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

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!

• Since $(X_n)_{n\ge1}$ is monotonic, the limit $Y:=\lim_{n\to\infty}X_n$ exists almost surely, and thus also in $\Bbb P$. Now $X_n\stackrel{\Bbb P}\to X$ forces $X=Y$ a.s. Commented May 19, 2023 at 17:06

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.