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.


Bernoulli measure is aperiodic?

  • $\begingroup$ Hint: can you show that there are countably many periodic points in $X$? Now all you need to show is that the measure of any countable set is zero for the Bernoulli measure. $\endgroup$
    – Dan Rust
    Aug 21, 2018 at 12:53
  • $\begingroup$ @DanRust any periodic sequence with a repeating an block of length n defines and periodic orbit. The set of all periodic sequences is countable.I can show that measure any countable set is zero but for Bernoulli measure,i do not know. $\endgroup$
    – Adam
    Aug 21, 2018 at 15:04

1 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?


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