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If one doesn't know a motivation, it's hard to memorize such a theorem. So do I.

Rudin RCA p.30

Let $(X,\Sigma,\mu)$ be a measure space such that $\mu(X)<\infty$ and $f\in L^1(\mu)$.

Let $S$ be a closed subset of $\mathbb{C}$

If [$\forall E\in\Sigma, \mu(E)>0 \Rightarrow \frac{1}{\mu(E)}\int_E f d\mu \in S$], then $f(x)\in S$ almost everywhere.

Maybe since it's deserves to be marked as a theorem in the text, Rudin provided it.

A problem is, he never gives a motivation.

I don't know any motivation for this theorem (which looks similar to the mean value theorem to me). And in the hypothesis, why $S$ has to be closed? The theorem still holds even when $S$ is a $F_\sigma$ set. And what is an example this theorem is applied?

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This follows from the Lebesgue differentiation theorem, which is one of the corner stones of integration theory, so you should really read and understand it. –  Your Ad Here Feb 15 at 9:50
    
Even so, this doesn't answer the OP's question (a possible response is "this is why this is a comment"). The point is not the proof, but rather why this might be important. Why would someone think this is relevant? I don't mean to be rude. –  Mark Fantini Feb 15 at 9:52

1 Answer 1

up vote 3 down vote accepted

Rudin later uses it to prove the Radon-Nikodym theorem (theorem 6.10 on pp. 121-123) as well as a couple of further results following Radon-Nikodym (theorems 6.12 and 6.16).

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