# ring generated by cylinder sets vs. filtration /sub-sigma algebras

I'm a little confused about the construction carried out in Shields' text (ergodic theory of discrete sample paths), which goes as follows. Consider an alphabet $A$, and let $A^{\infty}$ be the set of all infinite sequences $\mathbf x=x_1,x_2,\dots,x_i,\dots$ with $x_i \in A,~i \in \mathbb N$. We can define the $\sigma$-algebra on $A^\infty$ generated by the cylinder sets $[\mathbf a_{1..n}] ,~a_i \in A,~n \in \mathbb N$ where
\begin{equation*} [\mathbf a_{m..n}]:=\{\mathbf x \in A^{\infty}:x_i=a_i, i\in m..n\}. \end{equation*} For $n \in \mathbb N$, let $\mathcal C_n$ be the collection of cylinder sets $[\mathbf a_{1..n}]$ and let $\mathcal R_n$ denote the ring generated by $\mathcal C_n$. Let $\mathcal C=\cup_n \mathcal C_n$. Then $\cup_n \mathcal R_n$ is the ring generated by all the cylinder sets. I have two questions.

1. It is asserted that the sequence $\{\mathcal R_n\}$ is increasing. I don't know if I'm missing an obvious fact here, but I actually think it's decreasing in that $\mathcal R_i \subseteq \mathcal R_j$ for $i \geq j$.
2. How are these rings related to filtration? Specifically, is there an obvious way to turn $\{\mathcal R_n\}$ to sub-sigma algebras?

Many thanks!

Your alphabet $A$ is finite? Say it has $k$ elements.
Every cylinder in $\mathcal C_n$ is the union of $k$ cylinders in $\mathcal C_{n+1}$. And therefore it belongs to $\mathcal R_{n+1}$.
For example, say your alphabet is $A = \{\mathtt{a},\mathtt{b},\mathtt{c}\}$. Then cylinder $[\mathtt{a}\mathtt{c}\mathtt{a}]$ is the union of three cylinders $$[\mathtt{a}\mathtt{c}\mathtt{a}\mathtt{a}] \\ [\mathtt{a}\mathtt{c}\mathtt{a}\mathtt{b}] \\ [\mathtt{a}\mathtt{c}\mathtt{a}\mathtt{c}]$$
$\mathcal R_n$ already is a finite $\sigma$-algebra on $A^\infty$, and $\mathcal R_1 \subseteq \mathcal R_2 \subseteq \dots$. So we have a filtration.