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Let $X_1$, $X_2$, ... be independent random variables. Show that sup$X_n$ < ∞ almost surely if and only if $\sum_{n=1}^∞$Pr($X_n$ > A) < ∞ for some A > 0.

Here is my idea:

(Forward direction): Since sup$X_n$ is bounded almost surely, $X_n$ is bounded almost surely for each n. So there exists a positive A such that $X_n$ < A almost surely for each n, i.e. Pr($X_n$ > A)=0 for each n (which is stronger than Pr($X_n$ > A i.o.) = 0). Also, since $X_1$, $X_2$, ... are independent, by Borel-Cantelli lemmas, $\sum_{n=1}^∞$Pr($X_n$ > A) < ∞.

However, I failed to show the reverse direction. Given $\sum_{n=1}^∞$Pr($X_n$ > A) < ∞ for some A > 0, by Borel-Cantelli lemmas, I can only get Pr($X_n$ > A i.o.) = 0. But it doesn't necessarily tell us Pr(sup$X_n$ < A) = 1 since it's still possible for finitely many $X_n$ to exceed A, does it? Or doesn't it only mean Pr(limsup$X_n$ < A) = 1?

Could anyone please tell me where I got wrong?

Thanks in advance!

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  • $\begingroup$ A macroscopical non-sequitur is "$Y<\infty\text{ a.c.}\implies Y\text{ bounded a.c.}$". For instance, the function $Y(x)=e^x$ is finite a.c. (id est, $<\infty$) but there is no $A$ such that $Y\le A$ a.c. (by which I mean: considering on $\Bbb R$ a probability which has the same null sets as the Lebesgue measure). $\endgroup$ – user228113 Apr 17 '17 at 2:44
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($X_n : \Omega \to \bar{\mathbb{R}}$).

For the other direction, note that $$ \sum_{n\geq 1}\mathbb{P}(X_n > A)<\infty \implies \mathbb{P}\left(\{w : X_n(w) > A, i.o.\}\right) = 0. $$ Now, read this as, for almost every $w$ (i.e. except on a set of measure $0$); $X_n(w) > A$ happens finitely many times. For instance, suppose that for a particular $w$ the indices that makes this happen are $i_1,\dots,i_{N_w}$. For this particular $w$, $$ \sup_n X_n(w) \leq \max\{A,X_{i_1}(w),\dots,X_{i_{N_w}}(w)\} <\infty, $$ hence $sup_{n}X_n < \infty$ for almost every $w$.

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  • $\begingroup$ Got it! Thank you so much! I think the reason why I got confused is that I presume that $X_n$ could be ∞. But since $X_n$ is real-valued, it can't be ∞. $\endgroup$ – user435845 Apr 19 '17 at 2:11

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