If $\lim_{\alpha \to \infty}\alpha P[X > \alpha] = 0$ then $E[X] < \infty$? Let $X$ be a positive random variable. Suppose that $\lim_{\alpha \to \infty}\alpha P[X > \alpha] = 0$ Does this implies that $X$ has finite expectation? that is $E[X] < \infty $
I know that if $E[X] < \infty$ $\Rightarrow$ $\lim_{\alpha \to \infty}\alpha P[X > \alpha] = 0$ (For any positive random variable see:  Expected value as integral of survival function) , so I was wondering if the converse is true.
I have also tried to think in a counterexample but unfortunately I have not been successfull.
I would really appreciate any hints or suggestions with this problem.
 A: No!
Example: Let $\mathbb{P}(X>\alpha)=\alpha^{-1}(\log\alpha)^{-1+\delta}$ for all sufficiently large $\alpha$, with $\delta\in[0,1)$.  Then $\int^\infty\mathbb{P}(X>\alpha)\,\mathrm{d}\alpha=+\infty$ and hence $\mathbb{E}X$ cannot be finite.
A: Here is an another counter example based on a problem solved here
Consider the probability space $((0,1),\mathscr{B}((0,1)),\lambda)$ where $\lambda$ is Lebesgue's measure restricted to the unit interval, and consider the function $X:(0,1)\rightarrow\mathbb{R}$  defined by
$$X(t):=\frac{1}{t|\log t|}\mathbb{1}_{(0,e^{-1}]}+e\mathbb{1}_{(e^{-1},1)}(t)$$
It is not difficult to check that $X$ is continuous, and that  $\lim_{t\rightarrow0+}X(t)=\infty$, $X$ is continuous and strictly monotone decreasing on $(0,e^{-1}]$, and
$$
\int_{(0,1]}X\,d\lambda\geq-\int_{(0,e^{-1}]}\frac{dx}{x\log x}=-\log(-\log(x))|^{e^{-1}}_0=\infty
$$
For each $\alpha$ large enough, there is exactly one $a_\alpha<\tfrac1e$ such that $X(a_\alpha)=\alpha$. Then
$$\begin{align}
\lambda(X>\alpha)&=\lambda((0,a_\alpha))=a_\alpha\\
&=\frac{1}{\alpha}\frac{1}{|\log a_\alpha|}
\end{align}$$
Furthermore, as $a_\alpha\rightarrow0$ when $\alpha\rightarrow\infty$,
$$
\lim_{\alpha\rightarrow\infty}\alpha\lambda(X>\alpha)=\frac{1}{|\log a_\alpha|}\xrightarrow{\alpha\rightarrow\infty}0
$$
