Riesz representation theorem for $C([0,1])$ 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:
$\phi\left(f\right)=\int_{\left[0,1\right]}fd\mu$
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)
 A: 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 $t\geq b$. Assume that $t \in [a,b)$. Let $\varepsilon >0$ and choose $n$ large enough so that :
$$n > \max\left(2, \frac{\|\Gamma\|}{\varepsilon}\right)$$
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 :
$$\Gamma(\psi_n) \leq \Gamma(\phi_{t,n}) + \frac{1}{n}\|\Gamma\| \leq g(t) + 2\varepsilon$$
But, this yields that :
$$g(t) \leq g\left(t + \frac{1}{n^2}\right) \leq g(t) + 2 \varepsilon$$
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 $2\leq k\leq m$, 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 $\mu$ :
$$\Bigg| \Gamma(f_1) - \int_{[a,b]} f_2\mathrm{d}\mu \Bigg| \leq \varepsilon$$
But, it also is :
$$\int_{[a,b]} f_2\mathrm{d}\mu - \int_{[a,b]} f \mathrm{d}\mu \leq \varepsilon \mu([a,b])$$
Thus :
$$\Bigg|\Gamma(f) - \int_{[a,b]}f\mathrm{d}\mu\Bigg| \leq \varepsilon(2 \| \Gamma \| + \mu([a,b])$$
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.
A: Rudin = Rudin, W.: Real and Complex Analysis, 3rd ed., New York: McGraw-Hill, 1987.
This is a special case of Rudin, p.40, Theorem 2.14. Since Step I -Step IX of the theorem' proof serve to build the structure of a $\sigma$-algebra and a positive measure that $[0,1]$ already has. Therefore, only Step X is what one needs to read. For a complete discussion of the Riesz representation theorem, read Example 6.169 of
https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/mathematical-methods.
