How to prove $xP(|X_1|>x) \leq E(|X_1|1(|x_1|>x))$? It is from one proof of theorem in Probability: Theory and Examples by Durret.
The theorem is stated as below.

Let $X_1, X_2,\dots$ be i.i.d with $E|X_i|<\infty$. Let $S_n=X_1+\dots+X_n$ and let $u=E(X_1)$. Then $S_n/n \rightarrow u$ in probability.

The proof in the book shown as below: 
Proof:  Two applications of the dominated convergence theorem imply
$xP(|X_1|>x)\leq E(|X_1|1(|X_1|>x))\rightarrow 0$ as $x\rightarrow \infty$
$u_n= E(|X_1|1(|X_1|\leq n))\rightarrow E(X_1)=u$ as $n\rightarrow \infty$

My question is how to prove  $xP(|X_1|>x) \leq  E(|X_1|1(|x_1|>x))$ and $E(|X_1|1(|X_1|>x))\rightarrow 0$?

Can anyone explain it in details?  Many thanks!

Just compute the $P(|X_1|>x)$ and you will see things following. It is nothing but the Chebyshev's inequality. 
Notice that $$P(|X_1|>x)=\int \mathrm{1}_{|X_1|>x}\le \int\frac{|X_1|}{x}\mathrm{1}_{|X_1|>x}.$$
The last inequality above follows from the fact that on the set $\{|X_1|>x\}$ we have $\frac{|X_1|}{x}>1.$ Now all it takes is a multiplication by $x$ in both sides, and you get the first inequality in your question. 
For the second part we use the fact that $|X_1|=|X_1|\mathrm{1}_{|X_1|\le x}+|X_1|\mathrm{1}_{|X_1|>x},$ and then use the DCT to argue that $\int |X_1|\mathrm{1}_{|X_1|\le x}\to \int |X_1|$ as $x\to \infty.$ It follows therefore that $\int |X_1|\mathrm{1}_{|X_1|>x}\to 0$ as $x\to \infty.$
