This question is about the unanswered part of Big Rudin 1.40: Open Set is a countable union of closed disks?. Specifically, in Rudin's Real and Complex Analysis, we have
My question: It appears to me that the proof remains equally valid, if we let $\Delta$ be an open circular disk in $S^c$, doesn't it? For we only need to make one other change to the proof above:
$$\mid A_E(f)-\alpha\mid = \frac{1}{\mu(E)}\left\vert\int_E (f-\alpha)d\mu\right\vert \le \frac{1}{\mu(E)}\int_E \left\vert(f-\alpha)\right\vert d\mu < r,$$
(which is also impossible, since $A_E(f)\in S.$) Is this correct? If I missed something, I'd appreciate it if someone can point it out. Thanks a lot!
Yes, the proof also works with open disks. One point that may have made Rudin prefer to use closed disks is that
$$g(x) \leqslant r \text{ for all } x \implies \int_E g\,d\mu \leqslant r\cdot \mu(E) \text{ for all measurable } E$$
is a wee bit easier to prove than
$$g(x) < r \text{ for all } x \implies \int_E g\,d\mu < r\cdot \mu(E) \text{ for all measurable $E$ with } 0 < \mu(E) < +\infty\,.$$