Show $$\lim_{n\to\infty} \frac{1}{n}E\left(\frac{1}{X} I_{\{X&gt;1/n\}}\right) = 0.$$  I am totally lost on this problem. Any hint or reference to study will be greatly appreciated.
Let $X \ge 0$.  Show  \begin{align} \lim_{y\to\infty} y\,\Bbb{E}\left[ \frac1X \cdot1_{X&gt;y} \right] &amp;= 0 \quad\text{and} \\ \lim_{y \downarrow 0} y\,\Bbb{E}\left[ \frac1X \cdot1_{X&gt;y} \right] &amp;= 0. \end{align}  N.B.: The author doesn't assume $\Bbb{E}[1/X] &lt; \infty$.    My try:  Since the expectation appears on the RHS of Markov's inequality, I don't think this inequality can be applied to solve this question.  $$ \begin{aligned} y\,\Bbb{E}\left[ \frac1X \cdot1_{X&gt;y} \right] &amp;= \Bbb{E}\left[ \frac{y}{X} \cdot1_{X&gt;y} \right] \\ &amp;&lt; \Bbb{E}\left[ 1 \cdot1_{X&gt;y} \right] \\ &amp;= \Bbb{P}(X &gt; y) \\ &amp;\xrightarrow[y\to0]{} \Bbb{P}(X \ge 0) = 1 \end{aligned} \tag1 \label1 $$  For the first part $y \to \infty$, I apply Markov's inequality at the row $\Bbb{P}(X &gt; y)$ in \eqref{1}, so that $$\Bbb{P}(X &gt; y) \le \frac{\Bbb{E}[X]}{y} \xrightarrow[y\to\infty]{} 0?$$ But I don't have $\Bbb{E}[X] &lt; \infty$.  Thanks for any help.
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