Does an almost sure bounded sequence necessarily converge? Let $\{\xi_n\}$ be a sequence of random variable and let $\mathbf P (\lim\xi_n \to \xi) \ge 1 - \epsilon$ $\forall \epsilon \gt 0$. Is this implies that $\xi$ random varaible too? 
First I tried to use Borell-Cantelli lemma, https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma , to say things like: 
let $M_{n, \epsilon} = \{w \in \Omega: |\xi_n(w) - \xi(w)| \gt \epsilon\}$,  $\forall \epsilon \gt 0$ 
I know from the statement, that $\lim_{n \to \infty}\mathbf P(M_{n, \epsilon}) = 0$, but I can't apply Borel-Cantelli, since it may "slowly" goes to 0 (like $\frac{1}{n}$). And I stucked on this way. 
I also wanted to use Lusin's theorem from here: https://people.math.gatech.edu/~heil/6337/spring11/lusin.pdf , but I'm not sure how to do it right. Since in Luzin's theorem statement about single function, but here I have a sequence of functions (maybe somehow take $\eta_n = |\xi_n - \xi|$ and then use Lusin's theorem for each of $\eta_n$, conclude that all of them are measurable and hence $\xi$ measurable?) 
Or from original statement doesn't follow that $\xi$ measurable? (but I can't imagine/find counterexample to such problem) 
Thx!

Found a counterexample 
Statement is not true: 
Let $\Omega = {1, 2, 3}, F_\Omega = \{\Omega, \{1\}, \{2, 3\}, \emptyset\}$ and $\xi_n: \Omega \to \mathbb{R}$, $\xi_n(\omega) = 1$, $\forall \omega$ 
$P(\{1\}) = 1, P(\{2, 3\}) = 0$ 
$\xi(\omega) = \omega$, then: 
1. $\xi$ is not a random variable ($\xi^{-1}([\frac{1}{2}, \frac{5}{2}]) = \{1, 2\} \notin F_\Omega$) 
2. $\xi_n$ converge to $\xi$ in sense of main question (in probability) 
