Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
You might want to try this. –  Did May 24 '12 at 22:08
Thanks - but that nowhere mentions what the limsup of a sequence of measurable functions would be. It does give me a definition for the limsup of a function, but I'm pretty sure that something different in intended in the context of this question! –  Edward Hughes May 24 '12 at 22:12
Yes it does. $ $ –  Did May 24 '12 at 22:46
Where exactly? $\:$ –  Ricky Demer May 24 '12 at 23:37

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
I'm a little confused where all your $\infty$s came from, and how you used my result that $\mathbb{P}(\lim\sup X_n/\log n=1)=1$. Could you possibly clarify this? Also how does your pointwise limsup relate to my definition above involving sets? I think this is the critical point I don't understand! Thanks for clearing up that the limsup is defined pointwise. –  Edward Hughes May 24 '12 at 23:14
Three of those infinities were from me having misread your question (and have been fixed). $\:$ My pointwise limsup has no relation to your definition involving sets, because your definition is for a limsup of events while your lectures notes use the event of a limsup being 1. $\;\;$ –  Ricky Demer May 24 '12 at 23:36
Ah okay - got it now! –  Edward Hughes May 25 '12 at 11:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.