Let $(\Omega,\mathfrak{F})$ be a measurable space. When does there exist an injective measurable function $f:(\Omega,\mathfrak{F})\to (\mathbb{R}^n,B(\mathbb{R}^n))$ to some Euclidean space, here $B(\mathbb{R}^n)$ is the Borel $\sigma$-algebra.
Thoughts. Clearly, if $\Omega$ is a Riemannian manifold and $\mathfrak{F}$ is its Borel $\sigma$-algebra then this works but I'm thinking of more general, non-topological criteria. (If they exist)
(Basically the same as Michael's answer)
Theorem 6.5.7 of Measure Theory by V. Bogachev:
Theorem. The following are equivalent:
$\mathfrak{F}$ is countably separated (Bogachev Definition 6.5.1 (ii)): there exists an at most countable collection of sets $F_n \in \mathfrak{F}$ such that for every two distinct points $x,y \in \Omega$, there is some $F_n$ with $x \in F_n$, $y \notin F_n$;
there is an injective measurable function $f : \Omega \to [0,1]$ (or equivalently, to $\mathbb{R}^n$, since they are Borel isomorphic to $[0,1]$);
The diagonal $\Delta = \{(x,x) : x \in \Omega\}$ is measurable with respect to the product $\sigma$-algebra $\mathfrak{F} \otimes\mathfrak{F}$;
There exists a separable (i.e. countably generated) sub-$\sigma$-algebra $\mathfrak{F}_0 \subset \mathfrak{F}$ which contains all the singletons.
This is the case if and only if there exists a countable subfamily of $\mathfrak{F}$ that separates points. Necessity is straightforward; if such a function exists then the preimages of rectangles with rational coordinates at the endpoints will serve as the countable separating family.
For sufficiency, let $\mathcal{C}=\{C_1, C_2,\ldots\}$ be a countable separating family of measurable sets. Then the Marczewski function $f:\Omega\to [0,1]$ given by $f(\omega)=\sum_n 2/3^n1_{C_n}(\omega)$ will be measurable and injective.
