I'm reading this wikibook which states that, from a linear functional $\Lambda$, one can build a measure.
Setting: $X$ is a locally compact space.
Here are the steps. Let $\Lambda: C^0(X,R)\rightarrow R$ be a positive linear functional.
If $V$ is open in $X$ we define $$ \mu(V)=\sup\{ \Lambda(f)\, where f\leq 1 and\, supp(f)\subset V \}. $$ If $E$ is not open, we define $$ \mu(E)=\inf\{ \mu(V)\, where V\,\text{is open and } E\subset V \}. $$
Now let $K$ be compact in $X$ and (from Urysohn's lemma), consider $g\in C^0(X,R)$ such that $0\leq g\leq 1$ and $g(K)=\{1\}$.
I want to prove that $\mu(K)\leq \Lambda (g)$.
The wikibook says that this is true «by definition».
If $V$ is any open containing $K$, I can prove that $\mu(V)\geq \Lambda(g)$ and $\mu(K)\leq \mu(V)$.
The wikibook also plays with a intermediate open set $V_{\alpha}=\{x : f(x)>\alpha\}$ but I'm not sure how it helps ? Maybe with a limit $\alpha\rightarrow 1$ ? But there are no continuity for $\Lambda$.
EDIT: closely related to Rudin's Proof on Riesz Representation Theorem
EDIT: answer in "Riesz-Markov Representation Theorem" (Kumaresan) here
EDIT: two more references for a full proof:
- Theorem THOooTWZWooHqGDAx in giulietta (disclaimer: I'm the author)
- The Riesz Markov Kakutani Representation Theorem