Timeline for Atoms of a sequence of Sigma-algebras

6 events

when toggle format	what		by	license	comment
May 24, 2012 at 15:59	vote	accept	Michael Greinecker
May 24, 2012 at 15:59	comment	added	Michael Greinecker		@GE: Yes, every atom in $\Sigma_n$ is measurable in $\Sigma_m$ for all $m\geq n$. The second condition means that $\Sigma_{n+1}\backslash\Sigma_n$ contains no set that is a union of $\Sigma_n$ atoms (and by induction $\Sigma_m$-atoms for $m<n$).
May 24, 2012 at 15:34	comment	added	Pietro Majer		As I understand, "for all n" refers to the whole sentence. So if we denote $\Pi_n$ the complete set algebra generated by $\Sigma_n$, that is, unions of $\Sigma_n$-atoms, condition 2. should read: for any $n$, $\Sigma_{n+1}\cap\Pi_n=\Sigma_n$. This implies by induction that for any $n \le m$, $\Sigma_m\cap\Pi_n=\Sigma_n$. But, as shown by Nik Weaver's counterexample, $A\in\Pi_0$ and $A=\cup_n A_n$, with $A_n\in\Sigma_n$, already fail to imply $A\in\Sigma_0$.
May 24, 2012 at 14:35	answer	added	Nik Weaver		timeline score: 4
May 24, 2012 at 14:32	comment	added	Gerald Edgar		So, let's try to understand this. In (2), a countable union of $\Sigma_n$-atoms is already in $\Sigma_n$, so the new information here is for uncountable unions. Also, quantification of $n$ in (2) may need clarification. Since any $\Sigma_n$-atom is an element of $\Sigma_{n+1}$, can we conclude it must belong to $\Sigma_n$ for all $n$?
May 24, 2012 at 13:44	history	asked	Michael Greinecker	CC BY-SA 3.0