Atoms of a sequence of Sigma-algebras - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T17:48:40Zhttp://mathoverflow.net/feeds/question/97837http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/97837/atoms-of-a-sequence-of-sigma-algebrasAtoms of a sequence of Sigma-algebrasMichael Greinecker2012-05-24T13:44:55Z2012-05-24T14:35:17Z
<p>I'm trying for some time now to prove or disprove the following conjecture to no avail:</p>
<blockquote>
<p>Let $S$ be a set and let $(\Sigma _n)$
be a sequence of countably generated
$\sigma$-algebras on $S$ satisfying
the following two conditions:</p>
<ol>
<li>$\Sigma_n\subseteq\Sigma_{n+1}$ for all $n$.</li>
<li>If $A\in\Sigma_{n+1}$ is a union of $\Sigma_n$-atoms, then $A\in\Sigma_n$
for all $n$.</li>
</ol>
<p>Then for all $n$: If
$A\in\sigma\big(\bigcup_n\Sigma_n\big)$
is a union of $\Sigma_n$-atoms, then
$A\in\Sigma_n$.</p>
</blockquote>
<p>An <em>atom</em> 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.</p>
<p>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. </p>
http://mathoverflow.net/questions/97837/atoms-of-a-sequence-of-sigma-algebras/97839#97839Answer by Nik Weaver for Atoms of a sequence of Sigma-algebrasNik Weaver2012-05-24T14:35:17Z2012-05-24T14:35:17Z<p>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 ${\cal B}$.</p>
<p>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 <code>$A \times [n,\infty)$</code> for $A \in {\cal B}$ and sets of the form <code>$B\times \{k\}$</code> for $B \in {\cal B}'$ and $k < n$. Since ${\cal B}$ is countably generated, so is each $\Sigma_n$.</p>
<p>The atoms of $\Sigma_n$ are the singletons <code>$\{(x,k)\}$</code> for $x \in {\bf R}$ and $k < n$ and the sets <code>$\{x\}\times[n,\infty)$</code> 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 $\Sigma_0$.</p>