Given a dynamical system $(X,\mathcal{B},\mu,T) $ where $X$ is a space, $B$ is borel $\sigma$-algebra, $\mu$ is a probability measure and $T$ is a $\mu$ invariant transformation i.e. $\mu(T^{-1}(A)=\mu(A)$.
Now let $\tau_A(x)=\min\{n:T^nx \in A, n \in \mathbb{N} \} $ i.e the first hitting or return time.
In the proof of a theorem I cant quite follow the first few lines where a relationship between the hitting time and return time is constructed. Its probably very easy but I just cant see the reasoning. This is:
Let $n\geq 1$ be an integer. The disjoint union $$ \{\tau_A \leq n \}= \{\tau_A \circ T \leq n-1 \} \cup T^{-1}(A \cap \{\tau_A > n-1\}) $$ gives using the invariance of the measure $$ \mu(\tau_A=n)=\mu(A \cap\{\tau_A \geq n\}) $$
I am just a little confused about the disjoint union on the RHS is equal to the LHS and then how we obtain the second statement from this. Any help is greatly appreciated.
This is using the version of $\mathbb N$ that does not include $0$.
$\tau_A(x) \le n$ means $T^k x \in A$ for some $k$, $1 \le k \le n$.
First case: there is such a $k \ge 2$. That says $T^{k-1}(T x) \in A$, where $1 \le k-1 \le n-1$, i.e. $\tau_A(Tx) \le n-1$.
The other possibility is that there is no such $k \ge 2$, but only $k = 1$. That says $Tx$ itself is in $A$, but the next $n-1$ iterates are not in $A$, so $\tau_A(Tx) > n-1$, i.e. $x \in T^{-1} (A \cap \{\tau_A > n-1\})$.
We can also write $\{\tau_A \circ T \le n-1\}$ as $T^{-1}\{\tau_A \le n-1\}$
So we have
$$ \eqalign{\mu(\tau_A \le n) &= \mu(T^{-1}\{\tau_A \le n-1\}) + \mu(T^{-1} (A \cap \{\tau_A > n-1\})\cr &= \mu\{\tau_A \le n-1\} + \mu(A \cap \{\tau_A > n-1\})}$$ and thus $$ \mu(\tau_A = n) = \mu(\tau_A \le n) - \mu(\tau_A \le n-1) = \mu(A \cap \{\tau_A > n-1\})$$