Sigma-algebra generated by collection of cylinders.

Question

Let $X$ = {$0,1$}$^\mathbb{N}$ be the set of all infinite sequences of $0$’s and $1$’s. A typical element $x \in X$ is written as $x = x_1x_2x_3$···. A cylinder set is a subset of $X$ of the form {$x \in X; x_1 = a_1,x_2 = a_2,...,x_m = a_m$}, with $a_i \in$ {$0,1$} and $m \in \mathbb{N}$. The collection $S$ of cylinder sets is a semi-ring and is closed under finite intersections. We have also shown that the set function $\mu : S →[0,1]$ given by $\mu(${$x \in X; x_1 = a_1,x_2 = a_2,...,x_m = a_m$}$) = (1/2)^m$, can be uniquely extended to a measure on $\sigma(S)$, and we denote this measure also by $\mu$. Let $B$ be the set $B := ${$x \in X; x_2 = 0,x_4 = 0,x_6 = 0,...$}.

Show that $B \in σ(S)$.

My attempt is:

$S$ can be written as $S = \bigcup_{m \in \mathbb{N}}S_m$, where $S_m =$ {$x \in X; x_1 = a_1,x_2 = a_2,...,x_m = a_m$}. Then if $x_2 = 0$ in $S$, $B \in S$.

Since $\sigma(S)$ is a $\sigma$-algebra generated by $S$, and $S$ is itself a $\sigma$-algebra, it follows that $\sigma(S) = S$ and hence $B \in \sigma(S)$.

Is this correct?

Answer 1

It is not correct. $B$ is an intersection of cylinder sets, not a union. $B=\bigcap_N \{x: x_2=0,x_4=0,\cdots, x_{2N}=0\} \in \sigma (S)$. Note that $\{x: x_2=0,x_4=0,\cdots, x_{2N}=0\}$ is the union of the cylinder sets $\{x:x_1=a_1, x_2=0,x_3=a_3,x_4=0,\cdots, x_{2N-1}=a_{2N-1},x_{2N}=0\}$ over all possible values of $a_1,a_3,...a_{2N-1}$.

Answer 2

Let $X_N = \{x = x_1x_2 \dots x_{2N} \in \{0,1\}^{2N} \mid x_2=x_4= \dots = x_{2N} = 0\}$ and for $x \in X_N$, let $S_x$ be the cylinder

$$S_x = \{y=y_1y_2 \dots \in X \mid y_1= x_1, y_2=x_2, \dots , y_{2N} = x_{2N}\}.$$ Notice that $X_N$ is finite for every $N \in \mathbb N$.

Then you have

$$B = \bigcap_{N \in \mathbb N}\bigcup_{x \in X_N} S_x$$ proving that indeed, $B$ belongs to $\sigma(S)$.

What you wrote in the question is not correct, as you lack the way to formalize the fact that $S_x$, in general, is a finite union of cylinders.