how to prove almost surely If I have $X_i$ being iid, and $E(X_i)=\infty$, how do I show that $\limsup \frac{X_n}{n}=\infty$ almost surely?
I.e. how do I show $P(\limsup_{n\to\infty} (\frac{X_n}{n})= \infty)= 1$?
 A: I assume $X_{i}$ are all non-negative. (There is a way to treat general $X_{i}$, but then we need to delve into how we define expectations and such).
This can be shown using the second Borel-Catelli lemma. The lemma states that given $E_{1},E_{2},\ldots$ independent events such that $\sum_{n=1}^{\infty} P(E_{n}) = \infty$, then $P(\limsup_{n} E_{n}) = 1$.
How to apply the lemma
For a given $\alpha$, we denote $E_{n}^{(\alpha)} = \{X_{n} \ge \alpha n\}$. Thus, 
$$
\limsup_{n\to \infty} E_{n}^{(\alpha)} = \bigcap_{m=1}^{\infty} \bigcup_{n = m}^{\infty} E_{n}^{(\alpha)} = \left\{\forall m, \exists n>m, \frac{X_{n}}{n} \ge \alpha \right\} = \left\{ \limsup_{n\to \infty} \frac{X_{n}}{n} \ge \alpha \right\}
$$
Using the properties of propability measures, we have
$$
P\left(\limsup_{n\to \infty} \frac{X_{n}}{n} = \infty \right) = P\left( \bigcap_{\alpha=1}^{\infty} \left\{\limsup_{n\to \infty} \frac{X_{n}}{n} \ge \alpha \right\}\right) = \lim_{\alpha \to \infty} P\left( \limsup_{n\to \infty} \frac{X_{n}}{n} \ge \alpha \right)
$$
So, if we show $P(\limsup_{n\to \infty} E_{n}^{(\alpha)}) = 1$ for all $\alpha$, then the statement follows. 
How to prove the conditions of the lemma
The events $E_{n}$ are independent since $X_{i}$ are independent. Thus, we can prove the above using the second Borel-Cantelli lemma. We're left to show  $\sum_{n=1}^{\infty} P(E_{n}^{(\alpha)}) = \infty$. This can be derived using Darboux integration. Since $P(X \ge t)$ is monotincally decreasing in $t$ and $X_{i}$ are i.i.d,
$$
\mathbb{E}\left[X\right] = \int_{0}^{\infty} P(X \ge t) dt = \alpha\int_{0}^{\infty} P(X \ge \alpha s) ds \\ \le \alpha \sum_{s=0}^{\infty} P(X \ge \alpha s) = \alpha + \alpha \sum_{n=1}^{\infty} P(X_{n} \ge \alpha n)
$$
Now, we must have $\sum_{n=1}^{\infty} P(X_{n} \ge \alpha n) = \sum_{n=1}^{\infty} P(E_{n}^{(\alpha)}) = \infty$, otherwise $E[X]$ would be finite.
