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How do I show that the linear span of idempotents is dense in $L^{\infty}(\Omega,\mu)$ where $(\Omega,\mu)$ is a measure space? I don't really have any idea how to do this. Does it involve approximating some other class that is dense, say the simple functions?

cyc
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    The first thing is to identify the idempotents. The next thing is to note that the essential range of an $L^\infty$ function is (relatively) compact. – Daniel Fischer Nov 02 '13 at 19:29
  • Why do I need compactness of the essential range? – cyc Nov 02 '13 at 22:07
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    The total boundedness is the important thing. You want, for every $\varepsilon > 0$, a finite set so that each value is within a distance of $\varepsilon$ from some point in the finite set. – Daniel Fischer Nov 02 '13 at 22:29
  • @DanielFischer Am I right that $f\in L^\infty$ is idempotent if and only if $f=\chi_S$ (characteristic function) for $S\subseteq \Omega$ measurable? – Rudy the Reindeer Jun 24 '14 at 11:23
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    @MattN. Yes. Modulo $a.e. [\mu]$, if we're pedantic. – Daniel Fischer Jun 24 '14 at 11:26
  • @DanielFischer Thank you! And: Yes, you're right (we are pedantic : )) – Rudy the Reindeer Jun 24 '14 at 11:27
  • @DanielFischer Could you give me a hint on how to show that the simple functions are dense in $L^\infty$? What I have is that if $\varepsilon > 0$ then the goal is to find measurable sets $S_1,\dots, S_n$ and coefficients $c_1,\dots, c_n$ such that $$ |f- \sum_{k=1}^n c_k \chi_{S_k}|_\infty < \varepsilon$$

    Now I'm not sure how to actually construct the sets and determine the coefficients.

     – Rudy the Reindeer Jun 24 '14 at 11:41
  • @DanielFischer Oh, I see that you already gave this hint in your very first comment to this question! Thank you : ) – Rudy the Reindeer Jun 24 '14 at 11:42
  • @DanielFischer But why is the image of an (essentially) bounded function compact? I can only see that it's bounded (obviously). – Rudy the Reindeer Jun 24 '14 at 11:47
  • @MattN. The image of $f$ itself has no reason to be closed, nor actually bounded, if $\mu$ has infinite null sets. The essential range (${ c : (\forall \rho>0)(\mu(f^{-1}(B_\rho(c))) > 0)}$), however, is bounded and closed, so compact (but we don't need that it is closed; bounded, hence totally bounded [we're talking of subsets of euclidean space], is enough). – Daniel Fischer Jun 24 '14 at 12:03
  • @DanielFischer Oh, "essential range", I see (new word added to my vocabulary!). And now I also understand the second half of your comment. I thought we use compactness to cover the range using finitely many $\varepsilon$-balls. Then the inverse image of each ball yields a measurable set $S_k$. Pick $c_k$ to be any value in $f(S_k)$. If we choose $\varepsilon$ small enough it will make the error small enough. But boundedness is of course enough to do that. – Rudy the Reindeer Jun 24 '14 at 12:17
  • @DanielFischer Could you tell me if I got it? I posted a tentative proof here. – Rudy the Reindeer Jun 24 '14 at 13:01
  • @MattN. Answered here. – Daniel Fischer Jun 24 '14 at 13:08

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The simple functions are precisely the linear span of the idempotents.

Martin Argerami
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