Riesz representation for a countable family of functions

Question

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.

1. $\mathcal{C}_0$ is countable, made of nonnegative functions and closed under addition (i.e. $u,v \in \mathcal{C}_0 \Rightarrow u+v \in \mathcal{C}_0$);
2. 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

1. The Riesz representation theorem (for example Rudin, Real and Complex Analysis, Theorem 2.14),
2. 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.

Answer 1

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.