A measurable group is a group equipped with a $\sigma$-algebra making the multiplication and the inversion measurable.
For any measurable space $X$ and an equivalence relation $\sim$ on $X$, there is the unique $\sigma$-algebra on $X^*=X/\sim$ making the obvious projection $q:X\to X^*$ satisfy:
"For all $A\subseteq X^*$, $q^{-1}(A)$ is measurable if and only if $A$ is measurable."
Call this $\sigma$-algebra the quotient $\sigma$-algebra.
Let $G$ be a measurable group and let $N$ be its normal subgroup. Assume that $N$ is at most countable. Give $G/N$ the quotient $\sigma$-algebra. Is $G/N$ also a measurable group?
My Attempt
The analogous statement is true for topological groups. So I tried to mimic its proof. It goes as:
- Show that $q:G\to G/N$ is an open map;
- Show that, if $f:W\to Y$ and $g:X\to Z$ are open maps, then $f\times g$ is an open map; Conclude that $q\times q$ is open.
- The rest is easy.
If 1 - 3 are changed into the analogous statements, then 1 is also true for measurable groups (provided that $N$ is at most countable). But I could not prove 2. Once I prove 2, I am done.
For 2, let $\mathcal A$ be the algebra generated by the set $\{B\times C:\text{$B\subseteq W$ and $C\subseteq X$ are measurable}\}$. I proved that if $A\in\mathcal A$, then $(f\times g)(A)$ is measurable. But the $\sigma$-algebra on $W\times X$ is the $\sigma$-algebra generated by $\mathcal A$. I am stuck here.