Measure-preserving maps on probability spaces

A professor posed me a problem a few days ago, and I have not been able to find an answer to it.

Let $(X_1,\Sigma_1,\mu_1)$ and $(X_2,\Sigma_2,\mu_2)$ be probability spaces. Suppose the following conditions hold.

1. $\mu_1$ is atomless, i.e., $\not\exists A\in\Sigma_1$ such that $\mu(A)>0$ and for any measurable $B\subset A$, $\mu(B)=0$.

2. $X_2$ is a complete, separable and metrizable topological space. (I think these spaces are called Polish spaces, although I might be mistaken.)

3. $\Sigma_2$ is a Borel $\sigma$-algebra on $X_2$.

Does there exist a measure-preserving map $f:(X_1,\Sigma_1,\mu_1)\rightarrow(X_2,\Sigma_2,\mu_2)$?

I really appreciate it if someone is able to provide a rough proof sketch (with hints) and provide a reference for this result. Thank you!

• $\mu_1$ is atom-less, what about $\mu_2$? E.g. take $X_1=X_2=[0,1]$, $\mu_1$=Lebesgue on $[0,1]$, $\mu_2=\delta_0$... $f([1/3,1/2])$ has no way to have $\mu_2$-measure $1/6$... – Milly Oct 20 '14 at 17:21
• @Milly, I think you misunderstand what measure preserving means. It means, $\mu_1 (f^{-1}(B)) = \mu_2(B)$. In particular, for your example let $f$ be the constant 0 function. – Jason Rute Oct 21 '14 at 3:13
• oh, you're right. In particular in this case you would have $\mu_2=f_\sharp \mu_1$, i.e. $f$ is a transport map between $\mu_1$ and $\mu_2$, so the problem is existence of optimal maps... – Milly Oct 21 '14 at 3:16
• It is true. As for a hint, consider the case where the first space is [0,1]. Construct your $f$ in stages. At each stage partition the second space into parts extending the previous partition. Can you think how to construct $f$ to be measure preserving for this partition? I think it is due to Caratheodory, but I might be wrong. – Jason Rute Oct 21 '14 at 3:24
• Oh, sorry, I confused "transport plan" with "optimal transport plan", thanks for the remark. – Milly Oct 21 '14 at 3:43