$\lim_{x\rightarrow+\infty} x^{\alpha+\beta}P(|X|>x)=0\Leftrightarrow\lim_{x\rightarrow+\infty}x^{\alpha}E||X|^{\beta}\cdot1_{\{|X|>x\}}|=0$ I have met with a probability problem which I have no idea to deal with. It says:

Let $\alpha&gt;0$, $\beta\geq0$, prove: 

$\lim_{x\rightarrow+\infty} x^{\alpha+\beta}\mathbb{P}(|X|&gt;x)=0\Leftrightarrow\lim_{x\rightarrow+\infty}x^{\alpha}\mathbb{E}||X|^{\beta}\cdot1_{\{|X|&gt;x\}}|=0$.  

The sufficiency is easy to prove. However, I have no ideas how to prove the necessity. Can you give me some advice? Any hints would be appreciated.

Thanks for your time and patience.
You can simplify the statement by writing $Y$ for $|X|^{\beta}$ and repalcing $x^{\beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y&gt;t) \to 0$ iff $t^{s}EYI_{Y&gt;t} \to 0$ as $t \to \infty$.[Here $s=\frac {\alpha} {\beta}$. I will leave the case $\beta=0$ to you].  As noted by you the íf'part follows easily by Chebychev's Inequality. For the ónly if'part write $t^{s}EYI_{Y&gt;t} \to 0$ as $t^{s}\int_t^{\infty} P(Y&gt;u)du$ which is less than or equal to $t^{s}\int_t^{\infty} \epsilon u^{-1-s}du=\epsilon$ for $t$ sufficiently large.
Suppose 
$\lim_{x\to \infty} x^{\alpha+\beta}P(|X|&gt;x) = 0$.
Note that
\begin{align}
E[|X|^\beta \cdot 1_{|X|&gt;x}] 
&amp;= P(|X|&gt;x)E[|X|^\beta | |X| &gt; x] \\
&amp;= x^\beta P(|X|&gt;x) E[|X/x|^\beta | |X/x|&gt;1] \\
&amp;\le x^\beta P(|X|&gt;x) E[|X|^\beta], \quad \text{for large $x$}.
\end{align}
Note that for large $x$, $|X/x| &lt; |X|$, and thus
$|X/x|^\beta &lt; |X|^\beta$.

Assuming $E[|X|^\beta] &lt; \infty$, 
$$
\lim_{x \to \infty} x^\alpha E[|X|^\beta \cdot 1_{|X|&gt;x}]  = 0.
$$