I am trying to solve the following exercise :
Let $X_1,X_2, ...$ be independent. Show that $sup X_n < \infty$ a.s. if and only if $\sum P(X_n > A) < \infty$ for some A.
I want to show the "only if" part by showing that
$\sum P(X_n > A) = \infty$ for all A $\Rightarrow P(sup X_n = \infty)>0$
It's possible that my logic is already wrong but for me the opposite of "$sup X_n < \infty$ a.s" is $P(sup X_n = \infty)>0$
My attempt until now : By using the Borel Cantelli lemma for independent events I obtain, for all A $P(X_n>A$ infinitely often$) = 1$. I can use A for all the integer : for all $m \in \mathbb{N}, \exists N_m$ s.t $P(N_m)=0$ and $\forall \omega \in \Omega - N_m, X_n(\omega) > m$ infinitely often. I can rewrite this as $ \forall m \in \mathbb{N}, \exists N_m$ s.t $P(N_m)=0$ and $\forall \omega \in \Omega - N_m, \forall n \in \mathbb{N}, \exists N > n$ s.t $X_n(\omega)>m$
By using that the discrete union of negligible event is still negligible I obtain :
$\forall \omega \in \Omega - \cup_{m \in \mathbb{N}} N_m, \forall m, \forall n, \exists N>n$ s.t $X_n(\omega)>m$ i.e
$\forall \omega \in \Omega - \cup_{m \in \mathbb{N}} N_m, supX_n(\omega)=+\infty$ i.e
$P(supX_n=+\infty)=1$ which is stronger that what I really want to show so I was wondering if there is a mistake in my proof.