I was reading the solution of the following question here Bernoulli shift is a measure-preserving transformation :
Let (for simplicity) $(\{0,1\}^\mathbb{N},\mathcal{B},\mu,T)$ be a Bernoulli scheme, when $\mathcal{B}$ is the $\sigma$-algebra generated from the cylinder sets, $\mu$ is $(0.5,0.5)^\mathbb{N}$, and $T$ is "right shift", which means $\forall(x_n)_n\in\{0,1\}^\mathbb{N}: T((x_n)_n)=(x_{n+1})_n$
$T$ is a measure-preserving transformation, but why?
To prove that $T$ is measurable, let $A\in\mathcal{B}$, $T^{-1}(A)=\{x_n\mid (x_{n+1})_n\in A, x_1=0$ or $1\}$.
Is this set measurable? If it is, why $T$ is a measure-preserving transformation? I am struggling to prove it.
Thanks in advance.
My questions are:
1- I do not know what are the $\pi$- system and the $\lambda$-system and the Dynkin's $\pi-\lambda$ theorem the author used to solve this problem. Could someone explain this to me please?
2- Also, is there anyway to solve this question by a method where we do not need to use any of these terminologies?
Here is the answer I am referring to:
If $A \subseteq \{0,1\}^{\mathbb{N}}$ is a cylinder set, then $T^{-1}(A)$ is also a cylinder set. Define $\mathcal{G}$ by\begin{equation*}\mathcal{G} = \{A \subseteq \{0,1\}^{\mathbb{N}} \, \mid \, T^{-1}(A) \in \mathcal{B} \}.\end{equation*}An exercise shows that $\mathcal{G}$ is a $\sigma$-algebra. Since it contains the cylinder sets, $\mathcal{G} \supseteq \mathcal{B}$. That is, if $A \in \mathcal{B}$, then $T^{-1}(A) \in \mathcal{B}$, so $T$ is measurable.
If $A \subseteq \{0,1\}^{\mathbb{N}}$ is a cylinder set, then $\mu(T^{-1}(A)) = \mu(A)$ follows by a direct computation. The family of all cylinder sets if a $\pi$-system (since the intersection of two cylinders is a cylinder) and the family $\Lambda = \{A \in \mathcal{B} \, \mid \, \mu(A) = \mu(T^{-1}(A))\}$ is a $\lambda$-system. Therefore, Dynkin's $\pi-\lambda$ theorem implies that $\Lambda \supseteq \mathcal{B}$. In particular, $\mu(A) = \mu(T^{-1}(A))$ for all $A \in \mathcal{B}$, so $T$ is measure-preserving.