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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...

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

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  • $\begingroup$ 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

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