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I came across the following problem: Let $\xi\in L^{2}(\mathbb{P})$. Can we obtain that $\xi/T^{\epsilon}$ converges almost-sure to zero, for any $\epsilon$ as $T$ tends to infinite? Thus, is $\frac{\xi}{T^{\epsilon}}\overset{a.s.}{\rightarrow }0$ right? Thanks for your consideration.

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    The assumption $\xi\in L^2$ is too much. It is enough that $\xi$ is finite almost surely. Then the limit is zero on the set where $\xi$ is finite. – 2017-02-19
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    @zhoraster Thanks. $\xi$ is a random variable. – 2017-02-19
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    Yes, I thought so. If it is finite by according to your definition, then $\xi/T^\epsilon\to 0$ *for any* $\omega$, not only almost surely. – 2017-02-19

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$X=\lim_{T\to\infty}\xi/T^{\epsilon}$ exists and by the dominated convergence theorem $$0=\lim_{T\to\infty}E(\lvert\xi/T^{\epsilon}\rvert)=E(\lvert X\rvert)$$ so $X=0$ almost surely.

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    Thanks for your comments. I can obtain convergence $L^{1}$ from your equation $0=\lim_{T\to\infty}E(\lvert\xi/T^{\epsilon}\rvert)=E(\lvert X\rvert)$. However, this convergence does not imply almost surely. – 2017-02-19
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    @weilinhy Well $X$ is defined to be the limit (possibly taking values in $\{+\infty,-\infty\}$) and $X$ has been shown to be $0$ almost surely. – 2017-02-19
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    @ user375366 Thanks for your help. Is $\frac{\xi}{T^{\epsilon}}\overset{a.s.}{\rightarrow }0$ is right? – 2017-02-20
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    @weilinhy Yes it converges to $0$ almost surely. – 2017-02-21
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    @ user375366 Thanks for your kindly help. – 2017-02-21