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).

  • $\begingroup$ 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} $. $\endgroup$ – Transcendental Oct 26 '16 at 5:06
  • $\begingroup$ Yes, that's right. $\endgroup$ – Eric Wofsey Oct 26 '16 at 5:16

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