TITLE: Does anyone know an intuitive proof of the Birkhoff ergodic theorem?
QUESTION [47 upvotes]: For many standard, well-understood theorems the proofs have been streamlined to the point where you just need to understand the proof once and you remember the general idea forever.  At this point I have learned three different proofs of the Birkhoff ergodic theorem on three separate occasions and yet I still could probably not explain any of them to a friend or even sit down and recover all of the details.  The problem seems to be that they all depend crucially on some frustrating little combinatorial trick, each of which was apparently invented just to service this one result.  Has anyone seen a more natural approach that I might actually be able to remember?  Note that I'm not necessarily looking for a short proof (those are often the worst offenders) - I'm looking for an argument that will make me feel like I could have invented it if I were given enough time.

REPLY [6 votes]: I like an answer that was shown to me by Mate Wierdl.
You restate the maximal inequality this way:
Let
$$
Mf(x) = \sup_N (1/N)\sum_{n < N}f(T^nx).
$$
and first prove that $\|Mf\|_{1,\infty}\le\|f\|_1$.
Here by $\|\cdot\|_{1,\infty}$
I mean the "weak $L^1$ norm" (which is not actually a norm at all) defined by
$\|g\|_{1,\infty}=\sup_a a \cdot m( \{ x\colon |g|(x)\le a \}) $
(the area of the biggest rectangle that fits under $|g|$).
This has a beautiful (and simple and intuitive) proof if you are working on the integers and $T$ is the map $T(n)=n+1$. There is then a technique called transference that allows you to take the proof in your favourite system and transfer it to any other system.
You then show that this implies that when you have a maximal inequality, the set of functions for which you get almost everywhere convergence is closed in $L^1$. (Suppose the ergodic theorem works for $f_1,f_2,\ldots$ and $\|f_n-f\|_1\to 0$. Use the maximal inequality to deduce that the ergodic theorem holds for $f$).
At this stage you know that the set of $f$'s for which the ergodic theorem holds is a closed set, so you're done if you can find a dense set of $f$'s for which this works. Since you're looking for a dense set you're OK to work with $L^2$. The result is obvious for coboundaries and the orthogonal complement of the space of coboundaries is the set of invariant functions (for which the result is even more obvious). QED.