Show $\lim_{n \to \infty} n^{-1} E \left( \frac{1}{X}1_{[X>n^{-1}]} \right) =0$

Suppose $X$ is a non-negative random variable satisfying \begin{align*} P[0 \le X < \infty ]=1. \end{align*} Show

a) \begin{align*} \lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X>n]} \right) =0 \end{align*} b)

\begin{align*} \lim_{n \to \infty} n^{-1} E \left( \frac{1}{X}1_{[X>n^{-1}]} \right) =0 \end{align*}

What I did

I showed part a) already here is the proof:

\begin{align*} n E \left( \frac{1}{X}1_{[X>n]} \right) \le n E \left( \frac{1}{n}1_{[X>n]} \right)=n\cdot\frac{1}{n}P[X>n]=P[X>n] \end{align*} So $\lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X>n]} \right) \le \lim_{n \to \infty} P[X>n]=0$.

What about part b)? The same techniques no longer works.

I was thinking that \begin{align*} \lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X>n]} \right) \le \lim_{n \to \infty} n E \left( \frac{1}{X} \right) \le 0 \end{align*} but I am not sure that $E \left( \frac{1}{X} \right) < \infty$

Solution 1:

This was already asked on the site. Both results are consequences of Lebesgue dominated convergence theorem.

Define $X_n=nX^{-1}\mathbf 1_{X\gt n}$ and $Y_n=(nX)^{-1}\mathbf 1_{nX\gt1}$, then $X_n\to0$ almost surely because $X$ is almost surely finite and $Y_n\to0$ almost surely because $Y_n\leqslant n^{-1}Z$ with $Z=X^{-1}\mathbf 1_{X\gt0}$ almost surely finite.

Thus, $E(X_n)\to0$ and $E(Y_n)\to0$ as soon as a domination condition holds. But it happens that...

$$\qquad\qquad\qquad\qquad X_n\leqslant1\qquad Y_n\leqslant1$$