What is an intuitive explanation for Birkhoff's ergodic theorem? If I'm not familiar with measure theory, what is a good way to understand the idea behind the definitions involved, the interpretation of the theorem, and the proofs thereof? Particularly, it's not clear to me what is the significance of the map $T$ often referred to in the definitions, etc. Should it be thought of as a time shift? Even definitions of ergodicity don't mean much to me in terms of the notation. So, I'm looking for a tangible intuitive idea of what these quantities/sets/functions represent (maybe via an example?). A basic explanation with a concrete example or two explaining the definitions from scratch would be really helpful. Thank you!
 A: Formally, Birkhoff's ergodic theorem states that for an integrable function $f$, the time average equals the space average:
$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=0}^{n-1} f(T^i(x)) = \int_S f(x) d\mu(x).$$
Here, $(S,\mathcal{S})$ is a measurable space, $T : S \rightarrow S$ is an ergodic map for the measure $\mu$, and $f \in L^1$. 
Example 1
Suppose we flip a fair coin infinitely many times and encode the result as an infinite vector $\omega = (\omega_1, \omega_2,...)$, where $\omega_i = 0$ if the $i^{th}$ flip is heads and $1$ if it is tails. Let $T$ denote the shift: $T(\omega) = T((\omega_1,\omega_2,\dots)) = (\omega_2,\omega_3, \dots).$ Let $f(0) = 0$ and $f(1) = 10$. We expect the average of $f$ over all flips to be about 5. Birkhoff's ergodic theorem asserts this is true in a suitable sense. The law of large numbers is a special case of Birkhoff's ergodic theorem.
Example 2
Let $S = [0,1]$ be the unit circle. Let $\theta$ be an irrational number and for any $x \in S$, let $T(x) = x+\theta \pmod{1}$. $T$ corresponds to a rotation of the circle $S$. The trajectory of any point $x \in S$ is $\{x + k\theta : k = 1, 2, ... \}$. By plotting points, you will see that the trajectory becomes dense in the circle (because the 'angle' is irrational). The average of $f$ over its trajectory (the time average) is equal to the integral of $f$ over the circle (the space average).  
