# Uniform inverse-measure-preserving function for some class of measures

Working in the cantor space $2^\omega$. Giving two measurable spaces $(2^\omega, T, \mu)$ and $(2^\omega, T, \nu)$ an inverse-measure-preserving function $f:2^\omega \rightarrow 2^\omega$ is such that $\mu(A)=\nu(f^{-1}(A))$ for any borel set $A$.

Giving a class of measure $\mathcal{M}$, I wonder if one can find some conditions on $\mathcal{M}$ such that there exists a inverse-measure-preserving function $F$ between any measures on $\mathcal{M}$ and the Lebesgue measure :

i.e. there exists $F$ such that for any $\mu \in \mathcal{M}$ we have $\lambda(A)=\mu(F^{-1}(A))$

Does anyone know if there is some results in that direction, or any references I could look at ?

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Here $T$ is what, the natural Borel sigma-algebra? And the measures are probability measures? In my answer I assume these.
For this: $\mu(A)=\nu(f^{-1}(A))$, I might write $\mu = f(\nu)$ or maybe $\mu = f_*(\nu)$ and say that $\mu$ is the image of $\nu$ under $f$. Every probability measure on $2^\omega$ is an image of Lebesgue measure. Lebesgue measure is an image of a measure $\nu$ if and only if $\nu$ is atomless.
OK, we can change variables (except null sets) to get the following situation: $2^\omega$ is replaced by the square $[0,1] \times [0,1]$ and the map $f$ is the projection onto the first coordinate $[0,1]$. Now we want to know what are the measures on the square that project onto Lebesgue measure. Yes, indeed, there are lots of them.
Thank you very much for your answer, and sorry for the lack of precision. Yes I mean the natural Borel sigma-algebra, and yes I mean probability measures. The thing is that I would like to have the same function $f$ for several measures. As an example, I think the class of Bernoulli measures have that. If you take the function which spit a string in blocks of two bits and transform $01$ to $0$, $10$ to $1$ and $00$ and $11$ to the empty string, then I am pretty sure that lebesgue measure is the image of $\nu$ with the same function $f$ for any bernoulli measure $\nu$ – Archimondain Jun 13 '11 at 19:19
ps : What I call bernoulli measures are measures on $2^\omega$ such that $\mu(0)+\mu(1)=1$ and for any basic open $x_1 \dots x_n$, we have $\mu(x_1 \dots x_n))=\mu(x_1) \times \dots \times \mu(x_n)$ where each $x_i$ is either $0$ or $1$. – Archimondain Jun 13 '11 at 19:20