Show that convergence in probability implies convergence almost surely

Question

Let {$X_n : n ≥ 1$} and $X$ be random variables on a probability space $(\Omega,\mathcal{F},P)$. Suppose $X_n ≤ X_{n+1}$ for every $n ≥ 1$ and $X_n \to X$ in probability. Show that $ X_n → X $ a.s..

Can someone help me with this problem? I'm really struggling. Normally convergence in probability does not imply convergence almost surely, but why does $X_n ≤ X_{n+1}$ change the result?

Answer 1

Hint: Convergence in probability implies the existance of a subsequence $(X_{k_n})_{n\ge1}$ that converges almost surely. Now take monotonicity together with the converging subsequence (i.e. $X_{k_n} \le X_{k_{n+1}}$) to prove the result.