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?

  • 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

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.

  • $\begingroup$ 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

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