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.


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 $\lim_{n\to\infty}\mu(c_n)=0$.


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.


Let $\Pi=\bigcup_n\Pi_n$, and for $c,d\in\Pi$ write $c\le d$ iff $c\supseteq d$; $\langle\Pi,\le\rangle$ is then a forest in the usual sense, with levels $\Pi_n$.

For $n\in\omega$ let $\alpha_n=\max\{\mu(c):c\in\Pi_n\}$, and suppose that $\alpha=\inf\{\alpha_n:n\in\omega\}>0$. Let $C=\{c\in\Pi:\mu(c)\ge\alpha\}$; note that if $d\le c\in C$, then $d\in C$, so $C$ is a level-preserving sub-forest of $\Pi$. $C\cap\Pi_0$ is finite, so at least one of the component trees of the forest $C$ has infinite height; call it $C_0$. Now just apply König’s lemma to $C_0$ to conclude that $C$ has an infinite branch $\langle c_n:n\in\omega\rangle$: clearly $\lim_{n\to\infty}\mu(c_n)\ge\alpha>0$.

  • 1
    $\begingroup$ Thank you, that's great. I would have never thought to apply König's lemma. $\endgroup$ – Michael Greinecker Aug 29 '12 at 17:26

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