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? 



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


