Papa Rudin theorem $1.40$ There is the theorem:
Suppose $\mu(X) < \infty$, $f \in L^1(\mu)$, S is a closed set in the complex plane, and the averages $$A_E(f) = \frac{1}{\mu(E)} \int_E f \, d\mu$$. lie in S for every E $\in$
$\mathfrak M$ with $\mu(E)>0$. Then $f(x)\in S$ for almost all x $\in$ X.
There is the proof:
let $Δ$  be a closed circular disc ( with center at $\alpha$ and radius $r$ $\gt$ $0$, say) in the complement of $S$. Since $S^{c}$ is the union of countably many such discs, it is enough to prove that $\mu(E)$ $=$ $0$, where $E=f^{-1}(Δ)$ .
I don't understand why is it enough to prove that $\mu(E)$ $=$ $0$ and I also don't understand why we took $E=f^{-1}(Δ)$ .
Any help would be appreciated.

You want to show that $f(x) \in S$ for almost every $x \in X$. This is equivalent to saying that the set $$\{x \in X : f(x) \in S^c\} = f^{-1}(S^c)$$ has measure $0$. To show this, he decomposes $S^c$ in a countable union of disks $\Delta$ and shows that each of the $f^{-1}(\Delta)$ have measure $0$. Then, by $\sigma$-additivity, $f^{-1}(S^c)$ also has measure $0$.
