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In Halmos' Measure Theory, he says in problem 51.6 "The term 'Baire set' is suggested by the term 'Baire function' as used in analysis. If $\mathscr{B}$ is the smallest class of functions which contains all continuous functions and contains the limit of every pointwise (but not necessarily uniformly) convergent sequence of functions in it, then the functions of $\mathscr{B}$ are called the 'Baire functions' on $X$. A necessary and sufficient condition that a set be a Baire set is that it be a Borel set and that its characteristic function be a Baire function."

In this section of the book, $X$ is a locally compact Hausdorff space. For those not familiar with Halmos, in this context, he defines a Baire set to be a member of the $\sigma$-ring $\mathbf{S}_0$ generated by the class $\mathbf{C}_0$ of compact $G_\delta$ subsets of $X$, and a Borel set to be a member of the $\sigma$-ring $\mathbf{S}$ generated by the class $\mathbf{C}$ of compact subsets of $X$. One further piece of context is that in problem 51.3, he asks that you prove (and I have done so) that the $\sigma$-ring generated by the class of all bounded open sets coincides with $\mathbf{S}$.

I am able to prove that every Baire set is a Borel set whose characteristic function is a Baire function. It is the other direction that is giving me problems. I am desperately trying to avoid using transfinite induction to prove this, although I have no problem with using it to get an idea of how to proceed.

I have been trying to define various classes of sets and classes of functions in order to get the right type of containment, but I always seem to get the opposite containment from what I am looking for. My strategy has been the following. If $E$ is a bounded open set such that $\chi_E\in\mathscr{B}$, then if I can prove that $E$ is a Baire set, then I should be able to use 51.3 to show that it holds for any Borel set. So assuming $E$ is bounded and open (by the way, for Halmos, in this context, bounded means contained in a compact set), I know how to find a continuous real-valued function $f_E$ on $X$ with range contained in $[0,1]$, and compact sets $C$ and $K$, and open set $U$, such that $E\subseteq C\subseteq U\subseteq K$ and such that $f_E$ is $1$ on $C$ and $0$ on the complement of $U$. Now to the transfinite part. If $\chi_E\in\mathscr{B}$, then there is a transfinite collection of pointwise convergent sequences of functions in $\mathscr{B}$, starting with the continuous functions, that ends up at $\chi_E$. If I multiply every function involved by $f_E$, continuous functions stay continuous, pointwise limits stay pointwise limits (i.e., if $f_n\to f$, then $f_nf_E\to ff_E$), and $\chi_E$ stays as $\chi_E$, since $\chi_E=\chi_Ef_E$. I was hoping that this class $\mathscr{B}_E=\{ff_E:f\in\mathscr{B}\}$ would turn out to be a subclass of $\mathscr{B}_\sigma$ which I define to be the smallest class of functions which is closed under pointwise limits and contains all continuous functions with $\sigma$-compact support. For, I am able to show that every member of $\mathscr{B}_\sigma$ is Baire measurable. But these containments tend to work out the other way around; the smallest class of functions of some type tends to be contained in other classes of functions that you might create.

I have been racking my brains for a couple of weeks on this one, and could sure use some help. And I sure would appreciate it if the help stayed within the boundaries of Halmos Measure Theory. (Kelley Topology would be OK, too).

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Well, I've figured this out finally. My mistake was considering $\mathscr{B}_\sigma$. I should have considered $\mathscr{B}_\Sigma=\{f\in\mathscr{B}:f\text{ has $\sigma$-compact support}\}$. There are quite a few details, and I won't post them all unless someone requests it.

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  • $\begingroup$ Actually, that wasn't the proper fix. But I have worked out a solution. $\endgroup$
    – Jeff Rubin
    Jun 19, 2019 at 7:25

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