I'm trying for some time now to prove or disprove the following conjecture to no avail:



> 
> Let $S$ be a set and let $(\Sigma \_n)$
>  be a sequence of countably generated
>  $\sigma$-algebras on $S$ satisfying
>  the following two conditions:
> 
> 
> 1. $\Sigma\_n\subseteq\Sigma\_{n+1}$ for all $n$.
> 2. If $A\in\Sigma\_{n+1}$ is a union of $\Sigma\_n$-atoms, then $A\in\Sigma\_n$
>  for all $n$.
> 
> 
> Then for all $n$: If
>  $A\in\sigma\big(\bigcup\_n\Sigma\_n\big)$
>  is a union of $\Sigma\_n$-atoms, then
>  $A\in\Sigma\_n$.
> 
> 
> 


An *atom* is a minimal measurable set. In a countably generated $\sigma$-algebra, the atoms form a partition of the underlying space into points that can not be distinguished by measurable sets.


I have actually only little intuition for the problem. If $S$ is analytic and all the $\Sigma\_n$ are sub-$\sigma$-algebras of the Borel-$\sigma$-algebra, both condition 2. and the conjecture is automatically satisfied, due to a result of Blackwell, so counterexamples must be somewhat unnatural. 



Atoms of a sequence of Sigma-algebras
Counterexample. First, let ${\cal B}$ be the Borel $\sigma$-algebra on ${\bf R}$ and let ${\cal B}'$ be the $\sigma$-algebra generated by ${\cal B}$ together with one non-Borel set $E$. Note that $E$ is a union of atoms of ${\cal B}$.


Now for each $n$ let $\Sigma\_n$ be the $\sigma$-algebra of subsets of ${\bf R} \times {\bf N}$ generated by sets of the form $A \times [n,\infty)$ for $A \in {\cal B}$ and sets of the form $B\times \{k\}$ for $B \in {\cal B}'$ and $k < n$. Since ${\cal B}$ is countably generated, so is each $\Sigma\_n$.


The atoms of $\Sigma\_n$ are the singletons $\{(x,k)\}$ for $x \in {\bf R}$ and $k < n$ and the sets $\{x\}\times[n,\infty)$ for $x \in {\bf R}$. You don't get anything new in $\Sigma\_{n+1}$ that's a union of atoms of $\Sigma\_n$. However, $E\times{\bf N}$ appears in the $\sigma$-algebra generated by $\bigcup\_n \Sigma\_n$, and this is a union of atoms of $\Sigma\_0$.


