$\{X_1,X_2,...\}$ is uniformly integrable. Prove that $\mathbb{E}X_n \rightarrow \mathbb{E}X$ as $n \rightarrow \infty$. Suppose $X_n \rightarrow X$ a.s. as $n \rightarrow \infty$ and that $X_n \geq0$ for all $n, \omega$. Furthermore, $\{X_1,X_2,...\}$ is uniformly integrable. Prove that $\mathbb{E}X_n \rightarrow \mathbb{E}X$ as $n \rightarrow \infty$.
I know that a sequence of random variables converges almost surely to $X$ if for every $\epsilon > 0$ we have
\begin{equation}
P(\lim_{n \rightarrow \infty}|X_n - X|< \epsilon) = 1
\end{equation}
But I'm not sure how to proceed. Some hints would be greatly appreciated.
 A: You can always write
$$ \mathbb{E}\big[ \vert X_n-X \vert \big]= \int_F\vert X_n-X\vert dP+ \int_{F^c}\vert X_n-X\vert dP\leq \int_{F} \vert X_n\vert dP+ \int_F \vert X\vert dP+ \int_{F^c}\vert X_n-X\vert dP$$
Since $X$ is  integrable and $\{X_n\}$ is unifromly integrable, for every $\epsilon>0$ there exists $\delta_\epsilon>0$ such that
$$ \int_F\vert X_n\vert dP<\epsilon \quad \text{and} \quad \int_F\vert X\vert dP<\epsilon $$
if $P(F)<\delta_\epsilon$. Using almost sure convergence, we know that
$$ P( \vert X_n-X\vert>\epsilon )\to0. $$
Therefore, for $n$ large enough we have $ P( \vert X_n-X\vert>\epsilon )<\delta_\epsilon$ and therefore
$$ \mathbb{E}\big[ \vert X_n-X \vert \big] \leq  \int_{ \{  \vert X_n-X\vert>\epsilon \}} \vert X_n\vert dP+ \int_{ \{  \vert X_n-X\vert>\epsilon \} } \vert X\vert dP+ \int_{\{  \vert X_n-X\vert\leq\epsilon \}}\vert X_n-X\vert dP < $$
$$ < 2\epsilon+ \int_{\{  \vert X_n-X\vert\leq\epsilon \}} \epsilon\cdot dP\leq 3\epsilon.   $$
This is true for all $\epsilon>0$, so $\mathbb{E}[X_n]\to \mathbb{E}[X]$.
A: Another (similar) approach is as follows.
For every $\epsilon > 0$, choose $K(\epsilon)$ so that $E[|X_n|1_{X_n\geq K(\epsilon)}]<\epsilon$ and $E[|X|1_{X\geq K(\epsilon)}] < \epsilon$ (this is simply applying the definition of UI to $\{ X_n \}_n \cup \{X\}$.
Compute
$$
\begin{align}
E|X_n-X| \leq
& E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] \\
+ &E[|X_n|1_{X_n > K(\epsilon), X \leq K(\epsilon)}] \\
 +& E[|X_n|1_{X_n \leq K(\epsilon), X > K(\epsilon)}] \\
+& E[|X|1_{X_n > K(\epsilon), X \leq K(\epsilon)}]\\
+& E[|X|1_{X_n \leq K(\epsilon), X > K(\epsilon)}]\\
\leq &E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] \\
 + &  E[|X_n|1_{X_n > K(\epsilon)}] \\
+&E[|X|1_{X > K(\epsilon)}] \\
+&   E[|X_n|1_{X_n > K(\epsilon)}] \\
+&E[|X|1_{X > K(\epsilon)}] \\
\leq &E[|X_n-X|1_{X_n, X \leq K(\epsilon)}]   + 4\epsilon.
\end{align}
$$
Sending $n \to \infty$ we find
$$
\limsup_n E|X_n-X| \leq 4\epsilon.
$$
Now send $\epsilon \to 0$.
