Proving two sigma algebras coincide

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.

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

• 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 $\theta_n^{-1} (\mathcal{B}) \subseteq \mathcal{\tilde{B}}$
– Karl
Commented Dec 24, 2020 at 17:40
• The $B_i$ on the left hand side are the $D_i$ on the right hand side I imagine...
– Karl
Commented Dec 24, 2020 at 18:41
• @Karl, I made an addition. Proof by definition. Is there any questions? Commented Dec 24, 2020 at 19:12
• 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?
– Karl
Commented Dec 24, 2020 at 19:32
• @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? Commented Dec 24, 2020 at 20:13