Condition that doesn't implies integrability Let $X$ be a random variable, and $x\mathbb{P}\left(\left|X\right|>x\right)\to 0$ when $x\to\infty$. This condition doesn't implies integrability. Can anybody give an example of this? A random variable such that $\int_{0}^{\infty}\mathbb{P}(|X|>x)dx=\infty$ even when $x\mathbb{P}(|X|>x)\to 0$?
And what about the opposite: does integrability of $\left|X\right|$ implies $x\mathbb{P}(|X|>x)\to 0$?
 A: Suppose $X$ has pdf
\begin{align*}
f_X(x) = \frac{x + (x+e-1)\log(x+e-1)}{x^2(x+e-1)\log^2(x+e-1)} 
\end{align*}
defined for $x \ge 1$, and 0 for $x < 1$. We may compute
\begin{align*}
F_X(x) = \int_{1}^{x} f_X(t) dt = \left[1 - \frac{1}{t\log(t+e-1)}\right]_{t=1}^{t=x}= 1 - \frac{1}{x\log(x+e-1)}
\end{align*}
And hence
\begin{align*}
\mathbb{P}(|X| > x) = \mathbb{P}(X > x) = \frac{1}{x\log(x+e-1)}
\end{align*}
Hence $x \mathbb{P}(|X| > x) = \frac{1}{\log(x+e-1)} \rightarrow 0$ as $x \rightarrow \infty$. But, 
\begin{align*}
\int_{1}^{\infty}\mathbb{P}(|X| \ge x) dx &> \int_{1}^{\infty}\frac{1}{(x+e-1)\log(x+e-1)} dx \\
&= \left[\log(\log(x+e-1))\right]_{x=1}^{\infty} \\
&= \infty
\end{align*}
But, integrability implies $x \mathbb{P}(|X| > x) \rightarrow 0$. Indeed, integrability implies that $\mathbb{P}(|X| > x) = o(\frac{1}{x})$, where $o$ is the little $o$-notation. Hence $x\mathbb{P}(|X| > x) = o(1) \rightarrow 0$.
A: The condition $\lim_{x\to  +\infty}x\mathbb P\left\{X>x\right\}=0$ is equivalent to the convergence to zero of the sequence $\left(2^n\mathbb P\left\{X\geqslant 2^n\right\}\right)_{n\geqslant 1}  $. To get an example of non-integrable random variable $X$ such that $2^n\mathbb P\left\{X\geqslant 2^n\right\}$, consider $X$ discrete, taking the value $2^i$, $i\in\mathbb N^*$ with probability $p_i\in[0,1]$ and $\mathbb P\{X=l\}$ if $l$ is not a positive power of $2$. We should have
$$\sum_{i=1}^{ +\infty}p_i=1;\quad \sum_{i=1}^{ +\infty}2^i\cdot p_i=+\infty \mbox{ and } 2^n\sum_{i=n}^{+\infty}p_i\to 0.            $$
We can choose $p_i:=c^{-1}   2^{-i}/i$  with $c :=\sum_{i=1}^{ +\infty}      2^{-i}/i$. 
However, it is true that an integrable random variable $X$ satisfies $\lim_{x\to  +\infty}x\mathbb P\left\{X>x\right\}=0$. Take $Y$ a linear combination of indicator functions. Then by Markov's inequality, 
$$x\mathbb P\left\{X>x\right\}\leqslant x\mathbb P\left\{\left| X-Y\right|>x/2\right\}+x\mathbb P\left\{\left| Y\right|>x/2\right\}\\ \leqslant 2\mathbb E\left|X-Y\right| +x\mathbb P\left\{\left| Y\right|>x/2\right\}. $$
Since $Y$ is bounded, we have $\mathbb P\left\{\left| Y\right|>x/2\right\}$ for $x$ large enough hence 
$$\limsup_{x\to \infty}    x\mathbb P\left\{X>x\right\}\leqslant 2\mathbb E\left|X-Y\right| ,$$
and by definition of Lebesgue integral, the last term can be made arbitrarily small. 
