Given a sigma algebra $\mathcal{F}$ on a set $X$. Let be further given some sigma ideal $I$ https://en.wikipedia.org/wiki/Sigma-ideal. Then we can consider $\mathcal{F}/I=\{[A]\mid A\in \mathcal{F}\}$ where $[A]=\{Z:A\triangle Z\in I\}$.

Is $\mathcal{F}/I$ a sigma algebra on [X]?

I do not see how to show that $[A]\in \mathcal{F}/I$ implies $[A]^c \in \mathcal{F}/I$ since there is no information about $[A]^c=\{Z:A\triangle Z\in I\}^c$

  • $\begingroup$ It's for sure an 'abstract' $\sigma$-algebra, i.e. Boolean algebra with countable supremums. To represent abstract Boolean algebras as set algebras, one uses the set of (abstract) ultrafilters as base set. Maybe the same (or analogous) representation works for abstract $\sigma$-algebras.. $\endgroup$ – Berci Sep 11 at 12:42
  • $\begingroup$ @Berci thanks for that input. Is there no elementary way to show that this is an sigma algebra, i.e. without the notion of boolean algebra, ultrafilers etc.? $\endgroup$ – Roger Sweet Sep 11 at 12:50
  • $\begingroup$ I can't think of any simpler representation of the abstract $\sigma$-algebra $\mathcal F/I$. I guess ultrafilters (probably which are closed under countable intersections) will be needed anyway to find a 'base set'. $\endgroup$ – Berci Sep 11 at 12:57
  • $\begingroup$ You gave the definition of $[A]$ wrong. I fixed that. Also, $[A]%c$ should be the "complement" in the measure algebra, not the complement as a set. That is, $[A]^c$ should be defined to be $[A^c]$, not what you wrote. $\endgroup$ – David C. Ullrich Sep 11 at 14:26
  • $\begingroup$ @David C. Ullrich thanks for pointing that out but I thought both definitions of [A] are equivalent. Nevertheless it is still not possible to show the properties of sigma algebra elementary or? $\endgroup$ – Roger Sweet Sep 11 at 14:37

A $\sigma$-algebra is an algebraic structure (specifically, a Boolean algebra with a countably infinite operation of countable supremum (satisfying the axiom that it's the supremum of its arguments)), together with a representation of its elements as sets, making the operations the real set operations.

When forming the quotient, the 'algebra layer' (the operations and the axioms they satisfy) are kept, but in general, we lose the representation.

However, in some specific cases we can simply determine the new representation, like for example if $I=\{B\in\mathcal F:B\subseteq A\}$ for some $A\in\mathcal F$.

For the general case, we need to generalize the construction of the Boolean algebra case, and for that we need $\sigma$-ultrafilters, i.e. ultrafilters that are closed under countable intersections.

Let $Y:=\{U\ \,\sigma$-ultrafilter on $X: U\cap I=\emptyset\}$, and represent $[A]\in\mathcal F/I$ as $\{U\in Y: A\in U\}$.
Check that it's a well-defined injective $\sigma$-algebra morphism $\mathcal F/I\to\mathcal P(Y)$.

  • 1
    $\begingroup$ Unfortunately, a Boolean $\sigma$-algebra need not have any $\sigma$-ultrafilters. For example, the standard measure algebra (Borel subsets of $[0,1]$ modulo the ideal of measure-zero sets) has no $\sigma$-ultrafilters. So the morphism at the end of your answer won't be injective. $\endgroup$ – Andreas Blass Sep 11 at 19:37
  • $\begingroup$ Hmm.. I was not sure in that part. So, in general it can't be represented? $\endgroup$ – Berci Sep 11 at 20:10

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