# Halmos Measure Theory section 62 exercise 3

Is there a locally compact group $G$ and a Borel measure $\mu$ on $G$ such that \begin{equation*} H=\{g\in G\mid \mu(gE)=\mu(E) \: \text{for all measurable} \: E\} \end{equation*} is not a closed subgroup of $G$ ?

Theorem 62.G in Halmos' textbook states that if $\mu$ is a Baire measure then $H$ is always closed. So a counter example must involve a group $G$ which isn't first countable.

I don't know any such groups except those of the form $G^{I}$ with $I$ uncountable and $G$ some usual group. With those I tried to construct a measure so that $H$ is the set of elements with finitely many non trivial coordinates, but it wouldn't be Borel (finite on compact sets).

• One place I would look is at compact abelian groups with ordered dual. These were characterized by Hahn. One reference is a paper co-written by me: faculty.missouri.edu/~stephen/preprints/hahn.html. I simply have no idea whether these will work, but it is worth a try. Commented Jun 14, 2016 at 21:07
• Thank you, I'll try learning about them when I have a moment. Commented Jun 18, 2016 at 12:48