MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 4 down vote accepted

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).

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.