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?
Best Answer
Your second equality is wrong. You cannot take the limit inside the probability.
$P(|X_n| >\epsilon)\leq P(\cup_{m \geq n} (|X_m| >\epsilon)$ so $\lim P(|X_n| >\epsilon) \leq P(\lim \sup (|X_n| >\epsilon) =0$ since $\lim \sup P(|X_n| >\epsilon)$ has probability $0$ by assumption.