Equivalent conditions for almost sure convergence. 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:


$X_n \rightarrow X$ 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 $.
$X_n \rightarrow X$ 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 $P(A) &gt; P(M) =1 \Rightarrow P(A)=1$.

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?
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).
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).