Let $X \ge 0$. Show
\begin{align} \lim_{y\to\infty} y\,\Bbb{E}\left[ \frac1X \cdot1_{X>y} \right] &= 0 \quad\text{and} \\ \lim_{y \downarrow 0} y\,\Bbb{E}\left[ \frac1X \cdot1_{X>y} \right] &= 0. \end{align}
N.B.: The author doesn't assume $\Bbb{E}[1/X] < \infty$.
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.
$$ \begin{aligned} y\,\Bbb{E}\left[ \frac1X \cdot1_{X>y} \right] &= \Bbb{E}\left[ \frac{y}{X} \cdot1_{X>y} \right] \\ &< \Bbb{E}\left[ 1 \cdot1_{X>y} \right] \\ &= \Bbb{P}(X > y) \\ &\xrightarrow[y\to0]{} \Bbb{P}(X \ge 0) = 1 \end{aligned} \tag1 \label1 $$
For the first part $y \to \infty$, I apply Markov's inequality at the row $\Bbb{P}(X > y)$ in \eqref{1}, so that $$\Bbb{P}(X > y) \le \frac{\Bbb{E}[X]}{y} \xrightarrow[y\to\infty]{} 0?$$ But I don't have $\Bbb{E}[X] < \infty$.
Thanks for any help.
It's clear from the comments that you already got the first part. The second part is even easier than the first part. One has for $y>0$: \begin{align} \left|\frac{y}{X}\mathbf{1}_{y<X} \right| \leq 1 \end{align} Moreover: \begin{align} \lim_{y\to 0^+}\frac{y}{X}\mathbf{1}_{y<X} = 0 \end{align} By the Bounded Convergence Theorem: \begin{align} \lim_{y\to 0^+}\mathbb E\left[ \frac{y}{X}\mathbf{1}_{y<X} \right]=\mathbb E[0]=0 \end{align}