1
$\begingroup$

In "Ergodic theory with a view towards number theory" we are asked to show Rohlins lemma holds for aperiodic atomless invertible measure preserving systems.

Not only I can't find a proof, I even don't understand why the following is not a counterexample.

I will build a aperiodic atomless invertible measure preserving system satisfying that there is no nonempty $E$ with $E,T(E)$ disjoint which will cause a contradiction.

First I will give a failed counter-example that will fail being atomless-

A copy of $Z$ with the shift map, the only measureable sets being everything or nothing. Call this the trivial sigma algebra.

To upgrade this to being atomless we consider $[0,1]\times Z$, the sigma algebra is the product sigma algebra of the standard Lebesgue and the trivial sigma algebra. $T$ works by shifting the $Z$ part.

This clearly preserves measure, is atomless since we just think of this as $[0,1]$, is aperiodic, and clearly $E,T(E)$ can't be disjoint.

What am I missing?

$\endgroup$
  • 1
    $\begingroup$ You did not specify the measure. After that you need to verify that you have a Lebesgue space. $\endgroup$ – John B Mar 26 at 0:20
1
$\begingroup$

You are missing a hypothesis.

In Rokhlin's Lemma, the hypothesis is that you are working with a standard measure space, which your counterexample is not.

$\endgroup$
  • $\begingroup$ Thanks! I think this is not mentioned in the book, and indeed the lemma proven in the theorem (i.e not the exercise I'm talking about) doesn't need this- if you have a ergodic invertible measure preserving system that is atomless, it satisfies the lemma. $\endgroup$ – Andy Mar 26 at 3:26
  • $\begingroup$ @Andy I regret that this is exactly what I said in my comment, a bit before then the answer (which follows my comment). And it remains as I said: first you need to defined the measure, after which you need to show that you have a Lebesgue space. So, it is not possible to discuss what is in the answer before that. Is it obvious for any of you that it is not a Lebesgue space for any measure on the shift? :) $\endgroup$ – John B Mar 26 at 13:11
  • $\begingroup$ @JohnB I implicitly put the product measure in the example. I saw your comment and the answer at the same time (and upvoted both :) , thanks for it!). I plan on reading lebesgue spaces (which are the same as standard measure space if I understand correctly). I'll find a random internet source unless you have a good one $\endgroup$ – Andy Mar 26 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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