Preliminaries:
Definition of Bernoulli-shift
Let $X=\left\{0,1,2\right\}^{\mathbb{Z}^d}$ and $\mathcal{B}$ the Borel-$\sigma$-Algebra on it. Moreover, let $\mu_{p_0,p_1,p_2}$ denote a Bernoulli-measure on it, that is a measure on $\mathcal{B}$ that is defined via giving the cylinder sets $[x_i,x_{i+1},\ldots,x_{i+n}], x_j\in\left\{0,1,2\right\}$(this cylindersets generate $\mathcal{B}$) the measure $$ \mu_{p_0,p_1,p_2}([x_i,x_{i+1},\ldots,x_{i+n}])=\prod_{k=i}^{i+n}p(x_k),~~\text{where} \sum_{k=0}^{2}p(k)=1. $$ A translation-invariant measure $m$ on $X$ is a Bernoulli-shift if for some $p_0,p_1,p_2$ there is a bimeasurable measure-preserving transformation $$ f\colon(X,\mathcal{B},m)\to (X,\mathcal{B},\mu_{p_0,p_1,p_2}) $$ that commutes with all spatial translations.
Theorem 1 If $m_1$ is a Bernoulli-shift on $X$ and $$ f\colon (X,\mathcal{B},m_1)\to (X,\mathcal{B},m_2) $$ is measurable, measure preserving (but not necessarily invertible), and commutes with spatial translation, then $m_2$ is a Bernoulli-shift on $X$, too.
Now to my question, that is not too silly, I hope.
Let $\mu$ be a translation-invariant product measure on $X$, in which each state has positive probability. With other words: $\mu$ is a Bernoulli-measure with $p_0,p_1,p_2>0$.
Consider the following dynamics on $X$. If we have a $\eta\in X$, then $\eta(x)$ is the state on lattice point $x$ and $\eta_n(x)$ is the state on lattice point $x$ at time $n$.
From one time-step to the next, a 1 becomes a 2, a 2 becomes a 0, and a 0 becomes a 1 if at least on of its 2d neighbors is a 1. Denote the dynamics by $T\colon\eta_n\to\eta_{n+1}$.
This is a so called cellular automaton which is named Greenberg-Hastings model.
By $T^k\mu$ denote the distribution on $X$ at time $k$ when $\mu$ (as above) is the original distribution. Moreover, suppose that $\lim_{n\to\infty}\eta_{3n}$ exisits.
The aim is to show that $v:=\lim_{n\to\infty}T^{3n}\mu$ is a Bernoulli-shift. As a preliminary result, it is said, that for all $n$, $T^{3n}\mu$ is a Bernoulli-shift by Theorem 1. And that is what I do not understand.
I think this suggests to use Theorem 1: $\mu$ is a Bernoulli-shift, because in the definition of a Bernoulli-shift one might choose the $p_0,p_1,p_2$ of $\mu$ itself and $S\colon X\to X, (S(x))_i=x_{i+1}$. This shift $S$ is a bimeasurable (because it is invertible) measure preserving transformation that commutes with all spatial translations.
So, if $\mu$ is a Bernoulli-shift, why then are $$ T^{3n}\mu $$ Bernoulli-shifts? What function $f$ (in the Theorem 1) do I have to use for $$ f\colon (X,\mathcal{B},\mu)\to (X,\mathcal{B},T^{3n}\mu)? $$
Thought (maybe too much) about that. But do not know how to show that. Maybe $f=T^{3n}$, but I do not see that $T^{3n}$ is measure-preserving. Or maybe we can assume that $\mu$ is the measure on a period 3 state? I think then $T^{3n}$ would be measure-preserving? But... do not know if this really makes any sense at all. Maybe Theorem 1 has to be applied in a very different way...
Hope you can help me with that confusion.
Edit: Maybe $f=T^{3n}\circ S$ does it?