Q:

Durrett Ex.1.3.16

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.

A:

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}

