Show that $\frac{1}{n}X_n\to 0\mbox{ a.e.}$ 

Show that for any sequence $(X_n)$ of identically distributed integrable random variables it is 
    $$
\frac{1}{n}X_n\to 0\text{ a.e.}
$$


Hello! I am missing an idea...
I have to show that
$$
\mathbb{P}(\left\{\frac{1}{n}X_n\not\to 0\right\})=0.
$$
I do not know which tool I need to apply:


*

*weak law of large numers?

*strong law of large numbers?

*Ergoden?
Maybe you can give me a hint...
I am sorry that I cannot give more own ideas yet.
 A: Hint The inequality
$$\sum_{n =1}^{\infty} 1_{\{|X| \geq n\}} \leq |X|$$
entails
$$\sum_{n=1}^{\infty} \mathbb{P}(|X| \geq n) \leq \mathbb{E}(|X|) \tag{1}$$
(Thanks to @ByronSchmuland for the suggestion of this simplification.) Use this in order to prove
$$\sum_{n=1}^{\infty} \mathbb{P} \left( \left| \frac{X_n}{n} \right| > \varepsilon \right) < \infty.$$
Now apply Borel-Cantelli's lemma.
A: Here is my proof based on your help. Please tell me if it is ok.
Consider any $\varepsilon>0$.
Define $Y:=\frac{\lvert X_1\rvert}{\varepsilon}$.
$$
A_m:=\left\{\omega\in\Omega: m\leq Y(\omega)<m+1\right\}, m\geq 1.
$$
Consider an arbitrary fixed $\omega\in\Omega$. Then $Y(\omega)\in A_m$ for a $m\geq 1$.
Then it is
$$
Y(\omega)\geq \sum_{n=1}^m 1_{\left\{Y>n\right\}}=\sum_{n=1}^{\infty}1_{\left\{Y>n\right\}}.
$$
Because $\omega$ was chosen arbitrarily, this is for any $\omega\in\Omega$.
From this it follows
$$
\Rightarrow\mathbb{E}(Y)\geq\sum_{n=1}^{\infty}\mathbb{E}(1_{\left\{Y>n\right\}})=\sum_{n=1}^{\infty}\mathbb{P}(\left\{Y>n\right\})
$$
And (because of the integrability of $X_1$) it is
$$
\mathbb{E}(Y)=\frac{1}{\varepsilon}\int_{\Omega}\lvert X_1\rvert\, d\mathbb{P}<\infty.
$$
Define
$$
M_n:=\left\{\omega\in\Omega: \frac{1}{n}\lvert X_n\rvert>\varepsilon\right\}.
$$
From what is shown above, it follows from Borel-Cantelli, that
$$
\mathbb{P}(\limsup_{n\to\infty}M_n)=\mathbb{P}(\left\{\omega\in\Omega: \omega\in M_n\text{ for infinite many }n\right\})=0.
$$
Because of
$$
N:=\left\{\omega\in\Omega: \frac{1}{n}X_n(\omega)\not\to 0\right\}\subset\limsup_{n\to\infty}M_n,
$$
it follows
$$
\mathbb{P}(N)\leq\mathbb{P}(\limsup_{n\to\infty}M_n)=0\Rightarrow\mathbb{P}(N)=0.
$$
So $\frac{1}{n}X_n\to 0\text{ a.e.}~~\diamond$
