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Suppose $\left\{X_n\right\}$ are independent. If there exists A s.t. $\sum P(X_n>A) < \infty$, then $\sup_n X_n < \infty$ a.s.

My attemps is as follows:
By Borel-Cantelli lemma $P(\limsup X_n>A)=\lim_{n \rightarrow \infty}P(\cup_{m\ge n}X_m > A)=0$.

I don't know how to proceed.

It seems to me $\limsup$ is regarding the limiting tail part. If I choose $X_1=\infty$, and let the tail part still satisfy the condition. Then obviously $\sup_n X_n = \infty$ a.s

p.s.: The question is actually an "if and only if". The other direction is proved here: $X_n$ are independent rv's. $\sup X_n <\infty$ implies there exist $A$ such that $\sum_{n=1}^{\infty} P(X_n>A)<\infty$

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    It seems that you are having exactly the same problem as [this one](http://math.stackexchange.com/questions/2079640/is-it-possible-that-liminf-n-to-infty-nu-n0-where-u-n-sim-mathcal-u0-1/2080544#2080544). I do not mean it's a duplicate, just add this link for further reading. – 2017-01-02
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    I suspect you are supposed to assume the $X_n$ are real valued (not extended-real valued), so your proposed counterexample of $X_1 = \infty$ doesn't apply. – 2017-01-02
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    @NateEldredge Your are probably right. Two more questions regarding this comment: 1. Is there any case that $X_n$ are extended-real valued? Any examples?; 2. If $X_n$ are extended-real valued, does my counterexample apply? – 2017-01-03
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    Yes, of course your counterexample shows this cannot hold if the $X_n$ are allowed to be extended-real valued. You could still show $\limsup_{n \to \infty} X_n < \infty$ a.s., but I can't think of another similar statement that would still hold in this case. – 2017-01-03

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Let $E = \{X_n > A \text{ i.o.}\} = \bigcap_{m=1}^\infty \bigcup_{n=m}^\infty \{X_n > A\}$. Borel–Cantelli says $P(E)=0$. So for every $\omega \in E^c$, we have that there are only finitely many $n$ such that $X_n(\omega) > A$.

Now as a general fact, if you have an infinite set of numbers, and only finitely many of them exceed $A$, then the supremum of your set is finite. (This is basically the fact that a finite set is bounded). So this means that for every $\omega \in E^c$, we have $\sup_n X_n(\omega) < \infty$. Since $P(E)=0$, we have shown $\sup_n X_n < \infty$ almost surely.