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It's well know and easy to prove that given a finite number of events $A_1,\dots,A_n$ we can factorize the probability of their intersection as: $$\mathbb{P}(A_1\cap\dots\cap A_n)=\mathbb{P}(A_1)\mathbb{P}(A_2|A_1)\dots\mathbb{P}(A_n|A_1\cap \dots \cap A_n)$$ I was wondering if this remains true for a sequence of events, i.e. can we factorize a countable intersection in the same way?

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Yes, just take limit on both sides to get $P(\cap_n A_n)=P(A_1)\prod_{k=2}^{\infty} P(A_k|A_1,A_2,...,A_{k-1})$.

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  • $\begingroup$ So you are using sequential continuity of the probability the pass the limit inside P in the LHS? $\endgroup$ – StabiloBoss Jun 7 at 11:55
  • $\begingroup$ Right, the events $A_1\cap A_2\cap...\cap A_n$ decrease to the event $\cap_n A_n$ so we can pass to the limit. $\endgroup$ – Kavi Rama Murthy Jun 7 at 11:58

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