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Apr 3, 2021 at 4:06 comment added erz yes, $A$ is arbitrary measurable, but the whole point is that $h$ is independent of $A$. Check out Nishiura - Absolute Measurable Spaces, section 3.4
Apr 3, 2021 at 2:57 comment added Nate River Thank you, that seems like it could be relevant indeed. Just to clarify, here A is an arbitrary measurable set, and the choice of h in general depends on A?
Apr 3, 2021 at 2:43 comment added erz There is Ulam-Oxtoby theorem that says that if two measures $\mu$ and $\nu$ on $[0,1]$ are non-atomic, and every open set if of positive measure, then there is a homemorphism $h$ such that $\mu(A)=\nu(h(A))$. I cannot quite understand if there is any relation between this condition and topological conjugacy, but perhaps it is relevant
Apr 3, 2021 at 1:08 history asked Nate River CC BY-SA 4.0