Let $X$ be a measurable space whose $\sigma$-algebra is generated by a family $\mathcal{G}=\bigcup\_n \mathcal{G}\_n$ of subsets of $X$, where $(\mathcal{G}\_n)$ is a sequence of $\sigma$-algebras on $X$ of increasing fineness. I am not sure if the "$=\bigcup\dots$" is helpful, but I included it just in case.


Furthermore, let $\sim$ be an equivalence relation on $X$ and let $X/\sim$ be the quotient measurable space, i.e., elements of $X/\sim$ are $\sim$-equivalence classes and a set $Y \subseteq X/\sim$ is measurable if and only if $\bigcup\_{y\in Y} y$ is measurable in $X$.


The question: Is the $\sigma$-algebra on $X/\sim$ equal to the $\sigma$-algebra generated by the following family? $$\{ Y \subseteq X/\sim\ :\ \exists G \in \mathcal{G} \quad \bigcup\_{y\in Y} y = G\}$$


I suspect that the answer is no, but it would be nice to have confirmation either way. If the answer is yes, then I could save 2-3 pages.



Quotient sigma-algebra generated by quotient-measurable generating sets
No, it isn't.


Let $X = \omega = \{0,1,2,\dots\}$. Let $\mathcal{G}\_n = \sigma(\{\{0\}, \{1\}, \dots, \{n\}\})$ be the $\sigma$-field in which subsets of $\{0,1,\dots, n\}$, and their complements, are measurable. If $\mathcal{G} = \bigcup\_n \mathcal{G}\_n$, then $\sigma(\mathcal{G}) = 2^\omega$, but all sets in $\mathcal{G}$ are finite or cofinite.


Now take the equivalence relation $\sim$ where $x \sim y$ iff $x-y$ is even, so we have the quotient $Q = \{E, O\}$ where $E = \{0,2,4,\dots\}$ and $O = \{1,3,5,\dots\}$ are the sets of even and odd numbers. Clearly the quotient $\sigma$-algebra on $Q$ is $2^Q$. But since $E \notin \mathcal{G}$ and $O \notin \mathcal{G}$, we have
$$\sigma\left(\{ Y \subseteq X/\sim\ :\ \exists G \in \mathcal{G} \quad \bigcup\_{y\in Y} y = G\}\right) = \{\emptyset, \{E,O\}\} \ne 2^Q.$$


