Does convergence in probability imply almost sure convergence for a sequence of almost surely monotonically growing random variables?

Question

Let $ X_{1}, X_{2}, \ldots $ be a sequence of almost surely monotonically growing random variables. Show that the sequence converges almost surely to a random variable $ X $ (for $ n \rightarrow \infty $ ) if it converges in probability to $ X $.

Idea: I have encountered Theorem 3.1.4, which states that $X_n \stackrel{P}{\longrightarrow} X$ if and only if for all subsequences ${X_{n_i}}$ of ${X_n}$, there exists another subsequence ${X_{n_{i_j}}}$ such that $X_{n_{i_j}}$ converges almost surely to $X$.

Based on this theorem, if we have established $X_n \stackrel{P}{\longrightarrow} X$, can we conclude that the entire sequence $X_n$ converges almost surely to $X$? Or are there additional conditions or considerations needed to reach this conclusion?

Any insights or clarifications would be greatly appreciated. Thank you!

Answer 1

Suppose $X_n\to X$ in probability. Then there exists a subsequence $X_{n_j}$ such that $X_{n_j} \to X$ almost surely. Because the sequence is monotonic, it follows that $X_{n} \to X$ almost surely.

Indeed, Let $\omega$ be a member of the set of convergence. Given $\varepsilon>0$, there exists $N$ such that for $j\geq N$, $|X_{n_j}(\omega) -X(\omega)|<\varepsilon$. Since the sequence is montonic, for $n\geq n_{N}$, $|X_{n}(\omega) -X(\omega)|<\varepsilon$ as desired.