I've been looking at the proof of 'simple function is dense in $L^p$ space in the Rudin's book 'Real and complex analysis(RCA)', Page 69, theorem 3.13.

My question is in the assumption, there is a 'finite support simple function'. But the proof does not prove the simple function has finite support. Rudin might think this is obvious, but I'm not.

How could we verify this?


1 Answer 1


By definition a simple function $s$ has a finite range. In particular, if non-negative it has a minimum value $m>0$ on its support $\Omega$. If $L^p$-integrable (as in Rudin) we must have $$m^p\mu(\Omega)\leq \int s^p <+\infty$$ so the support must be finite.

  • $\begingroup$ I was too fool. Thanks! $\endgroup$
    – kayak
    Commented Aug 5, 2016 at 9:37

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