Borell-Cantelli lemma

Question

If $\sum_{n} P(|X_{n}|>n)<\infty$, then prove that the $\limsup_{n}$ $|X_{n}|/n \leq 1$ a.s.

My approach

Let $E_{n}=|X_{n}>n|.$

$\sum_{n}P(E_{n})<\infty$ implies $P(E_{n} \text{ i.o})=0$ which further implies $P(\limsup_{n} E_{n})=0$.

Can anyone throw some light on how to approach further!

Answer 1

Suppose $\omega \notin \limsup_n E_n$. This means that $\omega$ belongs to only finitely many $E_n$'s and in particular there exists some $N$ with the property that $$n \ge N \implies \omega \notin E_n \implies |X_n(\omega)| \le n \implies \frac{|X_n(\omega)|}{n} \le 1$$ which in turn implies $$\limsup_n \frac{|X_n(\omega)|}{n} \le 1.$$ Since $P(\limsup E_n) = 0$ this last inequality holds a.s.

Answer 2

Your choice of $E_n$ should be $E_n=(|X_n|>n)$. Since $P(E_{n}\, \text{i.o})=0$, it follows that with probability one $|X_n|/n\leq 1$ eventually whence $\limsup |X_n|/n\leq 1$.