Let $\{X_n\}$ and $X$ be random variables. If $X_n\rightarrow X$ a.e., then for every $r>0$, $$\mathscr{E}|X|^r\leq\underset{n\rightarrow\infty}{\underline{\lim}}\mathscr{E}|X_n|^r.$$If $X_n\rightarrow X$ in $L^r$, and $X\in L^r$, then $\mathscr{E}|X_n|^r\rightarrow\mathscr{E}|X|^r$.
$\bullet$ Proof.
(1) By Fatou's Lemma, we have $$\int_\Omega|X|^r\,d\mathscr{P}=\int_\Omega\underset{n\rightarrow\infty}{\lim}|X_n|^r\,d\mathscr{P}\leq\underset{n\rightarrow\infty}{\underline{\lim}}\int_\Omega|X_n|^r\,d\mathscr{P}=\underset{n\rightarrow\infty}{\underline{\lim}}\mathscr{E}|X_n|^r.$$
(2) Write $$X=X_n+(X-X_n)=X_n-(X_n-X).$$For $r>1$, we have by applying Minkowski's inequality on both equalities, $$\left(\mathscr{E}|X_n|^r\right)^{\frac{1}{r}}-\left(\mathscr{E}|X_n-X|^r\right)^{\frac{1}{r}}\leq\left(\mathscr{E}|X|^r\right)^{\frac{1}{r}}\leq\left(\mathscr{E}|X_n|^r\right)^{\frac{1}{r}}+\left(\mathscr{E}|X_n-X|^r\right)^{\frac{1}{r}}.$$Hence, as $n\rightarrow\infty$, we have $\mathscr{E}|X_n|^r\rightarrow\mathscr{E}|X|^r$. Next, for $0<r\leq1$, consider the inequality $$|x+y|^r\leq|x|^r+|y|^r\mbox{ for all }x,y\in\mathbb{R}.$$Which implies that $$\left(\mathscr{E}|X_n|^r\right)-\left(\mathscr{E}|X_n-X|^r\right)\leq\left(\mathscr{E}|X|^r\right)\leq\left(\mathscr{E}|X_n|^r\right)+\left(\mathscr{E}|X_n-X|^r\right).$$Thus, again we have $\mathscr{E}|X_n|^r\rightarrow\mathscr{E}|X|^r$ as $n\rightarrow\infty$.
$\Box$
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