$\limsup\limits_{n\to\infty}X_n / n = \infty$, if $X_n\geq 0$ is i.i.d. and has infinite mean. Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of i.i.d. real random variables with $X_1\geq 0$ a.s. and $\mathbb{E}(X_1)=\infty$ then it holds
$$
\limsup\limits_{n\to\infty}\frac{X_n}{n}=\infty \text{ a.s.}
$$
I tried to prove this statement, but I'm not quite sure, whether my proof is correct, since the usage of $\limsup$ with respect to random variables and with respect to sets confuses me a bit.
My attempt:
We know (e.g. by Kolmogorov's zero-one law) that $\limsup\limits_{n\to\infty}\frac{X_n}{n}$ is constant a.s., i.e. $\exists c \in\mathbb{R}\cup\{-\infty, \infty\}$ such that $P\bigl(\limsup\limits_{n\to\infty}\frac{X_n}{n}\leq c\bigr)=1$.
If we assume $P\bigl(\limsup\limits_{n\to\infty}\frac{X_n}{n}=\infty\bigr)<1$, then we see that $c<\infty$ and since $X_n\geq 0$ a.s., we can conclude, that $c\geq0$, yielding
$$P\Bigl(\limsup\limits_{n\to\infty}\frac{X_n}{n}> c\Bigr)=0,$$
for a $c\geq0$.
This however means that $P\bigl(\limsup\limits_{n\to\infty}\{\frac{X_n}{n}>c\}\bigr)=P\bigl(\frac{X_n}{n}>c \text{ i.o.}\bigr)=0$. Because if $P\bigl(\frac{X_n}{n}>c \text{ i.o.}\bigr)>0$, then $P\bigl(\frac{X_n}{n}>c \text{ i.o.}\bigr)=1$ (by the Lemma of Borel-Cantelli) and hence we find a subsequence $(n_k)_{k\in\mathbb{N}}\subset\mathbb{N}$ with $X_{n_k}>c \text{ a.s. } \forall k\in\mathbb{N}$. But then $\limsup\limits_{n\to\infty}\frac{X_n}{n}> c$ a.s. (To be more specific: Per definition $\limsup\limits_{n\to\infty}\frac{X_n}{n}=\inf\limits_{n\in\mathbb{N}}\sup\limits_{j\geq n}\{\frac{X_j}{j}\}$ and the existence of the subsequence $X_{n_k}$, ensures that $\sup\limits_{j\geq n}\{\frac{X_j}{j}\}>c$ a.s.
Having now established, that $P\bigl(\frac{X_n}{n}>c \text{ i.o.}\bigr)=0$ has to hold, we can use the Lemma of Borel-Cantelli to infer
$$
\sum_{n=1}^{\infty}P\Bigl(\frac{X_n}{n}>c\Bigr)<\infty,
$$
or equivalently
$$
\sum_{n=1}^{\infty}P\Bigl(\frac{X_n}{c}>n\Bigr)<\infty.
$$
This however is equivalent to $\mathbb{E}\bigl(\frac{X_1}{c}\bigr)<\infty$ (since $X_1\geq 0$ a.s.), which is a contradiction to $\mathbb{E}(X_1)=\infty$ and hence the result follows.
My question:
-Is the proof correct? I'm not 100% sure, whether my conclusion above $P\Bigl(\limsup\limits_{n\to\infty}\frac{X_n}{n}> c\Bigr)=0 \Longrightarrow P\bigl(\frac{X_n}{n}>c \text{ i.o.} \bigr)=0$ is valid. It seems valid to me, but dealing with $\limsup$ of functions and of sets confuses me. Is there a relation between both concepts?
Thanks in advance
 A: It looks valid to me. Maybe write with more care that your sequence $(n_k)_{k\in\mathbb N}$ depends on $\omega$, but this has no real impact on your proof. Also, just after the introduction of this sequence, you deduce that $\limsup\limits_{n\to\infty}\frac{X_n}{n}\ge c$, where the inequality is not necessarily strict because it can converge to $c$. But again, it does not invalidate the proof.
To recap, the key equality here is
$$
\left\{\limsup_{n\to+\infty}\frac{X_n}{n}=+\infty\right\}=\bigcap_{c\in\mathbb N^*}\limsup_{n\to+\infty}\left\{\frac{X_n}{n}>c\right\},
$$
which connects the limit superior of a random variable with that of sets and paves the way to Borel-Cantelli's lemma. As you mentioned, for all $c\in\mathbb N^*$, because $X_1$ is nonnegative we have
$$
\mathbb E\left[\frac{X_1}{c}\right]=\sum_{n\in\mathbb N}\mathbb P\left(\frac{X_1}{c}>n\right)=\sum_{n\in\mathbb N}\mathbb P\left(\frac{X_n}{n}>c\right)=+\infty,
$$
so by Borel-Cantelli's lemma, $\mathbb P\left(\limsup_{n\to+\infty}\left\{\frac{X_n}{n}>c\right\}\right)=1$. As a countable intersection of almost sure sets is itself almost sure, we deduce from the first equality above that $\mathbb P\left(\left\{\limsup_{n\to+\infty}\frac{X_n}{n}=+\infty\right\}\right)=1$. This is basically the proof you wrote, but presented in a more direct way.
