Let $f$ be an locally integrable function on the measure space $(\mathbb R^n,S,\mu)$, with $\mu$ be a radon measure proof that
\begin{align}\text{If }\int_{\mathbb R^n} f g\, d\mu \geq 0\text{ for all }g\in C_C^0(\mathbb R^n)\text{ s.t } g \geq 0 \text{ then }f \geq 0\text{ a.e.}\end{align}I see clear that the reasoning of these problem is by contradiction, then i suppose that $\mu(f^{-1}(- \infty ,0)) >0$ is easy to see that$$f^{-1}(- \infty ,0)=\bigcup_{n} f^{-1}(-\infty, -1/n)$$then i take $n_0$ s.t $\mu(f^{-1}(-\infty, -1/n_0))>0$ now my problem is how i can construct a continuous compact supported function $g$ s.t. $spt(g) \subset f^{-1}(-\infty, -1/n)$$g \geq0$ for find the contradictionAnother aproach to i think is using something like the density of $C_C^0(\mathbb R^n)$ but i not sure
EDIT
I think in a new approach:
By contradiction, sup. exists $A$ st. $ \mu(A)>0$ and $f<0$ in A, then like $\mu$ is Radon there exists $K$ compact such that $K \subset A$ and $\infty> \mu(K)>\frac{\mu(A)}{2}>0$ now like $\infty >\mu(K)$ then exists a sequence of compact supported continuous functions $( \psi _n )_n$ such that $0 \leq \psi_n \leq 1$ and $\psi_n \to 1_K$ then by dominated convergence theorem (I have doubts in these step)$$\int f \psi_n d \mu \to \int f 1_K d\mu = \int_K f d \mu <0$$But by hypothesis $0 \leq \int f\psi_n d \mu$, $\forall n \in \mathbb N$ then$$0 \leq \int f \psi_n d \mu <0$$And hence these is a contradiction.\
These proof is correct? Any hint or help I will be very grateful.