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Let X be a random variable defined on the probability space $(\Omega,\mathbf{F},P )$. If $E|X|<+\infty$, How do I prove that $$\lim_{n\to \infty}\int_{\left(|X|>n\right)} X \ dP=0 \,?$$

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    Please make the title more informative (and less subjective).2011-10-28
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    @Chris, this is **NOT** what is asked...2011-10-28
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    Oops, my mistake. Rolling back.2011-10-28

2 Answers 2

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Let $X_n=|X|\cdot[|X|\geqslant n]$. Since $\lim\limits_{n\to\infty} X_n=0$ almost surely, you ask why $\lim\limits_{n\to\infty}\mathrm E(X_n)=\mathrm E\left(\lim\limits_{n\to\infty}X_n\right)$.

Hint: Use Lebesgue's dominated convergence theorem with the domination condition $|X_n|\leqslant|X|$.

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Put $A_k:=\left\{\omega\in\Omega,k\leq |X(\omega)|