Rudin Real & Complex Analysis Thm 3.14

Question

In the proof, the author claims that by Lusin's theorem, $g(x) = s(x)$ except on a set of measure $< \epsilon$ and $|g| \leq \|s\|_\infty$, ($g(x) \in C_c(X)$, $s(x)$ simple and $\mu(\{x:s(x) \neq 0\}) < \infty$)). I don't understand how he can go from $|g| \leq \sup_X |s|$ to $|g| \leq \|s\|_\infty$ without any justification, because from what I understand $\sup_X |s|$ isn't necessarily equal to $\|s\|_\infty$ (suppose $s(x) := \chi_A$ where $\mu(A) = 0$ for example). I guess you can modify the proof of Lusin's to replace $\sup_X |s|$ with $\|s\|_\infty$, but then again I don't even understand why it's necessary to use $\|s\|_\infty$ here because $\sup_X |s|$ would've done exactly the same job in the proof. I guess what I'm asking is am I missing something major here?

Answer 1

In probability and measure theory almost (no pun intended) any purpose served by a simple function can be served equally well by a simple function whose supremum (which in the case of simple functions, is equal to its maximum) is equal to its essential supremum.

(The essential supremum of a measurable function $f$ on a domain $X$ is the smallest number $c$ for which the measure of $\{x\in X : f(x)>c\}$ is zero.)

Thus if Rudin's proof has a gap, you can fill it in in that way.

(OK, the pun may have been slightly intended.)