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Rudin RCA p.43

Riesz Representation for LCH:

Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a positive linear functional on $C_c(X)$. Then there exists a $\sigma$-algebra $\mathfrak{M}$ on $X$ which contains all Borel sets in $X$, and there exists a unique positive measure $\mu$ on $\mathfrak{M}$ which represents $\Lambda$ in the sense that $\Lambda f = \int_X f d\mu$ for every $f\in C_c(X)$.

Since Urysohn's lemma can be proven without choice when $X$ is locally compact Hausdorff and second countable, i thought Riesz-Representation theorem could be proven without choice if $X$ is, additionally, second countable.

All the other parts of the Rudin's proof works fine without choice, but there is one part i'm stuck.

Below is Rudin's argument:

(The notation $f\prec V$ will mean that $V$ is open, that $f\in C_c(X)$, $0≦f≦1$, and that $\text{supp}f \subset V$)

For every open set $V$ in $X$, define $\mu(V)=\sup\{\Lambda f : f\prec V\}$.

Then define $\mu(E)=\inf\{\mu(V):E\subset V, V \text{ open}\}$ for every subset $E$ of $X$. (It is consistent with the above definition. Then it can be shown that "$E\subset F \Rightarrow \mu(E)≦\mu(F)$"

Now, let $\{E_n\}$ be a sequence of subsets of $X$, and I need to prove "$\mu(\bigcup_{n\in\mathbb{N}} E_n)≦\sum_{n=1}^{\infty} \mu(E_n)$".

Since Urysohn's Lemma holds for locally compact second countable space, it can be proven that $\mu(V_1\cup V_2)≦\mu(V_1)+\mu(V_2)$ for every open sets $V_1,V_2$.

Suppose $\mu(E_n)<\infty$ for all $n\in\mathbb{N}$.

Fix $\epsilon>0$.

Then, by definition, there exist open sets $V_n$ such that $E_n\subset V_n$ and $\mu(V_n)<\mu(E_n) + 2^{-n}\epsilon$ (**Axiom of Countable Choice is used here)

Put $f\prec \bigcup_{n\in\mathbb{N}} V_n$. Since $\text{supp} f$ is compact there exists a finite subcover $\{V_1,...,V_m\}$ of $\text{supp} f$, hence $\Lambda f≦\mu(\bigcup_{i=1}^m V_i)≦\sum_{i=1}^m \mu(V_i)≦\sum_{i=1}^{\infty} \mu(E_i) + \epsilon$.

Thus $\mu(\bigcup_{n\in\mathbb{N}} V_n)≦\sum_{n=1}^{\infty} \mu(E_n)+\epsilon$. Since $\bigcup_{n\in\mathbb{N}} E_n \subset \bigcup_{n\in\mathbb{N}} V_n$, the proof is done.

I struggling to avoid that 'countable choice'.. Is there a clever way to avoid this choice?($X$ is assumed to be locally compact and second countable)

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Without the Axiom of Countable Choice, one cannot prove that the Lebesgue measure on $ \mathbb{R}^{n} $ is $ \sigma $-additive. Hence, one runs into such undesirable circumstances when trying to avoid the Axiom of Choice altogether. –  Haskell Curry Jan 20 '13 at 9:02

1 Answer 1

up vote 1 down vote accepted

You will want to find Fremlin's Measure Theory (available freely on his webpage), where he devotes a chapter in the fifth volume to analysis without the axiom of choice.

In particular you might be interested in theorem $\mathbf{564I}$.

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Thank you Asaf. I skimmed it up, but now i'm a bit worried.. I'm enjoying mathematics without choice but this seems completely different way to develop measure theory from the usual. Should i study measure theory 'with choice' first then come back and study Borel Hierarchy? Or is it ok to study directly this paper you suggested? –  Katlus Jan 21 '13 at 7:10
    
Well, it is said that it is better to start with simpler and easier theories, and then advance into the harder things. Measure theory with choice is much more technical and complicated; but can be very interesting. Fremlin's book has five volumes from which you can learn modern analysis and choiceless analysis. But I should say that your efforts are impressive, and you do a good job. Whether or not you wish to continue should be a decision based on what you want to do with it later. –  Asaf Karagila Jan 21 '13 at 7:28
    
Is there a negation missing in your comment "Measure theory with choice is much more technical"? Are you seriously suggesting that Borel codes are less technical than the usual approach to measure theory using at least countable choice? –  commenter Jan 21 '13 at 21:50
    
@commenter: No, the other way around. Without choice things are more complicated; not that with choice they are all that easy, but at least the basic levels can be dealt with quite smoothly (e.g. completion of a $\sigma$-additive measure is $\sigma$-additive). –  Asaf Karagila Jan 21 '13 at 22:59
    
Then we agree. Thanks. –  commenter Jan 21 '13 at 23:13

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