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 ?
Best Answer
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).