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Suppose that $(\Omega,\mathcal{A},\mu)$ is a $\sigma$-finite measure space of infinite measure and $T:\Omega\to\Omega$ a measure-preserving transformation with measurable inverse. Let be $\Omega_k\in \mathcal{A}$ an increasing sequence such that $\Omega_k\uparrow\Omega$ and $\mu(\Omega_k)<+\infty$ for all $k\in\mathbb{N}$.

Question 1: Given a set $A\in\mathcal{A}$, such that $\mu(A)>0$, is it true that the set $$E_k=\{\omega\in A; T^n(w)\notin A\ \forall n\in\mathbb{N}\ \text{and}\ T^{n_j}(w)\in \Omega_k \ \text{for some infinite sequence}\ (n_j(\omega)) \}$$ has zero measure ?

Question 2: If $T$ is not invertible is $\mu(E_k)=0$, in general ?

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  • $\begingroup$ Your set E depends on the value of k, right? Or do you want to merely require recurrence to some $\Omega_k$, where k which may depend on $\omega$? If E depends on k, so that we have sets $E_k$, do you want $E_k$ to have zero measure for some k? For all k? $\endgroup$
    – Vaughn Climenhaga
    Oct 30, 2010 at 2:09
  • $\begingroup$ Hi Vaughn, yes the set E depends on k. Perhaps I should mention in the question that I defined those sets because I am trying to use them to prove that almost all points are recurrent or goes to infinity, where a point $\omega$ goes to infinity it means that for any $k\in\mathbb{N}$, there is an $n_0\equiv n_0(k)$ such that for all $n\geq n_0$ we have $T^n(\omega)\notin \Omega_k$. $\endgroup$
    – Leandro
    Oct 30, 2010 at 2:25
  • $\begingroup$ I don't see it in the question now, but presumably $\mu$ should be an invariant measure? $\endgroup$
    – Vaughn Climenhaga
    Oct 30, 2010 at 2:35
  • $\begingroup$ Yes the measure is invariant. $\endgroup$
    – Leandro
    Oct 30, 2010 at 3:03

1 Answer 1

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If I remember my infinite ergodic theory correctly, any measure-preserving transformation $T$ of a $\sigma$-finite measure space $(\Omega,\mathcal{A},\mu)$ leads to a decomposition of $\Omega$ into a dissipative part $\Omega_d$ and a conservative part $\Omega_c$. (This notation is probably non-standard.) The dissipative part is the union of the wandering sets for $T$, where a set $E$ is wandering if it is disjoint from all its preimages $T^{-n}(E)$, $n\geq 0$. The conservative part is thee complement of the dissipative part.

On the dissipative part, everything "goes to infinity" in some sense (eventually leaves $\Omega_k$, for instance), while on the conservative part, the Poincare recurrence theorem holds. (Conservative measures are exactly those $\sigma$-finite measures for which the Poincare recurrence theorem still works.) I believe this establishes the dichotomy you want.

This may even work for transformations that don't preserve $\mu$, but only preserve the collection of null sets. I'd check a reference on that, though -- details of all of this are in Aaronson's book An Introduction to Infinite Ergodic Theory.

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  • $\begingroup$ You're very welcome -- it took me a little while to remember this, and it's worth remembering. $\endgroup$
    – Vaughn Climenhaga
    Oct 30, 2010 at 3:16

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