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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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    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)|. We have for all $k\geq 0$: $kP(A_k)\leq \int_{k\leq X hence $\sum_{k=0}^{+\infty}kP(A_k)\leq E|X|\leq\sum_{k=0}^{+\infty}(k+1)P(A_k).$ Since $E|X|$ is finite, the series $\sum_{k=0}^{+\infty}kP(A_k)$ is convergent and since $(k+1)P(A_k)\sim kP(A_k)$, the series $\sum_{k=0}^{+\infty}(k+1)P(A_k)$ is also convergent. Now, we have $\int_{|X|\geq n}|X|dP=\sum_{k=n}^{+\infty}\int_{A_k}|X|dP\leq \sum_{k=n}^{+\infty}(k+1)P(A_k),$ which is the remainder of a convergent series.