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?
Best Answer
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 $a_1,a_3,...a_{2N-1}$.