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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 ?

Thanks in advance.

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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.

[added Jun 14]

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.

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  • $\begingroup$ 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 $\nu$ $\endgroup$
    – Archimondain
    Commented Jun 13, 2011 at 19:19
  • $\begingroup$ 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 $1$. $\endgroup$
    – Archimondain
    Commented Jun 13, 2011 at 19:20

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