I was seeking a solution to a textbook's problem to check if mine is correct and found one that got me a doubt.
We'd like to show that:
\begin{align}\sum_{n=1}^{\infty}\mathbb{P}\left( \frac{|X_{n}|}{n}>1 \right) < \infty \implies \lim \sup_{n}\frac{|X_{n}|}{n}\leq 1\end{align}
My first thought was applying Borel-Cantelli lemma, so we show that it will be higher than 1 infinitely often with probability $0$.
The author did the above and proceeded with the following step:
$\mathbb{P}\left(\lim \sup \left[ \frac{|X_{n}|}{n}>1 \right]\right) = 0 \implies \mathbb{P}\left(\lim \sup \left[ \frac{|X_{n}|}{n} \leq 1\right]\right) = 1$
Is this because we know the following from the series?
$\mathbb{P}\left( \frac{|X_{n}|}{n}>1 \right) \downarrow 0 \implies\mathbb{P}\left( \left[\frac{|X_{n}|}{n}>1\right]^{c} \right) = \mathbb{P}\left( \frac{|X_{n}|}{n}\leq1 \right) \uparrow 1$
And thus a series based upon the complement sequence won't converge atributing probability $1$ to $\lim \sup_{n}\frac{|X_{n}|}{n}\leq 1$.
if the sum is zero, then each term is zero, which implies that $P(|X_n|/n > 1) = 0$ for any $n$. So $P(|X_n|/n \leq 1) = 1$. Since the statement holds for all $n$, the countable intersection of these probability $1$ events still has probability $1$, and so \begin{align*} P\left(\bigcap_n \{|X_n|/n \leq 1\}\right) = 1 \end{align*} Hence $\limsup |X_n|/n \leq 1$ with probability $1$.