Measure-preserving mapping

Question

Let $(X, \mu, T)$ be a mesure-preserving mapping. Let $A \subset X$ be a measurable subset such that any point in $A$ eventually comes back to $A$. We define space $(A, \mu_A)$, $ \mu_A ( B) = \mu (B) / \mu (A)$ and mapping $T_A: A \mapsto A$ such that $T_(x) = T^{n(x)} (x)$ where $n(x) = \min \{ n >0: T^n (x) \in A \}$. Show that $(A, \mu_A , T_A)$ is a measure- preserving mapping.

I have no idea how to deal with it. Please help in any way.

Answer 1

An intuitive way to think about this is via the pointwise ergodic theorem. The trajectory of a point $x$ that is typical according to $\mu$ visits $A$ with frequency $\mu(A)$. Among the time steps $\{n: T(x)\in A\}$ that the system is in $A$, the system will be in $B\subseteq A$ with frequency $\mu(B)/\mu(A)=\mu_A(B)$. This implies that for a point in $A$ that is typical according to $\mu$ (in other words, a typical point according to $\mu_A$), the trajectory of $T_A$ visits $B$ with frequency $\mu_A(B)$.

This can be made into a proof, but there is also an elementary argument. In a probabilistic language, we simply condition on the first iterate of $T$.

Namely, for $B\subseteq A$, we can write \begin{align} & \hspace{-2em}\bigcup_{n=1}^\infty \Big[T^{-n}(B)\cap\bigcap_{i=1}^{n-1}T^{-i}(A^c)\Big] \\ &= T^{-1}(B) \,\cup\, T^{-1}\Big( \bigcup_{m=1}^\infty \Big[T^{-m}(B)\cap\bigcap_{j=1}^{m-1}T^{-j}(A^c)\Big] \cap A^c \Big) \\ &= T^{-1}(B) \,\cup\, T^{-1}\Big( \bigcup_{m=1}^\infty \Big[T^{-m}(B)\cap\bigcap_{j=1}^{m-1}T^{-j}(A^c)\Big] \setminus T_A^{-1}(B) \Big) \;. \end{align} The left-hand side is simply the set of points in $X$ whose trajectories enter $A$ for the first time through $B$. We have decomposed this set into the set of points that go into $B$ in one step and the rest. Using the invariance of $\mu$ under $T$ we get \begin{align} & \hspace{-2em}\mu\left(\bigcup_{n=1}^\infty \Big[T^{-n}(B)\cap\bigcap_{i=1}^{n-1}T^{-i}(A^c)\Big]\right) \\ &= \mu(B) \,+ \, \mu\Big( \bigcup_{m=1}^\infty \Big[T^{-m}(B)\cap\bigcap_{j=1}^{m-1}T^{-j}(A^c)\Big] \setminus T_A^{-1}(B) \Big) \\ &= \mu(B) \,+\, \mu\Big( \bigcup_{m=1}^\infty \Big[T^{-m}(B)\cap\bigcap_{j=1}^{m-1}T^{-j}(A^c)\Big] \Big) \,-\, \mu\big(T_A^{-1}(B)\big) \;. \end{align} which implies $\mu\big(T_A^{-1}(B)\big)=\mu(B)$.

Answer 2

Write $A_n=\{x\in A: n(x)=n\}$ for $n\in\mathbb N$. Then note that $A_n$ are disjoint and exhaustive in $A$, because every point in $A$ returns to $A$ in finite time.

Then for any $B$ in the $\sigma$ algebra restricted to $A$ space, write $B=\cup_n (B\cap A_n)$, thus $\mu(B)=\sum_n \mu(B\cap A_n)$.

Does it help in showing $\mu\circ T_A^{-1}=\mu$? Notice that showing this is enough as $\mu_A=\mu/\mu(B)$.