Inducing an ergodic system gives an ergodic system

Let $U={\bigcup }_{n\ge 0}{T}^{-n}{E}_{\mathrm{\infty }}$. It is measurable and conull by positive-set sweep-out. It is strictly invariant: if Tx eventually enters the core then x does; if x eventually enters at a positive time then Tx does, while if x is already in the core, its next positive return belongs to the core. Thus $T^{-1}U=U$. On U let $q(x)=\min \{n\ge 0:T^{n}x\in E_{\mathrm{\infty }}\}$. The fibers $\{q=n\}=T^{-n}{E}_{\mathrm{\infty }}\setminus {\bigcup }_{j<n}{T}^{-j}{E}_{\mathrm{\infty }}$ are measurable.

Take any finite real measurable f on the core with $f\circ T_E=f$ everywhere. Define $F(x)=f(T^{q(x)}x)$ for $x\in U$ and $F(x)=0$ otherwise. The measurable q-fibers make F measurable. If $x\in U\setminus E_{\mathrm{\infty }}$, then $q(Tx)=q(x)-1$, hence F(Tx)=F(x). If $x\in E_{\mathrm{\infty }}$, then $q(Tx)=r_E(x)-1$: before the first return there is no E visit, and the first return is in the core. Therefore $F(Tx)=f(T_{E}x)=f(x)=F(x)$. On $X\setminus U$, both x and Tx are outside U and F is zero.

Ergodicity of T and the everywhere invariant-function criterion give a constant c with F=c almost everywhere. Since F=f on the core, f=c for ${\mu }_E$-almost every point there. The same criterion applied to the probability-preserving T_E now proves its ergodicity. Working with everywhere invariant functions avoids choosing representatives for almost-everywhere invariance.