Q:

Why is $\mu(\bigcap_{n=1}^{\infty}T^nX)=1$? 

Let $\mu$ be an $T$-invariant measure on $(X,\mathcal{B})$. Then it is
$$
\mu(T^nX)=\mu(T^{-n}T^nX)=\mu(X)=1. 
$$
Why is then
$$
\mu(\bigcap_{n=1}^{\infty}T^n(X))=1? 
$$

A:

For all $n$ you have $\mu(X \setminus T^nX)=0$, so
$$\mu \left( \bigcup_n (X \setminus T^nX) \right) \le \sum_n\mu (X \setminus T^nX) = \sum_n 0 = 0$$
But now, 
$$0=\mu\left( \bigcup_n (X \setminus T^nX) \right) = \mu \left(X \setminus \left( \bigcap_n T^nX \right) \right) = 1- \mu\left( \bigcap_n T^nX \right)$$

