Showing that $\mu(\phi^{-1}(V)) = \lambda (V).$

Question

In a process of showing that the operator $T$ we gave here Showing that a set has measure zero? is ergodic, I want to prove that $\mu(\phi^{-1}(V)) = \lambda (V).$ where I have the following givens to prove it:

Suppose $V$ is a measurable subset of $[0,1).$ Let $\lambda$ be Lebesgue measure on $[0,1).$ Prove that $\mu(\phi^{-1}(V)) = \lambda (V).$ we have the following hint:

"It is enough to show this for $V = [\frac{m}{2^n}, \frac{m+1}{2^n})$ for each $n,$ as such sets generate the $\sigma$-algebra for $[0,1).$ what is the preimage of $V$ under $\phi$?"

Here is so far what we know about the setting of the problem:

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.

We now know 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.

We also, know that the following map is bijective:

$\phi: X\setminus X_{\infty} \to [0,1)$ 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.

And we may assume that $\phi$ is measurable with measurable inverse.

my questions are:

1- How does this will help us in proving ergodicity?

2- How do we knew that the suggested $V$ in the hint will generate the $\sigma$-algebra for $[0,1)$?

3- Finally, how can the preimage of $V$ under $\phi$ help me in the proof?

Could someone clarify this to me please?

Answer 1

I'll answer your questions in order:

1. Consider what you know and what you want to eventually show. You have $\phi$ is a bijection between $X\setminus X_\infty$ and $[0,1)$. If you have that $\phi$ and its inverse are measureable and (as you want to show) the measures are preserved under the map, then the measure spaces are in essence the same (or isomorphic). Thus, you could prove that the map $\phi\circ T\circ \phi^{-1}$ is ergodic on the measure space $[0,1)$ instead, and thus conclude $T$ is ergodic on the space $X\setminus X_\infty$. This is useful because the action of the map might be much easier to understand on $[0,1)$.

2. There may be some confusion here over what the "measurable sets" should be for $[0,1)$. I believe the question might be referring to Borel measurable sets (otherwise, I don't think that statement is correct). Showing that sets $[\frac m {2^n}, \frac{m+1}{2^n})$ generate the Borel measurable sets is just a matter of showing they generate all open intervals. This isn't too hard to show: take $(a,b)$ and prove for every $x\in(a,b)$ there is a set $[\frac m {2^n}, \frac{m+1}{2^n}) \subset (a,b)$ containing $x$, which can be done by choosing $n$ large enough. (There is also the case of intervals of the form $[0,b)$, but that can be done similarly.)

3. It helps to think of $\phi^{-1}(V)$ as the binary expansion of numbers in $V$. Note that for elements in $[\frac m {2^n}, \frac{m+1}{2^n})$, the binary expansion of the first $n$ numbers is fixed. This should make $\mu(\phi^{-1}(V))$ easy to calculate. And it should correspond to the Lebesgue measure $\lambda(V) = 1/2^n$.

Edit: Expanding a bit on the last point. If $V = [\frac m {2^n}, \frac{m+1}{2^n})$ then $\phi^{-1}(V)$ will consist of all binary expansions of this interval (each binary expansion is unique since we disallow repeated 1's on the set $X\setminus X_\infty$). In particular, the preimage of $V$ will be all $(a_1,a_2,\ldots)$, where the first $n$ terms are fixed so that $a_1 2^{-1} + \cdots + a_n 2^{-n} = \frac m {2^n}$. Thus we can define sets $A_i = \{0\}$ or $A_i = \{1\}$ (depending on what value we have for $a_i$) so that $\phi^{-1}(V) = \{\mathbf x \mid x_i \in A_i \text{ for } i=1,\ldots,n\}$. Thus by the definition of $\mu$ we have $\mu\left(\phi^{-1}(V)\right) = \prod_{i=1}^n \frac{|A_i|}{2} = \frac 1 {2^n}.$ This agrees with $\lambda(V) = \frac 1 {2^n}$, as expected.