# The space of Borel function modulo comeager sets is Dedekind complete

Let $$(X,\tau)$$ be a topological space. Denote by $$Bor(X)$$ the space of Borel functions $$f:X\to\mathbb{R}$$ where we identify two functions whenever they agree on the complement of a meager set. We endowed $$Bor(X)$$ with the pointwise partial order. I have read somewhere that this space is Dedekind complete. Is that true? Where can I find the proof worked out?

• (I suspect there is some relation, because the Boolean algebra of regular open sets, again under certain assumptions I'm not too sure about, coincides with the Boolean algebra of Borel sets modulo meager sets, cf. Halmos, "Lectures on Boolean Algebras" (1963), §13. But I'm afraid I'm unable to fill the dots.) – Gro-Tsen Mar 7 at 18:43
• @Gro-Tsen An answer on math.SE seems to be relevant – მამუკა ჯიბლაძე Mar 7 at 20:36
• I should mention the following fact: If $(X,\Sigma)$ is a measure space equipped with a $\sigma$-ideal $\mathcal{N} \subseteq \Sigma$, then the poset of measurable functions $f : X \rightarrow [0,1]$ measurable with respect to the Borel $\sigma$-algebra on [0,1], identifying functions that differ only on a set from $\mathcal{N}$, is a complete lattice iff $\Sigma/\mathcal{N}$ is a complete Boolean algebra (instead of just $\sigma$-complete). With this result, the question is equivalent to the question of whether the Borel sets modulo the meagre sets is always a complete Boolean algebra.. – Robert Furber Mar 7 at 21:01
• @Littlefield The set of measurable functions to $\mathbb{R}$ is not literally a complete lattice - for instance the sequence of constant functions $n : X \rightarrow \mathbb{R}$ has no least upper bound, because it has no upper bound. The actual statement is that every bounded set of functions has a least upper bound. By translating and rescaling, this is equivalent to $[0,1]$-valued functions being a complete lattice, and this is a little easier to prove technically. Unfortunately I do not know a reference, but I can outline the proof in an answer. – Robert Furber Mar 8 at 8:35
• @Littlefield You are right that Borel sets modulo meagre sets is always a complete Boolean algebra - I just couldn't see the right way of proving this at first, which is to enlarge the Borel $\sigma$-algebra to the $\sigma$-algebra of sets having the Baire property. – Robert Furber Mar 8 at 8:36

Fremlin's measure theory textbook is a good reference for these things. I am splitting things up into the Boolean algebra part and the real-valued functions part.

## Complete Boolean algebras:

The way to show that $$\mathcal{B}or(X)/\mathcal{M}(X)$$ (using $$\mathcal{M}(X)$$ for the meagre sets) is a complete Boolean algebra is to reformulate it a bit first. The $$\sigma$$-algebra of sets with the Baire Property, $$\mathcal{BP}(X)$$, consists of the sets $$S \subseteq X$$ such that there exists an open set $$U \subseteq X$$ such that $$S \vartriangle U$$ is meagre ($$\vartriangle$$ meaning symmetric difference). As it is a $$\sigma$$-algebra and contains the open sets, $$\mathcal{B}or(X) \subseteq \mathcal{BP}(X)$$, and in fact $$\mathcal{BP}(X)$$ is the completion of $$\mathcal{B}or(X)$$ with respect to the $$\sigma$$-ideal $$\mathcal{M}(X)$$ in the same sense that the $$\sigma$$-algebra of Lebesgue measurable sets is the completion of the Borel sets with respect to the $$\sigma$$-ideal of Lebesgue null sets. For short, I will write $$\mathrm{BP(X)} = \mathcal{BP}(X) / \mathcal{M}(X)$$.

Because of this, the inclusion map $$\mathcal{B}or(X) \hookrightarrow \mathcal{BP}(X)$$ induces an isomorphism $$\mathcal{B}or(X)/\mathcal{M}(X) \cong \mathrm{BP}(X)$$, so we only need to prove that the latter is a complete Boolean algebra. This is done in 514I in volume 5, part 1 of Fremlin's Measure Theory.

The regular open sets $$\mathrm{RO}(X)$$ form a complete Boolean algebra. Each regular open set has the Baire property, and this defines a complete Boolean homomorphism $$\mathrm{RO}(X) \rightarrow \mathrm{BP}(X)$$. This is surjective, and is injective iff $$X$$ is a Baire space, i.e. if every meagre open set is empty. This is discussed in the same place in Fremlin's book, along with the fact that if $$X$$ is the Stone space of a complete Boolean algebra $$A$$, regular open sets are the same thing as clopens, so $$\mathrm{RO}(X) \cong A$$.

## Real-valued functions

The way we get to real-valued measurable functions from the above has to do with Dana Scott's construction of the reals in a Boolean-valued model of set theory, as described in section 4.3 of Solovay's article Real-valued Measurable Cardinals.

Although expositions of these things often restrict to the case of a probability measure space, every proof I've seen goes through without using any particulars about the $$\sigma$$-ideal of null sets other than the fact that $$\mathcal{B}or(X)/\mathcal{N}$$ is a complete Boolean algebra. So in the following, we will say that a triple $$(X,\Sigma,\mathcal{N})$$ is a negligibility space if $$\Sigma$$ is a $$\sigma$$-algebra on $$X$$ and $$\mathcal{N}$$ is a $$\sigma$$-ideal in $$\Sigma$$ such that $$\Sigma/\mathcal{N}$$ is a complete Boolean algebra. (I know no standard terminology for this, so this is my own term)

We define $$L^0(X,\Sigma,\mathcal{N})$$ (or $$L^0(X)$$ for short) to be the set of $$\mathbb{R}$$-valued measurable functons on $$(X,\Sigma)$$ modulo the equivalence relation of being equal almost everywhere, i.e. outside a set from $$\mathcal{N}$$. Similarly $$L^\infty(X)$$ consists of bounded functions, $$L(X;[0,1])$$ of $$[0,1]$$-valued functions.

The key point is that we can, in a suitable sense, define the Borel sets of $$\mathbb{R}$$ and $$[0,1]$$ by generators and relations, and then define measurable functions in terms of this and the complete Boolean algebra $$\Sigma/\mathcal{N}$$. This way of doing things is also explained in Fremlin's section 364, and I think 364B, C and M are enough to establish what you want, but here is my explanation anyway.

For each $$q \in \mathbb{Q}$$, define $$b_q = (q,\infty)$$, and for each $$q \in \mathbb{Q}\cap[0,1]$$ define $$c_q = (q,1]$$ (we will write $$[0,1]_{\mathbb{Q}}$$ for $$[0,1] \cap \mathbb{Q}$$ from now on). Now, $$(b_q)_{q \in \mathbb{Q}}$$ generates the Borel sets of $$\mathbb{R}$$, subject to the relations:

1. $$b_q = \bigvee\limits_{q' > q} b_{q'}$$ for all $$q \in \mathbb{Q}$$.
2. $$\bigvee_{q \in \mathbb{Q}} b_q = 1$$
3. $$\bigwedge_{q \in \mathbb{Q}} b_q = 0$$,

while $$(c_q)_{q \in [0,1]_{\mathbb{Q}}}$$ generates the Borel sets of $$[0,1]$$ subject to the relations: $$c_q = \bigvee\limits_{q' > q} c_{q'}$$

For a complete Boolean algebra $$A$$, we define $$\mathcal{C}(\mathbb{Q},A)$$ to be the set of functions $$f : \mathbb{Q} \rightarrow A$$ subject to the relations described above, (so for the first one $$f(q) = \bigvee_{q' > q}f(q')$$, and so on). We define $$\mathcal{C}([0,1]_{\mathbb{Q}}, A)$$ in a similar manner. We define the ordering on these pointwise, i.e. $$f \leq g$$ iff for all $$q \in \mathbb{Q}. f(q) \leq g(q)$$ (respectively with $$[0,1]_{\mathbb{Q}}$$ in that case). Then we can define, for a family $$(f_i)_{i \in I}$$ a function $$f$$ $$f(q) = \bigvee_{i \in I}f_i(q).$$ It is easy to prove that in the case that all $$f_i \in \mathcal{C}([0,1]_{\mathbb{Q}}, A)$$, then $$f$$ is too. But in the case of $$f_i \in \mathcal{C}(\mathbb{Q},A)$$, relation 3 does not hold automatically, but if there is an upper bound $$g \in \mathcal{C}(\mathbb{Q},A)$$ for $$(f_i)_{i \in I}$$ then it does. It then follows that $$f$$ is a least upper bound for $$(f_i)_{i \in I}$$.

To use this for $$L^0(X)$$ and $$L(X;[0,1])$$ we build poset isomorphisms. So define $$F_{X,\mathbb{Q}} : L^0(X) \rightarrow \mathcal{C}(\mathbb{Q},\Sigma/\mathcal{N})$$ as follows: $$F_{X,\mathbb{Q}}([a])(q) = [a^{-1}(b_q)]$$ and $$F_{X,[0,1]_{\mathbb{Q}}} : L(X;[0,1]) \rightarrow \mathcal{C}([0,1]_{\mathbb{Q}},\Sigma/\mathcal{N})$$ similarly: $$F_{X,[0,1]_{\mathbb{Q}}}([a])(q) = [a^{-1}(c_q)].$$ Then prove that this is well-defined, an order embedding, injective and surjective, the last two relying heavily on the countability of $$\mathbb{Q}$$ and the fact that $$\mathcal{N}$$ is a $$\sigma$$-ideal. These isomorphisms then establish that $$L^0(X)$$ is Dedekind complete, and $$L(X;[0,1])$$ is a complete lattice, and so $$L^\infty(X)$$ is Dedekind complete.

Regarding the question from Gro-Tsen: I unfortunately do not have the time to work out if this really works, but an isomorphism between $$L^\infty(X,\mathcal{B}or(X),\mathcal{M}(X))$$ and the normal lower semicontinuous functions would follow for completely regular Baire spaces if normality of a lower semicontinuous function $$a : X \rightarrow [0,1]$$ were the same as $$a^{-1}((q,1])$$ being a regular open set, rather than just open. This condition certainly implies that $$a$$ is a normal lower semicontinuous function, by reinterpreting Dilworth's Theorem 3.2, but I don't know if it goes the other way.

• This paper uses the term "localizable measurable space". (It also defines measurable spaces to be equipped with a $\sigma$-ideal of negligible sets, which is unfortunately not standard, but "localizable measurable space" is still unambiguous.) – Dap Mar 12 at 14:09
• @Dap I'm aware of Dmitri Pavlov's terminology, but I don't agree with it. I don't like the term "measurable space", because otherwise there's no name left for a set with a $\sigma$-algebra on it. The other problem is that the name implies that it will be possible to put a measure on a "measurable space", but this is not so - for any Polish space $X$ without isolated points, the only countably additive measure on $(X,\mathcal{B}or(X), \mathcal{M}(X))$, that vanishes on the $\sigma$-ideal $\mathcal{M}(X)$ is 0. – Robert Furber Mar 13 at 15:13
• In some sense, it is the usual thing for complete Boolean algebras not to admit countably additive measures, but any complete Boolean algebra can be presented as a $\sigma$-algebra modulo a $\sigma$-ideal. – Robert Furber Mar 13 at 15:14

After some googling, I managed to track a reference that answers the question, under the assumption that $$X$$ is a Baire space (i.e., every nonempty open subset is non-meager):

Kusraev & Kutateladze, Boolean-Valued Analysis (Springer 1999), §5.1.7 example (6) on page 206

— explicitly states that, if $$X$$ is a Baire space, the set of Borel functions $$X\to\mathbb{R}$$ modulo equality on the complement of a meager set is a “$$K$$-space”, or Kantorovich space, meaning (op. cit., §5.1.4) a Dedekind-complete vector lattice.

Furthermore, this provides some hint of a link to the space I was referring to in the comments, since the immediately following example (7) states that the space in question can also be identified with the set of lower semicontinuous functions $$X \to \overline{\mathbb{R}} := \mathbb{R}\cup\{\pm\infty\}$$ such that $$f^{-1}(-\infty)$$ and the closure of $$f^{-1}(+\infty)$$ are nowhere dense, again modulo equality on the complement of a meager set.

Meta: I hope someone can post a better answer than this one and clarify the exact relation (under suitable hypotheses on $$X$$, maybe Baire + T3½) between “real-valued Borel functions modulo meager sets” (as above) and “normal lower-semicontinuous real-valued functions” (or “continuous real-valued functions on the Stone space of the Boolean algebra of regular open sets”) as in Dilworth's 1950 paper “The Normal Completion of the Lattice of Continuous Functions” that I mentioned in the comments. (I suspect the only difference is in boundedness or something like this.) I must say, I find it maddening that the word “meager” (or “first-category”) does not appear anywhere in Dilworth's paper and that no reference to Dilworth's paper appears in the Kusraev & Kutateladze book (even though they seem tantalizingly close). If the answer is not staring at me in the face, I might ask a different question about this.