Prove $X_n$ fullfills Weak Law of Large Numbers Use following facts:


*

*$\forall_{\epsilon>0} \mathbb{P}(X \geq \epsilon) \leq \frac{\mathbb{E}X}{\epsilon}$ (Chebyshev's inequality) 

*When $\lim_n n \mathbb{P}(|X| > n) =0$ and $X_n^* = X \mathbb{1}_{\{|X|\leq n\}}$ then $\lim_n \frac{\mathrm{Var(X_n^*)}}{n} = 0$ (proven here)


to prove:
Suppose $\lim_n n \mathbb{P}(|X| > n) =0$ and $X_n$ are i.i.d. with same distribution as X. Prove that $\{X_n\}$ fullfills the Weak Law of Large Numbers by showing:
$$
\lim_n \frac{S_n - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n} \to 0 \text { in probability }
$$
where $S_n = \sum_{i=1}^k X_i$.
So far I have found out:


*

*$0 = \lim_n n \mathbb{P}(|X|\geq n) \leq \mathbb{E}X$ but that gives me nothing useful

*If we define  $X_n^* = X \mathbb{1}_{\{|X|\leq n\}}$ then
$$
\lim_n \frac{S_n - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n} = \frac{S_n - n\mathbb{E}X_n^*}{n} \geq \frac{S_n - n\cdot n\mathbb{P}(|X| \leq n)}{n} = \frac{S_n - n^2}{n}
$$
which is also rather pointless.
I appreciate any hints or solutions.
 A: \begin{eqnarray}
\begin{split}
& \mathbb P\left(\left| \frac{S_n - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n}\right| \geq  \varepsilon\right)\cr  & =  \mathbb P\left(\left| \frac{S_n^* - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n}\right|\geq \varepsilon, |X_1|\leq n,\ldots, |X_n|\leq n\right) \cr & + \mathbb P\left(\left| \frac{S_n - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n}\right|\geq \varepsilon, |X_i|> n \text{ for some } i=1,\ldots, n\right) \cr
& \leq  \mathbb P\left(\left| \frac{S_n^* - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n}\right|\geq \varepsilon\right)+n\mathbb P\left(|X_1|> n\right). 
\end{split}\tag{1}
\end{eqnarray}
The second term tends to zero. Consider the first term and apply Chebyshev's inequality. 
$$
\mathbb P\left(\left| \frac{S_n^* - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n}\right|\geq \varepsilon\right) \leq \frac{\sum_{i=1}^n \text{Var}(X_i^*)}{n^2\varepsilon^2} = \frac{\text{Var}(X_1^*)}{n\varepsilon^2} \to 0.
$$
Note also, that you are proved WLLN in the conditions that expectation $\mathbb EX$ need not to exist. The only thing that is given to you is the condition $n\mathbb P(|X| >n)\to 0$, which is more weak condition than $\mathbb E|X|<\infty$. 
The hint to use Chebyshev's inequality did not mean that expectation exists. This is just a tool that can be applied in this problem to those random variables that have a expectation. 
Addition:
The inequality in (1) follows from the inequality $\mathbb P(A\cap B)\leq \mathbb P(A)$ (as well as $\mathbb P(B)$). So 
$$
\mathbb P\left(\left| \frac{S_n^* - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n}\right|\geq \varepsilon, |X_1|\leq n,\ldots, |X_n|\leq n\right)\leq \mathbb P\left(\left| \frac{S_n^* - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n}\right|\geq \varepsilon\right)
$$
and 
$$
\mathbb P\left(\left| \frac{S_n - n\mathbb{E}X\mathbb{1}_{\{|X|\leq n \}}}{n}\right|\geq \varepsilon, |X_i|> n \text{ for some } i=1,\ldots, n\right)
$$
$$\leq \mathbb P\left(\bigcup_{i=1}^n\{|X_i|> n\}\right)\leq \sum_{i=1}^n P\left(|X_i|> n\right)=n\mathbb P(|X_1|>n).
$$
