Recall the notion of a dense subcategory $\mathcal{D}$ of a category $\mathcal{C}$. It means that the restricted Yoneda functor $\mathcal{C} \to \mathrm{Hom}(\mathcal{D}^{op},\mathbf{Set})$, $A \mapsto \mathrm{Hom}(-,A)|_{\mathcal{D}}$ is fully faithful. Roughly, it means that $\mathcal{D}$ "detects morphisms" in $\mathcal{C}$.

One can show that $\mathbf{Meas}$, the category of measurable spaces$^1$, has no small dense subcategory. Trivially, $\mathbf{Meas}$ is a dense subcategory of $\mathbf{Meas}$, but that is not very interesting.

**Question.** What is an example of a "quite small" proper dense full subcategory of $\mathbf{Meas}$?

By "quite small" I mean that we are not just removing a bunch of measurable spaces, but rather that the objects of the dense subcategory are parametrized by a very simple structure. Imagine, very informally, there was a measure on $\mathbf{Meas}$, then I want the dense subcategory to be of measure $0$.

We can assume that the one-point measurable space belongs to the subcategory. If $\mathcal{K}$ denotes the rest, we have the following characterization of density: If $X,Y$ are measurable spaces, then a map $f : X \to Y$ is measurable iff for every measurable map $a : A \to X$ for $A \in \mathcal{K}$ the composition $f \circ a : A \to Y$ is measurable. (This is what I meant above with "detecting morphisms"). The question asks for such a class of measurable spaces.

At first you might think that this is completely impossible. I had the same suspicion for $\mathbf{Top}$, but it turns out that for $\mathbf{Top}$ it *is* possible: take the one-point-space and the topological spaces of the form $P \cup \{\infty\}$ for directed sets $P$, where the sets $P_{\geq p} \cup \{\infty\}$ form a local base at $\infty$. This subcategory is dense: This is just a fancy way of saying that a map is continuous iff it preserves net convergence. Maybe there is some similar theory of "net convergence" for measurable spaces? I found the related discussion What properties are preserved under a measurable mapping?, but I am not sure if Eric Wofsey's answer settles my question, because convergent filters cannot be seen as maps.

$^1$ Since Dmitri Pavlov's notion of a measurable space has become quite prominent on mathoverflow, let me mention that I use the "classical" definition here. It's just a set with a $\sigma$-algebra. However, if there was a very good answer for Pavlov's measurable spaces, I would be happy to hear about that too.

Measis a complete subcategory of the category of uniform spaces. So one could ask the same question for that larger category as well. $\endgroup$