Kac return-time formula without invertibility

Put $C_0=X$ and $C_j={\bigcap }_{i=0}^{j-1}{T}^{-i}{E}^c$ for $j\ge 1$. Since $T^{-1}{C}_j={\bigcap }_{i=1}^j{T}^{-i}{E}^c$, its disjoint split at E consists of $E\cap \{r_E>j\}$ and $C_{j+1}$. Indicator invariance and finite additivity therefore give $\mu (E\cap \{r_E>j\})=\mu (C_j)-\mu (C_{j+1})$ for $j\ge 0$. In particular at j=0 this is $\mu (E)=1-\mu (E^c)$.

For $N\ge 1$, the simple function ${\sum }_{j=0}^{N-1}{\mathbf{1}}_{E\cap \{r_E>j\}}$ equals $\min (r_E,N)$ on E and zero elsewhere. Its integral telescopes to $1-\mu (C_N)$. The sets C_N decrease to points avoiding E at all nonnegative times, a null set by sweep-out. Since $\mu (C_0)=1$, continuity from above yields $\mu (C_N)\to 0$.

The tail sums increase pointwise to $r_E{\mathbf{1}}_E$, including infinite return times. Monotone convergence gives ${\int }_E{r}_{E}d\mu =1$. The complement of the infinitely returning core in E is measurable null, so its nonnegative integral is zero, even where $r_E=\mathrm{\infty }$. Restriction and normalization therefore give ${\int }_{E_{\mathrm{\infty }}}{r}_{E}d{\mu }_E=1\mathrm{/}\mu (E)$; multiplying by the positive denominator reverses the equivalence.