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?

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    $\begingroup$ You seem to assume we all have that book in front of us and know what $g$ and $s$ are. ${}\qquad{}$ $\endgroup$ Apr 24 '15 at 15:12
  • $\begingroup$ Given a $g(x) \in C_c(X)$, there exists a $s(x)$ simple and $\mu({x:s(x)\neq0}) < \infty$ such that etc. That's exactly all you know about $g$ and $s$. I guess I should also mention that $X$ is locally Hausdorff. $\endgroup$ Apr 24 '15 at 22:11
  • $\begingroup$ Locally compact. $\endgroup$ Apr 24 '15 at 22:29
  • $\begingroup$ I agree that $\sup |s|$ would have done same job. Have you found any reason why Rudin uses $||s||_{\infty}$? $\endgroup$
    – Error 404
    Oct 16 '18 at 7:23

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.)

  • $\begingroup$ Sorry, what about the case where $s(x)$ is the indicator of $\{0\}$ under the $\mathbb{R^1}$ Lebesgue measure? In which case $sup_{x \in \mathbb{R^1}} s(x) = 1$, whereas $\mu({x\in X: s(x) > 0.5}) = 0$? Thank you. $\endgroup$ Apr 24 '15 at 22:20
  • $\begingroup$ The essential supremum of that function of a singleton is $0$. ${}\qquad{}$ $\endgroup$ Apr 24 '15 at 22:25
  • $\begingroup$ Sorry, didn't read your comment properly, I understand what you want to claim now. But still, isn't that sort of claim kind of really big and absolutely non-trivial? If possible, please give some justification thanks. $\endgroup$ Apr 24 '15 at 22:27
  • $\begingroup$ Exactly so the essential supremum is not always equal to the supremum. $\endgroup$ Apr 24 '15 at 22:28
  • $\begingroup$ A typo afflicts my comment above: I wrote "of the indicator function of a singleton" and then intended to change it to "of that function", but I neglected to delete the words "of a singleton". ${}\qquad{}$ $\endgroup$ Apr 25 '15 at 1:31

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