claim : let $\{X_n, \, n \geq 1\}$ be a monotonic sequence of random variables
$X_n\rightarrow0\quad \text{in probability} \implies X_n\rightarrow 0\quad \text{a.s}$
proof attempt :
we have for every $\epsilon > 0$ $$\lim_{n \to +\infty}\mathbb{P}(|X_n| \geq \epsilon) = 0$$
which intuitively means to me that : (since it holds for every $\epsilon, \, $ and $\epsilon > 0$ and $|X_n |\geq 0$)
$$\lim_{n \to +\infty}\mathbb{P}(|X_n| > 0) =\lim_{n \to +\infty}\mathbb{P}(|X_n| \neq 0) = 0$$
I'm not sure concerning this step.... anyways
by the monotone convergence theorem : $$\mathbb{P}(\omega \mid\lim_{n \to +\infty}|X_n(w)| \neq 0) = 0$$
then :
$$\mathbb{P}(\omega \mid\lim_{n \to +\infty}|X_n(w)| = 0) = \mathbb{P}(\omega \mid\lim_{n \to +\infty} X_n(w) = 0) = 1$$
thus the almost sure convergence, I need some feedback concerning this attempt at proof, thanks !