Limits of Expectations 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.

Hint: For any $k > 1$, $\int_y^\infty \frac{y}{x} \ dP(x) \le \int_y^{ky}\  dP(x) + \int_{ky}^\infty \frac{1}{k} \ dP(x) \le \ldots$
