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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.

Thank you in advance!

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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.

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