Probability of a supremum of a sequence of independent random variables I'm trying to prove the next:
Suppose $\{X_n\}$ is an independent sequence of random variables. Show that $$P(\sup X_n<\infty)=1$$ if and only if $$\sum_n P(X_n>M)<\infty$$ for some $M.$
I'm having problems with this. Here is my attempt:
We have that if :$$\sum_n P(X_n>M)<\infty$$ then $P(X_n>M i.o)=0$ because of Borel-Cantelli lemma. Then $$P(\limsup \{X_n\leq M\})=1=P(\liminf \{X_n\leq M\}),$$ but here I don't find how to relate this with the event $\{\sup X_n<\infty\}$ because not all $X_n$ satisfy $M$ is upper bound for them even though many infinity of them do it. 
For other direction I was trying a proof based on contradiction and using a Borel $0-1$ law without much success.
Any kind of help is thanked in advanced.
 A: If $\omega\in\liminf\{X_n\leq M\}$ then there is $n_0\in\mathbb{N}$ such that $X_n(\omega)\leq M$ for all $n\geq n_0$. Hence the sequence $X_n(\omega)$ is bounded from above, which means its supremum is finite. So $\liminf\{X_n\leq M\}\subseteq\{\sup X_n<\infty\}$, and hence $P(\liminf\{X_n\leq M\})=1$ implies that $P(\sup X_n<\infty)=1$. 
Now to the other direction. Suppose for all $M$ we have $\sum_{n=1}^\infty P(X_n>M)=\infty$. Since the random variables are independent the second Borel-Cantelli lemma tells us that $P(X_n>M i.o)=1$. Note that by assumption this is true for all $M>0$, hence it is true for all $M\in\mathbb{N}$. A countable intersection of sets with probability $1$ is a set with probability $1$, so $P(\cap_{M=1}^\infty\{X_n>M i.o\})=1$. And of course if $\omega\in\cap_{M=1}^\infty\{X_n>M i.o\}$ then the sequence $X_n(\omega)$ has an infinite supremum, since for each $M\in\mathbb{N}$ there is an element (and even infinitely many elements) that are greater than $M$. So $\cap_{M=1}^\infty\{X_n>M i.o\}\subseteq\{\sup X_n=\infty\}$, and hence $P(\sup X_n=\infty)=1$. This implies $P(\sup X_n<\infty)=0\ne 1$. 
A: If $\sum_{n} P(X_n>M)<\infty$ for some $M$, then as you mentioned $P(X_n>M\, \text{i.o})=0$. So $X_n\leq M$ eventually with probability one (formally for almost every $\omega$, there exists $N=N(\omega)$ such that $n\geq N$ implies that $X_n(\omega)\leq M$. So for almost every $\omega$
$$
X_n(\omega)\leq \sup_{n<N}X_n(w)+\sup_{n\geq N} X_{n}(\omega)<\infty
$$
whence $\sup X_n<\infty$ with probability one.
For the converse suppose that $\sum_{n} P(X_n>M)=\infty$ for each $M$. Since the $X_n$ are independent, for each $k\geq 1$, $X_n>k$ i.o with probability one. Consider the intersection $A=\cap _{k} (X_n>k\,\text{i.o})$ which is also a sure set. In particular with probability one there is a subsequence of $X_n$ which goes to $\infty$. So $\sup_{n} X_n=\infty$ w.p1.
