Learning about Lebesgue-Rohlin spaces is prominently on my to-do-list, so I'm reading Fundamentals of measurable dynamics by Daniel Rudolph, were I'm stuck on an exercise.
Framework
There is a nonempty set $X$ and a sequence $T=(\Pi_0,\Pi_1,\ldots)$ of partitions of $X$ such that each partition is finite, for every two points in $X$ there is an index $n$ such that the two points lie in different cells of $\Pi_n$, and $\Pi_{n+1}$ is finer than $\Pi_n$ for all $n$. We call this sequence a tree. That $\Pi_{n+1}$ is finer than $\Pi_n$, means that every cell in $\Pi_n$ is a union of cells in $\Pi_{n+1}$.
There is a finitely additive probability measure $\mu$ on the algebra generated by $\bigcup_n\Pi_n$. A chain in the tree is a sequence $(c_0,c_1,\ldots)\in\prod_n \Pi_n$ such that $c_n\supseteq c_{n+1}$ for all $n$. We say that $\mu$ is atomless if for every chain $(c_0,c_1,\ldots)$, we have $\lim_{n\to\infty}\mu(c_n)=0$.
Question
The exercise I have trouble with requires us to show the following:
If $\mu$ is atomless, then $\lim_{n\to\infty}\mu(c_n)=0$ uniformly over all chains.
It is relatively easy to see that this is equivalent to showing that for each $\epsilon>0$, there exists an $n$ such that for all $c\in\Pi_n$ we have $\mu(c)<\epsilon$, but I don't know what more I can do.