Almost sure convergence implies convergence in probability

Remarks

Proof: fix $\varepsilon > 0$ and let $A_n = \bigcup_{m \ge n} \{|X_m - X| > \varepsilon\}$. The events $A_n$ decrease, and their intersection has probability $0$ by almost sure convergence. By the continuity of probability, $P(A_n) \to 0$. Since $\{|X_n - X| > \varepsilon\} \subseteq A_n$, the claim follows. The converse is false. Let $X_n$ be independent with $P(X_n = 1) = 1/n$ and $P(X_n = 0) = 1 - 1/n$. Then $X_n \to 0$ in probability. But by the second Borel–Cantelli lemma, $X_n = 1$ for infinitely many $n$ almost surely, so $X_n$ does not converge to $0$ almost surely.