Let $\{X_n\}$ be i.i.d integrable r.v.s, show that $\frac{1}{n}\max_{1\leq j\leq n}|X_j|\to 0 \quad \mbox{a.e.}$ This problem is an exercise in Probability theory,independence,interchangeable, martingale(Chow), exercise 4.1.10.
Let $\{X_n,n\geq 1\}$ be independent identical distributed integrable random variables, then
$$\frac{1}{n}\max_{1\leq j\leq n}|X_j|\to 0 \quad \mbox{a.e.}$$
 A: Show and use the following facts:


*

*If $\{Y_n\}$ is a sequence of random variables such that for each $\varepsilon>0$, 
$$\sum_{n=1}^{\infty}P(|Y_n|\geqslant \varepsilon)<\infty,$$
then the sequence $\{Y_n\}$ converges to $0$ almost surely. 

*If $X$ is an integrable random variable, then the series $\sum_n P(|X|>n)$ is convergent.
A: Davide provided a good hint, I just give some supplementary, since this is a question I want to know long ago.
For any fixed $\epsilon>0,$
$$
\sum_{n=1}^{\infty}P(|X_n|>n\epsilon)=\sum_{n=1}^{\infty}P(|X|>n\epsilon)=\sum_{n=1}^{\infty}P(\frac{|X|}{\epsilon}>n)<\infty
$$
because $\mathbb{E}[\,\frac{|X|}{\epsilon}]<\infty$. Apply Borel-Cantelli lemma,
$$ P(|X_n|>n\epsilon\quad i.o.)=0
$$
which implies $P(|X_n|\le n\epsilon\quad e.v.)=1$. With probability $1$, $\exists\; N(\omega)>0$ such that
$$
 |X_n|\le n\epsilon\quad \text{for}\ \;n>N
$$
Hence there exists an even greater integer $N_1(\omega)>0$ (because $\max|X_j|$ grows less faster than $n\epsilon$) such that
$$
 \max_{1\le j\le n}|X_j|\le n\epsilon\ \Longleftrightarrow \frac{1}{n}\max_{1\le j\le n}|X_j|\le \epsilon\quad \text{for}\ \;n>N_1
$$
