I am proving the following :
Suppose $\{X_n : n\in\mathbb{N}\}$ are i.i.d. random variables. Then $P(\sup_{n\in\mathbb{N}}X_n < \infty) = 1$ if and only if $ \sum_{n\in\mathbb{N}}{P(X_n > M)} < \infty$ for some $M>0.$
I guess by the form of the problem, that Borel-Canteli lemma will be used at some point, but I could not thought of a good way to connect it to the problem.
Any idea or hint will be really helpful. Thanks.
(edit)
Assume $\sum_{n\in\mathbb{N}}{P(X_n > M)} < \infty.$ In fact, as they are i.i.d., $P(X_n > M) = P(X_1>M)$ for all $n.$ Thus $P(X_1>M) = 0$ and $P(\sup_{n\in\mathbb{N}}X_n > M) \le \sum_{n\in\mathbb{N}}{P(X_n > M)} = 0.$
Conversely, I've realized that $\{sup_{n\in\mathbb{N}}{X_n} < \infty\} = \bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}\{X_n \le m\}.$
Suppose $P(\sup_n X_n < \infty) = 1$. Since $\{\sup_n X_n < \infty\} = \bigcup_{M=1}^\infty \{\sup_n X_n \le M\}$, we have $\lim_{M \to \infty} P(\sup_n X_n \le M) = 1$. But if $P(X_i > M) > 0$, $P(\sup_n X_n > M) = 1$. So for some $M$ we must have $P(X_i > M) = 0$.
Conversely, if $P(\sup_n X_n < \infty) < 1$, then for all $M$ we have $P(\sup_n X_n > M) > 0$, and $P(X_i > M) > 0$.