Is a Measure algebra a sigma algebra

Question

Given a sigma algebra $\mathcal{F}$ on a set $X$. Let be further given some sigma ideal $I$. 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]?

Answer 1

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 $\mathcal F/I\to\mathcal P(Y)$.