TITLE: Limits of a distribution property
QUESTION [6 upvotes]: I've been cracking my brains since several days so far trying to solve this exercise. I'll firstly claim the problem and then will clarify what I know and what I've tried to do.
So, we are given a real-valued r.v. on some probability space s.t. $\mathbb E(|X|)<\infty$ and by $F$ we denote the distribution function.
What we need to show is
$$\lim_{z \ \rightarrow \ - \ \infty}z F(z) = \lim_{z \ \rightarrow \ + \ \infty}z (1-F(z)) = 0$$
We know, that $\lim_{z \ \rightarrow \ - \ \infty}F(z) = 0$ and $\lim_{z \ \rightarrow \ + \ \infty}F(z) = 1$ and what I though of was to somehow show that the distribution function increases to $1$ faster than $z$ goes to infinity. Also, on this step we still don't know whether a density exists (it will be mentioned later in the exercise).
I realize that the solution might be quite simple but I'm really stuck with it.
Any help would be highly appreciated.

REPLY [6 votes]: It suffices to prove that $$\tag{*}\lim_{z\to +\infty}z\Pr\left(\left|X\right|\gt z\right)=0,$$
since the two wanted limits are up to a constant that of $z\Pr\left(X\gt z\right)$ and $z\Pr\left(-X\gt z\right)$ as $z$ goes to $+\infty$. 
The convergence (*) follows from Markov's inequality: 
$$z\Pr\left(\left|X\right|\gt z\right) \leqslant \mathbb E\left[\left|X\right|\mathbf 1\left\{\left|X\right| \gt z\right\}\right]$$
and then use monotone convergence.