Sequence of iid RVs with infinite expectation Let $X_n$ be a sequence of iid RVs with $E|X_n|=+\infty$. Show that for every positive number $A$, $P[|X_n|>nA, \; \mbox{i.o.}]=1$ and $P[|S_n|<nA,\;\mbox{ i.o.}]=1$.
Here, $S_n=\sum_{k=1}^nX_k, k=1,2,\ldots$ and i.o. means "infinitely often," i.e. $P(\limsup_{n\to\infty}A_n)=P(A_n,\; \mbox{i.o.})$.
The only place I've ever seen the i.o. terminology used is the Borel-Cantelli Lemma, so I tried to apply this portion of the lemma:
If {$A_n$} is an independent sequence of events such that $\sum_{n=1}^\infty PA_n=\infty$, then $PA=P(A_n,\;\mbox{i.o.})=1$.
Is there a way to apply that here?
 A: *

*\begin{align*} \infty &= E [ |X_1| ] \\& = \int_0^\infty P(|X_1|>t) dt \\
& \le A+\sum_{k=1}^\infty A P(|X_1|>k A)\\
& =A + A\sum_{k=1}^\infty P(|X_k|>k A). \end{align*}
Now use Borel-Cantelli (II) to obtain $P(|X_k|>k A\mbox{  i.o.})=1$.

*The second statement is wrong, at least as expressed here (I apologize if I misunderstood).  Take $X_1$ be any nonnegative random variable taking values in $[1,\infty)$ with infinite expectation (e.g. density $c x^{-2}$ for $x\ge 1$). Then $S_n \ge \frac{1}{2}n$ for all $n$, so statement fails for $A=\frac 12$ (or any $A\in (0,1)$ for that matter). 
A: Yes, there is. We can write the inequality $$ \frac{|X_n|}{n} \le 1+\sum_{i=1}^\infty I\left(\frac{|X_n|}{n}\ge i\right)$$ where $I$ is the indicator function. This makes sense if you stare at it for awhile... the RHS just adds one until it hits $|X_n|/n.$
Then take the expected value of both sides to get $$\frac{1}{n}E(|X_n|) \le 1 +\sum_{i=1}^\infty P(|X_n|\ge ni)$$ which implies $$ \sum_{i=1}^\infty P(|X_n|\ge ni) = \infty$$ Since the $X$'s are IID, this means we can replace $X_n$ with $X_i$ and write $$ \sum_{i=1}^\infty P(|X_i|\ge ni) = \infty.$$
So you can apply the Borel-Cantelli lemma and conclude that $P(|X_i| > ni,\;\mbox{i.o})=1$ is true for any positive integer $n$. Since $P(|X_i|>ni)$ is decreasing in $n$, it follow that it holds true for any real $n>0$.
EDIT
As FnaCool showed, the statement that $P(|S_n|<nA,\;\mbox{i.o.})=1$ for all $A>0$ is wrong. However, it is true that $P(|S_n|>nA,\;\mbox{i.o.})=1.$ This follows from the first statement $P(|X_n|>nA,\;\mbox{i.o.})=1$ $\forall A>0.$
To see how, note the first statement is equivalent to $$\limsup_n \frac{|X_n|}{n} = \infty$$ almost surely, and the second is $$\limsup_n \frac{|S_n|}{n} = \infty$$ almost surely.
To prove the second from the first, observe $$ \frac{|X_n|}{n} = \frac{|S_n-S_{n-1}|}{n} \le \frac{|S_n|}{n}+\frac{|S_{n-1}|}{n}$$ so that $$\limsup_n \frac{|S_n|}{n} \ge \frac{1}{2}\limsup_n \frac{|X_n|}{n} $$
