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Let $X=\{0,1\}^\mathbb{Z}$ with measure $\mu=(p,1-p)^{\mathbb{Z}}$.

Let $(\phi(x))_i=(x_i+x_{i+1})$mod$2$.

If $p=1/2$, then $\phi(X)=X$. If $p \not = 1/2$, then $\phi(X)$ is not a Bernoulli scheme (i.i.d.).

For $x \in X$, define $x^*$ so that $x^*_i=(x_i+1)$mod$2$. Then $\phi(x)=\phi(x^*)$.

A factor map $\psi$ is finitary if for almost every $x \in X$ there exists integers $m \leq n$ such that the zero coordinates of $\psi(x)$ and $\psi(x')$ agree for almost all $x' \in X$ with $x[m,n]=x'[m,n]$.

In the case that $p \not = 1/2$, is it possible to construct a finitary map $\psi:(X, \mu) \to (X, \mu)$, such that for almost all $x \in X$, $\psi(x)=\psi(x^*)$?

Thank you.

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    $\begingroup$ You didn't say this, but I presume you also want that $\psi$ should preserve the measure $\mu$? (Otherwise your previous map $\phi$ will do the job). $\endgroup$ Jan 25, 2011 at 22:32
  • $\begingroup$ Yes, I'd like $\psi$ to preserve $\mu$. Thank you Anthony. $\endgroup$ Jan 25, 2011 at 23:35
  • $\begingroup$ Question has been edited to clarify that $\psi$ should preserve $\mu$. $\endgroup$ Jan 30, 2011 at 12:05
  • $\begingroup$ I am probably missing something here. Forget about finitary, is it even possible to get a measurable map? Won't the push forward of any such map be (1-p, p)^\Z? $\endgroup$ Feb 11, 2018 at 15:26

1 Answer 1

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Let $G\subset X$ be the set of generic points for $\mu$, which is of measure 1. Since $p\neq1-p$, we have that for $G^*=\{x^*:x\in G\}$, $$G\cap G^*=\emptyset.$$ Now define $\psi$ as folllows: For $x\in G^*$, $$\psi(x)=x^*,$$otherwise, $$\psi(x)=x$$We get for all $x\in G$, the map is finitary (with $m=n=0$), and satisfies $$\psi(x)=\psi(x^*).$$

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