Limit Superior of Random Variables Suppose $X_n$ are iid random variables with $\mathbb{P}(X_n\le x)=1-e^{-x}$. By using the Borel Cantelli Lemmas it's fairly easy to show that $\mathbb{P}(\lim\sup X_n/\log n=1)=1$. My lecture notes go on to claim that hence $\lim\sup X_n/\log n=1$ almost surely and hence $\sup X_n=\infty$ almost surely. I don't understand this at all. As far as I know $$\{\lim\sup X_n/\log n=1\}=\cap_n\cup_{m\geq n}\{\omega \in \Omega:X_m(\omega)=\log n\}$$
I have no definition for what $\lim\sup X_n$ is, however. How does one define the $\lim\sup$ of a sequence of  measurable functions? And moreover surely it is nontrivial to show that this  precisely 'factors out' of the event in the required way? I'm clearly missing some critical background information here, so a clear explanation would be greatly appreciated! Many thanks.
 A: The $\: \limsup \:$ of a sequence of measurable functions is defined pointwise. $\;\;$ (I could show how

to get from "pointwise limits of sequences of measurable functions are always measurable" to

"the limsup of a sequence of measurable functions is always measurable" if you want me to.)

$\displaystyle\lim_{n\to \infty} \: \log n \;\; = \;\; \scriptsize+\normalsize\infty$

For all members $\omega$ of the domain of your random variables,
$\displaystyle\limsup_{n\to \infty} \frac{X_n(\omega)}{\log n} \: = \: 1 \;\; \implies \;\; \displaystyle\limsup_{n\to \infty} \frac{X_n(\omega)}{\log n} \: \not\leq \: 0$
$\implies \;\; \displaystyle\limsup_{n\to \infty} X_n(\omega) \: = \: \scriptsize+\normalsize\infty \;\; \implies \;\; \sup_n X_n(\omega) \: = \: \scriptsize+\normalsize\infty$

$\left\{\omega \;\; : \;\; \displaystyle\limsup_{n\to \infty} \frac{X_n(\omega)}{\log n} \: = \: 1 \right\} \;\;\;\; \subseteq \;\;\;\; \left\{\omega \;\; : \;\; \displaystyle\sup_n X_n(\omega) \: = \: \scriptsize+\normalsize\infty \right\}$

$1 \;\; = \;\; \mathbb{P}\left(\displaystyle\limsup_{n\to \infty} \frac{X_n(\omega)}{\log n} \: = \: 1 \right) \;\; \leq \;\; \mathbb{P}\left(\displaystyle\sup_n X_n(\omega) \: = \: \scriptsize+\normalsize\infty\right) \;\; \leq \;\; 1$

$\mathbb{P}\left(\displaystyle\limsup_{n\to \infty} X_n(\omega) \: = \: \scriptsize+\normalsize\infty\right) \;\; = \;\; 1$

Therefore $\;\;\;\; \displaystyle\sup_n \: X_n(\omega) \;\; = \;\; \scriptsize+\normalsize\infty \;\;\;\;$ almost surely.

A: I should say that 
$$\{\lim\sup X_n/\log n=1\} \neq \cap_n\cup_{m\geq n}\{\omega \in \Omega:X_m(\omega)=\log n\}$$
NOTE: The left side is $\limsup$ first and then equal.
while I guess what you want to say is:
$$\big\{\lim\sup \{X_n/\log n=1\}\big\} = \cap_n\cup_{m\geq n}\{\omega \in \Omega:X_m(\omega)=\log n\}$$
which is of course correct by definition.
Also you should know that 
$$\mathbb{P}\big\{\lim\sup \{X_n/\log n=1\}\big\} = 0$$
which is not the question ask for.
To solve this problem you can consult here($\lim\inf$ and $\lim\sup$ of events and random variables). See @BCLC's four theorems.
then the whole solution is given here(Lim sup of Random Variables and Events). See my answer and note that the NOTE at the bottom are of the most importance.
