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Consider the definition of almost sure convergence:

The sequence of random variables ${(X_n)}_{n \in \mathbb{N}}$ defined on the probability space $(\Omega, \mathcal{F}, P)$ converges almost surely to a random variable $X$ defined on the same probability space, if $$ P(\{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}) = 1. $$

Set $A := \{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}$. Clearly, a.s. convergence implies that $P(A^{c}) = 1 - P(A) = 0$.

I am wondering if the definition of a.s. convergence is equivalent to any of the following statements:

  1. $X_n \rightarrow X$ a.s. $\iff$ $\exists M \in \mathcal{F}$ with $P(M) = 1$, such that $\lim_{n \rightarrow \infty} X_n(\omega) = X(\omega) \quad \forall \omega \in M $.
  2. $X_n \rightarrow X$ a.s. $\iff$ $\exists N \in \mathcal{F}$ with $P(N) = 0$, such that $\lim_{n \rightarrow \infty}X_n(\omega) = X(\omega) \quad \forall \omega \in N^c$

The first and the second statements follow from one another, provided that either of them is correct. To show "$\Rightarrow$" of the first statement, set $M := A$. To show "$\Leftarrow$", observe that $M \subset A$ and thus $P(A) > P(M) =1 \Rightarrow P(A)=1$.

I would like to know if the reasoning is correct or there is something that I neglect. Is the completeness of the probability space relevant?

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  • $\begingroup$ Yes, the above looks fine, except a minor error: If $M \subset A$ then $P[M]\leq P[A]$ (we cannot claim strict inequality). $\endgroup$ – Michael Dec 5 '17 at 7:52
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With that definition of almost sure convergence then you will indeed need to assume completeness for your statements to be equivalent. Indeed if the set $\Omega - \{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}$ is not measurable but negligible (i.e. can be included in a measurable set with measure 0), than its probability is not even defined, i.e. it does not converge a.s. as per the first definition. It does however converge a.s. as per your proposed definitions 1. and 2.

Moreover, on a probability space, almost sure convergence will be equivalent with almost uniform convergence (by Egoroff's theorem, see for instance: https://en.wikipedia.org/wiki/Egorov%27s_theorem#Statement_of_the_theorem).

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  • $\begingroup$ Could you give some details on why exactly one needs the completeness? $\endgroup$ – Holden Oct 31 '17 at 23:08
  • $\begingroup$ I added some more details. Hope this helps. $\endgroup$ – Julien Nov 1 '17 at 21:42
  • $\begingroup$ But that set is always measurable, see math.stackexchange.com/questions/2501235/… $\endgroup$ – Holden Nov 2 '17 at 13:06
  • $\begingroup$ Indeed, I missed the fact that you had assumed $X$ to be measurable as well (in this case completeness is not required; it is required though if you relax that hypothesis, because then you can't assume the a.s. limit to be measurable). $\endgroup$ – Julien Nov 2 '17 at 18:04

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