Constructing measure preserving maps between non-atomic measures Suppose $(\mu, X,\Sigma)$ and $(\mu^\prime, X^\prime, \Sigma^\prime)$ are non-atomic probability measures. Is it always possible to construct a measure preserving map between the two spaces? 
(If not, suggestions of other simple conditions that would guarantee the possibility would be welcome.)
 A: Of course this is not true in general. I think that that a necessary and sufficient condition is the following:
There exist partitions $\{X_n : n < \omega\}$, $\{Y_n : n < \omega\}$ of $X, Y$ respectively such that for each $n$, $\mu(X_n) = \mu'(Y_n)$, and the measure algebras of $\mu \upharpoonright X_n, \mu' \upharpoonright Y_n$ are homogeneous and have the same Maharam type.
If you do not understand this answer, try reading about Maharam's theorem.
A homogeneous example where this doesn't hold is obtained by looking at product measure spaces of the form $2^X$ where $2 = \{0, 1\}$ has uniform measure. If you take $X, Y$ of different cardinalities then these space aren't measure isomorphic. Although it is possible that both $2^X$ and $2^Y$ have same cardinality.
Greg Martin's comment isn't correct. There is a measure preserving bijection from $[0, 1]$ to $[0, 1]^2$ since are both isomorphic to $2^{\omega}$.
A: If there is an inverse measure preserving function from $X$ to $Y$, then the measure algebra of $Y$ completely embeds into that of $X$ so that the Maharam type (density of the measure algebra) of the associated measure algebra of $Y$ is at most that of $X$. Recall that the Maharam type/density of a measure algebra $(B, \mu)$ is just the least size of a dense set in the associated metric space $(B, d)$ where $d(x, y) = \mu(x \Delta y)$.
Suppose $\kappa_1 < \kappa_2$ are two infinite cardinals. Then there is no inverse measure preserving map $f: (2^{\kappa_1}, \Sigma_1, \mu_1) \to (2^{\kappa_2}, \Sigma_2, \mu_2)$ ($(2^{\kappa_i}, \Sigma_i, \mu_i)$ is product measure space with uniform measure on 2) - Inverse measure preserving means for every $A \in \Sigma_2$, $\mu_{1}(f^{-1}[A]) = \mu_2(A)$. Suppose not. Let $N_i$ denote the null ideal over $2^{\kappa_i}$. Define $F: \Sigma_2 / N_2 \to \Sigma_1 / N_1$ by $F(A / N_2) = f^{-1}[A] / N_1$. $F$ is an injective boolean homorphism which means that the density of the measure algebra $\Sigma_2 / N_2$ is no more that that of $\Sigma_1 / N_1$. But this is impossible since $\kappa_2 > \kappa_1$. 
Remark: In case the target space is $[0, 1] \cong 2^{\omega}$, inverse measure preserving maps can always be constructed.
