Almost sure convergence to 0 implies probability convergence to 0 I've seen proof of almost sure convergence implying probability convergence but I want to ask whether or not the "proof" for the 0 case is correct:
$X_n$ converges almost surely to $X$ if $\mathbb{P}(\lim_{n\to\infty}X_n = X) = 1$. 
Given that $\lim_{n\to\infty}X_n = 0$ then $X_n$ converges to $X = 0$ almost surely, then I prove that it converges in probability to 0 by considering for $\epsilon > 0$,
$$\lim_{n\to\infty}\mathbb{P}(|X_n - 0| > \epsilon) \\ = \lim_{n\to\infty}\mathbb{P}(|X_n| > \epsilon)\\ = \mathbb{P}(\lim_{n\to\infty} |X_n| > \epsilon)\\
= \mathbb{P}(0 > \epsilon) \\ = 0
$$
since $\epsilon > 0$. So since I've shown the probability goes to zero, it converges in probability. Is this how it's done?

Here's a way to salvage your proof. Note that $P(|X_n|>\epsilon) = E(1_{|X_n|>\epsilon})$. Since $X_n \to 0$ a.s,  we have $1_{|X_n|>\epsilon} \to 0$ a.s.
Since $1_{|X_n|>\epsilon}\leq 1$, the dominated convergence theorem applies and yields $\lim_n E(1_{|X_n|>\epsilon}) = E(0) = 0$.
