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.

  • $\begingroup$ 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 $n$? $\endgroup$ May 24 '12 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 $A\in\Sigma_0$. $\endgroup$ May 24 '12 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 $m<n$). $\endgroup$ May 24 '12 at 15:59

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 $\Sigma_0$.


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.