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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.

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  • $\begingroup$ 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 $n$? $\endgroup$
    – Gerald Edgar
    Commented May 24, 2012 at 14:32
  • $\begingroup$ 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 $A\in\Sigma_0$. $\endgroup$
    – Pietro Majer
    Commented May 24, 2012 at 15:34
  • $\begingroup$ @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 $m<n$). $\endgroup$
    – Michael Greinecker
    Commented May 24, 2012 at 15:59

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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 $\Sigma_0$.

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