# Measurable almost everywhere right inverse when almost every range element has unique preimage

Let $$X$$ and $$Y$$ be compact metric spaces, $$\pi: X \to Y$$ a continuous onto map, and $$\nu$$ a Borel probability measure on $$Y$$ such that $$\nu$$-almost every $$y \in Y$$ has unique $$\pi$$-preimage. Does there exist a measurable map $$s: Y \to X$$ such that for $$\nu$$-almost every $$y \in Y$$, $$(\pi \circ s)(y) = y$$?

For the application I'm interested in, $$X$$ and $$Y$$ are shift spaces with finite alphabet over a countable group, if that helps at all.

Every continuous onto map $$\pi: X \to Y$$ between compact metric spaces has a Borel measurable section, i.e. there is a Borel measurable map $$s: Y \to X$$ such that $$\pi \circ s (y) = y$$ for all $$y$$. The assumptions about the measure $$\nu$$ and the preimages of $$\pi$$ are superfluous.
For a proof of this, you can look up Theorem 6.9.7 in Bogachev's Measure Theory, Volume 2. The idea of the proof is to start with the case where $$X \subset [0,1]$$. In this case, for a given $$y \in Y$$ you can select a preimage of $$y$$ by taking the smallest one, i.e. by setting $$s(y) = \inf \; \pi^{-1}(y).$$ In the general case, the trick is to use the fact that for any compact metric space $$X$$ there is some compact set $$K \subset [0,1]$$ and a surjective continuous map $$f: K \to X$$ (in fact you can take $$K$$ to be the Cantor set). Then you can apply the previous case to the composition $$\pi \circ f$$.
• Thanks! The assumptions about the measure $\nu$ and the preimages of $\pi$ are needed for my application, but it's good to know they aren't needed for the section to exist. – Sophie MacDonald Mar 27 at 16:57
• My pleasure! Of course under these assumptions any two sections will agree $\nu$-almost everywhere, whereas in general one can't say much. – Dominique R.F. Mar 27 at 17:12
• Yes, and in fact my situation is that $\nu = \pi_* \mu$ where $\mu$ is a fully supported probability measure on $X$; this means that $s$ is also a left inverse of $\pi$ $\mu$-almost everywhere. Moreover, $\mu$ and $\nu$ are both invariant under the continuous action of a countable group $G$, and $\pi$ is $G$-equivariant; the assumptions thus imply that $s$ is also $G$-equivariant almost everywhere. – Sophie MacDonald Mar 27 at 17:28