The conditions for $\lim_{x\to\infty}{g(x)\bar{F}(x)}=0$ Let $X$ be a nonnegative random variable with distribution $F$, and $g$ is a continuous function such that $E[g(X)]< \infty$. Is
$$
\lim_{x\to\infty}{g(x)\bar{F}(x)} = 0
$$
always true, where $\bar{F}(x) = 1-F(x)$ is the survival function? If not, what other conditions should $g$ have?
 A: *

*It holds if  $g\colon[0,\infty)\rightarrow[0,\infty)$ is monotone nondecreasing and $g(x)\xrightarrow{x\rightarrow\infty}\infty$. To wit,
$$P[X>x]\leq P[g(X)\geq g(x)]\leq\frac{1}{g(x)}E\big[g(X)\mathbb{1}_{\{g(X)\geq g(x)\}}\big]$$
to equivalently
$$
g(x)E[X>x]\leq g(x)E[g(X)\geq g(x)]\leq E\big[g(X)\mathbb{1}_{\{g(X)\geq g(x)\}}\big]$$
If $g(X)$ is integrable, then by dominated convergence, $$\lim_{x\rightarrow\infty}E\big[g(X)\mathbb{1}_{\{g(X)\geq g(x)\}}\big]=0$$
whence we conclude that $g(x)E[X>x]\xrightarrow{x\rightarrow\infty}0$.


*If $g\colon[0,\infty)\rightarrow[0,\infty)$ is bounded (not necessarily monotone nondecreasing) the conclusion of the statement in the OP holds trivially.


*It is left to the OP to decide whether the statement holds if $g\colon[0,\infty)\rightarrow[0,\infty)$ is unbounded but not necessarily monotone nondecreasing.

Edit: Here is a hint for what I leave to the OP:
Starting from $g(x)=\sum^\infty_{n=1}e^x\mathbb{1}_{(n,n+\frac1{n^2}]}(x)$ and $X\sim\operatorname{Exp}(1)$, one may construct a continuous function $\phi:[0,\infty)\rightarrow[0,\infty)$ such that $\phi(x) E[X>x]$ diverges as $x\rightarrow\infty$ even though $E[\phi(X)]<\infty$.
