Prove $\lim_{\varepsilon\to 0^+}\frac{1}{\varepsilon}\int\limits_{X\leqslant \varepsilon}X\mathrm{d}\mathbb{P}=0$ 
Let $X$ be a random variable taking values in $[0,+\infty]$. Prove that:
  $$\lim_{\varepsilon\to 0^+}\frac{1}{\varepsilon}\int\limits_{X\leqslant \varepsilon}X\mathrm{d}\mathbb{P}=0 ~~~~\mathrm{and~~~} 
\lim_{x \to +\infty}\frac{1}{x}\int\limits_{X\leqslant x}X\mathrm{d}\mathbb{P}=0.$$

Attempt. Of course we have 
$\displaystyle \frac{1}{\varepsilon}\int\limits_{X\leqslant \varepsilon}X\mathrm{d}\mathbb{P}\leqslant \frac{1}{\varepsilon} \varepsilon =1,$
but we don't get anything interesting by that (the same arguments holds for the other integral also).
Thank you in advance.
 A: Observe that
$$
\int_{X\leq \epsilon}X\ d\mathbb P=\int_{0<X\leq \epsilon}X\ d\mathbb P\leq \int_{0<X\leq \epsilon}\epsilon\ d\mathbb P=\epsilon \mathbb P(0<X\leq \epsilon),
$$
and therefore
$$
0\leq \frac{1}{\epsilon}\int_{X\leq \epsilon}X\ d\mathbb P\leq \mathbb P(0<X\leq \epsilon).
$$
As $\epsilon\to 0$ the right side tends to $\mathbb P(0<X\leq 0)$ which is zero (regardless of whether $\mathbb P(X=0)>0$), so we get the desired limit, by the squeeze theorem.
The second limit has a simple proof if $\mathbb EX<\infty$, in which case
$$
\int_{X\leq x}X\ d\mathbb P\leq \int_{\Omega}X\ d\mathbb P=\mathbb EX<\infty,
$$
and therefore
$$
\frac{1}{x}\int_{X\leq x}X\ d\mathbb P\leq\frac{\mathbb EX}{x}
$$
which tends to zero as $x\to\infty$.
Here is a more involved argument which works when $\mathbb EX=\infty$ as well. Apply the tail formula for the expectation to the random variable $X\cdot 1_{X\leq x}$ to obtain the formula
$$\int_{X\leq x}X\ d\mathbb P=\int_0^x\mathbb P(t\leq X<\infty)\ dt.$$
As $t\to\infty$ the quantity $\mathbb P(t\leq X<\infty)$ tends to zero (regardless of whether $\mathbb P(X=\infty)$ is positive). Thus, for every $\epsilon>0$ we can find an $N<\infty$ such that $\mathbb P(t\leq X<\infty)<\epsilon$ for all $t>N$. Consequently for all $x>N$ we have the upper bound
$$
\int_0^x\mathbb P(t\leq X<\infty)\ dt\leq N+\int_{N}^x\mathbb P(t\leq X<\infty)\ dt\leq N+(x-N)\epsilon.
$$
Dividing both sides by $x$ and sending $x\to\infty$ yields that
$$\limsup_{x\to\infty}\frac{1}{x}\int_{X\leq x}X\ d\mathbb P\leq\epsilon.$$
Since $\epsilon>0$ was arbitrary, we therefore obtain that
$$
\limsup_{x\to\infty}\frac{1}{x}\int_{X\leq x}X\ d\mathbb P=0,
$$
and hence the desired limit.
