Let $\{X_n:n\geq 1\}$ be a sequence of r.v. with $X_n\to X$ a.s., then $X$ is r.v I'm trying to understand the "almost sure convergence" concept. I've seen two definitions,
Let $\{X_n:n\geq 1\}$ be a sequence of r.v. defined on the same probability space $(\Omega,\mathcal{A},P) $.
Definition 1: Let $X:\Omega\to\mathbb{R}$ be a function. We will say that $X_n\to X$ a.s. (almost sure convergence) if $\exists N\subset\Omega : P(N)=0 $ with $\lim_n X_n(w)=X(w), \forall w\in N^c $
Definition 2: Let $X$ be a r.v. on $(\Omega,\mathcal{A},P) $, then we will say that $X_n\to X$ a.s. if $P(\lim_n X_n = X )=1$
In the second one, we are defining the limit as a r.v., but in the first one, it is not said that $X$ is a r.v. so I have searched if it is (trying not to use the second one).
What I found is to define $X^*(w)=\begin{cases} X(w)  & w\in N^c \\ 0  &w\in N \end{cases} $, then it is said that $X_n\to X^*$. As measurable functions preserves limits, $X^*$ is r.v., and because $X=X^*$ a.s, we have $X$ is r.v. too.
I'm not seeing why $\lim_n X_n(w)= 0$ when $w\in N$. Is this proof correct? I don't think so. The argument satisfies me, but we don't know if $X_n\to X^*$.

If $X_n\to X$ on $A\in\mathcal{A}$ with $\mathsf{P}(A^c)=0$, $X$ need not be measurable unless $\mathsf{P}$ is complete. On the other hand, $X_n\to X^*$ on $A$, and $X^*$ is measurable, i.e., it is a random variable.
