Htykuut

i’m trying to prove the special case of Riesz representation theorem:
Every positive (non-negative on non-negative functions) linear continuous functional $phi$ on the normed space $C([0,1])$ is given by some measure $mu$ by the rule:
$phileft(fright)=int_{left[0,1right]}fdmu$

I want to do it with using measure extension theorem:
First I need to build $mu$ on elementary sets. But I don't know what it should be like.
Can you help me with this? (for open interval, for example)

Proving it generally for $C[a,b]$ :

First, assume that $Gamma$ is positive. For $a leq t < b$ and for $n$ large enough so that $t + frac{1}{n} leq b$, let :

$$phi_{t,n}(x) = begin{cases} 1 & text{if} ; x in [a,t] \ 1- n(x-t) & text{if} ; x in (t,t + frac{1}{n}] \ 0 & text{if} ; x in (t + frac{1}{n}, b]end{cases}$$

If $n leq m$, then $ 0 leq phi_{t,m} leq phi_{t,n} leq 1$. It follows that ${Gamma(phi_{t,n})}$ is decreasing and bounded below by $0$. Therefore, we can define :

$$g(t) = begin{cases} 0 & text{if} ; t<a \ lim_{n to infty} Gamma(phi_{t,n}) & text{if} ; t in [a,b) \ Gamma(1) & text{if} ; t geq b end{cases}$$

Moreover, if $t_1 > t$, we have : $phi_{t,m} leq phi_{t_1,n}$.

Since $Gamma$ is positive, $g(t)$ is monotonically increasing. It is clear that $g(t)$ is right continuous if $t<a$ or if $tgeq b$. Assume that $t in [a,b)$. Let $varepsilon >0$ and choose $n$ large enough so that :

$$n > maxleft(2, frac{|Gamma|}{varepsilon}right)$$

and also that : $g(t) leq Gamma(phi_{t,n}) leq g(t) + varepsilon$.

Let :

$$psi_n(x) = begin{cases} 1 & text{if} ; x in [a, t + frac{1}{n^2}] \ 1 - frac{n^2}{n-2}left(x-t-frac{1}{n^2}right) & text{if} ; x in (t + frac{1}{n^2}, t + frac{1}{n} - frac{1}{n^2}] \ 0 & text{if} ; x in (t + frac{1}{n} - frac{1}{n^2}, b] end{cases}$$

It then is : $| psi_n - phi_{t,n}|_infty leq 1/n$. That means :
$$Gamma(psi_n) leq Gamma(phi_{t,n}) + frac{1}{n}|Gamma| leq g(t) + 2varepsilon$$

But, this yields that :

$$g(t) leq gleft(t + frac{1}{n^2}right) leq g(t) + 2 varepsilon$$

Since $g(t)$ is increasing, it is sufficient to show that $g(t)$ is right continuous. The Hahn-Banach Extension Theorem gives a Borel measure $mu$ such that $mu((alpha,beta]) = g(beta) - g(alpha)$. In particular, if it is $a leq c leq b$, then it is :

$$mu([a,c]) = mu((a-1,c]) = g(c)$$

Let $f in C([a,b])$ and let $varepsilon >0$. Let $delta$ be such that if $|x-y| < delta$ and $x,y in [a,b]$, then :
$$|f(x) - f(y)| < varepsilon$$

Now, let $P ={a=t_0,t_1,dots,t_m=b}$ be a partition with $sup(t_k - t_{k-1}) < delta/2$. Then choose $n$ to be large enough so that :
$$frac{2}{n} < inf(t_k-t_{k-1})$$
$$text{and}$$
$$g(t_k) leq Gamma(phi_{t,n}) leq g(t_k) + frac{varepsilon}{mu|f|_infty}$$
Next, let :
$$f_1(x) = f(t_1) + phi_{t_1,n} + sum_{k=1}^m f(t_k)(phi_{t_k-n} - phi_{t_{k-1},n})$$
$$text{and}$$
$$f_2(x) = f(t_1)_{mathcal{X}[t_0,t_1]} + sum_{k=2}^m f(t_k)_{mathcal{X}[t_{k-1},t_k]}$$
It can be seen that $f_1$ is continuous and piecewise linear, while $f_2$ is a step function. Both $f_1$ and $f_2$ agree with $f(x)$ at each point $t_k$ for $k geq 1$. Moreover the function $f_1$ takes values between $f(t_{k-1})$ and $f(t_k)$ on the interval $[t_{k-1},t_k]$ of course.

It is :

$$|f_1-f|_infty leq varepsilon$$
$$text{and}$$
$$sup{|f_2(x)-f(x)| : x in [a,b]} leq varepsilon$$

From the above, we can conclude that :
$$|Gamma(f) - Gamma(f_1)| leq varepsilon|Gamma|$$

Now, for $2leq kleq m$, it is :

$$|Gamma(phi_{t_k,n} - phi_{t_{k-1},n}) - (g(t_k)-g(t_{k-1}))| leq frac{varepsilon}{m|f|_infty}$$

Now, applying $Gamma$ to $f_1$ and integrating $f_2$ with respect to $mu$ :
$$Bigg| Gamma(f_1) - int_{[a,b]} f_2mathrm{d}mu Bigg| leq varepsilon$$
But, it also is :
$$int_{[a,b]} f_2mathrm{d}mu - int_{[a,b]} f mathrm{d}mu leq varepsilon mu([a,b])$$
Thus :
$$Bigg|Gamma(f) - int_{[a,b]}fmathrm{d}muBigg| leq varepsilon(2 | Gamma | + mu([a,b])$$
But $varepsilon$ is arbitrary and we can yield :

$$Gamma(f) = int_{[a,b]} f mathrm{d}mu$$
for every $f in C[a,b]$. It also is $|Gamma| = Gamma(1) = |mu|([a,b])$.

Now, recall the Jordan Decomposition Theorem, which states that :

Let $Gamma in C([a,b])^*$. Then there exist positive linear functionals $Gamma^+, Gamma^- in C([a,b])^*$ such that :
$$Gamma = Gamma^+ - Gamma^-$$
$$text{and}$$
$$|Gamma| = Gamma^+(1) + Gamma^-(1)$$

The general result now follows from that theorem and the proof is completed.