About Ergodic Theorems Is there any demonstration purely dynamic of the Birkhoff Ergodic Theorem, i.e, without the Maximal Ergodic Theorem??
I ask this question because I never understood the intuition behind The Maximal Ergodic Theorem.
 A: I know a proof of aubadditive ergodic theorem, without using the Birkhoff ergodic theorem. Thus, the Birkhoff ergodic theorem can be regarded as a Corollary of subadditive ergodic theorem. 
Given the system $(M,f,\mu)$, we call a sequence $\varphi_n: M \to \Bbb R$ subadditive if $\varphi_{m+n} \leq \varphi_m + \varphi_n \circ f^m$.
The statement of the subadditive ergodic theorem is as follows:

Let $\varphi_n : M \to \Bbb R$, $n\ge 1$ be a subadditive sequence of
  measurable functions such that $\varphi_1^{+}\in L^1(\mu)$. Then the
  sequence $(\varphi_n/n)_n$ converges for $\mu$-almost every point to a
  measurable function $\varphi:M \to [-\infty,+\infty)$. Moreover,
  $\varphi^{+}\in L^1(\mu)$ and 
$$ \int \varphi\,d\mu  = \lim_n \frac{1}{n} \int \varphi_n\,d\mu  = \inf_{n}\frac{1}{n} \int \varphi_n \,d\mu\in [-\infty , + \infty). $$

One easily sees that the Birkhorf ergodic theorem follows from the above theorem, since the sequence $\varphi_n=\sum_{i=0}^{n-1}\varphi\circ f^i$ is a subadditive function.
The idea of the proof is to consider the number sequence $a_n=\int \varphi_nd\mu$ and thus, by subadditivity, define 
$$L=\lim_n\frac{a_n}{n}=\inf_n\frac{a_n}{n}$$
Then define 
$$\varphi_-(x)=\liminf \frac{a_n}{n}; \varphi_+(x)=\limsup \frac{a_n}{n}$$
Finally, one tries to show that 
$$\int\varphi_-(x)d\mu\geq L\geq \int\varphi_+(x)dmu$$
and finishes the proof.
The proof following the above idea is given by A.Avila and J.Bochi, Proof of the subadditive ergodic theorem.
http://www.mat.puc-rio.br/~jairo/docs/kingbirk.pdf
