(Borel-Cantelli) If $X_i$ is a sequence of identical and independent variables. Using Borel-Cantelli, prove that... 
If $X_i$ is a sequence of identical and independent variables. Using Borel-Cantelli, prove that
$$E|X_1| \lt \infty \to P(|X_n| \gt n\ \text{infinite times}) = 0.$$
(uses that $E(x) = \sum_{n=0}^{\infty}P(X \gt n)$).

Let $(A_n)_{n \in \mathbb N}$ be a sequence of events.
1- If $\sum_{n=1}^{\infty} P(A_n) \lt \infty$, then:
$$P(\lim_{n\to \infty} \sup A_n) = 0$$
2-If $(A_n)_{n \in \mathbb N}$ are independent and $\sum_{i=1}^{\infty} P(A_i) = \infty$, then:
$$P(\lim_{n\to \infty} \sup A_n) = 1$$
what i managed to do
$$P(\lim_{n\to \infty} \sup A_n)= P(\cap_{n_0}(\cup_{n \gt n_0}A_n)) \le P(\cup_{n \gt n_0}A_n)\le   \lim_{n_0 \to \infty}\sum_{n \ge n_0}P(A_n)=0$$
Is my answer to the exercise acceptable?
Thanks.
 A: $\Bbb{E}(X)=\sum_{n=0}^{\infty}\Bbb{P}(X>n)$ is true for integer valued non-negative random variables.
Instead you can argue like this:-
Let $A_{n}=\{|X_{n}|> n\}$
Then $\Bbb{P}(A_{n})=\Bbb{P}(\{|X_{n}|>n\})=\Bbb{P}(|X|>n)$.
You have :-
$$\Bbb{E}[|X|]=\int_{0}^{\infty}\Bbb{P}(|X|>t)\,dt$$
(I am deliberately not using the notation for Lebesgue Integral as you might not be familiar with it).
The above is due to Tonelli's Theorem:-
$$\Bbb{E}[|X|]=\Bbb{E}[\int_{0}^{\infty}\,\mathbf{1}_{(0,|X(\omega)|]}dt]=\Bbb{E}[\int_{0}^{\infty}(\mathbf{1}_{\{0<t<|X(\omega)|\}})\,dt]$$
$$=\int_{0}^{\infty}\Bbb{E}[\mathbf{1}_{\{0<t<|X(\omega)|\}}]\,dt = \int_{0}^{\infty}\Bbb{P}(|X|>t)\,dt$$. For more detailed proof see here. The proof for the integer valued random variables is almost the same . You just have to replace the Lebesgue measure on $\Bbb{R}$ with the counting measure on $\Bbb{Z_{\geq 0}}$.
Now $\{\Bbb{P}(|X|>n)\}_{n=0}^{\infty}$ is a decreasing sequence of real numbers and since the integral $\displaystyle\int_{0}^{\infty}\Bbb{P}(|X|>t)\,dt<\infty$ we have $\displaystyle\sum_{n=0}^{\infty}\Bbb{P}(|X|>n)<\infty$ due to Integral test
Now  you have $\sum_{n=0}^{\infty}\Bbb{P}(A_{n})=\sum_{n=0}^{\infty}\Bbb{P}(|X|>n)<\infty$ and hence by Borel Cantelli you can conclude that $\Bbb{P}(|X_{n}|>n\,\text{infinitely often})=\Bbb{P}(\lim\sup A_{n})= 0$.
Now if indeed $X_{1}$ was an integer valued random variable then you can use the indentity $\Bbb{E}(|X|)=\sum_{n=0}^{\infty}\Bbb{P}(|X|>n)$ and conclude that $\sum_{n=0}^{\infty}\Bbb{P}(A_{n})<\infty$ and hence by Borel-Cantelli you have your required result.
