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I just started looking through the proof of the Riesz Representation Theorem (in Rudin's Real and Complex Analysis), and I am still very confused about several things. I'll just write the statement of the theorem as given in the book:

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 $M$ in $X$ which contains all Borel sets in $X$, and there exists a unique positive measure $\mu$ on $M$ which represents $\Lambda$ in the sense that,

(a) $\Lambda f = \int_Xf \ d\mu$ for every $x\in X$

(b) $\mu(K)<\infty$ for every compact set $K\subset X$

(c) For every $E\in M$, we have $\mu(E)=\inf\{\mu(V):E\subset V, V \text{ open}\}$

(d) The relation $\mu(E) = \sup\{\mu(K): K\subset E, K \text{ compact}\}$ holds for every open set $E$, and for every $E\in M$ with $\mu(E)<\infty$.

(e) If $E\in M$, $A\subset E$, and $\mu(E)=0$, then $A\in M$. "

The proof of this statement spans 7 pages, and while I was able to follow the logic step-by-step, I feel like I have left with very little intuition regarding the theorem as a whole. I am currently trying to solve the problem:

Show that

$ \Lambda f = \int_{-\infty}^\infty f(x) |x| dx, \ \ \ f\in C_c(\mathbb{R}) $

is a positive linear functional on $C_c(\mathbb{R})$. Denoting by $\mu$ the measure representing $\Lambda$, compute $\mu([0,\pi])$.

So I suppose my specific question is how do the restrictions on the measure given in $(b)-(d)$ allow us to explicitly calculate the value of $\mu$ for $[0,\pi]$ for the $\Lambda$ above? More generally, it would be nice to see another proof of the Riesz Representation Theorem, as Rudin's has not really sunk in yet. If you have any reference recommendations that give a more intuitive (or just different) proof, it'd be much appreciated.

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The measure of a set is the integral of its indicator function. So you can approximate the indicator function from above by continuous functions that converge pointwise to the indicator function. You can do this by (b) in a way such that the integrals of this sequence are eventually finite, so you obtain the measure as the limit integral by the dominated convergence theorem. –  Michael Greinecker Nov 8 '12 at 14:44

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up vote 2 down vote accepted

Let $$ f(t)=\begin{cases} 1,&\text{ if } t\in[0,\pi]\\ nt+1,&\text{ if }t\in[-1/n,0]\\ -nt+n\pi+1,&\text{ if }t\in[\pi,\pi+1/n]\\ 0,&\text{ elsewhere} \end{cases} $$ Then, as $f_n\searrow 1_{[0,\pi]}$, Dominated Convergence gives us $$ \mu([0,\pi])=\int 1_{[0,\pi]}\,d\mu=\lim_n\int f_n\,d\mu=\lim_n\int_{-\infty}^\infty\,f_n(x)|x|\,dx=\int_0^\pi\,x\,dx=\frac{\pi^2}2 $$ (the point is that we are approximating $1_{[0,\pi]}$ with continuous functions, where we can revert to the functional to calculate the integrals)

As for an "easier" proof of Riesz-Markov, I haven't seen one. But I think that, once you get some experience on the subject, Rudin's proof is not that bad.

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That is a decreasing sequence of functions. –  Michael Greinecker Nov 8 '12 at 14:50
    
Good point! ${ }$ –  Martin Argerami Nov 8 '12 at 15:09
    
Okay, this makes sense. Is there a way to determine what $\mu$ is in general, not just on a specific interval? –  Steve W Nov 8 '12 at 15:14
    
If by "in general" you expect a formula, and you mean on any other interval, then the same trick works. Otherwise, I guess the best you can say is that $$\mu(E)=\int_E\,|t|\,dm(t),$$ where $m$ is the Lebesgue measure. –  Martin Argerami Nov 8 '12 at 15:52

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