TITLE: Non-decreasing sequence of random variable convergence in probability implies it also converges almost surely.
QUESTION [10 upvotes]: 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.

REPLY [2 votes]: 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,
$$
\begin{align}
P(\{ |X_n - X| \geq \epsilon \} \,\, \text{i.o.}) &= P(\cap_{n=1}^{\infty} \cup_{i=n}^{\infty} \{|X_i - X| \geq \epsilon \}) \\
&= P(\cap_{n=1}^{\infty} \{ |X_n - X| \geq \epsilon \}) \\
&= \lim_{n \to \infty} P(|X_n - X| \geq \epsilon) \\
&= 0 .
\end{align} 
$$