Induced transformations preserve restricted finite measure

Fix a trace set $B\subseteq E_{\mathrm{\infty }}$; it is ambient measurable. Write $H_n=E\cap T^{-n}B\cap {\bigcap }_{j=1}^{n-1}{T}^{-j}{E}^c$ and $R_N=T^{-N}B\cap {\bigcap }_{j=0}^{N-1}{T}^{-j}{E}^c$. Pulling B back once and splitting at E gives $\mu (B)=\mu (H_1)+\mu (R_1)$. Pulling $R_N$ back and splitting at E gives $\mu (R_N)=\mu (H_{N+1})+\mu (R_{N+1})$. Thus for each $N\ge 1$, $\mu (B)={\sum }_{n=1}^N\mu (H_n)+\mu (R_N)$.

The H_n are disjoint first-return pieces. Each differs from $H_n\cap E_{\mathrm{\infty }}$ by a subset of the measurable null set $E\setminus E_{\mathrm{\infty }}$; these differences are themselves measurable. Consequently the finite-sum identity implies $\mu ({\bigcup }_{n=1}^N(H_n\cap E_{\mathrm{\infty }}))\le \mu (B)$. Passing to the increasing union gives $\mu (T_E^{-1}B)\le \mu (B)$.

Apply the same inequality to $C=E_{\mathrm{\infty }}\setminus B$. Since $T_E$ maps its entire domain to itself, $T_E^{-1}C=E_{\mathrm{\infty }}\setminus T_E^{-1}B$. All measures here are finite, so $\mu (E_{\mathrm{\infty }})-\mu (T_E^{-1}B)\le \mu (E_{\mathrm{\infty }})-\mu (B)$ gives the reverse inequality. Equality follows for every B; division by $0<\mu (E)<\mathrm{\infty }$ proves invariance of ${\mu }_E$.