Davide Giraudo Apr 02 '15 at 20:03
  • I'm just stuck on showing that the sets in which they agree are sigma algebra. – MathLover123 Apr 02 '15 at 22:05
  • Actually, it is not necessarily true: http://math.stackexchange.com/questions/201164/do-probability-measures-have-to-be-the-same-if-they-agree-on-a-generator-of-bore . However, this collection is a monotone class and we can conclude. – Davide Giraudo Apr 03 '15 at 08:06

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