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.
Best Answer
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.