# Is every Baire subset of a locally compact Hausdorff topological group $\sigma$-finite with respect to a Haar measure?

In his book Measure Theory, Halmos defines a Baire subset of a locally compact Hausdorff space $X$ as an element of the smallest $\sigma$-ring (not $\sigma$-algebra) on $X$ containing the $G_{\delta}$ compact subsets of $X$, i.e., compact subsets that are the intersection of countably many open subsets.

My question is as follows:

Question. If $G$ is a locally compact Hausdorff topological group, and $\mu$ is a Haar measure on $G$ (defined on the Borel $\sigma$-ring on $G$, i.e., the smallest $\sigma$-ring on $G$ containing the compact subsets of $G$), then is every Baire subset of $G$ (which is a Borel subset of $G$ also) $\sigma$-finite w.r.t. $\mu$?

Halmos’s discussion of measurable groups in his book seems to take this as fact, but I was unable to find a proof anywhere.

This is basically immediate from the fact that $\mu$ is finite on compact sets. The collection of sets $A$ such that $\mu$ is $\sigma$-finite on $A$ is a $\sigma$-ring (since it is closed under taking countable unions and subsets), and thus contains all Baire sets (indeed, it contains all Borel sets if you define those as the $\sigma$-ring generated by compact sets).
• Thanks! Did you mean the following? The collection $\mathscr{R}$ of all $\sigma$-finite elements of the Borel $\sigma$-ring $\mathscr{B}$ is a sub-$\sigma$-ring of $\mathscr{B}$, but as $\mathscr{R}$ contains the compact subsets of $G$, it follows that $\mathscr{R}$ must be all of $\mathscr{B}$. – Transcendental Oct 26 '16 at 5:06