# A strong Borel selection theorem for equivalence relations

In Kechris' book "Classical Descriptive Set Theory" there is the following theorem (12.16):

Let $$X$$ be a Polish space and $$E$$ an equivalence relation such that every equivalence class is closed and the saturation of any open set is Borel. Then $$E$$ admits a Borel selector.

So under the hypotheses of the theorem, one gets a set of representatives that is Borel.

There are also stronger forms of this theorem that weaken the assumptions for instance to the classes being $$G_\delta$$ instead of closed (see Miller (1980)).

I have the feeling that a theorem with an even weaker assumption as follows should be true.

Question: Let $$X$$ be a Polish space and $$E$$ an equivalence relation such that every equivalence class is Borel and the saturation of any open set is Borel. Does $$E$$ always admit a Borel selector?

I am not an expert in this field and therefore also happy if you can just give a reference.

Let $$X$$ be the Cantor space $$2^\omega$$, and let $$E$$ be the relation of "equivalence mod $$\mathrm{Fin}$$" -- i.e., $$xEy$$ if and only if $$\{n \in \omega :\, x(n) \neq y(n) \}$$ is finite. The equivalence classes for this relation are countable (hence Borel, and even $$F_\sigma$$). If $$U \subseteq 2^\omega$$ is open, then $$U$$ contains a basic clopen subset of $$2^\omega$$, which (by how these are defined) contains a representative of every equivalence class of $$E$$. So the saturation of any nonempty open set with respect to $$E$$ is $$2^\omega$$ itself. However, $$E$$ does not admit a Borel selector: any selector for $$E$$ is non-measurable with respect to the usual Haar measure on $$2^\omega$$, by Vitali's classical argument.