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I'm quite embarrassed not to know the answer to this question, but I think someone else will.

Suppose that $(X, T)$ is a topological dynamical system, and $\mathcal{E}$ is the collection of ergodic $T$-invariant Borel probability measures.

It's well-known that the measures in $\mathcal{E}$ are "pairwise mutually singular" in the sense that $\mu \perp \nu$ for all $\mu \neq \nu \in \mathcal{E}$. But must $\mathcal{E}$ be "strongly mutually singular" in the sense that there exists $f: \mathcal{E} \rightarrow \mathcal{B}(X)$ where $\mu(f(\mu)) = 1$, and $\nu(f(\mu)) = 0$ for all $\mu \neq \nu \in \mathcal{E}$?

This immediately holds if $\mathcal{E}$ is countable (upon taking intersections), and slightly less immediately holds for all $\mathcal{E}$ if $X$ has a countable basis (the proof of Ian Morris from "Strongly mutually singular" families of measures, and the set of ergodic measures basically does this, and it also appears in the textbook of Oliveira and Viana).

But I actually don't immediately see a proof for the most general case of arbitrary $\mathcal{E}$ and $X$. My feeling is that it should hold, since it seems related to ergodic decomposition, on which I've never heard of a necessary restriction like countable basis for $X$.

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    $\begingroup$ If $C(X)$ is separable, say then you can take $(g_k)$ to be a dense subset and set $f(\mu)=\{x\colon \frac 1N\sum_{n=0}^{N-1}g_k(T^nx)\to\int\g\,d\mu\}$. But this is essentially repeating what Ian Morris said. What do you mean by a topological dynamical system? Are you just assuming that $X$ is compact and $T$ is continuous? or is there a Hausdorff assumption also? (in that case $C(X)$ is separable). BTW: isn't there a standard assumption that $(X,\mu)$ is a Lebesgue space for ergodic decomposition? $\endgroup$
    – Anthony Quas
    Commented Nov 5, 2022 at 7:45
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    $\begingroup$ I won't be surprised if it is something basic like this; I'm bad at remembering these basic topology things since I always work in nice spaces. 1. Of course I think you're right about Lebesgue; I didn't consider the fact that this implicitly restricts the $\sigma$-algebra as well! (It has a countable generating set) 2. That being said, I'm still curious if this fails in the non-Lebesgue case. I guess you need compact, Hausdorff and second countable for $C(X)$ separable, right? So maybe the question is: if $X$ is not necessarily 2nd ctable, does strong mutual singularity still hold? $\endgroup$
    – Ronnie Pavlov
    Commented Nov 5, 2022 at 14:05
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    $\begingroup$ Followup since I ran out of room: I guess (from Googling) that $[0, \omega_1]$ with the order topology is compact, Hausdorff, but not 2nd countable. I think $[0,1]^I$ for uncountable $I$ and the product topology is another one, but didn't check details. $\endgroup$
    – Ronnie Pavlov
    Commented Nov 5, 2022 at 14:11
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    $\begingroup$ Oops! I guess that's what happens if I think I remember "facts" rather than proofs. $\endgroup$
    – Anthony Quas
    Commented Nov 5, 2022 at 17:24
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    $\begingroup$ Is this an answer or an intuition, i.e. do you have a counterexample for a non-Lebesgue space? I've been unable to construct one. $\endgroup$
    – Ronnie Pavlov
    Commented Dec 8, 2022 at 15:29

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I think the answer is "no" for silly reasons. Consider $X = \{0,1\}^{[0,1]}$ with the product topology and $T$ the identity. Then for every $x \in X$, the "delta-measure" $\delta_x$ is ergodic. I used quotes because singletons are not measurable in the product $\sigma$-algebra, but the measures can still be defined and are all distinct.

But now there are $2^c$ measures and the product $\sigma$-algebra only has (I think?) $c$ sets. So a function $f$ as in my question can't even be injective, trivially precluding strong mutual singularity.

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    $\begingroup$ Do you want the product $\sigma$-algebra or the Borel $\sigma$-algebra? $\endgroup$
    – John Griesmer
    Commented Nov 6, 2022 at 0:51
  • $\begingroup$ Yeah I literally just came here to post that I just realized the Borel $\sigma$-algebra for the product topology is actually not the same as the product $\sigma$-algebra here (I hate weird spaces). So I guess this is an example to show that the strong mutual singularity need not hold for an arbitrary $\sigma$-algebra, but it's not an answer to my original question. So I won't accept it :) $\endgroup$
    – Ronnie Pavlov
    Commented Nov 6, 2022 at 0:56

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