Riesz representation for a countable family of functions In reading this I have found the following result that I don't know how to prove precisely:

Theorem: Let $(X,d)$ be a compact metric space and let $C(X)$ be the Banach space of real valued continuous functions with the supremum
  norm. There exists $\mathcal{C}_0 \subset C(X)$ s.t.
  
  
*
  
*$\mathcal{C}_0$ is countable, made of nonnegative functions and closed under addition (i.e. $u,v \in \mathcal{C}_0 \Rightarrow u+v \in \mathcal{C}_0$);
  
*For every $L: \mathcal{C}_0 \to [0, + \infty)$ additive (i.e. $L(u+v)=Lu+Lv \,\, \forall \, u,v \in \mathcal{C}_0$) there exists a
  unique finite positive Borel measure $\mu$ on $X$ s.t. 
  $$Lu = \int_X u
 \text{ d} \mu \quad \forall \, u \in \mathcal{C}_0.$$

Here with "finite positive Borel measure" I mean a countably additive set function from $\mathcal{B}(X)$ (the Borel sigma algebra of $X$) to $[0,+\infty)$.
I think the ingredients for the proof should be


*

*The Riesz representation theorem (for example Rudin, Real and Complex Analysis, Theorem 2.14),

*The separability of $C(X)$.


I really don't know how to fill the dots. For sure there exists $\mathcal{C} \subset C(X)$ countable, dense and closed under addition (even $\mathbb{Q}$-vector space) but not made of nonnegative functions. For sure a bounded additive functional $L: \mathcal{C} \to [0, + \infty)$ extends to a functional $\tilde{L}: C(X) \to [0, + \infty)$ which is additive and positive (in the sense of Rudin i.e. $\tilde{L}(u) \in [0, + \infty)$ if $u(X) \subset [0, + \infty)$). But this is not linear, hence I can't use the RRT and conclude that $\tilde{L}$ is represented by a unique finite positive Borel measure $\mu$ on $X$, which would also represent $L$ on $\mathcal{C}$ as required.
I show that $\tilde{L}: \mathcal{C}_0 - \mathcal{C}_0$ as in the comments to the answer of harfe is bounded.
Take $u \in \mathcal{C}_0 - \mathcal{C}_0 \subset C(X)$ and fix $\epsilon \in \mathbb{Q}^+$. We know that $u=u_1-u_2$ for some $u_1, u_2 \in \mathcal{C}_0$. Moreover, by density, there exists $\tilde{u}_{\epsilon} \in \mathcal{C}$ s.t. 
$$ -\epsilon/2 \le u(x)-\tilde{u}_{\epsilon}(x) \le \epsilon/2 \quad \forall \, x \in X \Rightarrow 0 \le u(x) - (\tilde{u}_{\epsilon} -\epsilon/2) \le \epsilon \quad \forall \, x \in X.$$
Define $u_{\epsilon} := \tilde{u}_{\epsilon} - \epsilon/2$ so that $u_{\epsilon} \in \mathcal{C}$ (it is closed under addition and $-\epsilon/2 \in \mathbb{Q}$) and $0 \le u - u_{\epsilon} \le \epsilon$. Then $u_{\epsilon}^+, u_{\epsilon}^- \in \mathcal{C}_0$. Hence $u_{\epsilon} = u_{\epsilon}^+ - u_{\epsilon}^- \in \mathcal{C}_0- \mathcal{C}_0$ and
$$ u-u_{\epsilon} = u_1 - u_2 - u_{\epsilon}^+ + u_{\epsilon}^- = (u_1+ u_{\epsilon}^-) - (u_{\epsilon}^+ + u_2) \text{ with } u_1+ u_{\epsilon}^- \ge u_{\epsilon}^+ + u_2 $$
so that $u-u_{\epsilon} \in \mathcal{C}_0$.Then
\begin{equation} \tag{1}
\left | \tilde{L}u \right | - \left | \tilde{L}u_{\epsilon} \right | \le \left | \tilde{L}u - \tilde{L}u_{\epsilon} \right | = \left |\tilde{L} (u-u_{\epsilon}) \right |  = \left | L(u-u_{\epsilon}) \right | \le C|u-u_{\epsilon}| \le C\epsilon
\end{equation}
being $L$ bounded on $\mathcal{C}_0$. Moreover
$$ \|u_{\epsilon} \| = \max \left \{ \|u_{\epsilon}^+\|, \|u_{\epsilon}^-\| \right \} $$
and then
\begin{align*}
\tilde{L}u_{\epsilon} &= Lu_{\epsilon}^+ - Lu_{\epsilon}^- \le Lu_{\epsilon}^+ \le C \| u_{\epsilon}^+\| \le C \|u_{\epsilon}\| \\
-\tilde{L}u_{\epsilon} &= Lu_{\epsilon}^- - Lu_{\epsilon}^+ \le Lu_{\epsilon}^- \le C \|u_{\epsilon}^-\| \le C \|u_{\epsilon}\| 
\end{align*}
so that 
\begin{equation} \tag{2}
\left | \tilde{L}u_{\epsilon} \right | \le C \|u_{\epsilon}\|. 
\end{equation}
Joining (1) and (2), we obtain
$$ \left | \tilde{L}u \right | \le  \left | \tilde{L}u_{\epsilon} \right |+ C\epsilon \le C\|u_{\epsilon}\| + C\epsilon \le C \|u\| + 2C\epsilon. $$
Passing to the limit as $\epsilon \downarrow 0$ we obtain that $\tilde{L}$ is bounded.
 A: You pointed out that there are three gaps to fill in your argument:


*

*make $\mathcal C_0$ such that it is made of nonnegative functions,

*show that $L$ is bounded,

*show that $\tilde L$ is linear.


Let me provide sketches for each of those.
make $\mathcal C_0$ such that it is made of nonnegative functions:
Choose $\mathcal C$ as you suggest, i.e. dense, countable, and closed under addition.
Then you can construct $\mathcal C_0$ by taking the positive and negative part
$\max(f,0)$ and $\max(-f,0)$ of each function $f\in\mathcal C$.
In order to make $\mathcal C_0$ closed under addition, you probably need to add
additional functions to $\mathcal C_0$. However, it should still be countable.
Note that $\mathcal C_1:= \mathcal C_0-\mathcal C_0$ is dense in $C(X)$.
show that $L$ is bounded:
We first make some modifications to $C_0$.
First, we add the constant functions $q$ for each $q\in\mathbb Q$ to $\mathcal C_0$.
Then, we also add all functions $f-g$ to $\mathcal C_0$
if $f\geq g$ and $f,g\in \mathcal C_0$.
In order to make $\mathcal C_0$ closed under addition, you probably need to add
additional functions to $\mathcal C_0$. However, it should still be countable.
After all these additions, the set $\mathcal C_0$ should still be countable.
Let $f\in \mathcal C_0$ be bounded,
i.e. $\|f\|\leq q$ for some $q\in\mathbb Q$.
Then we have $f(x)\leq q\cdot1$ for all $x\in X$.
Then we have $q-f\in C_0$.
It follows that
$$
Lf \leq Lf+L(q-f) = L(f+q-f)=Lq=q L1
$$
and therefore $L$ is bounded (with the constant $L1$)
show that $\tilde L$ is linear:
You can construct $\tilde L$ by extending $L$ from $\mathcal C_0$ to
$\mathcal C_1$ and then to $C(X)$ by extending it continuously.
Now, you can choose $\mathcal C_0$ such that
$qf\in\mathcal C_0$ for all $q\in\mathbb Q_+,f\in\mathcal C_0$,
see your linked answer.
Let $a\in\mathbb R$ and $f\in C(X)$.
Suppose that $a_k\to a$ and $f_k\to f$ with $f_k\in\mathcal C_1$,
$a_k\in\mathbb Q$.
Then $L(a_kf_k)=a_kL(f_k)$ can be shown using the additivity of $L$.
We obtain
$$
\tilde L(af)
=\lim \tilde L(a_k f_k)
=\lim L(a_k f_k)
=\lim a_kL(f_k)
=a\lim L(f_k)
=a\tilde L(f)
$$
and $\tilde L$ is linear.
