Primitive system is ergodic if original system is ergodic

Question

Suppose that $(X,\mathscr{A},\mu)$ is a probability space. Let $T\colon X\to X$ be a measure preserving ergodic transformation . Let $f\in L^{1}(X,\mathscr{A},\mu)$ be such that $f(X)\subset\mathbb{N}$. We define the primitive (or integral) system of $T$ under $f$ by:

1. set: $X_{f}:=\{(x,k) \ | \ x\in X, \ k\in\mathbb{N}, \ k\leq f(x)\}$,
2. $\sigma$-algebra: $\mathscr{A}_{f}$ generated by sets of the form $A_{f}^{k}:=\{(x,k) \ | \ x\in A, \ k\leq f(x)\}$, where $A\in\mathscr{A}$ and $k\in\mathbb{N}$.
3. measure: $$\mu_{f}(A_{f}^{k}):=\frac{\mu(A)}{\int_{X}f \ \text{d}\mu},$$ and extend $\mu_{f}$ to all of $\mathscr{A}_{f}$.
4. transformation: $T_{f}\colon X_{f}\to X_{f}$ defined by $$T_{f}(x,k):=\begin{cases}(x,k+1)&k<f(x)\\(Tx,1)&k=f(x)\end{cases}.$$

I have been able to prove that $T_{f}$ is measure preserving. But how do I show that $T_{f}$ is ergodic? Any suggestions are greatly appreciated. For instance, it enough to prove it for generators of the $\sigma$-algebra $\mathscr{A}_{f}$?

Answer 1

Note that $(X,\mu,T)$ is measure theoretically isomorphic to $(X^f_0,\mu^f_0, R)$, where $\mu^f_0=\mu^f(\bullet\,|\, X^f_0)$ and $R=R(T^f;X^f_0):X^f_0\to X^f_0$ is the first return map of $T^f$ to $X^f_0$.

Let $A\subseteq X^f$ be a measurable $T^f$-invariant subset with $\mu^f(A)>0$. Then $A\cap X^f_0\subseteq X^f_0$ is a measurable $R$-invariant subset with $\mu^f_0(A)>0$. Indeed, denoting by $X^f_m$ the height-$m$ slice of $X^f$, we have $A=\biguplus_{m\in\mathbb{Z}_{\geq0}} (A\cap X^f_m)$. Then for some $m$, $\mu^f(A\cap X^f_m)>0$, and since $(T^f)^{-m}(A\cap X^f_m)\subseteq X^f_0$ and $A$ is $T^f$-invariant, $(T^f)^{-m}(A\cap X^f_m)\subseteq A\cap X^f_0$, so that

$$0< \mu^f(A\cap X^f_m)=\mu^f((T^f)^{-m}(A\cap X^f_m))\leq \mu^f(A\cap X^f_0) = \mu^f_0(A)\,\mu^f(X^f_0).$$

Further, $R^{-1}(A\cap X^f_0)$ is by definition a subset of $X^f_0$ ($\mu^f_0$-a.e.) and since $R$ is the first return map of $T^f$, $R^{-1}(A\cap X^f_0)\subseteq_{\mu^f_0} A\cap X^f_0$. Reversing the time direction we get that $R^{-1}(A\cap X^f_0)=_{\mu^f_0} A\cap X^f_0$.

Ergodicity is a measure theoretical property; thus $R$ is ergodic. Consequently $A\cap X^f_0=_{\mu^f_0} X^f_0$, whence $A=_{\mu^f} X^f$, so that $T^f$ is ergodic w/r/t $\mu^f$.