Intersection of sigma algebras generated by shifts

Question

EDIT: Iosif's answer showed that my motivation for this question was mislead. To keep this question interesting for a broader readership, let us forget about sequence spaces and tail algebras and focus this question on the more abstract problems in questions 1 and 2.

Question 1': Given a Borel subset $A \subseteq Y \times W$ of a product of Polish spaces that satisfies $A = \pi^{-1}(\pi(A))$ with the projection $\pi \colon Z \rightarrow Y$, has it been studied under which conditions there exists a Borel set $B \subseteq Y$ with $A = \pi^{-1}(B)$?

Question 2': Under which conditions is a Borel-sub-sigma algebra $\mathcal{F}$ on a Polish space generated by $\{A \in \mathcal{F} | A \text{ is closed}\}$?

Original formulation below

Let $X$ and $Y$ be Polish spaces. Equip $\bar{X} := X^{\mathbb{N}}$ and $Z := Y \times \bar{X}$ with the product topologies. Elements of the sequence space will be denoted $\bar{x} = (x_0,x_1,\dots) \in \bar{X}$. All spaces are endowed with their Borel sigma-algebras, which we denote $\mathcal{B}_Y$, $\mathcal{B}_Z$ etc. Consider the 'extended' shift map $f \colon Z \rightarrow Z$ defined by $(y,x_0,x_1,\dots) \mapsto (y,x_1,x_2,\dots)$. I am trying to understand the sigma-algebra $$\Sigma := \bigcap_{k \geq 1} f^{-k}(\mathcal{B}_Z).$$ My first instinct was to argue '$f^k(y,\bar{x})$ does not depend on $x_0,\dots,x_{k-1}$ and, hence, $\Sigma = \pi^{-1}(\mathcal{B}_Y)$ should be generated by the projection $\pi \colon Z \rightarrow Y$.' Trying to turn this argument into something rigorous exposed some technical difficulties. (NB: the inclusion $\pi^{-1}(\mathcal{B}_Y) \subseteq \Sigma$ is trivial.)

My approach is as follows. Take a closed set $A \in \Sigma$, and fix any $(y,\bar{x}) \in A$ and $\bar{x}' \in \bar{X}$. Denote $\bar{x}_k = (x_0',\dots,x_{k-1}',x_k,x_{k+1},\dots)$ so that $\bar{x}_k \rightarrow \bar{x}'$. The fact that $f^k(y,\bar{x}_k) = f^k(y,\bar{x})$ and $A \in f^{-k}(\mathcal{B}_Z)$ implies $(y,\bar{x}_k) \in A$. Since $A$ is closed, it contains the limit $(y,\bar{x}')$. We have shown $A = \pi^{-1}(\pi(A))$. If it so happens that $\pi(A) \in \mathcal{B}_Y$, then $A \in \pi^{-1}(\mathcal{B}_Y)$. But of course, in general, $\pi(A)$ is not Borel.

One possibility to proceed is to add the assumption that $X$ and $Y$ be locally compact. In this case, $A$ can be written as a countable union of compact sets, which would guarantee that $\pi(A)$ is Borel.

Question 1: Can the local compactness assumption be relaxed? Even if $\pi(A)$ is not itself Borel, are there weaker assumptions that imply the existence of a Borel set $B \in \mathcal{B}_Y$ with $A = \pi^{-1}(B)$?

Moving on, suppose we have shown (with or without local compactness) that $$\mathcal{C} := \{ A \in \Sigma \,|\, A \text{ is closed} \} \subseteq \pi^{-1}(\mathcal{B}_Y).$$

Can we argue that $\mathcal{C}$ generates $\Sigma$? This is definitely the case if $\Sigma$ happens to be countably generated. But it is well known that a sub-sigma-algebra of $\mathcal{B}_Z$ need not be countably generated itself. At the same time, being countably generated is certainly not necessary: the standard counterexample to the above, namely the countable/cocountable sigma-algebra is still generated by its closed sets.

Question 2: Under which conditions does $\mathcal{C}$ generate $\Sigma$? More generally, when does $\{ A \in \mathcal{F} \,|\, A \text{ is closed} \}$ generate a Borel-sub-sigma-algebra $\mathcal{F}$?

Cheers!

Answer 1

$\Sigma = \pi^{-1}(\mathcal{B}_Y)$ should be generated by the projection $\pi \colon Z \rightarrow Y$.

This is not true in general. E.g., suppose that $Y=\{0\}$ and $X=\{0,1\}$. Then $\pi^{-1}(\mathcal{B}_Y)$ is the trivial $\sigma$-algebra $\{\emptyset,Z\}$ over $Z$. On the other hand, $$\Big\{(y,x_0,x_1,\dots)\in Z\colon\lim_{n\to\infty}\frac1n\,\sum_{i=1}^n x_i=\frac12\Big\}\in\Sigma\setminus\{\emptyset,Z\}.$$ So, $\Sigma \ne \pi^{-1}(\mathcal{B}_Y)$.