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? 
 A: 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:


*

*$b_q = \bigvee\limits_{q' > q} b_{q'}$ for all $q \in \mathbb{Q}$.

*$\bigvee_{q \in \mathbb{Q}} b_q = 1$

*$\bigwedge_{q \in \mathbb{Q}} b_q = 0$,


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.
A: 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.
