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Let $(X_{n})$ be a sequence of nonnegative uniformly integrable random variables. Is it true that $\limsup X_{n} \in L_{1}$? Thanks!

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First, the question should be rephrased to avoid giving orders. Second, the result is false. –  Did Mar 12 '11 at 21:01
    
Can you give a counterexample? –  user7762 Mar 12 '11 at 21:15
    
Yes, I can. But did you read the first part of my comment? –  Did Mar 12 '11 at 21:17
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Now that you modified your question, let us turn to the second part of my comment, saying that the result is false. Hint 1: any i.i.d. sequence of integrable random variables is uniformly integrable. Hint 2: what is the limsup of such a sequence? –  Did Mar 12 '11 at 23:52
    
what is the limsup of such a sequence? –  user7762 Mar 13 '11 at 0:50
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up vote 1 down vote accepted

Take $\{X_n\}$ a sequence of i.i.d. , with a density of unbounded support, and integrable.

  • The supremum in the definition of uniform integrability $\sup_{n\in\Bbb N}\int_{\{|X_n|\geq R}|X_n|dP$ is actually $\int_{\{|X_1|\geq R\}}|X_1|dP$ which converges to $0$ when $R\to +\infty$, using monotone convergence for example.
  • Let $j$ a fixed integer, and $E_{j,n}:=\{X_n\geq j\}$. Since the series $\sum_{n\geq 1}P(E_{j,n})$ is divergent and the events $\{E_{j,n}\}$ are independent, by Borel-Cantelli lemma, $P(\limsup_n E_{j,n})=1$. Hence for almost all $\omega$, there exist $S_{\omega,j}\subset \Bbb N$ infinite such that for each element $n$ of this set $X_n(\omega)\geq j$. Hence $\limsup_{n\to +\infty}X_n(\omega)\geq j$. Since it's true for almost all $\omega$, we conclude that $\limsup_{n\to +\infty}X_n$ is not integrable.
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