(1) Suppose that I have a compact Hausdorff space $X$ with a countable base. Why are the Borel algebra $\mathcal{B}(X)$ (the $\sigma$-field generated by the open sets) and the Baire algebra $\mathcal{B}a(X)$ (the $\sigma$-field generated by the compact $G_\delta$ sets) equal? Where can I find a proof of this?

(2) Suppose now that $X$ has an uncountable base. In that case, $\mathcal{B}(X)$ and $\mathcal{B}a(X)$ do not coincide anymore, and I know that considering the Baire sets avoids some pathologies of the Borel sets. What are those pathologies? Also, what would be an example of a Borel set which is not Baire?

  • $\begingroup$ In (2) $X$ is still compact Hausdorff? $\endgroup$ Aug 16, 2020 at 10:42

1 Answer 1


To see in the first case that Baire sets and Borel sets coincide, it suffices to note that the generating sets for the Baire sets (compact $G_\delta$) are always Borel (compact implies closed in Hausdorff spaces) so that Baire $\subseteq$ Borel easily. And if $O$ is open we can write it as a countable union of compact $G_\delta$ sets, so all open sets are in the Baire $\sigma$-field, so all Borel sets are too. (Second countable Hausdorff compact implies perfectly normal etc.)

To see what might go wrong more generally, check out $X=\omega_1 + 1$ which is compact Hausdorff but not second countable. In it, $\{\omega_1\}$ is closed (so Borel) but not Baire (Halmos proves in his Measure Theory that a compact set is Baire if it is a $G_\delta$ and this singleton isn’t). The Dieudonné measure on $X$ is a Borel measure that is not regular, but is regular when we work on Baire sets. See Halmos’ book, or Fremlin’s extensive work in topological measure theory. Taking Baire sets gives us more than enough sets to do integration stuff etc and gives better behaved measures in terms of regularity properties.


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