Let $X_1,X_2,\ldots$ be iid random variables, then we are trying to prove that: $$\limsup_{n\to\infty} \frac{1}{n} \max_{k\leq n} \lvert X_k \rvert = \begin{cases}0 &\textrm{if } E\lvert X_1\rvert < \infty \\ \infty &\textrm{if } E\lvert X_1\rvert = \infty \\ \end{cases}$$
In the proof given in class, we proved that for all $a>0$: $$P(\lvert X_n \rvert > an \ i.o.) = \begin{cases}0 &\textrm{if } E\lvert X_1\rvert < \infty \\ 1 &\textrm{if } E\lvert X_1\rvert = \infty \\ \end{cases}$$ Using the above result, if $E\lvert X_1\rvert = \infty$, then it is obvious that $\limsup \frac{1}{n} \max_{k\leq n} \lvert X_k \rvert > a$ for all $a> 0$ which gives us the conclusion. But, if $E\lvert X_1 \rvert <\infty$, i don't understand how the previous result guarantees that $\limsup \frac{1}{n} \max_{k\leq n} \lvert X_k \rvert < a$, since we're always taking the maximum over all elements.
I think I have it figured out (can someone please verify this? or provide a simpler way if there is?):
If $E\lvert X_n\rvert < \infty$, then $P(\lvert X_n \rvert = \infty) = 0$. Consider the set $$\left(\bigcap_{n\geq 1} \{\lvert X_n \rvert < \infty\}\right) \cap \{ \lvert X_n \rvert > an \ i.o\}^c$$ This set has probability $1$ since it is the countable intersection of sets with probability $1$. Now, let $w$ be an element of the set above, then there exists $n_a$ such that $\forall n\geq n_a: \lvert X_n(w) \rvert \leq an$, and since $X_1(w),\ldots, X_{n_a-1}(w) < \infty$, there exists $n_0$ such that $$\forall n\geq n_0: \frac{1}{n}\max \{ X_1(w),\ldots, X_{n_0-1}(w) \} \leq a.$$ This gives us the required result by considering $n\geq \max\{n_0,n_a\}$.