I'm specifically concerned in the case where $\mu$ is a finite measure, but that hasn't shown up anywhere yet. I've also been working in the case where $L^1$ is separable just because that makes the topology easier to handle (I don't know anything about nets other than that they're supposed to be useful).
The extreme points of the unit ball are exactly those functions $\chi_A-\chi_{A^c}$, where $A$ is some measurable set. Denote such a function $f_A$. Then for each infinite sequence $f_{A_n}$, the fact that $L^\infty$ is weak star compact implies that $f_{A_n}$ has a limit point, so there is a subsequence $f_{A_m}\xrightarrow{w*}f$, where $f\in L^\infty$, $\|f\|_\infty\le 1$. To show that $f$ is an extreme point, normally I would assume that it is not and produce a set $B$, $\mu(B)>0$, such that $|f|<1-\epsilon$ on $B$ for some $\epsilon$. Unfortunately, this doesn't seem to play well with the weak star convergence.
In general, one would probably want to find a weak open neighborhood containing any non-extreme point disjoint from each extreme point to show closure of the set of extreme points. Again, if $|f|<1-\epsilon$ on $B$ for some $B$, $W(f;\chi_B)$ seems like a natural choice that hasn't yet panned out.
Thanks in advance.
This does not hold if $\mu$ is not atomic. If you even have that $\mu$ does not contain any atom, every point in the unit ball can be approximated by extreme points, see Approximate partition of unity by characteristic functions.
In the case that $\mu$ is atomic, it is quite easy to prove.