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Let $(X, \mathscr B_X, \mu, T)$ be a measure-preserving system such that the condition in the title is satisfied for all $n$ larger than $N \in \Bbb N$. How can we conclude that $A$ is trivial? Any help\hints are appreciated.

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  • $\begingroup$ What happens if $B=X$? $\endgroup$
    – user133281
    Aug 30, 2016 at 22:00
  • $\begingroup$ Doesn't that just read "$\mu(A) = \mu(A)$"? $\endgroup$ Aug 30, 2016 at 22:06
  • $\begingroup$ So then the conclusion does not follow, right? $\endgroup$
    – user133281
    Aug 30, 2016 at 22:09
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    $\begingroup$ If $B=X$, then the given condition reads $\mu(A) = \mu(A)$, which is satisfied for any $A$. It does not necessarily hold that $\mu(A) \in \{0,1\}$. $\endgroup$
    – user133281
    Aug 30, 2016 at 22:52
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    $\begingroup$ The notation $\mu(B)$ suggests that $B$ is an element of the $\sigma$-algebra, while the notation $(X, B, \mu, T)$ suggests that $B$ is the $\sigma$-algebra. Clarification is needed here. $\endgroup$
    – Math1000
    Aug 30, 2016 at 23:18

2 Answers 2

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Here is a more elementary proof.

Take any $B$ with $0 < \mu(B) < 1$. Let $N_1$ be such that $\mu(B \cap T^{-N_1} B) = \mu(B)^2$. With $N_1,\dots,N_k$ chosen, choose $N_{k+1} > N_k$ such that $$\mu\left(\bigcap_{j \in \Sigma_k} T^{-N_j}B \cap T^{-N_{k+1}} B\right) = \mu(B)^{1+|\Sigma_k|}$$ for each $\Sigma_k \subseteq \{0,N_1,\dots,N_k\}$. Let $$A = \bigcup_{k \ge 1} [B \cap T^{-N_1} B \cap \dots \cap T^{-N_{k-1}}B \cap T^{-N_k}B^c \cap T^{-N_{k+1}}B].$$ Then, since the union is a disjoint one, $$\mu(A) = \sum_{k \ge 1} \mu(B)^{k+1}(1-\mu(B)) = \mu(B)^2,$$ so $$\mu(A)\mu(B) = \mu(B)^3,$$ while for $j \ge 3$, $$\mu(A \cap T^{-N_j} B) = \sum_{k=1}^{j-2} \mu(B)^{k+2}(1-\mu(B))+\mu(B)^j(1-\mu(B))+\sum_{k=j+1}^\infty \mu(B)^{k+1}(1-\mu(B))$$ $= \mu(B)^3+(1-\mu(B))^2\mu(B)^j.$

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    $\begingroup$ Will look at this over the weekend! Appreciate the response $\endgroup$ Sep 7, 2019 at 3:31
  • $\begingroup$ Could you explain the intuition behind this calculation? I can check that each step is correct but I am totally lost and have no idea why we are doing this. $\endgroup$
    – Bach
    Jan 30, 2020 at 0:53
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    $\begingroup$ @Bach well, first I should say that this solution took me a long time to find. I first saw this problem around 3 years ago now in ergodic theory with a view towards number theory and thought about it on and off for a year until I looked it up and was a bit annoyed that there was no elementary solution. and I've thought about it every once in a while for the past 2 years since then. in any event, my initial goal was to prove it for Bernoulli shifts. Consider the measure space modelling a bi-infinite coin flip with $B$ being the event that the $0^{th}$ flip is heads. Let me know if $A$ has $\endgroup$ Jan 30, 2020 at 1:06
  • $\begingroup$ some intuition now (if I recall correctly, it should). $\endgroup$ Jan 30, 2020 at 1:06
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    $\begingroup$ @mathworker21 Thank you! It's very helpful to learn this! $\endgroup$
    – Bach
    Jan 30, 2020 at 4:12
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Actually, there is no non-trivial dynamical system $(X,\mathcal B,\mu)$ for which $\mu(A\cap T^{-n}B)=\mu(A)\mu(B)$ holds eventually for any $A$ and $B$. This is a consequence of Baire theorem. Indeed, fix a measurable set $B$ and consider the complete pseudo-metric space $(\mathcal B,\rho)$ where $\rho(A,B)=\mu(A\Delta B)$. The map $F_n\colon A\mapsto \mu(A\cap T^{-n}B)-\mu(A)\mu(B)$ is continuous. Defining $F_N:=\bigcap_{n\geqslant N}F_n^{-1}\{0\}$, we obtain the existence of $N_0$, a positive $\delta$ and a measurable set $A_0$ such that if $\rho(A,A_0)\lt \delta$ then for each $n\geqslant N_0$, $F_n(A)=0$. Using successively this property with $A\cup A_0$ and $A_0\setminus A$ instead of $A$, we obtain that if $\mu(A)\lt \delta$, then for each $n\geqslant N_0$, $F_n(A)=0$.

We now use a proposition in Bogachev's book: if $\varepsilon$ is a positive number and $(X,\mathbb B,\mu)$ is a finite measure space, then there exists a finite partition $(B_i)_{1\leqslant i\leqslant N}$ such that either $\mu(B_i)\lt\varepsilon$ or $B_i$ is a $\mu$-atom of measure greater than $\varepsilon$. Consequently, we obtain that $\mu(T^{-n}B\cap T^{-n}B)=\mu(B)^2$, hence $\mu(B)$ is $0$ or $1$.

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  • $\begingroup$ I'm having trouble seeing the use of Baire's theorem here. From what I understand Baire's theorem asserts that the intersection of countably many open dense sets is dense. How does that relate to the intersection $F_{N}$? $\endgroup$
    – user135520
    Apr 12, 2019 at 21:48
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    $\begingroup$ By considering complements, Baire theorem can also be stated as follows: if a countable union of closed subsets sets of a complete metric space has a non-empty interior, then one of the sets has a non-empty interior. $\endgroup$ Apr 13, 2019 at 7:55

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