I've been fighting with this homework problem for a while now, and I can't quite see the light. The problem is as follows,
Assume random variable $X \ge 0$, but do NOT assume that $\mathbb{E}\left[\frac1{X}\right] < \infty$. Show that $\lim_{y \to 0^+}\left(y \, \mathbb{E}\left[\frac{1}{X} ; X > y\right]\right) = 0$
After some thinking, I've found that I can bound
$ \mathbb{E}[1/X;X>y] = \int_y^{\infty}\frac1{x}\mathrm dP(x) \le \int_y^{\infty}\frac1{y}\mathrm dP(x) $
since $\frac1{y} = \sup\limits_{x \in (y, \infty)} \frac1{x}$ resulting in
$ \lim_{y \to 0^+} y \mathbb{E}[1/X; X>y] \le \lim_{y \to 0^+} y \int_y^{\infty}\frac1{y}\mathrm dP(x) = P[X>0]\le1 $
Of course, $1 \not= 0$. I'm not really sure how to proceed...
EDIT: $\mathbb{E}[1/X;X>y]$ is defined to be $\int_y^{\infty} \frac{1}{x}\mathrm dP(x)$. This is the notation used in Durret's Probability: Theory and Examples. It is NOT a conditional expectation, but rather a specifier of what set is being integrated over.
EDIT: Changed $\lim_{y \rightarrow 0^-}$ to $\lim_{y \rightarrow 0^+}$; this was a typo.