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Suppose $\Omega := [0,1]^{\mathbb Z}$ is equipped with the product topology and endowed with the Borel $\sigma$-algebra $\mathcal B(\Omega)$ and there is a probability measure $\mathbb P$ on $(\Omega,\mathcal B(\Omega))$ such that the shift $T:\Omega \to \Omega$, $$T(\omega)(k) := \omega(k+1),\quad \omega\in\Omega,k\in \mathbb Z$$ is measure preserving, i.e. $\mathbb P = \mathbb P \circ T^{-1}$ on $\mathcal B(\Omega)$, and ergodic, i.e. $A=T^{-1}(A)$ implies $\mathbb P (A)\in\{0,1\}$ for any $A\in\mathcal B(\Omega)$. Now let $f:[0,1]^3\to[0,1]$ a measurable function and $U:\Omega \to \Omega$ the transformation defined by $$ U(\omega)(k) := f(\omega(2k-1),\omega(2k),\omega(2k+1)),\quad \omega\in\Omega,k\in\mathbb Z.$$ We consider the probability measure $\widetilde {\mathbb P}:= \mathbb P\circ U^{-1}$ where $U^{-1}$ denotes the preimage.

Then, by $T\circ U= U\circ T^2$, it holds that $(\Omega,\mathcal B(\Omega), \widetilde {\mathbb P},T)$ is still a measure-preserving dynamical system. Is it also ergodic?

Edit: What are examples of probability measures $\mathbb P$ on $\mathcal B(\Omega)$ and sets $A\in\mathcal B(\Omega)$ such that $T^{-2}(A)=A$ but $\mathbb P(A)\notin \{0,1\}$ (and hence necessarily $T^{-1}(A)\neq A$)?

maliesen
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  • Try to show that $T^{-1}(E)=E$ iff $T^{-1}(U(E))=U(E)$. So, using the ergodic of the previous system we have $\widetilde{\mathbb{P}}(E) = \mathbb{P}(U^{-1} (E))=0$ or $1$ – JPMarciano Nov 25 '20 at 01:21
  • @JPMarciano From $T^{-1}(E)=E$ it follows $T^{-2}(U^{-1}(E)) = U^{-1}(E)$. If it was true, that yet $T^{-1}(U^{-1}(E)) = U^{-1}(E)$, it would follow $\mathbb P(U^{-1}(E))\in{0,1}$ by ergodicity of the previous system. Do I get the idea right? From $T^{-1}(E)=E$ it also follows $T^{-1}(U(E))=U(E)$ but the converse does not hold in general. From $T^{-1}(U(E))=U(E)$, how do you deduce $T^{-1}(U^{-1}(E)) = U^{-1}(E)$? – maliesen Nov 25 '20 at 16:00
  • You are right, sorry! – JPMarciano Nov 25 '20 at 17:31

1 Answers1

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The answer is negative: Let \begin{align*} \mathbb P &:= \frac 1 2 \left(\delta_{\left(\mathbb 1_{2 \mathbb Z}(k)\right)_{k\in \mathbb Z}} + \delta_{\left(\mathbb 1_{2 \mathbb Z+1}(k)\right)_{k\in \mathbb Z}} \right), \\ A &:= \left\lbrace (1)_{k\in\mathbb Z} \right\rbrace ,\\ f(x,y,z) &:= y. \end{align*}

The probability $\mathbb P$ corresponds to the irreducible Markov chain on state space $\{0,1\}$ with transition matrix $P = \begin{pmatrix} 0 & 1\\ 1& 0\end{pmatrix}$ which has unique stationary distribution $(\frac 1 2 \,\,\, \frac 1 2)$. In light of the answer to this math.SE question the dynamical system $(\Omega,\mathcal B(\Omega),\mathbb P, T)$ is measure-preserving and ergodic (but not mixing). Now, $T^{-1}(A)=A$ but $$U^{-1}(A) = \prod_{k\in \mathbb Z} \begin{cases} \{1\},&\quad k\in 2 \mathbb Z \\ [0,1],&\quad k\in 2\mathbb Z+1\end{cases}, $$ whence $\widetilde{\mathbb P}(A) = \frac 1 2 $.

maliesen
  • 582