I've been trying to prove the following version of Poincaré recurrence with a weaker hypothesis (finite additivity in place of countable additivity for the measure) and with a stronger conclusion (a bound on the return time). here is the problem:

Let $(X,\mathcal{B}, μ, T)$ be a measure-preserving system with $μ$ only assumed to be a finitely additive measure, and let $A ∈ \mathcal{B}$ have $μ(A)> 0$. Show that there is some positive $n \leq \frac{1}{\mu(A)}$ for which $μ(A ∩ T^{-n}A) > 0$.

I think I could prove this, but I have some questions.
Here is my proof:

At first, I used this lemma:
Lemma Suppose $D$ is an algebra of subsets of $X$, $\mu$ is a finitely additive measure on $D$, $A_1,\dots,A_n\in D$ and $\sum_{j=1}^n\mu(A_j)>\mu(X)$. Then there exist $j,k$ with $j\ne k$ such that $\mu(A_j\cap A_k)>0$.
Proof : If $\mu(A_j\cap A_k)=0$ for all $j\ne k$ then additivity shows that $\mu(\bigcup A_j)>\mu(X)$, contradiction.

Say $T:X\to X$ is measurable if $T^{-1}(A)\in D$ for every $A\in D$. The result we're calling Poincare recurrence follows easily:

Cor Suppose $D$ is an algebra of subsets of $X$ and $\mu$ is a finitely additive measure on $D$ with $\mu(X)<\infty$. Suppose $T:X\to X$ is measurable and measure-preserving. If $A\in D$ and $\mu(A)>0$ there exists $n\ge 1$ with $\mu(A\cap T^{-n}(A))>0$.
proof: If on the other hand $\mu(A\cap T^{-n}A)=0$ for every $n\ge0$ then I want to see that $\mu(T^{-n}(A)\cap T^{-m}(A))=0$ for all $n\ne m$; hence the corollary follows from the lemma.

My questions :

1 - How can I conclude this in the proof of Cor :
If $\mu(A\cap T^{-n}A)=0$ for every $n\ge 0$ then $\mu(T^{-n}(A)\cap T^{-m}(A))=0$ for all $n\neq m$

2- Does this proof really work? I mean Does the Cor provide all hypothises of lemma ? For example, $A_1,\dots,A_n\in D$ and $\sum_{j=1}^n\mu(A_j)>\mu(X)$.

3-In the main question I have to show that $ n \leq \frac{1}{\mu(A)}$, How can I do this ?


Note that if $\mu(A\cap T^{-n}A)=0$ for all $n\le 1/\mu(A)$, then also $$ \mu(T^{-m}A\cap T^{-n}A)=\mu(T^{-m}(A\cap T^{-n+m}A))=\mu(A\cap T^{-n+m}A)=0 $$ whenever $m<n\le1/\mu(A)$ (because $\mu$ is invariant). Therefore, $$ \mu\left(\bigcup_{n\le1/\mu(A)}T^{-n}A\right)=\sum_{n\le1/\mu(A)}\mu(T^{-n}A)=\sum_{n\le1/\mu(A)}\mu(A)>1, $$ again because $\mu$ is invariant (we are also including $n=0$ and thus why the sum is larger than $1$). This contradiction gives the claim.

A minor comment is that "with a stronger conclusion (a bound on the return time)" is not correct since the same holds in the usual recurrence theorem (and nothing changes in the former argument).

  • $\begingroup$ Would you mind giving me a counterexample with $ n \leq \frac{1}{\mu(A)}$ ? $\endgroup$ – Reza Dec 13 '19 at 22:42
  • $\begingroup$ Sorry, in the beginning I was forgetting $0$. Since it is also there, the statement is correct. I have updated my answer accordingly. $\endgroup$ – John B Dec 13 '19 at 23:28
  • $\begingroup$ Thank you, but the last question. What is the contradiction? Shall I take the space a probabiliy space ? I didn't see this in the main question. $\endgroup$ – Reza Dec 14 '19 at 3:55
  • $\begingroup$ And why is the sum larger than 1 ??? $\endgroup$ – Reza Dec 14 '19 at 4:09
  • 1
    $\begingroup$ PS: That's why it is irrelevant whether it is $1/\mu(A)$ or something like $k+1/\mu(A)$ for some $k$. $\endgroup$ – John B Dec 14 '19 at 16:18

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