Let ($A_n : n \in \mathbb{N} $) be a sequence of events in some probability space $( \Omega, \mathcal{F}, \mathbb{P} )$. Set

$A = \{ \omega \in \Omega : \omega \in A_n \text{ infinitely often} \} $ , $B = \{ \omega \in \Omega : \omega \in A_n \text{ for all sufficiently large } n \} $

Show that $ B = \cup_{n=1}^{\infty} \cap_{k=n}^{\infty} A_k $

I tried getting my head round what this question means, or is even asking, but this is too wacky...

  • $\begingroup$ Just out of curiosity, since $A$ doesn't feature in the problem, are you also supposed to show that you get $A$ if you reverse the union and intersection? $\endgroup$ – Arthur Feb 10 '17 at 9:25
  • $\begingroup$ The next wants you to show $ \mathbb{P}(A) \leq \sum_{k=n}^{\infty} \mathbb{P}(A_k) $, and if $ \sum_{n=1}^{\infty} $ converges then that $ \mathbb{P}(A) = 0$. Hmmm... $\endgroup$ – Christopher Turnbull Feb 10 '17 at 14:51

Whatever the sets $\{U_n\}$ are, the following is true (by definition)

$$ \bigcup_{n=1}^{\infty} U_n =\{\omega\in\Omega: \text{ there is an } n \text{ for which } \omega \in U_n\}.$$

At the same time, if

$$U_n=\bigcap_{k=n}^{\infty}A_k=\{\omega\in\Omega:\omega \in U_k \text{ for all } k\geq n \text{ } \ \}.$$

then $$\bigcup_{n=1}^{\infty} U_n=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k=$$ $$=\{\omega\in\Omega: \text{ there is an } n \text{ for which } \omega \in A_k\text{ for all } k\geq n\}=$$ $$ =\{\omega\in\Omega: \text{ for all sufficiently large } n, \ \omega \in A_n \}=B.$$

  • $\begingroup$ Wow very clear, thanks! I had the same idea but couldn't formalise it like you have. I guess practice makes perfect... $\endgroup$ – Christopher Turnbull Feb 10 '17 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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