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On a standard Borel (or Polish) space $X$, any probability measure $\mu$ is the push-forward of the uniform measure on $[0, 1]$ under some $f : [0, 1] \to X$.

This $f$ is not unique in general. Suppose that $f, f' : [0, 1] \to X$ have the same push-forward measure. Is there a measure-preserving $\alpha : [0, 1] \to [0, 1]$ such that $f = f' \circ \alpha $?

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  • $\begingroup$ Of course, this cannot work pointwise, so you probably wanted the equality to hold a.e. on $[0,1]$ ? $\endgroup$
    – Christian Remling
    Commented Oct 19, 2022 at 19:53
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    $\begingroup$ Atomless probability spaces where this is possible a very special. The main examples are uncountable products of the uniform measure and hyperfinite Loeb spaces. The book "Model Theory of Stochastic Processes" by Fajardo and Keisler discusses the property in detail. $\endgroup$
    – Michael Greinecker
    Commented Oct 19, 2022 at 21:16

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No: take $X = [0,1]$, $f$ the identity, and $f'(x) = 1-|2x-1|$. One must have $\alpha(y) \in \{y/2, 1-y/2\}$ for (almost) every $y$, but if we set $A = \{y\,:\, \alpha(y) = y/2\}$, $\alpha$ pushes Lebesgue measure onto the measure with density $2(1_{A/2} + 1_{(1-A^c)/2}) \not\equiv 1$.

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