Kac's lemma

Remarks

The return time is finite almost everywhere on $A$ by the Poincaré recurrence theorem. For invertible $T$, let $A_n = \set{x \in A : r_A(x) = n}$. The sets $T^j A_n$ with $n \ge 1$ and $0 \le j < n$ are pairwise disjoint, and their union is invariant up to a null set and contains $A$, so by ergodicity it has measure $1$. Hence $1 = \sum_{n \ge 1} n \, \mu(A_n) = \int_A r_A \, d\mu$. The non-invertible case follows by passing to the natural extension. Example: for the rotation $x \mapsto x + 1/q \bmod 1$ on $\set{0, 1/q, \dots, (q-1)/q}$ with the uniform measure and $A$ a single point, the return time is $q$ and $\int_A r_A \, d\mu = q \cdot \frac{1}{q} = 1$. Without ergodicity the integral equals the measure of the set of points whose orbit meets $A$, which can be less than $1$. For the identity map, $r_A = 1$ on $A$ and $\int_A r_A \, d\mu = \mu(A)$.

Going deeper

For a student

Kac's lemma explains why Poincaré recurrence does not contradict the second law of thermodynamics. For a macroscopic system, the set of states near a given low-entropy state has astronomically small measure. So the mean return time is astronomically long. It is also the basis of the induced transformation $T_A(x) = T^{r_A(x)} x$ on $A$, which preserves the normalized measure on $A$ and is ergodic when $T$ is. For an ergodic Markov chain started at a state $s$ with stationary probability $\pi(s) > 0$, it gives the classical formula $1/\pi(s)$ for the expected return time to $s$.