$\lim_{x\rightarrow+\infty} x^{\alpha+\beta}P(|X|>x)=0\Leftrightarrow\lim_{x\rightarrow+\infty}x^{\alpha}E||X|^{\beta}\cdot1_{\{|X|>x\}}|=0$ I have met with a probability problem which I have no idea to deal with. It says:
Let $\alpha>0$, $\beta\geq0$, prove: 
$\lim_{x\rightarrow+\infty} x^{\alpha+\beta}\mathbb{P}(|X|>x)=0\Leftrightarrow\lim_{x\rightarrow+\infty}x^{\alpha}\mathbb{E}||X|^{\beta}\cdot1_{\{|X|>x\}}|=0$.  
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.
 A: You can simplify the statement by writing $Y$ for $|X|^{\beta}$ and repalcing $x^{\beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y>t) \to 0$ iff $t^{s}EYI_{Y>t} \to 0$ 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 $t^{s}EYI_{Y>t} \to 0$ as $t^{s}\int_t^{\infty} P(Y>u)du$ which is less than or equal to $t^{s}\int_t^{\infty} \epsilon u^{-1-s}du=\epsilon$ for $t$ sufficiently large.
A: Suppose 
$\lim_{x\to \infty} x^{\alpha+\beta}P(|X|>x) = 0$.
Note that
\begin{align}
E[|X|^\beta \cdot 1_{|X|>x}] 
&= P(|X|>x)E[|X|^\beta | |X| > x] \\
&= x^\beta P(|X|>x) E[|X/x|^\beta | |X/x|>1] \\
&\le x^\beta P(|X|>x) E[|X|^\beta], \quad \text{for large $x$}.
\end{align}
Note that for large $x$, $|X/x| < |X|$, and thus
$|X/x|^\beta < |X|^\beta$.
Assuming $E[|X|^\beta] < \infty$, 
$$
\lim_{x \to \infty} x^\alpha E[|X|^\beta \cdot 1_{|X|>x}]  = 0.
$$
