0

Does anyone know where I find the proof of the following version of Poincaré recurrence:

Let $(X, \mathcal{B}, \mu, T)$ be a measure-preserving system with $\mu(X)=1$ only assumed to be a finitely additive measure ($\mu(∅) = 0$ and $\mu(A ∪ B) = \mu(A) + \mu(B)$ for any disjoint elements $A$ and $B$), and let $A \in \mathcal{B}$ have $\mu(A) >0$. Then there is some positive $n \leq \dfrac{1}{\mu(A)} $ for which $\mu(A ∩ T^{−n}A) > 0$

Ilovemath
  • 2,921
  • You are missing a statement on the measure of the full space. It should be $μ(X)=1$. Now how large can a sequence $A,TA,..., T^kA$ be so that all its elements are pairwise "disjoint in measure"? The argument for the pre-images should be similar. – Lutz Lehmann May 02 '19 at 22:03
  • $\mu$ is finite measure, it does not have to be probability. – Ilovemath May 02 '19 at 22:07
  • Then it should be $n \le \frac{μ(X)}{μ(A)}$, as rescaling the measure does not influence the geometry of the problem. – Lutz Lehmann May 02 '19 at 22:28
  • got it. At some point in the proof must use Kac's Theorem – Ilovemath May 02 '19 at 23:21
  • more importantly do you know a book that has proof of this theorem? – Ilovemath May 02 '19 at 23:23
  • Try by contradiction: suppose that $$\mu(A\cap T^{-n}A)=0\text{ for all }n\leq 1/\mu(A).$$ Then, because $\mu$ is $T$-invariant, $$\mu(T^{-m}A\cap T^{-n}A)=\mu(A\cap T^{-n+m}A)=0,$$ for any $m<n$. Hence, $$\mu\left(\bigcup_{k=0}^{1/\mu(A)}T^{-k}A\right)=\sum_{k=0}^{1/\mu(A)}\mu(A)>1.$$ – defacto Aug 10 '23 at 16:05

0 Answers0