Why is the first return map measure preserving? $\newcommand{\set}[1]{\{#1\}}$
$\newcommand{\mc}{\mathcal}$
Definitions and Context
Let $(X, \mc X, \mu, T)$ be an invertible measure preserving system and let $A$ be a measurable subset of $X$ with $\mu(A)>0$.
By Poincare recurrence we know that for almost all points $x\in A$ there is a positive integer $k$ such that $T^kx\in A$.
Thus we may define a map $r_A:A\to \mathbb N$ by writing
$$
r_A(x) = \min\set{k\geq 1:\ T^kx\in A}
$$
So $r_A$ is well-defined almost everywhere in $A$.
Also define a map $T_A:A\to A$ by writing
$$
T_Ax= T^{r_A(x)}x
$$
Let $\mu_A$ be a measure on $A$ obtained by restricting $\mu$ to $A$ and normalizing, that is, $\mu_A(B)=\mu(B)/\mu(A)$ for any measurable subset $B$ of $A$.
The Problem
Lemma 2.43 in Einsiedler and Ward's Ergodic Theory with a view towards Number Theory states that $T_A$ preserves the measure $\mu_A$.
What the proof given in the book shows is that for a measurable subset $B$ of $A$ we have $\mu_A(T_A(B))=\mu_A(B)$.
I do not see how this shows that $T_A$ is measure preserving.
What one needs to show is that for any measurable subset $B$ of $A$ we have $\mu_A(T_A^{-1}(B))=\mu_A(B)$, which is equivalent to showing that $\mu(T_A^{-1}(B))=\mu(B)$.
Here is what I tried.
Define, as in the book, $A_n=r_A^{-1}(n)$.
Thus the $A_i$'s partition $A$.
Let $B$ be an arbitrary measurable subset of $A$.
Then we have
$$
T_A^{-1}(B) = \bigsqcup_{n\geq 1} \set{x\in A_n:\ T_Ax\in B} = \bigsqcup_{n\geq 1} \set{x\in A_n:\ T^nx\in B}
$$
Thus
$$
T_A^{-1}(B) = \bigsqcup_{n\geq 1} A_n\cap T^{-n}(B)
$$
So we have
$$
\mu(T_A^{-1}(B)) = \sum_{n\geq 1} \mu(A_n\cap T^{-n}(B))
$$
I am stuck here.
 A: This fact is general for any invertible measure preserving system:

Let $(X,\mathcal{B},\mu,T)$ be such that $T$ is invertible and measurable with $T^{-1}$ also measurable. Then $\mu(T^{-1}(A))=\mu(A)$ for all $A\in\mathcal{B}$ if and only if $\mu(TA) = \mu(A)$ for all $A\in\mathcal{B}$.

In fact we only need one direction of this theorem, that $\mu(TA)=\mu(A)$ for all $A\in\mathcal{B}$ implies that $\mu(T^{-1}(A))=\mu(A)$ for all $A\in\mathcal{B}$. This direction alone holds even if $T$ is not invertible, but is surjective. We will prove that in a minute, but this is what we need for your question.
Back to your question: Let $B$ be a measurable subset of $A$. Then since $T$ and $r_A$ are measurable, we have that $C:=T_A^{-1}(B)$ is also a measurable subset of $A$.
The proof given in the book shows that $\mu_A(T_A C) = \mu_A(C)$. Since $T_A$ is surjective (see proof below) we have $$T_A(C) = T_A T_A^{-1}(B)=B$$ which implies that $\mu_A(B)=\mu_A(T_A^{-1}(B))$ as required.
The proof that $T_A$ is surjective goes as follows: Consider the measure-preserving system $(X,\mathcal{B},\mu,T^{-1})$ and the same measurable set $A$. By poincare recurrence theorem for almost all $a\in A$ there exists $n$ such that $T^{-n}(a)\in A$. Let $b:=T^{-n}(a)$ for the minimal $n$, then $T_A(b)=a$. This shows that $T_A$ is essentially surjective which is enough for our purposes.
