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Problem: Let $\mu$ be a positive (finite?) measure on a space $X$. Show that every $f\in L^\infty(\mu)$ is a uniform limit of simple functions $f_i$.

This question came while reading a proof in Rudin's Real and Complex Analysis (Theorem 6.16). It is mentioned without any explanation. Is this a simple corollary of another theorem? I can't see why this is true. I know that every non-negative measurable function is the pointwise limit a simple functions, but that doesn't help.

Cachalot
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    follow the proof of the statement you've mentioned (pointwise limit) and see how good the approximation is at each step, at least for bounded functions. If you want to you can first assume $0 \leq f \leq 1$ to see how it works and then generalize. Oh, and if you need extra help be sure to post what you got from following my hint. – mm-aops Feb 02 '15 at 23:32
  • @mm-aops Since $f:X\to \mathbb{C}$ is essentially bounded, I guess that we can only find simple measurable functions $f_i$ that uniformly converges to $f$ on $X-N$ with $\mu(N) = 0$. – flyingwith Nov 25 '19 at 07:03
  • $\mu$ is a positive $\sigma$-finite measure in that theorem. – stoic-santiago Jul 09 '21 at 09:40

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