Why are abelian groups amenable?
Here is a simpler argument, combining 1--6 into one step.
Let $G$ be a countable abelian group generated by $x_1,x_2,\ldots$. Then a Følner sequence is given by taking $S_n$ to be the pyramid consisting of elements which can be written as
$a_1x_2+a_2x_2+\cdots+a_nx_n$ with $\lvert a_1\rvert\leq n,\lvert a_2\rvert\leq n-1,\ldots,\lvert a_n\rvert\leq 1$.
The invariant probability measure is then defined by $\mu(A)=\underset{\omega}{\lim}\lvert A\cap S_n\rvert / \lvert S_n\rvert$ as usual.
A more natural way to phrase this argument is:
- The countable group $\mathbb{Z}^\infty$ is amenable.
- All countable abelian groups are amenable, because amenability descends to quotients.
But I would like to emphasize that there is really only one step here, because the proof for $\mathbb{Z}^\infty$ automatically applies to any countable abelian group. This two-step approach is easier to remember, though. (The ideas here are the same as in my other answer, but I think this formulation is much cleaner.)
2016 Edit: Here is an argument to see that $S_n$ is a Følner sequence. It is quite pleasant to think about precisely where commutativity comes into play.
Fix $g\in G$ and any finite subset $S\subset G$. We first analyze the size of the symmetric difference $gS\bigtriangleup S$. Consider the equivalence relation on $S$ generated by the relation $x\sim y$ if $y=x+g$ (which is itself neither symmetric, reflexive, or transitive). We will call an equivalence class under this relation a "$g$-string". Every $g$-string consists of elements $x_1,\ldots,x_k\in S$ with $x_{j+1}=x_j+g$.
The first key observation is that $\lvert gS\bigtriangleup S\rvert$ is at most twice the number of $g$-strings. Indeed, if $z\in S$ belongs to $gS\bigtriangleup S$, then $z$ must be the "leftmost endpoint" of a $g$-string; if $z\notin S$ belongs to $gS\bigtriangleup S$, then $z-g$ must be the "rightmost endpoint" of a $g$-string; and each $g$-string has at most 2 such endpoints (it could have 1 if the endpoints coincide, or 0 if $g$ has finite order).
Our goal is to prove for all $g\in G$ that $\frac{\lvert gS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$ as $n\to \infty$. Since $\lvert abS\bigtriangleup S\rvert\leq\lvert abS\bigtriangleup bS\rvert+\lvert bS\bigtriangleup S\rvert= \lvert aS\bigtriangleup S\rvert+\lvert bS\bigtriangleup S\rvert$, it suffices to prove this for all $g_i$ in a generating set.
By the observation above, to prove that $\frac{\lvert g_iS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$, it suffices to prove that $\frac{\text{# of $g_i$-strings in $S_n$}}{\lvert S_n\rvert}\to 0$. Equivalently, we must prove that the reciprocal $\frac{\lvert S_n\rvert}{\#\text{ of $g_i$-strings in $S_n$}}$ diverges, or in other words that the average size of a $g_i$-string in $S_n$ diverges.
We now use the specific form of our sets $S_n=\{a_1g_1+\cdots+a_ng_n\,|\, \lvert a_i\rvert\leq n-i\}$. For any $i$ and any $n$, set $k=n-i$ (so that $\lvert a_i\rvert\leq k$ in $S_n$). The second key observation is that every $g_i$-string in $S_n$ has cardinality at least $2k+1$ unless $g_i$ has finite order. Indeed given $x\in S_n$, write it as $x=a_1g_1+\cdots+a_ig_i+\cdots+a_ng_n$; then the elements $a_1g_1+\cdots+bg_i+\cdots+a_ng_n\in S_n$ for $b=-k,\ldots,-1,0,1,\ldots,k$ belong to a single $g_i$-string containing $x$. If $g_i$ does not have finite order, these $2k+1$ elements must be distinct. This shows that the minimum size of a $g_i$-string in $S_n$ is $2n-2i+1$, so for fixed $g_i$ the average size diverges as $n\to \infty$.
When $g_i$ has finite order $N$ this argument does not work (a $g_i$-string has maximum size $N$, so the average size cannot diverge). However once $N<2k+1$, the subset containing the $2k+1$ elements above is closed under multiplication by $g_i$. In other words, once $n\geq i+N/2$ the set $S_n$ is $g_i$-invariant, so $\lvert g_iS_n\bigtriangleup S_n\rvert=0$.
I'm grateful to David Ullrich for pointing out that this claim is not obvious, since the quotient of a Følner sequence need not be a Følner sequence (Yves Cornulier gives an example here).
There is a slightly quicker approach than that outlined by Mariano, using the Markov-Kakutani fixed point theorem. I first learned of this from Rudin's Functional Analysis -- more accurately, I remember skimming over that part as a struggling undergraduate, and then years later, once I'd heard of amenability, realizing the connection. (As far as I know, the word "amenable" is never mentioned in the book.)
The proof can be found via Theorems 5.23, 5.24 and 5.25 of the aforementioned book (2nd ed. if that makes any difference) and I don't think I can improve on the exposition there.
(Regarding the approach Tom outlines: it might be worth observing that amenability passes to quotient groups (this is particularly obvious using the invariant mean, but probably isn't too hard in most of the other formulations). Therefore, going from 3 to 4 doesn't need the classification theorem.)
A vague remark on possible alternative routes. From one point of view, the "invariant mean" formulation of amenability is just a convenient way to avoid epsilon-delta arguments with Folner nets. Now, Folner sequences can be easily constructed in ${\mathbb Z}^n$ for any finite $n$ -- just take cubes centred at zero of ever increasing width -- but at present I can't quite nail down how to transfer the construction over to arbitrary abelian groups. (Of course, things would be easier if we could pass to the formulation with an invariant mean; but then one might as well work with the invariant mean throughout.)
I believe that the neatest proof for amenability of abelian topological groups, is via the Markov-Kakutani fixed point theorem: see the half-page proof of this result as Theorem G.2.1 in: MR2415834 (2009i:22001) Bekka, Bachir; de la Harpe, Pierre; Valette, Alain Kazhdan's property (T). New Mathematical Monographs, 11. Cambridge University Press, Cambridge, 2008. xiv+472 pp. ISBN: 978-0-521-88720-5