Halmos Measure Theory has a problem (51.6) which goes as follows: 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 Halmos's setting, $X$ is a locally compact Hausdorff space, and he works with (generated) $\sigma$-rings rather than $\sigma$-algebras. He defines Baire sets as the $\sigma$-ring generated by the compact $G_\delta$ sets and the Borel sets as the $\sigma$-ring generated by the compact sets.

I can prove one direction (Baire set is Borel and its characteristic function is a Baire function), but have had no luck with the other direction. My questions are: 1) how do I go about proving the reverse implication? 2) What would be the more modern equivalent statement of the problem using $\sigma$-algebras? 3) Would the proof of that statement be substantially different than the proof of the original problem?

I've tried all sorts of constructions to no avail. Initially, I wanted to stay away from using an equivalent transfinite definition of $\mathscr{B}$, but I've relaxed that self-imposed restriction and played with those types of construction, too, but without success. Any help would be greatly appreciated.

  • $\begingroup$ The terminology is amazingly confusing: there is something known as “sets having the property of Baire” but apparently it's not the same thing as the “Baire sets” Halmos talks about. Incredible! ⁂ Anyway, there's a related theorem 24.3 in Kechris's Classical Descriptive Set Theory (Springer GTM 156, 1995), stated in a different context (essentially Polish spaces), but maybe trying to carry out the proof in the Hausdorff locally compact context still gives something of use? $\endgroup$
    – Gro-Tsen
    Jun 19 '19 at 10:04
  • $\begingroup$ Is this motivated by a concrete application or is it just for the love of theory? In lots of applications, e.g., to probability, one works with Polish spaces which are Radon. Namely, you can forget about Baire and no longer have to worry about that. $\endgroup$ Jun 19 '19 at 15:16
  • $\begingroup$ @Gro-Tsen Agreed. The terminology is confusing. Combine that with Halmos' definitions for Borel and Baire sets (with $\sigma$-rings rather than $\sigma$-algebras), and it is even more confusing. $\endgroup$
    – Jeff Rubin
    Jun 20 '19 at 6:53
  • $\begingroup$ @AbdelmalekAbdesselam You know, it was (originally) motivated by a prelude to probability, and Halmos was the most complete measure theory text I had, so I thought I'd pursue it through all he had to say about Baire functions (which was in the stated problem in the original post). But then it turned out to be difficult (for me) to prove and so the "love of theory" part took over, hence the posting. Once I'm through with this topic, I'll try to move on to probability theory. $\endgroup$
    – Jeff Rubin
    Jun 20 '19 at 6:58

I gave a simple counterexample in the comments yesterday but I guess it got buried. Let $X$ be the disjoint union of two copies of the long line. The characteristic function of either of them is continuous, and therefore Baire according to the stated definition, but neither is a Baire set.

To make the result true you need to define "Baire functions" to be the smallest class containing all continuous functions which vanish at infinity and closed under pointwise sequential limits. (The proof is given in another of my comments.)

  • 1
    $\begingroup$ I'm guessing that this is not a counterexample in the Halmos setting. I would guess (I can't prove it right now) that one copy (of the two copies) of the long line is not a Borel set according to Halmos' definition of the Borel sets, $\mathbf{S}=$ the $\sigma$-'ring' generated by the compact sets. Theorem 51.A says that with Halmos' definition, every Borel set is $\sigma$-bounded (contained in a countable union of compact sets), and since a single copy of the long line is not compact or even $\sigma$-compact, I'm guessing that it isn't contained in a countable union of compact sets either. $\endgroup$
    – Jeff Rubin
    Jun 20 '19 at 6:50
  • $\begingroup$ Yes according to that (unusual) definition these sets aren't Borel. How does that affect my counterexample? What specifically is wrong? $\endgroup$
    – Nik Weaver
    Jun 20 '19 at 10:54
  • $\begingroup$ Because the problem statement is “A necessary and sufficient condition that a set be a Baire set is that it be a Borel set and that it’s characteristic function be a Baire function.” Nevertheless, it is an interesting space to consider that I hadn’t heard about before (I.e., it isn’t in my copy of “Counterexamples in Topology”), so I do appreciate your mentioning it. $\endgroup$
    – Jeff Rubin
    Jun 20 '19 at 16:01

After more consideration, I have come up with what I think is an answer to my first question. I won't bore people with all the details, but for me the Aha! moment was when I considered the class $\mathscr{B}_{\Sigma}=\{f\in\mathscr{B}: f\text{ has $\sigma$-compact support}\}$. I can prove $\mathscr{B}_\Sigma$ contains the continuous functions with $\sigma$-compact support and is closed under pointwise limits. I can further prove that $\mathscr{B}\mathscr{B}_\Sigma=\{fg: f\in\mathscr{B},g\in\mathscr{B}_\Sigma\}\subseteq\mathscr{B}_\Sigma$ (which depended on my previously having shown that $\mathscr{B}$ is actually an algebra of functions (in particular, it is closed under multiplication of two functions). Finally, I can prove that every $f\in\mathscr{B}_\Sigma$ is Baire measurable. I'll omit the rest of the proof unless someone asks me to post it.


Since the above version doesn't work, let me present, in detail, a version that I think does work. I'll start with some terminology. Let $\mathbf{F}_0$ represent the class of all continuous real valued functions on $X$. Let $\mathscr{F}$ be the subclass of $\mathbf{F}_0$ consisting of continuous functions whose range is contained in $[0,1]$. If $\mathbf{T}$ is any class of real valued functions on $X$, and $\alpha$ is an ordinal, then we define $\mathscr{B}_\alpha(\mathbf{T})$ by transfinite induction as follows:

Let $\mathscr{B}_0(\mathbf{T})=\mathbf{T}$. If $\alpha$ is a successor ordinal, let $\mathscr{B}_\alpha(\mathbf{T})$ be the class of all pointwise limits of functions in $\mathscr{B}_{\alpha-1}(\mathbf{T})$. If $\alpha\leq\omega_1$ is a limit ordinal (where $\omega_1$ is the first uncountable ordinal), let $\mathscr{B}_\alpha(\mathbf{T})=\bigcup_{\beta<\alpha}\mathscr{B}_\beta(\mathbf{T})$.

Next, we define $\mathscr{B}(\mathbf{T})$ to be the smallest class of real valued functions on $X$ which is closed under pointwise limits and which contains $T$. Thus, the $\mathscr{B}$ of the problem statement is just $\mathscr{B}(\mathbf{F}_0)$.

Proposition 1. $\mathscr{B}(\mathbf{T})=\mathscr{B}_{\omega_1}(\mathbf{T})$.
Proof. Suppose $f_n\in\mathscr{B}_{\omega_1}(\mathbf{T})$ for $n=1,2,\dots$ and that the limit $f=\lim f_n$ exists. Then for each $n=1,2,\dots$, there exists an ordinal $\alpha_n<\omega_1$ such that $f_n\in\mathscr{B}_{\alpha_n}(\mathbf{T})$. Let $\alpha<\omega_1$ be an ordinal greater than each $\alpha_n$. Then $f_n\in\mathscr{B}_\alpha(\mathbf{T})$ for each $n=1,2,\dots$ and $f\in\mathscr{B}_{\alpha+1}(\mathbf{T})\subseteq\mathscr{B}_{\omega_1}(\mathbf{T})$. That is, $\mathscr{B}_{\omega_1}(\mathbf{T})$ is closed under pointwise limits. $\mathbf{T}=\mathscr{B}_0(\mathbf{T})\subseteq\mathscr{B}_{\omega_1}(\mathbf{T})$ so $\mathscr{B}_{\omega_1}(\mathbf{T})$ contains $\mathbf{T}$, hence contains all of $\mathscr{B}(\mathbf{T})$.

Conversely, $\mathscr{B}_{\omega_1}(\mathbf{T})\subseteq\mathscr{B}(\mathbf{T})$, for if not, there exists a function in $\mathscr{B}_{\omega_1}(\mathbf{T})$ which is not in $\mathscr{B}(\mathbf{T})$. Hence the class of countable ordinals $\alpha$ such that $\mathscr{B}_\alpha(\mathbf{T})$ contains a function not in $\mathscr{B}(\mathbf{T})$ is non empty. Let $\alpha$ be the least such ordinal. Then if $\beta<\alpha$, $\mathscr{B}_\beta(\mathbf{T})\subseteq\mathscr{B}(\mathbf{T})$. If $\alpha=0$ we have a contradiction since $\mathscr{B}_0(\mathbf{T})=\mathbf{T}\subseteq\mathscr{B}(\mathbf{T})$. $\alpha$ cannot be a limit ordinal, since then $\mathscr{B}_\beta(\mathbf{T})$ would contain a function not in $\mathscr{B}(\mathbf{T})$ for some $\beta<\alpha$, contradicting the minimality of $\alpha$. So $\alpha=\beta+1$ for some countable ordinal $\beta$, and there is a function $f\in\mathscr{B}_\alpha(\mathbf{T})$ which is not in $\mathscr{B}(\mathbf{T})$. By definition $f=\lim f_n$ for some sequence $\{f_n\}$ of functions in $\mathscr{B}_\beta(\mathbf{T})\subseteq\mathscr{B}(\mathbf{T})$, so by the definition of $\mathscr{B}(\mathbf{T})$, $f\in\mathscr{B}(\mathbf{T})$, a contradiction. $\quad\Box$

Proposition 2. Let $g$ be an arbitrary real valued function on $X$. Let $g\mathscr{B}(\mathbf{T})=\{gf:f\in\mathscr{B}(\mathbf{T})\}$ and let $g\mathbf{T}=\{gf: f\in\mathbf{T}\}$. Then $g\mathscr{B}(\mathbf{T})\subseteq\mathscr{B}(g\mathbf{T})$.
Proof. Suppose not. Then, using Proposition 1, there exists a least countable ordinal $\alpha$ such that $g\mathscr{B}_\alpha(\mathbf{T})$ contains a function not in $\mathscr{B}(g\mathbf{T})$. As in Proposition 1, $\alpha$ cannot be $0$ or a limit ordinal, so let $\alpha=\beta+1$. Then there exists $f\in\mathscr{B}_\alpha(\mathbf{T})$ such that $gf\notin\mathscr{B}(g\mathbf{T})$ and there exists a sequence $\{f_n\}$ in $\mathscr{B}_\beta(\mathbf{T})$ such that $f=\lim f_n$. By the minimality of $\alpha$, $g\mathscr{B}_\beta(\mathbf{T})\subseteq\mathscr{B}(g\mathbf{T})$, so $gf=\lim gf_n\in\mathscr{B}(g\mathbf{T})$, a contradition. $\quad\Box$

Halmos Measure Theory Theorem 50.B. If $C$ is a compact set, $F$ is a closed set, and $C\cap F=\varnothing$, then there exists a function $f$ in $\mathscr{F}$ such that $f(x)=0$ for $x$ in $C$ and $f(x)=1$ for $x$ in $F$.

Halmos Measure Theory Theorem 50.D. If $C$ is compact, $U$ is open, and $C\subseteq U$, then there exist sets $C_0$ and $U_0$ such that $C_0$ is a compact $G_\delta$, $U_0$ is a $\sigma$-compact open set, and $$C\subseteq U_0\subseteq C_0\subseteq U.$$

Halmos Measure Theory Theorem 51.A. Every Borel set is $\sigma$-bounded; every $\sigma$-bounded open set is a Borel set.

Halmos Measure Theory Theorem 51.B. If a real valued continuous function $f$ on $X$ is such that the set $N(f)=\{x:f(x)\neq 0\}$ is $\sigma$-bounded, then $f$ is Baire measurable.

Proposition 3. Let $g$ be a continuous real valued function on $X$ having $\sigma$-compact support. Then every function in $\mathscr{B}(g\mathbf{F}_0)$ is Baire measurable.
Proof. Let $\mathbf{M}=\{h\in\mathscr{B}(g\mathbf{F}_0):\text{ $h$ is Baire measurable}\}$. $\mathbf{M}$ is closed under pointwise limits since both $\mathscr{B}(g\mathbf{F}_0)$ and the class of Baire measurable functions are. If $f\in\mathbf{F}_0$ then $h=gf$ is continuous and has $\sigma$-compact support, so $N(h)$ is $\sigma$-bounded, so by Theorem 51.B, $h$ is Baire measurable. Hence $g\mathbf{F}_0\subseteq\mathbf{M}$ and therefore $\mathscr{B}(g\mathbf{F}_0)\subseteq\mathbf{M}\subseteq\mathscr{B}(g\mathbf{F}_0)$, so every $h\in\mathscr{B}(g\mathbf{F}_0)$ is Baire measurable. $\quad\Box$.

Proposition 4. If $E$ is a Borel set and if $\chi_E\in\mathscr{B}$ then $E$ is a Baire set.
Proof. By 51.A, $E$ is $\sigma$-bounded, say $E\subseteq\cup C_n$ where $C_n$ is compact, $n=1,2,\dots$. Since a finite union of compact sets is compact, without loss of generality, we may assume $C_1\subseteq C_2\subseteq\cdots$. By 50.D applied to $C_n\subseteq X$, there exist open $U_n$ and comapct $K_n$ such that $C_n\subseteq U_n\subseteq K_n$. By 50.B applied to $C_n$ and $F=X-U_n$, there exists $f_n\in\mathscr{F}$ such that $(1-f_n)(C_n)=\{1\}$ and $(1-f_n)(X-U_n)=\{0\}$. In addition, $(1-f_n)(X-K_n)=\{0\}$ since $X-K_n\subseteq X-U_n$. For $x\in E$, there is an $n_0$ such that $x\in C_n$ for all $n\geq n_0$, so $(1-f_n)(x)\to 1$. For $x\notin\cup K_n$, $x\in X-K_n$, $n=1,2,\dots$, so $(1-f_n)(x)=0$, $n=1,2,\dots$, hence $(1-f_n)(x)\to 0$. Therefore, $g_n=1-f_n$ is continuous with compact support. By Proposition 2, $g_n\mathscr{B}=g_n\mathscr{B}(\mathbf{F}_0)\subseteq\mathscr{B}(g_n\mathbf{F}_0)$. Then, by Proposition 3, since $\chi_E\in\mathscr{B}$, we have that $g_n\chi_E$ is Baire measurable, $n=1,2,\dots$. Since $g_n(x)\to 1$ for $x\in E$, and since $\chi_E(x)=0$ for $x\notin E$, it is clear that $\chi_E=\lim g_n\chi_E$, and hence $\chi_E$ is Baire measurable. For Halmos Measure Theory, which works with $\sigma$-rings, a function $f$ is measurable if and only if $N(f)\cap f^{-1}(M)$ is measurable for each Borel set $M$ of the real line. Taking $f=\chi_E$ and $M$ to be the entire real line, this says that $E=E\cap X=N(\chi_E)\cap X=N(\chi_E)\cap\chi_E^{-1}(\mathbb{R})$ is measurable; i.e., a Baire set. $\quad\Box$

  • $\begingroup$ What about this: consider the family of Borel sets whose characteristic function is Baire, apply monotone class lemma. Voila, such sets constitute a $\sigma$-algebra which contains every compact $G_\delta$. $\endgroup$
    – Nik Weaver
    Jun 16 '19 at 9:16
  • $\begingroup$ ... and therefore contains the Baire sets.That's fine for the other (easy) direction of the implication. What I was having trouble with is to show that a Borel set whose characteristic function is a Baire function is a Baire set. The difficulty (for me) comes about because the family you describe is not a $\sigma$-algebra (or -ring) generated by anything that I can find and therefore it's hard to prove that something else contains it. $\endgroup$
    – Jeff Rubin
    Jun 17 '19 at 2:08
  • $\begingroup$ I no longer believe the answer that I posted above is correct, as there was a flaw in my "proof" that every $f\in\mathscr{B}_\Sigma$ is Baire measurable. So I am back to square one. $\endgroup$
    – Jeff Rubin
    Jun 17 '19 at 3:02
  • $\begingroup$ Oh, sorry, I misread you. Maybe the key words you need here are "prove that every Baire function, in your sense, is Baire-to-Borel measurable (or Baire to Baire, there's no difference on $\mathbb{R}$). So, every continuous function that vanishes at infinity is measurable, as are pointwise limits of sequences of measurable functions. $\endgroup$
    – Nik Weaver
    Jun 17 '19 at 7:07
  • $\begingroup$ Is the issue continuous vs. continuous and vanishing at infinity? I think you need that condition. $\endgroup$
    – Nik Weaver
    Jun 17 '19 at 7:08

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