Here is a remaining part of the question I asked here Showing that a set has measure zero? :
Let $S : [0,1) \to [0,1) $ be the doubling map, $S(x) = 2x \pmod{1}.$ I want to show that $\phi(T(\textbf{x})) = S(\phi(\textbf{x})).$
I am just wondering how can I do this? What is $S(\phi(\textbf{x}))$?
Note that the map $\phi$ is defined as: Consider the map $\phi: X\setminus X_{\infty} \to [0,1)$ defined by $$\phi(a_1, a_2, \dots , a_n) = \sum_{n=1}^{\infty}a_n2^{-n}$$ so $\phi$ sends each sequence $(a_1, a_2,\dots)$ to the point $x$ in $[0,1)$ with that binary decimal expansion.
I also, want to know how can I use the fact that $S$ is ergodic to prove that $T$ is ergodic?
Any hints for the above two questions will be greatly appreciated.
Here is the general setting of the problem that is given in the link mentioned above:
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 map $\phi: X\setminus X_{\infty} \to [0,1)$ defined by $$\phi(a_1, a_2, \dots , a_n) = \sum_{n=1}^{\infty}a_n2^{-n}$$ so $\phi$ sends each sequence $(a_1, a_2,\dots)$ to the point $x$ in $[0,1)$ with that binary decimal expansion.We know that $\phi$ is a bijection. We may assume that $\phi$ is measurable with measurable inverse. Also, we know that if $V$ is a measurable subset of $[0,1)$ and if $\lambda$ is Lebesgue measure on $[0,1).$Then $\mu(\phi^{-1}(V)) = \lambda (V).$
uspackagetypo, other errors prevent me building your diagram, so I can't help you translate it. – Eric Towers Oct 01 '24 at 19:55\arrowwon't take"T" redas an argument. Is there any version of your commutative diagram that builds somewhere (not necessarily here at M.SE.com)? – Eric Towers Oct 01 '24 at 21:23