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Let $(X, \mathcal{B}, \mu, S)$ and $(Y, \mathcal{C}, \nu, T)$ be invertible probability-measure-preserving systems, with a measurable factor map $\pi: X \to Y$, i.e. $\pi \circ S = T \circ \pi$. Suppose that there is some $N \in \mathbb{N}$ such that $\nu$-almost every $y \in Y$ has at most $N$ preimages under $\pi$, i.e. $\# \pi^{-1}(\{y\}) \leq N$. If necessary, I'm happy to assume that $X$ and $Y$ are compact metric spaces, $\mu$ and $\nu$ are Borel, and $\pi, S, T$ are continuous.

I'm trying to show that $\pi$ preserves measure-theoretic entropy, i.e. $h(\nu) = h(\mu)$. I'd like to apply the Abramov-Rokhlin entropy formula, which expresses the entropy of a skew product transformation as the sum of the entropy of the base and the entropy of the fibre. To do this, I'd like to realize $X$ as the direct product $Y \times S$ (or a subset of this), with $F = \{ 1, \dots, N \}$, together some probability vector $\mathbf{p} \in \mathbb{R}^N$, such that $\mu = \nu \times \mathbf{p}$. I'd then try to realize $S$ as $S(y,k) = (T(y), \alpha(y)(k))$ where for each $y$, $\alpha(y)$ is an invertible measure-preserving transformation of $(F, \mathbf{p})$.

The problem is that if the $\alpha$'s give a transitive action of $S_N$, this would require that $\mathbf{p}$ be uniform. In particular, it seems that the fiber cardinality would have to be constant, not just bounded. Even forgetting about the measure, this is an issue, because the $\alpha$'s would have to be bijections.

Two ideas I've played with:

  • The number of preimages $\# \pi^{-1}(\{y\})$ is constant along the orbit of $y$, so I could try partitioning $X$ and $Y$ into at most $N$ disjoint subsystems on which $\pi$ is constant-to-one, then applying the Abramov-Rokhlin formula separately on each one, but I don't know how to show that this decomposition would be measurable.

  • Bögenschutz and Crauel (1990) have a generalization of the Abramov-Rokhlin formula for a skew product transformation (with the notation above) in which we no longer require a fixed measure on the fiber, but only that $\mu$ be invariant with fixed $Y$-marginal $\nu$. This might solve the problem by letting me add a null set of fixed points to $X$, for instance, to make $\pi$ constant-to-one.

I'm also hoping to apply this in the case where $S$ and $T$ are really actions of a countable amenable group $G$, using Ward and Zhang's generalization of the Abramov-Rokhlin formula. I haven't worked through the proofs to see if the Bögenschutz-Crauel and Ward-Zhang generalizations are compatible with each other.

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    $\begingroup$ Jisang Yoo's paper: arxiv.org/abs/1612.08648 has a lemma (Lemma 3.1) showing measurability of the decomposition by number of pre-images. $\endgroup$ – Anthony Quas Mar 31 '20 at 17:18
  • $\begingroup$ Thanks! It looks like this decomposition may not quite be measurable, but it is universally measurable (in particular, it becomes measurable if you replace $\mathcal{B}$ and $\mathcal{C}$ by their completions with respect to $\mu$ and $\nu$, respectively), which is enough. Universal measurability will actually help elsewhere in this project, too, so I'm glad to have learned about it here. $\endgroup$ – Sophie MacDonald Mar 31 '20 at 17:53
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One should always keep in mind that the natural substrate of ergodic theory is Lebesgue-Rokhlin (aka Lebesgue or standard) measure spaces which enjoy a lot of properties not necessarily present in general measure spaces. One of these properties is an explicit description of homomorphisms of such spaces obtained by Rokhlin in his 1949 paper, which, in particular, implies measurability of the cardinality of the preimages. Therefore, in your situation the number of preimages is constant along the ergodic components, and it is an easy exercise to check that the entropy does not decrease for quotient maps of this kind.

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  • $\begingroup$ Thank you for the suggestion to look at Rokhlin's paper! The idea I was missing was similar to the one in my reply to Anthony Quas's comment on the question: work with a complete $\sigma$-algebra, rather than with Borel sets only. $\endgroup$ – Sophie MacDonald Apr 1 '20 at 23:24

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