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It is fairly well known that if $\mu$ and $\nu$ are nonatomic measures on the standard Borel spaces $(X,B)$ and $(Y,C)$ such that $\mu(X)=\nu(Y)$. If $X$ and $Y$ are uncountable, then there exists a Borel isomorphism $f : X \to Y$ such that $f_\# \mu=\nu$.

Keep in mind that if $(X,B)$ and $(Y,C)$ are measurable spaces, $f: X \to Y$ is a measurable map, and $\mu$ is a measure on $B$, then we write $f_\#\mu$ for the push forward measure on $C$, which is defined by $f_\#\mu(A) = \mu(f^{-1}(A))$ for all $A \in C$.

Is there a continuous analog for this? Specifically, considering that Borel isomorphisms need not be continuous, is there anything similar to the result above, but where $f$ is continuous?

Any insights or sources would be valuable.

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    $\begingroup$ If $X$ and $Y$ are Polish spaces with their Borel $\sigma$-algebra, then there exists always a finer Polish topology on $X$ that makes $f$ continuous. I can't think of more one could do. $\endgroup$
    – Michael Greinecker
    Commented Dec 28, 2021 at 9:37
  • $\begingroup$ True if $X$ and $Y$ are intervals (both closed or both open) on the real line and the support of $\mu$, $\nu$ is $X$, $Y$, respectively. For higher dimensional analogues one should look in geometric measure theory. $\endgroup$
    – user95282
    Commented Dec 28, 2021 at 13:21
  • $\begingroup$ @MichaelGreinecker Thank you for your comment. Do you have a source for that statement? $\endgroup$
    – O-Schmo
    Commented Dec 28, 2021 at 22:03
  • $\begingroup$ @O-Schmo This follows from Theorem 4.59 in the 2006 book "Infinite Dimensional Analysis" by Aliprantis and Border. $\endgroup$
    – Michael Greinecker
    Commented Dec 28, 2021 at 22:18
  • $\begingroup$ @user95282 I think I can prove this on my own, but do you have a source for that statement and/or a good geometric measure theory book? $\endgroup$
    – O-Schmo
    Commented Jan 17, 2022 at 23:50

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This is a very good (and also well studied) question, especially for homeomorphisms of measures.

For example, the Haar measures on the zero-dimensional compact groups $\mathbb Z_2^\omega$ and $\mathbb Z_3^\omega$ are not homeomorphic because their sets of values on clopen sets are distinct. More information on homeomorphisms of measures can found in this paper of Akin.

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