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I'm having my first tiny little bits of ergodic theory so please forgive the probable naivete of the question. So, I'm looking at the first page of chapter 8 of Cornfeld, Fomin, Sinai's "Ergodic Theory". They state:

Now let us show that an arbitrary automorphism $T'$ of the measure space $(M', \mathscr{F}, \mu)$ naturally gives rise to stationary random processes.

Consider some countable partition $$\xi=(C_1,\cdots,C_k),1\leq k\leq\infty$$

For the state space choose $Y=(1,2,\cdots,k)$ and put $M=\prod_{n=-\infty}^{\infty}Y^{(n)}$ ; $Y^{(n)}=Y$. Consider the map $\phi:M'\rightarrow M$ defined as follows: the $n$th coordinate of the point $\phi x'$ equals $j$ if and only if $(T')^nx'\in C_j$. The map $\phi$ is measurable. It transforms the measure $\mu'$ on $M'$ onto a certain measure $\mu$ on $M$: $ \mu(A)\equiv\mu'(\phi^{-1}A), A\in\mathscr A$, while the transformation $T'$ is the shift on $M$. From the fact that $T'$ is an automorphism it follows that the random process $(M,\mathscr A, \mu)$ is stationary (i.e. its measure is shift-ivariant (e.d.)).

Now, I have problems with the bold text: first, $T'$ does not act on $M$, so I guess they mean: "$\phi T'\phi^{-1}$ is the shift on $M$"; then, I can't see how the fact that $T'$ preserves the measure $\mu'$ of $M'$ should guarantee me that $\phi T'\phi^{-1}$ preserves the $M$-measure $\mu$.

I actually tried to sketch a proof: $$\mu(B) = \mu'(\phi^{-1}(B))= \mu'(T'\phi^{-1}(B))$$ And also: $$\mu(\phi T'\phi^{-1}B) = \mu'(\phi^{-1}\phi T'\phi^{-1}B)$$ Now, if I got things right I would want the two to be equal; yet, it seems to me that this would be the case only if $\phi$ itself was an isomorphism, so that $\phi^{-1}\phi\equiv id_{M'}$.

What am I missing here?

Thanks in advance!

EDIT: I also found the same statement in Sinai's "Metric entropy of dynamical systems":

Take a finite partition $\xi = {C_1 , C_2 , \cdots , C_r }$ of $M$ . It generates a stationary random process of probability theory with values $1, 2, \cdots , r$ if one uses the formula:

$w_k(x) = j\quad$ if $\quad x ∈ T^{−k} C_j , −∞ < k < ∞ .$

...just to be more confident that the statement is, in itself, correct!

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