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)



Existence of measurable "inclusion" into Euclidean space
(Basically the same as Michael's answer)


Theorem 6.5.7 of *Measure Theory* by V. Bogachev:



> 
> **Theorem.** The following are equivalent:
> 
> 
> 1. $\mathfrak{F}$ 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 $x \in F\_n$, $y \notin F\_n$;
> 2. there is an injective measurable function $f : \Omega \to [0,1]$ (or equivalently, to $\mathbb{R}^n$, since they are Borel isomorphic to $[0,1]$);
> 3. The diagonal $\Delta = \{(x,x) : x \in \Omega\}$ is measurable with respect to the product $\sigma$-algebra $\mathfrak{F} \otimes\mathfrak{F}$;
> 4. There exists a separable (i.e. countably generated) sub-$\sigma$-algebra $\mathfrak{F}\_0 \subset \mathfrak{F}$ which contains all the singletons.
> 
> 
> 


