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Let $(S, \mathcal{S})$ be a measurable space and $\Omega= \prod_{i=0}^{\infty} S$, let $X_i:\Omega \to S$ be the "coordinates" and $\theta_n$ the shift operators on $\Omega$: $w=(x_0,x_1,...) \mapsto (x_n,x_{n+1},...)$. Finally let $\mathcal{A}=\sigma(X_n, n\geq 0)$ and $ \mathcal{G}_n=\sigma(X_n,X_{n+1},...)$. How would you prove that $\mathcal{G}_n=\theta_n^{-1}(\mathcal{A})$? I am puzzled because it seems to me that the antiimage of an element of $\mathcal{A}$ could very well be in $\mathcal{A}$ and not in $\sigma(X_n,X_{n+1},...)$. This claim is in the context of canonical Markov Chains, Ionescu-Tulcea theorem.

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It's known (see, e.g. math.stackexchange.com/questions/7881) that $$f^{-1}(\sigma(C))=\sigma(f^{-1}(C))$$ and that $A$ is generated by cylinder sets $\mathcal{B}$.

Hence $\theta_n^{-1}(\mathcal{A}) = \theta_n^{-1} (\sigma(\mathcal{B})) = \sigma(\theta_n^{-1} (\mathcal{B})) = \sigma( \mathcal{\tilde{B}}) = \mathcal{G}_n $,

where $\tilde{B}$ are cylinder sets, that generate $\mathcal{G}_n$.

Addition: elements of $\mathcal{B}$ have the next form: $F_{t_1, \ldots, t_N, D_1, \ldots, D_N} = \{(y_0, y_1, \ldots): y_{t_1} \in D_1, y_{t_2} \in D_2, \ldots, y_{t_N} \in D_N \}$, where $D_i$ are Borel sets of real line and $t_i \ge 0$.

Elements of $\tilde{\mathcal{B}}$ have the next form: $\tilde{F}_{s_1, \ldots, s_N, D_1, \ldots, D_N} = \{(y_0, y_1, \ldots): y_{s_1} \in D_1, y_{s_2} \in D_2, \ldots, y_{s_N} \in D_N \}$, where $D_i$ are Borel sets of real line and $s_i \ge n$.

Thus $\theta_n^{-1} (\mathcal{B}) = \mathcal{\tilde{B}}$ because $$\theta_n^{-1} (\mathcal{B}) = \theta_n^{-1} ( \{ F_{t_1, \ldots, t_n, D_1, \ldots, D_N}\}) = \{ \theta_n^{-1} (F_{t_1, \ldots, t_N, D_1, \ldots, D_N}) \} = \{ \tilde{F}_{t_1+n, \ldots, t_N+n, D_1, \ldots, D_N} \} = \mathcal{\tilde{B}}.$$

Addition2: $$\theta_n^{-1} ( {F}_{t_1, \ldots, t_N, D_1, \ldots, D_N} ) = \{ (y_0, y_1, \ldots) : \theta_n (y_0, y_1, \ldots) \in {F}_{t_1, \ldots, t_N, D_1, \ldots, D_N} \}$$ $$= \{ (y_0, y_1, \ldots) : (y_n, y_{n+1}, \ldots) \in {F}_{t_1, \ldots, t_N, D_1, \ldots, D_N} \} = $$ $$ = \{ (y_0, y_1, \ldots) : y_{n+t_1} \in D_1, y_{n+t_2} \in D_2, \ldots \} = \tilde{F}_{t_1+n, \ldots, t_N+n, D_1, \ldots, D_N}.$$

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    $\begingroup$ Thanks but since this is not really an exercise, I would prefer a more elementary approach that doesn't use the fact you quote and could perhaps be more intuitive. If that answer doesn't come, I will accept yours. In any case I don't understand why if $\mathcal{\tilde{B}}$ generate $\mathcal{G}_n$ then $\theta_n^{-1} (\mathcal{B}) \subseteq \mathcal{\tilde{B}}$ $\endgroup$
    – Karl
    Commented Dec 24, 2020 at 17:40
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    $\begingroup$ The $B_i$ on the left hand side are the $D_i$ on the right hand side I imagine... $\endgroup$
    – Karl
    Commented Dec 24, 2020 at 18:41
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    $\begingroup$ @Karl, I made an addition. Proof by definition. Is there any questions? $\endgroup$ Commented Dec 24, 2020 at 19:12
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    $\begingroup$ I really appreciate your effort, but there are some things that confuse me, my $X_i$ take value in a measurable space $S$, not in $\mathbb{R}$. Are you assuming $S=\mathbb{R}$ for simplicity? $\endgroup$
    – Karl
    Commented Dec 24, 2020 at 19:32
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    $\begingroup$ @Karl, it's true that it may be useful for simplicity. But I just forgot that $X_n$ are not usual random variables) In general case, of course, we should take $D_i$, which are measurable in $S$. Is there any questions? $\endgroup$ Commented Dec 24, 2020 at 20:13

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