Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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 $\lim_{n\to\infty}\mu(c_n)\ge\alpha>0$.

share|improve this answer
    
Thank you, that's great. I would have never thought to apply König's lemma. –  Michael Greinecker Aug 29 '12 at 17:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.