Non-decreasing sequence of random variable convergence in probability implies it also converges almost surely. The problem stated as follow:

Suppose $X_1 \leq X_2 \leq \cdots$ and $X_n \xrightarrow[]{p} X$. Show that $X_n \to X$ a.s.

I'm think about may be use the continuity of probability measure, but I don't know if that's correct.
 A: You don't need to use the subsequence argument.  For each $\omega$, $\{X_n(\omega)\}$ is an increasing sequence of real numbers, and so it has a limit $Y(\omega) \in (\infty, +\infty]$.  Being a pointwise limit of measurable functions, $Y$ is measurable, i.e. a random variable.  It remains to show that $Y = X$ a.s.  But since $X_n \to Y$ pointwise, we also have $X_n \to Y$ in probability and limits in probability are unique up to null sets, so indeed $Y=X$ a.s.
A: Alternative proof:
Fix $\epsilon > 0$.  Since the sequence $\{ X_n \}_{n=1}^{\infty}$ is monotone decreasing and $X_ n \to X$ in probability, then the events $\{ |X_n - X| \geq \epsilon \}$ are monotone decreasing.  Therefore,
$$
\begin{align}
P(\{ |X_n - X| \geq \epsilon \} \,\, \text{i.o.}) &= P(\cap_{n=1}^{\infty} \cup_{i=n}^{\infty} \{|X_i - X| \geq \epsilon \}) \\
&= P(\cap_{n=1}^{\infty} \{ |X_n - X| \geq \epsilon \}) \\
&= \lim_{n \to \infty} P(|X_n - X| \geq \epsilon) \\
&= 0 .
\end{align} 
$$
A: Since $X_n \to X$ in probability, there is a subsequence $\{X_{n_k}\}_{k=1}^\infty$ of $\{X_n\}_{n=1}^\infty$ such that $X_{n_k} \to X$ almost surely.  That $X_n \to X$ almost surely now follows from the fact that if a subsequence of a monotone sequence converges, then the original sequence converges to the same limit.
