Suppose $K$ is a compact metric space and let $F$ be a continuous linear functional on $C(K)$ (here $C(K)$ denotes the set of continuous functions on $K$). One version of the Riesz representation theorem (see e.g theorem 7.3 of https://assets.press.princeton.edu/chapters/s9627.pdf) states that there exists a unique finite signed Borel measure $\mu$ such that $F(f) = \int f d\mu$ for all $f \in C(K)$.
Under the assumption that $F(f) \geq 0$ for every positive $f \in C(K)$, I want to prove that the set $A =\{E \in \mathcal{B}: \mu(E) \geq 0\}$ is a $\sigma$-algebra.
The fact that $K \in A$ follows from $\mu(K) = F(\chi_K) \geq 0$ since $\chi_K \geq 0$ is continuous. However, I am stuck at proving that $A$ is closed under countable unions and complements.
Do you have any suggestions on how one could proceed?
If I'm not mistaken, then $A$ equals the Borel $\sigma$-algebra $\mathcal{B}$.
Because $K$ is compact, each element in $C(K)$ has a compact (and therefore bounded) image and thus is a bounded function. So $C(K)$ equals $C_b(K)$, the space of continuous bounded functions on $K$, and $F$ is a continuous linear functional on $C_b(K)$. By your assumption $F$ is also positive. So you can apply Theorem 7.4 in https://assets.press.princeton.edu/chapters/s9627.pdf which tells you that the measure $\mu$ is positive, i.e., $\mu(E)\geq 0$ for each $E$ in $\mathcal{B}$. So $A$ equals $\mathcal{B}$.