Bernoulli measure is aperiodic

Question

Let $\Sigma=\{s_1,\dots,s_m\}$ be a finite list of symbols, and put $X=\Sigma^\mathbb{Z}$.

Consider the left two-sided shift $T:X\to X$ given by $T(x_n)=(x_{n+1})$. Given an $m$-dimensional vector $\vec{p}=(p_1,\dots,p_m)$, we can construct a measure on $\Sigma$ by $\sum_{i=1}^m p_i \delta_{s_i}$, which then generates an infinite product measure on $X$. Such a measure is called a Bernoulli shift, and is ergodic for $T$.

A measure is called aperiodic if set of periodic points has zero measure.

Question

Bernoulli measure is aperiodic?

Answer 1

Let $(x_i)_{i \geq 1}$ be a countable subset of $X$ and let $$B_{i,k} = \{x \in X \mid (x)_{[-k,k]} = (x_{i})_{[-k,k]}\}$$ be the cylinder set of length $k$ which contains $x_i$.

Show that $$B(n) := \bigcup_{i=0}^\infty B_{i,n2^i}$$ is a nested sequences of subsets that contains every $x_i$ for any value of $n$ and then show that the measure of $B(n)$ converges to $0$ as $n$ increases. Hence, what must the measure of $\{x_i \mid i \geq 0\}$ be?