How to show $S_n(x)\geq(n-M)(\overline{A}(x)-\epsilon)$? Question
Suppose $T$ is a measure-preserving transformation of the probability space $(X,{\cal A},\mu)$, and let $B\in{\cal A}$. Set $$S_n(x)=\sum_{i=0}^{n-1}1_B(T^ix)$$ and $$A_n=\frac{1}{n}S_n(x),$$ for each $n\in\mathbb{N}$. Let $$\overline{A}(x)=\limsup_{n\to\infty}A_n(x),$$ and let $$\tau(x)=\min\{n\in\mathbb{N}:A_n(x)\geq\overline{A}-\epsilon\}.$$ Suppose that there exists an integer $M$ such that for $\mu$-almost every $x$, $\tau(x)\leq M$.
Why $$S_n(x)\geq(n-M)(\overline{A}(x)-\epsilon)?$$
Context
This is part of a proof in Bedford et al (1996)'s "Ergodic Theory, Symbolic Dynamics, and Hyperbolic Spaces". The theorem says that for $\mu$-almost every $x$, $\lim_{n\to\infty}A_n(x)$ exists (this is a general version of the Ergodic Theorem). He says that the inequality

follows by using $\tau$ to decompose the orbit of $x$ up to time $n$ into pieces on each of which the average number of visits to $B$ is at least $\overline{A}-\epsilon$ (remember that also $\overline{A}$ is $T$-invariant), and the piece left over has length bounded by $M$.

So far, I have figured out that
$$\forall n\in[1,\tau(x))~~A_n(x)<\overline{A}(x)-\epsilon,$$
but I don't understand any of the above comment or see any way to get the mentioned inequality.
 A: Define $H_M = \{x\in X: \tau(x) \leq M\}$, from the assumptions $\mu(H_M) =1$. Let 
$$
H= \bigcap_{n\geq 0} \left(T^{-n}(H_M)\cap H_M \right),
$$ and observe that by the invariance of the measure $\mu(H)=1$. The good property of $H$ is that for any $x\in H$ it is verified that $T^n(x) \in H_M$, for every $n\geq 0 $.
For $x\in H$, define inductively $n_1(x) = \tau(x)$, $n_2(x)= \tau(T^{n_1(x)}(x))$, $\cdots$ , $n_k(x) = \tau( T^{n_1(x) + \cdots + n_{k=1}(x)}(x))$. For each $x\in H$, the numbers $n_i(x)$ divide the natural numbers on the times with "good" estimate. By construction $n_{i}(x) \leq M$. For each $k\geq 1$, write $N_k(x) = n_1(x) + \cdots + n_k(x)$.
For each $x\in H$ and $n>0$, let $K=K(x,n) = \max \{i\in \mathbb{N}: n_1(x) + \cdots + n_i(x) \leq n \}$, we have that 
$$
\frac{1}{n}S_n(x) = \frac{1}{n}\left( S_{n_1(x)}(x) + \cdots + S_{n_{K}(x)}(T^{N_{K-1}(x)}(x)) \right) + \frac{1}{n} S_{n-N_{K}(x))} (T^{N_{K}(x)}(x)).
$$
Of course, $\frac{1}{n} S_{n-N_{K}(x))} (T^{N_{K}(x)}(x))\geq 0$. On the other hand, by the definitions of the times $n_i(x)$
$$
S_{n_1(x)}(x) + \cdots + S_{n_{K}}(T^{N_{K-1}(x)}(x)) \geq n_1(x)(\overline{A}(x)-\varepsilon) + \cdots + n_K(x)(\overline{A}(T^{N_{K-1}(x)}(x))-\varepsilon).
$$
By the $T$ invariance of the function $\overline{A}$, we conclude that
$$
S_{n_1(x)}(x) + \cdots + S_{n_{K}}(T^{N_{K-1}(x)}(x)) \geq (n_1(x) + \cdots n_K(x)) (\overline{A}(x)-\varepsilon).
$$
Since the times $n_i(x) \leq M$ and by the definition of $K=K(x,n)$, we obtain $n_1(x) + \cdots n_K(x) \geq n-M$. Hence
$$
\frac{1}{n}S_n(x) \geq \frac{n_1(x) + \cdots + n_K(x)}{n}.( \overline{A}(x)-\varepsilon) \geq \frac{n-M}{n}( \overline{A}(x) -\varepsilon).
$$
So for every $n$ large enough, we have
$$
\frac{1}{n}S_n(x) \geq (\overline{A}(x) - 2\varepsilon).
$$
From this, one concludes that 
$$
\liminf_{n\to +\infty} \frac{1}{n} S_n(x) \geq \overline{A} - 2\varepsilon.
$$
Since $\varepsilon$ is arbitrary, one obtains that for $\mu$-almost every point the limit $\displaystyle\lim_{n\to + \infty} \frac{1}{n} S_n(x)$ exists.
