# Convergence of Expectation for $X_n=\frac nY 1_{\{Y>n\}}$ for any $Y$ such that $P(0\le Y<\infty)=1$

Here is another self-study exercise that I am struggling mightily with:

$X_n=\frac nY 1_{\{Y>n\}}$ for any $Y$ such that $P(0\le Y<\infty)=1$

I am told that $X_n\to X$ a.s for some $X$, and am to show whether $E(X_n)\to E(X)$ as $n\to\infty$

I do not need to explicitly calculate the expectation, but just show its convergence, if applicable.

As I get more and more familiar with dominated convergence, monotone convergence, Fatou, etc. I may not need as much explicit help, but in this exercise if you could help me identify which of the convergence theorems is necessary (and hints at the justification for such), it would be of great help.

I'm having trouble visualizing the series in this form.

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Using $nI_{\{Y > n\}} \leq Y$, conclude that $|X_{n}| \leq 1$. Hence, you can take $Z = 1$ as the integrable function which dominates $|X_{n}|$ and conclude from the Dominated Convergence Theorem that, since $X_{n}$ converges a.s. to $X$, $E[X_{n}]$ converges to $E[X]$.
You can actually find out what $X$ is. Observe that $\{\omega: X_{n}(\omega) \neq 0\} = \{\omega: Y(\omega) > n\}$. Using this, you can work out that $X = 0$.
Hint: $n/Y < 1$ when $Y > n$, so $0 < X_n < 1$ almost surely. Thus $E(X_n) \le P(Y_n > n)$.