Measure preserving transformation that makes two partitions independent

Question

I am looking for a reference for the following result. I think it is well known but I haven't seen it written down anywhere.

Let $(X, \mathcal{B}, \mu)$ be a standard measure space and let $\mathcal{P}, \mathcal{P'}$ be two finite measurable partitions of $X$. Then there is a map $\varphi: X \to X$ which preserves the measure $\mu$ and such that the partitions $\mathcal{P}$ and $\varphi^{-1}\mathcal{P'}$ are independent with respect to $\mu$.

(Two partitions $\mathcal{P}$ and $\mathcal{Q}$ are independent with respect to $\mu$ if for any two cells $A \in \mathcal{P}$, $B \in \mathcal{Q}$, $\mu(A \cap B) = \mu(A)\mu(B)$.)

Answer 1

The property does not hold if the measure $\mu$ is atomic, so we assume that the given standard probability space $(X, \mathcal{B}, \mu)$ is nonatomic, whence it is isomorphic to he unit interval. Thus we may assume that $(X, \mathcal{B}, \mu)$ is the unit interval, endowed with the Lebesgue $\sigma$-algebra and measure. Let $\mathcal{P}=\{A_1,\ldots,A_n\}$ and $\mathcal{P'}=\{B_1,\ldots,B_m\}$. Denote $a_i:=\mu(A_i)$ for $i=1,\ldots, n$ and $s_k:=\sum_{i=1}^k a_i$ for $k=1,\ldots,n$. Now each $A_k$ is a standard measure space when endowed with the restrictions of the Lebesgue $\sigma$-algebra and of $\mu$ (see [1]). Thus, there exists a measure preserving transformation $f:[0,1] \to [0,1]$ such that $f^{-1}[s_{k-1}, s_k)=A_k$ up to measure zero. More precisely, $f:[0,1] \to [0,1]$ may be defined by $$\forall k \in\{1,\ldots n\}, \quad \forall x \in A_k, \quad f(x)=s_{k-1}+\mu\Bigl(A_k \cap [0,x] \Bigr) \,.$$

Next, define $\psi:[0,1) \to [0,1]$ $$\forall k \in\{1,\ldots n\}, \quad \forall x\in [s_{k-1},s_k), \quad \psi(x)= (x-s_{k-1})/a_k \,. $$ Observe that $\psi$ preserves Lebesgue measure. Finally, define $\varphi: [0,1] \to [0,1]$ by $\varphi:=\psi \circ f$. Then $\varphi$ preserves Lebesgue measure $\mu$, and the partitions $\mathcal{P}$ and $\varphi^{-1}\mathcal{P'}$ are independent with respect to $\mu$.

Some further details: The isomorphism $f$ converts $\mathcal P$ into a partitions into intervals. Then $\psi$ is a piecewise affine map, taking each interval $f(A_k)=[s_{k-1,}s_k]$ onto $[0,1]$, so for each $B_j \in \mathcal P'$, the preimage $\psi^{-1}(B_j)\cap [s_{k-1,}s_k]$ is a copy of $B_j$, translated and shrunk by a factor of $a_k$. This yields the required independence.

[1] Bogachev, Measure Theory, Vol 2, Springer, Proposition 9.4.10 page 284 https://diendantoanhoc.org/index.php?app=core&module=attach&section=attach&attach_id=12513

Answer 2

Here is a variant of the answer by Yuval Peres. His answer has the advantage that the same map $\varphi$ works for any choice of $\mathcal{P}'$ (the map $\varphi$ erases the information about $\mathcal{P}$). The variant below has the advantage that the $\varphi$ it provides is bijective.

Suppose $\mathcal{P}=\{A_1,A_2,\ldots,A_n\}$ and $\mathcal{P}'=\{B_1,B_2,\ldots,B_n\}$. Following Yuval's answer, there exist:

• An isomorphism $f\colon X\to[0,1)$ and $0=s_0\leq s_1\leq \cdots \leq s_n\leq 1$ such that $f^{-1}[s_{i-1},s_i)=A_i$ (modulo null sets) for $i=1,2,\ldots,n$.
• An isomorphism $g\colon X\to[0,1)$ and $0=t_0\leq t_1\leq \cdots \leq t_m\leq 1$ such that $g^{-1}[t_{j-1},t_j)=B_j$ (modulo null sets) for $j=1,2,\ldots,m$.

(Note: To be clear, an isomorphism refers to a measure-preserving bijective bi-measurable $f\colon X_0\to I_0$ where $X_0\subseteq X$ and $I_0\subseteq[0,1)$ are measurable, and $X\setminus X_0$ and $[0,1)\setminus I_0$ are null sets.)

Let $Q$ and $Q'$ be the partitions of $[0,1)$ given by $(s_i)_i$ and $(t_j)_j$ respectively. There exists an isomorphism $\beta:[0,1)\to[0,1)$ such that $Q$ and $\beta^{-1}Q'$ are independent.

The construction is based on the standard method of sampling two independent random variables with prescribed discrete distributions using a single uniformly distributed random number from $[0,1)$.

Namely, consider the two partitions $Q_0:=\big(s_{i-1}+(s_i-s_{i-1})t_{j-1}\big)_{i,j}$ and $Q'_0:=\big(t_{j-1}+(t_j-t_{j-1})s_{i-1}\big)_{i,j}$. Note that $Q_0$ is a refinement of $Q$, and $Q'_0$ is a refinement of $Q'$. Furthermore,

• There is a natural one-to-one correspondence between the blocks of $Q_0$ and $Q'_0$. Namely, \begin{gather} \big(s_{i-1}+(s_i-s_{i-1})t_{j-1},s_{i-1}+(s_i-s_{i-1})t_j\big) \\ \updownarrow\\ \big(t_{j-1}+(t_j-t_{j-1})s_{i-1}, t_{j-1}+(t_j-t_{j-1})s_i\big) \end{gather} Note that these two parts have the same measure $(s_i-s_{i-1})(t_j-t_{j-1})$.

It is easy to use the latter one-to-one correspondence to construct an isomorphism $\beta$ with the desired property.

Define $\varphi:X\to X$ as $\varphi:=g^{-1}\circ\beta\circ f$.