Let $G$ be a locally compact Hausdorff group with a Haar measure $\mu$, let $H$ be a closed normal subgroup of $G$, and let $q: G \to G/H$ be the quotient homomorphism. Let $\nu$ be a Haar measure on the locally compact Hausdorff group $G/H$.

My question is this: if $E$ is a union of cosets of $H$ in $G$ that is $\mu$-measurable (if necessary, $E$ can be assumed to be a Borel set); must then $q(E)$ be $\nu$-measurable?

I have seen work in descriptive set theory which shows that $q(E)$ is actually Borel whenever $E$ is a Borel $H$-invariant set, even when $H$ is not normal, but such work always assumes that $G$ is second countable. I am interested in locally compact Hausdorff groups that may not be second countable, and would be happy with $q(E)$ being just $\nu$-measurable.

If the answer is negative in general, does it hold at least if $G$ is abelian or compact?


It seems that the answer to this problem is affirmative for compact groups. Indeed, if the union $E$ of $H$-cosets is measurable in $G$, then for every $\epsilon>0$ we can find two compact subsets $A\subset E$ and $B\subset G\setminus E$ such that $\mu(A)+\mu(B)>1-\epsilon$. It follows that $q(A)\subset q(E)\subset q(G)\setminus q(B)$ and the projections $q(A)$, $q(B)$ have Haar measures $\lambda(q(A))=\mu(AH)\ge\mu(A)$ and $\lambda(q(B))\ge\mu(BH)=\mu(B)$, which implies that $\lambda(q(A))+\lambda(q(B))>1-\epsilon$ and hence $\lambda(q(A))\le\lambda(q(E))<\lambda(q(G)\setminus q(B))=1-\lambda(q(B))<\lambda(q(A))+\epsilon$. This means that the set $q(E)$ is measurable in the quotient group $G/H=q(G)$.

In the proof we used the fact that the Haar measure of $G$ projects onto the Haar measure of $G/H$. This follows from the uniqueness of the Haar measure (as a unique probability Borel invariant measure on the compact group).


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