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Consider the probability space $X = \{0,1\}^{\mathbb N}$ of sequences $x = (a_1, a_2, a_3, \dots)$ where each $a_i$ is either $0$ or $1.$ Take $\sigma$ to be the smallest $\sigma$-algebra that contains all cylinder sets $$C(n,A) = \{\textbf{x}| x_n \in A, A \subset \{0,1\}\}$$ that is, those sequences whose $n^{th}$ term belongs to some subset $A$ of $\{0,1\}.$ Let $\mu$ be the measure so that, given a finite collection $A_1, \dots , A_n \subset \{0,1\},$ we have $$\mu (\{\textbf{x} \,|\, x_i \in A_i, i = 1, \dots , n \}) = \Pi_{i=1}^{n}\frac{|A_i|}{2}$$ Consider the shift map $T: X \to X$ defined by $$T(a_1, a_2, a_3, \dots ) = (a_2, a_3, \dots )$$

Assuming that $T$ is measurable and preserves $\mu.$ We want to show that $T$ is ergodic.

$(a)$ Show that the set $X_{\infty} = \{ (a_1, a_2, a_3, \dots )\,|\,\exists N s.t. \forall n \geq N, a_n = 1 \}$ has measure zero.

Hint: Write $X_{\infty}$ in terms of nested cylinder sets with arbitrary small measure.

My question is:

How can I write $X_{\infty}$ in terms of nested cylinder sets with arbitrary small measure?

Is this a valid way of writing it $$X_{\infty} = \cup_{n=1}^{\infty} B_n$$ with $$B_N = \cap_{n = N}^{\infty} C(n, \{1\})$$? Should my sequence of $B_n$ be increasing or decreasing?

Any help will be greatly appreciated!

Hope
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1 Answers1

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Let us prove

The set $X_{\infty} = \{ (a_1, a_2, a_3, \dots )\,|\,\exists N s.t. \forall n \geq N, a_n = 1 \}$ has measure zero.

Proof: For each $N \in \Bbb N$, $N \geqslant 1$, let us define $X_N = \{ (a_1, a_2, a_3, \dots )\,|\,\forall n \geq N, a_n = 1 \}$. It is immediate that $$ X_\infty = \bigcup_{N=1}^\infty X_N$$ So, it is enough to prove that each $X_N$ has measure zero.

Given any $N, p \in \Bbb N$, let us define $C_{N,p} = \{ (a_1, a_2, a_3, \dots )\,|\, a_n = 1 \text{ for } n \in \{N, N+1, \dots, N+p \}\}$. Then, we have that $$ X_N = \bigcap_{p=0}^\infty C_{N,p}$$ and, $ \mu ( C_{N,p})= \frac{1}{2^{p+1}}$. So $\mu(X_N)=0$.

Ramiro
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  • $X_N$ is countable right? – Hope Oct 01 '24 at 13:53
  • why did you define the $C_{N,p}$ like this? Are they still cylinder sets? – Hope Oct 01 '24 at 13:54
  • I also edited my question, I am wondering if you have an answer to my last question? – Hope Oct 01 '24 at 13:56
  • @Hope The family ${X_N}{N \geqslant 1}$s is countable. So $X\infty$ is the union of a countable family of sets $X_N$. – Ramiro Oct 01 '24 at 15:24
  • So your sequence of $X_N$ is increasing or decreasing? – Hope Oct 01 '24 at 16:50
  • @Hope , I defined $C_{N,p}$ the way I did, to ensure that $ X_N = \bigcap_{p=0}^\infty C_{N,p}$ and that $\lim_{p \to \infty} \mu ( C_{N,p}) =0$. Yes, the $C_{N,p}$ are cylinder sets. – Ramiro Oct 01 '24 at 16:56
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    @Hope , The sequence ${X_N}{N \geqslant 1}$ is increasing. But, each $X_N$ has measure zero, and to prove it, we used the sequence ${C{N,p}}_{p\in \Bbb N}$, which is a decreasing sequence of nested cylinders. – Ramiro Oct 01 '24 at 17:03
  • @Hope , Regarding the last question you added to your question: IF your $C(n,{1}) = {\textbf{x}| x_n \in {1}}$, then you can write $X_{\infty} = \cup_{n=1}^{\infty} B_n$ with $B_N = \cap_{n = N}^{\infty} C(n, {1})$. Those $B_N$ coincides to the $X_N$ in my answer. – Ramiro Oct 01 '24 at 17:12