finding global bound for measurable partitions

Question

Let $(X, \mathcal{F},\mu)$ be a probability space and $\xi , \eta \subset \mathcal F$ be two measurable partitions, i.e. $\xi$ is finite, $\mu (A\cap B)=0$ for all $A,B \in \xi$, and $$\mu\left( X \setminus \bigcup_{C \in \xi} C \right) =0$$and analogously for $\eta$. Also suppose that $\xi$ and $\eta$ have the same cardinality.
Let $\sigma : \xi \to \eta$ be a bijection chosen in a way such that $$\chi(\sigma) := \sum_{C \in \xi} \mu\big(C \cap \sigma(C)\big)$$is maximal under all possible bijections.
Is there a number $r>0$, only dependent on the cardinality of $\xi$, such that $$\sum_{C \in \xi} \mu\big(C \cap \sigma(C)\big)=\chi(\sigma)>r?$$

Answer 1

Yes. Suppose the cardinality of $\xi$ is equal to $k$. Let $A\in\xi$ be the set maximizing $\mu(A)$, so that $\mu(A)\ge 1/k$. Consider all the sets $B_i\in\eta$ such that $B_i\cap A\neq\varnothing$. Clearly, there are at most $k$ such sets. Since $\eta$ is a partition, we have $$ \mu(A)=\sum_i \mu(B_i\cap A).$$ Let $B_{i_0}$ be the set maximizing $\mu(B_i\cap A)$, so that we have $$\mu(B_{i_0}\cap A)\ge \frac{\mu(A)}{k}\ge \frac{1}{k^2}.$$ Let $\sigma$ be any bijection such that $\sigma(A)=B_{i_0}$. Then, $\chi(\sigma)\ge 1/k^2$.

This bound is probably far from optimal.