= Solution
The union bound gives $\mathbb P(\bigcup_{n\geq m}A_n)\leq\sum_{n\geq m}\mathbb P(A_n)\to0$. Taking the decreasing intersection proves the first <Borel-Cantelli lemmas>[Borel-Cantelli lemma].
If $X_n\to X$ almost surely, then $1_{\{|X_n-X|>ε\}}\to0$ almost surely; <dominated convergence theorem> gives convergence of its expectation, hence convergence in probability.
If convergence in probability holds, any subsequence has a further one with $\mathbb P(|X_{n(k_r)}-X|>2^{-r})<2^{-r}$. Borel-Cantelli then gives almost-sure convergence. Conversely, failure in probability supplies a subsequence with probabilities bounded below by some positive constant, and no further subsequence can converge almost surely because that would imply convergence in probability.
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