$X_1,X_2,...$ converges to $0$ almost surely if and only if $P(|X_n|>\epsilon \text{ i.o.})=0$ for all $\epsilon > 0$

I feel like I have a general idea of this proof, but could I please check in terms of mathematical grammar? I'm looking to write the proof in a short manner as I'll be given a time limit.

"$\implies$"

$X_n\rightarrow0 \text{ a.s.} \implies P(\lim_{n\rightarrow\infty}|X_i-0|>\epsilon)=0 \implies P(\limsup_{n\rightarrow\infty}|X_i|>\epsilon)=0 \implies P(|X_i|>\epsilon \text{ i.o.})=0$

"$\impliedby$"

$P({|X_n|>\epsilon} \text{ i.o.})=0 \implies P(\limsup_{n\rightarrow\infty}|X_n|>\epsilon)=0\implies P(\lim_{n\rightarrow\infty}|X_n|>\epsilon)=0 $

$\implies P(\lim_{n\rightarrow\infty}|X_n-0|>\epsilon)=0 \implies X_n \rightarrow0$ a.s.

Since $X_n$ converges pointwise, I assume that it's limsup and limit are equal.

Solution 1:

The "$\impliedby$" direction can be instead done as follows: We know that for every $m\in\mathbb{N}$, the event

$$\{|X_n|> 1/m\text{ for infinitely many }n\}$$

has probability zero. Thus, the event

$$A_m:=\{|X_n|\leq 1/m\text{ for all sufficiently large }n\}$$

has $\mathbb{P}(A_m)=1$ for every $m\in\mathbb{N}$. A countable intersection of events with probability one is still an event with probability one, so $A:=\bigcap_{m=1}^{\infty}A_m$ has $\mathbb{P}(A)=1$. Moreover, the $X_n$ converges to $0$ on the event $A$. Therefore, $X_n\to 0$ almost surely (on the event $A$).