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Rudin RCA p.51

Not only this, i really don't like this book since it gives really terse proofs..

This is the argument how Rudin defines Lebesgue Measure:

Define $P_n=\{a\in\mathbb{R}^k|\text{For any} 1≦i≦k, a_i \text{ is an integral multiple of } 2^{-n}\}$ for every $n\in\mathbb{Z}^+$.

Then, he defines $\Lambda_n f=2^{-nk}\sum_{x\in P_n} f(x)$ for any complex function $f$ on $\mathbb{R}^k$ with compact support and $n\in\mathbb{Z}^+$.

Surely $P_n$ is countable, but is $\sum_{x\in P_n} f(x)$ unconditionally convergent? I seems this works because $f$ has a compact support, but i don't know how.. Thank you in advance

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Continuous images of compact sets are compact. Compact sets are bounded. –  Andres Caicedo Jan 23 '13 at 19:57
@Andres Well, $f$ is not continuous in the hypothesis. –  Katlus Jan 23 '13 at 20:03
Compact support implies that the terms of the sum are non-zero only finitely often. –  Thomas Andrews Jan 23 '13 at 20:07
Almost - you also need the spacing of the set $P_n$. See my proof below and ask questions if confused. –  pre-kidney Jan 23 '13 at 20:12

1 Answer 1

Let $S$ be the support of $f$. Note that the points in the set set $P_n$ are spaced some distance $\epsilon$ away from each other. Thus, the set $S\cap P_n$ is finite.

It follows that $\sum_{x\in P_n} f(x)=\sum_{x\in S \cap P_n} f(x)$ is a finite sum. No issues of convergence to worry about.

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You don't even need the spacing everywhere. It is sufficient to know that $P_n\cap [-a,a]^k$ has at most something like $(2^na+1)^k$ elements. –  Hagen von Eitzen Jan 23 '13 at 20:17
@pre-kidney There's something wrong with my iphone. I keep trying to accept your answer, but then notification appears saying "You can accept an answer in 2 minutes". I'll accept your answer later. –  Katlus Jan 23 '13 at 23:18

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