Let $\tilde\Omega_\mathbb{R}$ denote the collection of all functions $\tilde\omega : \mathbb{R} \to \mathbb{R}$ and $\mathcal{B}_\mathbb{R}$ the $\sigma$-algebra generated by all cylindrical subsets of $\tilde\Omega_\mathbb{R}$. Similarly, let $\tilde\Omega_\mathbb{Z^+}$ denote the collection of all functions $\tilde\omega : \mathbb{Z^+} \to \mathbb{R}$ and $\mathcal{B}_\mathbb{Z^+}$ the $\sigma$-algebra generated by all cylindrical subsets of $\tilde\Omega_{\mathbb{Z^+}}$.
My question concerns the following characterization of the sets in $\mathcal{B}_\mathbb{R}$:
$S \in \mathcal{B}_\mathbb{R}$ if and only if there exists $B \in \mathcal{B}_{\mathbb{Z^+}}$ and an infinite sequence of real numbers $t_1, t_2, \dots$ such that
$$
S = \left\{ \tilde\omega \in \tilde\Omega_\mathbb{R} \, : \, \big( \tilde\omega(t_n) \big)_{n \in \mathbb{N}} \in B \right\}
$$
I found this question that asks how to prove this characterization, and in the accepted answer an outline for the proof is provided. The first step is to show that the collection of sets $\Sigma$ defined by
$$
\Sigma := \Big\{ S \subseteq \tilde\Omega_\mathbb{R} \, : \, \exists B \in \mathcal{B}_\mathbb{Z^+}, \, \{t_n\}_{n \in \mathbb{N}} \subset \mathbb{R} \text{ such that } S = \left\{ \tilde\omega \in \tilde\Omega_\mathbb{R} \, : \, \big( \tilde\omega(t_n) \big)_{n \in \mathbb{N}} \in B \right\} \Big\}
$$
is a $\sigma$-algebra. I am having difficulty in proving that $\Sigma$ is closed under countable unions.
The following is my attempt. I thought it would be helpful to think in terms of projections. Specifically, for a sequence of real numbers $\{t_n\}_{n \in \mathbb{N}}$, define $\pi_{\{t_n\}_{n\in\mathbb{N}}} : \tilde\Omega_\mathbb{R} \to \tilde\Omega_\mathbb{Z^+}$ by $\pi_{\{t_n\}_{n\in\mathbb{N}}}(\tilde\omega):=\big( \tilde\omega(t_n) \big)_{n\in\mathbb{N}}$. Then $S \in \Sigma$ iff there exists $B \in \mathcal{B}_{\mathbb{Z}^+}$ and a projection $\pi_{\{t_n\}_{n\in\mathbb{N}}}$ such that $S = \pi_{\{t_n\}_{n\in\mathbb{N}}}^{-1}(B)$. Moreover, $\Sigma$ is the family of all such pre-images where $B$ and the sequence of times range over all possibilities:
$$
\Sigma := \Big\{ S \subseteq \tilde\Omega_\mathbb{R} \, : \, \exists B \in \mathcal{B}_\mathbb{Z^+}, \, \{t_n\}_{n \in \mathbb{N}} \subset \mathbb{R} \text{ such that } S = \pi_{\{t_n\}_{n\in\mathbb{N}}}^{-1}(B) \Big\} = \bigcup_{B \in \mathcal{B}_{\mathbb{Z}^+}} \bigcup_{\{t_n\}_{n\in\mathbb{N}} \subset \mathbb{R}} \{\pi_{\{t_n\}_{n\in\mathbb{N}}}^{-1}(B)\}
$$
Let $\{S_n\}_{n\in\mathbb{N}} \subset \Sigma$, so there exists $\{t_{n,k}\}_{n,k \in \mathbb{N}} \subset \mathbb{R}$ and $\{B_n\}_{n \in \mathbb{N}} \subset \mathcal{B}_{\mathbb{Z}^+}$ such that
$$
\forall n \in \mathbb{N}, \qquad S_n = \pi_{\{t_{n,k}\}_{k\in\mathbb{N}}}^{-1}(B_n)
$$
We aim to show $\bigcup_{n\in\mathbb{N}} S_n \in \Sigma$. I thought about trying to diagonalize the array of times to form a sequence $\{\tau_n\}_{n\in\mathbb{N}}$
$$
\tau_1 := t_{1,1}, \, \tau_2 := t_{1,2}, \, \tau_3 := t_{2,1}, \, \tau_4 := t_{3,1}, \, \dots
$$
and writing $\bigcup_{n\in\mathbb{N}} S_n = \pi_{\{\tau_n\}_{n\in\mathbb{N}}}^{-1}\left( \bigcup_{n\in\mathbb{N}} B_n \right)$, but this doesn't sit right with me. Does anyone have any suggestions?
Thank you very much for your help! Also, please give me feedback on how I can improve my questions on MSE as this is my first post!
Best Answer
Let $\mathbb{R}$ be equipped with Borel $\sigma$-algebra $\mathcal{B}$.
For every subset $S\subseteq\mathbb{R}$ let $\bar{\Omega}_{S}$ denote the set of functions $S\to\mathbb{R}$ and let it be equipped with the smallest $\sigma$-algebra on $\bar{\Omega}_{S}$ such that for every $s\in S$ function $\pi_{s}:\bar{\Omega}_{S}\to\mathbb{R}$ prescribed by $f\mapsto f\left(s\right)$ is measurable.
We denote this $\sigma$-algebra as $\mathcal{B}_{S}$ and it can also be described as the $\sigma$-algebra generated by cylindrical subsets of $\bar{\Omega}_{S}$.
For $K,L\in\mathcal{P}\left(\mathbb{R}\right)$ with $L\subseteq K$ let $\pi_{L}^{K}:\bar{\Omega}_{K}\to\bar{\Omega}_{L}$ be prescribed by $\bar{\omega}\mapsto\bar{\omega}\upharpoonright L$ and abbreviate $\pi_{L}^{\mathbb{R}}$ as $\pi_{L}$.
It is easily verified that every function $\pi_{L}^{K}$ is measurable.
Let $\mathcal{C}$ denote the collection of countable subsets of $\mathbb{R}.$
Then $A\in\Sigma\iff\exists K\in\mathcal{C}\exists B\in\mathcal{B}_{K}\left[A=\pi_{K}^{-1}\left(B\right)\right]$
Let $\left(A_{n}\right)_{n}$ denote a sequence of elements of $\Sigma$.
Then there is a sequence $\left(K_{n}\right)_{n}$ in $\mathcal{C}$ together with a sequence $\left(B_{n}\right)_{n}$ where $B_{n}\in\mathcal{B}_{K_{n}}$ such that $A_{n}=\pi_{K_{n}}^{-1}\left(B_{n}\right)$
Here $K:=\bigcup_{n=1}^{\infty}K_{n}\in\mathcal{C}$ and writing $\pi_{K_{n}}=\pi_{K_{n}}^{K}\circ\pi_{K}$ we find that $A_{n}=\pi_{K}^{-1}\left(\left(\pi_{K_{n}}^{K}\right)^{-1}\left(B_{n}\right)\right)$.
Then: $$\bigcup_{n=1}^{\infty}A_{n}=\bigcup_{n=1}^{\infty}\pi_{K}^{-1}\left(\left(\pi_{K_{n}}^{K}\right)^{-1}\left(B_{n}\right)\right)=\pi_{K}^{-1}\left(\bigcup_{n=1}^{\infty}\left(\pi_{K_{n}}^{K}\right)^{-1}\left(B_{n}\right)\right)$$
Here $\bigcup_{n=1}^{\infty}\left(\pi_{K_{n}}^{K}\right)^{-1}\left(B_{n}\right)$ is a countable union of elements of $\sigma$-algebra $\mathcal{B}_{K}$ hence is itself an element of $\mathcal{B}_{K}$ justifying the conclusion that $\bigcup_{n=1}^{\infty}A_{n}$ is an element of $\Sigma$.