# Show that $\lim\limits_{y\downarrow 0} y\mathbb{E}=0$.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $X$ be a nonnegative random defined on this space. For any $A\in \mathcal{F}$, let $\mathbb{E}[X;A]:=\mathbb{E}[X\mathbb{1}_{A}]$, where $\mathbb{1}_{A}$ denotes the indicator random variable on $A$. Without assuming $\mathbb{E}\left[\frac{1}{X}\right]<\infty$, show that
\begin{eqnarray}
\lim\limits_{y\downarrow 0} ~~y~\mathbb{E}\left[\frac{1}{X};X>y\right]=0.
\end{eqnarray}
Any initial ideas will be greatly appreciated.

#### Solutions Collecting From Web of "Show that $\lim\limits_{y\downarrow 0} y\mathbb{E}=0$."

Hint: For any $\epsilon>0$ there exists $y_0>0$ such that $\mathbb{P}(0<X\le y_0)\le\epsilon$ (why?). For $y<y_0$, we have
\begin{align} y\mathbb{E}\left[\frac{1}{X};X>y\right] &= y\left(\mathbb{E}\left[\frac{1}{X};X>y_0\right]+\mathbb{E}\left[\frac{1}{X};y<X\le y_0\right]\right) \\
&\le y\left(\frac{1}{y_0}+\frac{1}{y}\mathbb{P}(y<X\le y_0)\right) \\
&\le \frac{y}{y_0}+\epsilon.
\end{align}
What can you say as $y\rightarrow 0$?