# a version of Poincaré recurrence

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 $$m (because $$\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.
• Would you mind giving me a counterexample with $n \leq \frac{1}{\mu(A)}$ ? – Reza Dec 13 '19 at 22:42
• Sorry, in the beginning I was forgetting $0$. Since it is also there, the statement is correct. I have updated my answer accordingly. – John B Dec 13 '19 at 23:28
• PS: That's why it is irrelevant whether it is $1/\mu(A)$ or something like $k+1/\mu(A)$ for some $k$. – John B Dec 14 '19 at 16:18