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This is a step used in proving Riesz representation theorem. However I cannot follow his short proof. For every compact set $K$, he construct an $f\in C_{c}(X)$ such that $0\le f\le 1$, and $f(x)=1,x\in K$. This is fine. But then in the end he claim $\int_{X}2fd\mu<\infty$. This means we need to show $$\int_{X}fd\mu<\infty$$ But why this is true? $f$ can really be any function. So more specific if $X=\mathbb{R}^{*}$, $K=[0,1]$, $x=1$ on $K$ and as $x-2$ outside of $K$, then $f$ is definitely not in $L_{1}(X)$. I think I must be confused with something somewhere.

A proof clearly borrowed from his book can be found at here, page 77, Step 3.

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    I have written a proof for Riesz representation Theorem, two months ago when I teahing Real analysis 1 problem solving that there I explore it in 10 pages in details, but it is in persian, can you read persian? to I send it for you.2012-12-23
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    You are wrong because you didn't pay attention to definition of your measure by this theorem.2012-12-23
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    that function you definef is not continuos too, so it is not in $C_{C}(X)$.2012-12-23
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    I see where I wrote become nonsense. The function is supposed to have compact support. But here is the problem - why integrating a function with compact support must yield a finite number? Consider $\frac{1}{x}$, for example.2012-12-23
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    Integrating $\frac{1}{x}$ on a compact subset of its domain is finite. what is your problem with it?2012-12-23
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    This is clear on the real line, but for abstract locally Hausdauff space I do not see why $f$ must be absolutely integrable.2012-12-23
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    For instance don't think on your question and look at continuos function on $\mathbb{R}$, then if you select a compact subset of $\mathbb{R}$ like $K$ then we know that $f$ suppose its maximum and its minimum on $K$ and then you cas see easily by defenition of integral that $\int_{K}f<\infty$.2012-12-23
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    So your question is exactly about the proof of Riesz Represantation theorem, where the measure defined, bring a compact set onto finite number? or somthing else?2012-12-23
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    No, that's not the problem. On the real line the compact sets are bounded, and we know $f$ is bounded from above by $1$ already. But in general arbitrarily locally Hausdauff space I am wondering why $\int_{K}\chi(K)<\infty$.2012-12-23
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    Pay attention that the map of compact set $K$ by a continuos function like $f$ is compact, now if your codomain will be $\mathbb{R}^{n}$ then it is bounded by Hinie-Borel theorem.2012-12-23
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    For the real line this is fine, but for abstract topological space this is not obvious at all. We do not have the measure yet, it has to come from $f$'s value on the domain.2012-12-23
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    How you define integral without measures?2012-12-23
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    In Riesz representation theorem, we do not have the measure yet; we use the function to construct the measure. If we have the measure then we do not need the theorem at all.2012-12-23
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    In this theorem we have a positive linear function from $C_{C}(X)$, then we define a $\sigma$-algebra and a measure on it, then we prove this measure have all we want.2012-12-23
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    So at the time we prove that the integral...., we have a measure there!2012-12-23
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    I see. So the confusion stems from treating the linear functional's value on $f$ as infinite, which cannot happen.2012-12-23
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    Yes, exactly. And if you want I will send my notes for you and if you cann't translate them, tell me to I translate them in English for you if you want.2012-12-23
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    No need for the notes, but thanks for the help and kind of ashamed that I am totally confused.2012-12-23
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    In page 75 of the linked .pdf it is stablished that $\mu$ is locally finite wich is exactly the same thay say that $\mu(K)\lt\infty$ for any subset $K$ compact. In order to have compact sets with finite measure it's enough for the measure to be $\sigma-$finite.2012-12-24
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    No, they did not show it in page 75. They did show it in page 77. You did not read carefully enough.2012-12-24

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