Show that the shift map is measurable and measure-preserving

Question

Show that the shift map $\theta$ of Definition 6.3 is measurable and measure-preserving.

Not sure how to represent $\theta^{-1}$ which I believe is where I am stopped in solving this problem.

Answer 1

We can prove more general fact. Let $(\Omega,F,p)$ be a probability space where are defined iidrv's $X_k(k=1,2, \cdots)$(in our case $(\Omega,F,p)=(R^N,\cal{E},\mu)$) and $\{ X_k : k \in N\}$ are coordinate projections). The $\theta$-shift invariance means the following: For each Borel subset $A \subset \cal{B}(R^N)$ the following equality $p(\{\omega : (X_k(\omega))_{k \ge 1} \in A\})= p(\{\omega : (X_k(\omega))_{k \ge 2} \in A\})$ holds true. By using Charatheodory measure extension theorem from the algebra to minimal sigma algebra and a uniqueness property of the extended measure, it is sufficient to prove the validity of this equality for elementary subsets of $R^N$ having the form $A_1\times \cdots \times A_n\times R^{N \setminus \{1,\cdots, n\}}$, where $n \in N$ and $A_k \in \cal{B}(R)$ for $k=1,\cdots,n$, because such sets constitute an algebra which exactly generates a Borel class of subsets of $R^N$. Since $(X_k)_{k \ge 1}$ is iidrv's so is $(X_k)_{k \ge 2}$. The latter relation implies $$p(\{\omega : (X_k(\omega))_{k \ge 1} \in A_1\times \cdots \times A_n\times R^{N \setminus \{1,\cdots, n\}}\})=\prod_{k=1}^n p(\{ \omega : X_k(\omega) \in A_k\})=\prod_{k=1}^n m(A_k)=\prod_{k=1}^{n} p(\{ \omega : X_{k+1}(\omega) \in A_k\})=p(\{\omega : (X_k(\omega))_{k \ge 2} \in A_1\times \cdots \times A_n\times R^{N \setminus \{1,\cdots, n\}}\}). $$

Answer 2

Some confusion might stem from neglecting the fact that $\theta^{-1}$ is not defined as a function, at least not defined on the image set of $\theta$. Rather, for every function $\theta:E\to F$ (whether $\theta$ is injective or surjective or not) and every $A\subseteq F$, one defines $\theta^{-1}(A)=\{x\in E\mid\theta(x)\in A\}$. Thus, $\theta^{-1}$ is not defined on $F$ but on $2^F$, with values in $2^E$.

Likewise, when $E=F$ is a measure space $(E,\mathcal E,\mu)$, to say that $\theta$ preserves the measure $\mu$ means that $\mu(\theta^{-1}(A))=\mu(A)$, for every $A$ in $\mathcal E$.