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Given a funtion $f \in L^p([0,1])$ (take $p=\infty$ if you'd like), and also a measure preserving map $s:[0,1] \to [0,1]$ (meaning $s$ pushes Lebesgue measure forward to itself) I would like to know if there exists some $f^{\ast} \in L^p([0,1])$ such that $f^{\ast} \circ s = f$. If $s$ is invertible this is of course obvious but measure preserving maps need not be invertible (although must be onto).

Recall that given $f$ there exists a monotone rearrangement of $f$, and a measure preserving map $t:[0,1] \to [0,1]$ which yields this rearrangement.

However my question is in some sense the reverse question (with no monotonicity added).

Somehow it seems intuitive that such a map should exist but I'm not able to prove it directly. It seems like something which may be well known however.

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Consider $$s(x) = \begin{cases}2x, & x \in [0,1/2] \\\ 2x-1,&x \in ]1/2,1]\end{cases}.$$ Then for an arbitrary $f^{\ast}$, we have $$(f^{\ast}\circ s)(x+1/2) = (f^{\ast}\circ s)(x)$$ for all $x \in ]0,1/2]$. Certainly not all functions $f \in L^p([0,1])$ possess this property.

Also, measure preserving maps need not be strictly onto (although the image has to have full measure).

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You can characterize the functions that are of the form $g\circ s$. They are precisely the $s^{-1}\mathcal F$-measurable functions.

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