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!

The first inequality $x\mathbb P(|X_1|>x) \le \mathbb E\left[|X_1|1_{(|X_1|>x)}\right]$ is just a use of Markov inequality:
$$\mathbb E\left[|X_1|1_{(|X_1|>x)}\right] \ge \mathbb E\left[x\cdot 1_{(|X_1|>x)}\right] \ge x\mathbb P(|X_1|>x)\qquad (\because |X_1| >x\text{ on the support}).$$
The dominated convergence theorem implies:


*

*Since $|X_1| 1_{(|X_1|>x)}$ is bounded by an integrable function $|X_1|$, 


$$\lim_{x\to\infty}\mathbf E\left[|X_1| 1_{(|X_1|>x)}\right] = \mathbf E\left[ \lim_{x\to\infty} |X_1| 1_{(|X_1|>x)}\right] = \mathbf E\left[ 0\right]=0.$$
The proof of $u_n\to u$ is similar to the above.
