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A consequence of Birkhoff ergodic theorem tells us that ergodicity is equivalent to:

$\forall A,B \in \mathcal{B} \quad \frac{1}{N}\sum_{n=0}^{N-1}\mu(A\cap T^{-n}(B))\stackrel{N\to \infty}{\longrightarrow} \mu(A)\mu(B)$.

In other words, a system is ergodic if every pair of measurable sets are asymptotically independent in Cesàro sum. With this definition of ergodicity is natural to define the mixing systems that are the systems in which every pair of measurable sets are asymptotically independent.

I was wondering if exists a class of system that for every pair of measurable sets $A,B$ exists $N > 0$ such that $\forall n \geq N$ $\mu(A \cap T^{-n}B)=\mu(A)\mu(B)$.

That class of systems are clearly smaller than the mixing class, but I don't know if there are examples or anything about them. I think that the Bernoulli and Markov shifts presents that kind of behavior.

Any help or reference will help :)

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Even for Bernoulli shifts this property is not true, as the following example shows.

Let $(X,\mu)$ be the $(\frac 12,\frac 12)$ Bernoulli shift on two symbols, 0 and 1. That is, $X = \{0,1\}^\mathbb{N}$ is the set of all one-sided infinite sequences of 0s and 1s, and $\mu$ gives weight $2^{-n}$ to every $n$-cylinder.

Let $A=[1]$ be the set of sequences beginning with the symbol 1. Interpreting $x\in X$ as a series of coin flips, with 0 as heads and 1 as tails, $A$ corresponds to the event that the first flip is tails.

Let $B$ be the set of sequences $x\in X$ with the property that for every $k=1,2,3,\dots$, there is some index $i\in [2^{k-1}, 2^k) \cap \mathbb{N}$ such that $x_i = 1$. Then $B$ corresponds to the event that the first flip is tails, then (at least) one of the next two flips is tails, then (at least) one of the next four is tails, and so on.

Note that $[0] \cap B = \emptyset$ and so $\mu(B) \leq \frac 12$. On the other hand, $X \setminus B \subset [0] \cup \sigma^{-1}[00] \cup \sigma^{-3}[0000] \cup \cdots$, and so $\mu(X\setminus B) \leq \sum_{k=0}^\infty 2^{-2^k} < 1$. Thus $0 < \mu(B) < 1$.

Now consider $B \cap \sigma^{-n}(A)$. This is the event that $B$ holds (first flip tails, one of next two tails, etc.), and also that the $n$th flip is tails. Let $k'$ be the unique integer with $n\in [2^{k'},2^{k'+1})$. Considering conditional probabilities, we have $$ \frac{\mu(B\cap \sigma^{-n}(A))}{\mu(A)} = \mathbb{P}(B \mid \sigma^{-n}(A)) = \prod_{k\neq k'} (1-2^{-2^k}) > \prod_{k} (1-2^{-2^k}) = \mathbb{P}(B) = \mu(B). $$ In particular, $$ \mu(B\cap \sigma^{-n}(A)) > \mu(A) \mu(B), $$ and this holds for every $n$.

There may be simpler examples. I don't know how to generalise this to arbitrary measure-preserving transformations, but my expectation would be that for every mpt there are measurable sets $A,B$ such that you do not get equality in the mixing condition for any finite $n$.

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  • $\begingroup$ it s my idea, or is it true, that indeed there is not any measure non supported in finitely many points for which the statement can be true. Because, the assumption implies that the measure is Bernoulli, and your argument applies after. $\endgroup$ – user39115 Apr 22 '14 at 20:59
  • $\begingroup$ In a system where every set has measure 0 or 1 the property is true, but that doesn't said much. $\endgroup$ – user90803 Apr 23 '14 at 0:10
  • $\begingroup$ @Vaughn Climenhaga +1. Very nice. I was able to get any measure preserving system using similar ideas. See my solution $\endgroup$ – mathworker21 Sep 6 at 13:55
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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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  • $\begingroup$ +1. Very clever!! See my solution below for an alternative method. $\endgroup$ – mathworker21 Sep 6 at 13:55
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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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