$\sup_nX_n<\infty$ almost surely iff $\sum_nP(X_n>A)<\infty$ 
Let $(X_n)_{n\in\mathbb N}$ be a sequence of independent random variables. Show that $\sup_nX_n<\infty$ almost surely iff there exists $A>0$ such that, $\sum_nP(X_n>A)<\infty$

By Borel-Cantelli we have
$$\sum_nP(X_n>A)=\infty\Longleftrightarrow P(\limsup_nX_n>A)=1$$
Hence $\forall A>0$;
$\sum_nP(X_n>A)=\infty\Rightarrow 1=P(\limsup_nX_n>A)<P(\sup_nX_n>A)$ 
Can you give a hint for the other direction, Thanks.
 A: I'll assume the random variables are real-valued.
'direction 1'
Let $A_n^c := \{X_n > A\}$. By BCL1, we have
$$P(\limsup A_n^C) = 0$$
$$\to P(\liminf A_n) = 1$$
$$\to P(\lim A_n) = 1$$
$$\to \lim P(A_n) = 1$$
$$\to P(\bigcap_{n=1}^{\infty} A_n) = 1$$
$$\to \prod_{n=1}^{\infty} P(A_n) = 1 \ \text{Why?}$$
$$\to \forall n \in \mathbb N, P(A_n) = 1$$
$$\to \forall n \in \mathbb N, P\{X_n \le A\} = 1$$
$$\to P( \sup_{n \ge 1} (X_n) < \infty) = 1$$
'direction 2'
Show the contrapositive:
By BCL2, we have
$$P(\limsup A_n^C) = 1 \ (\forall A > 0)$$
$$\to P(\liminf A_n) = 0$$
$$\to \lim_{m \to \infty} P(\bigcap_{n=m}^{\infty} A_n) = 0$$
$$\to \lim_{m \to \infty} \prod_{n=m}^{\infty} P(A_n) = 0$$
$$\to \sup_{m \ge 1} \prod_{n=m}^{\infty} P(A_n) = 0$$
$\to \forall m \ge 1, \exists n \ge m$ s.t. $P(A_n) = 0$
$$\to P(X_n \le A) = 0 \tag{*}$$
Now suppose on the contrary that $P[\sup X_n < \infty] = 1$.
$$\to \sup[X_1, X_2, ...] < \infty \ \text{a.s.}$$
$\to \exists N \ge 0$ s.t.
$$P(\bigcap_{n=1}^{\infty} X_n \le N ) = 1$$
$$\to \prod_{n=1}^{\infty} P(X_n \le N ) = 1$$
$$\to P(X_n \le N ) = 1 \forall n \ge 1$$
Choose $A = N$ in $(*)$. ↯
$$\therefore, \forall A > 0, \sum_{n=1}^{\infty} P(X_n > A) =\infty \to P[\sup X_n < \infty] < 1$$
QED
