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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)

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  • $\begingroup$ Possibly, you could to define $\mu^\star(A) = \phi(1_A)$ where $A$ is an interval on $[0,1]$. Did you try that? $\endgroup$ – Yanko Dec 9 '18 at 11:09
  • $\begingroup$ it would be great, but $\chi_{A}$ is not continuous ($\notin C([0,1])$) $\endgroup$ – Ilya Dec 9 '18 at 11:19
  • $\begingroup$ i think $\mu$ has to somehow be consistent with the norm on $C([0,1])$ $\endgroup$ – Ilya Dec 9 '18 at 11:21
  • $\begingroup$ The idea would be to find a sequence of continuous functions $f_n$ that approximates $1_A$ (say, converging pointwise and boundedly), and then define $\mu(A) := \lim_{n \to \infty} \phi(f_n)$. $\endgroup$ – Nate Eldredge Dec 9 '18 at 16:51
  • $\begingroup$ @NateEldredge It's already done below. $\endgroup$ – Rebellos Dec 9 '18 at 16:53
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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.

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