Under what conditions on a metric space $X$, equipped with the Borel $\sigma$-algebra, does there exist a measurable total ordering of the elements of $X$?


By "measurable total ordering" we mean that any initial segment $I\_y:=\{x: x<y\}$ is Borel-measurable.


Edit: We know that separability is sufficient for a measurable total order to exist.


Edit II: Vladimir Pestov (private communication) has shown that a measurable total order always exists; will post answer soon with link to full paper.



Measurable total order
As mentioned in the OP, Vladimir Pestov has answered the question affirmatively. See Appendix D here:
<https://arxiv.org/pdf/1906.09855.pdf>


