Show that $\mu \geq 0$ on Borel subsets of $X.$ 
Let $X$ be a compact Hausdorff space and $\mu$ be a complex  Borel measure on $X$ such that $\int_{X} f\ d\mu \geq 0$ for every $f \in C(X)$ with $f \geq 0.$  Then show that $\mu$ is non-negative.

Could anyone give some suggestion in this regard?
 A: Let $E\subseteq X$ Borel. Let $\epsilon > 0$. By regularity, we can choose a compact subset $K\subseteq X$ and an open subset $U\subseteq X$ with $K \subseteq E \subseteq U$ and $|\mu|(U\setminus K) < \epsilon$. By Urysohn's lemma, we can pick $f\in C(X,[0,1])$ with $\chi_K\le f \le \chi_U$. Then
$$\int_X|\chi_E-f|d|\mu| \le |\mu|(U\setminus K) < \epsilon.$$
This shows that the function $\chi_E$ can be approximated by positive continuous functions in the $L^1(|\mu|)$-norm.
Next, let $E\subseteq X$ be a Borel subset. Choose a sequence of positive continuous functions $\{f_n\}_{n=1}^\infty\subseteq C(X, [0, \infty[)$ such that
$$\int_X|f_n-\chi_E|d|\mu|\stackrel{n \to \infty}\longrightarrow 0.$$
Hence,
\begin{align*}\left|\int_X f_n d\mu-\int_X \chi_Ed\mu\right| \le \int_X |f_n-\chi_E|d|\mu| \stackrel{n \to \infty} \longrightarrow 0\end{align*}
and we deduce that
$$\mu(E) = \int_X \chi_Ed\mu = \lim_n \underbrace{\int_X f_n d \mu}_{\ge 0}\ge 0.$$
Since $E$ was an arbitrary Borel set, we conclude that $\mu$ is a positive (finite) measure.
