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Let the sequence $\left\{ \left| X_n - X \right|^r \right\}$ be uniformly integrable for $r > 0$. This means that $E \left[ \left| X_n - X \right|^r 1 \left\{ \left| X_n - X \right|^r > a \right\} \right] \xrightarrow{a \rightarrow \infty} 0$, uniformly in $n$.

I would like to understand a proof that $X_n \xrightarrow{L_r} X$ as $n \rightarrow \infty$, meaning that $E \left| X_n - X \right|^r \xrightarrow{n \rightarrow \infty} 0$.

Fix $\varepsilon > 0$ \begin{eqnarray*} E \left| X_n - X \right|^r & = & E \left[ \left| X_n - X \right|^r 1 \left\{ \left| X_n - X \right|^r \leqslant \varepsilon \right\} \right] + E \left[ \left| X_n - X \right|^r 1 \left\{ \left| X_n - X \right|^r > \varepsilon \right\} \right]\\ & \leqslant & \varepsilon + E \left[ \left| X_n - X \right|^r 1 \left\{ \left| X_n - X \right|^r > \varepsilon \right\} \right] \end{eqnarray*} Now if I could conclude from here that $$ \limsup_{n \rightarrow \infty} E \left| X_n - X \right|^r \leqslant \varepsilon $$ then that would prove it. The problem is the second term on the left hand side of the inequality goes to zero only as $\varepsilon$ gets large. How is it possible to proceed from here?

Edit: What are the consequences of adding the assumption $X_n \xrightarrow{a.s.} X$ here? (It is a bit stronger than the suggestion in the comment below.)

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You need the additional hypothesis that $X_n \to X$ in measure for this implication to be true. It is certainly not the case that any family of uniformly integrable functions converges. –  Chris Janjigian Dec 6 '12 at 2:40
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1 Answer 1

up vote 2 down vote accepted

As the question is tagged (probability-theory), I assume the underlying space of finite measure. Let $Y_n:=|X-X_n|^r$; we assume that $Y_n\to 0$ in probability. Then $$E[Y_n]\leqslant E[Y_n\chi_{\{Y_n\leqslant R\}}]+\sup_{k\in\Bbb N}E[Y_k\chi_{\{Y_k\geqslant R\}}].$$ A consequence of the dominated convergence theorem is that $\lim_{n\to +\infty} E[Y_n\chi_{\{Y_n\leqslant R\}}]=0$ for each $R$, which gives, for each $R>0$, $$\limsup_{n\to +\infty}E[Y_n]\leqslant \sup_{k\in\Bbb N}E[Y_k\chi_{\{Y_k\geqslant R\}}].$$ We conclude using uniform integrability assumption.

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Thanks for this short, elegant and simple proof. It's great! –  Learner Dec 6 '12 at 11:12
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