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.



Existence of rearrangements of functions in $L^p([0,1])$ when given a measure preserving map
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).


