# Show that $E(|X|)<\infty$ and $E(X_n)\rightarrow E(X)$

After a few weeks off I am back at my self-study of Measure-Theoretic probability. As always, I thank the community for any detail and answers they can provide as I try to work myself through these exercises.

Perhaps this is an application of Levi?

The question is:

Suppose $X_1,X_2,...$ is a sequence of random variables, not necessarily nonnegative, and $X_n\uparrow X a.s.$. Also assume $\sup_n E(|X_n|)<\infty$.

Show that $E(|X|)<\infty$ and $E(X_n)\rightarrow E(X)$

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Can't you just apply monotone convergence to $Y_n = X_n - X_1$? – Nate Eldredge Jul 28 '12 at 16:16

First, Fatou's lemma gives $E (\liminf_n |X_n|) \leq \liminf_n E |X_n| \leq \sup_n E |X_n| < \infty$. Since $\liminf_n |X_n| = |X|$ a.e., this shows that $E |X| < \infty$.
We have $0 \leq X - X_n \leq X - X_1$ a.e. We know that $E |X-X_1| \leq E |X| + E |X_1| < \infty$, so apply the DCT to get $\lim_n E (X - X_n) = 0$, from which the result follows.