Please correct the following proof. I am trying to show that a linear functional $T$ must be identically $0$ if its dual representation satisfies a particular property.
Let $C_0(\mathbb{R},\mathbb{C})$ denote the complex-valued continuous functions on $\mathbb{R}$ that vanish at infinity, endowed with the uniform norm. Say $T : C_0(\mathbb{R},\mathbb{C}) \to \mathbb{C}$ is a continuous $\mathbb{C}$-linear functional. By Riesz representation, we know there exists some complex-valued Radon measure on $\mathbb{R}$ with finite total variation such that $$ Tu = \int_\mathbb{R} u(t)\ d\mu(t) \quad \forall u \in C_0(\mathbb{R},\mathbb{C}). $$ Now, let us suppose $\mu$ is such that we can show $$ \int_\mathbb{R} v(t)\ d\mu(t) = 0 $$ for all $v \in C_c(\mathbb{R},[0,\infty))$, meaning $v$ is a continuous, compactly supported function on $\mathbb{R}$ that take values in $[0,\infty)$. Then, we can conclude that $\mu$ is an identically zero measure. It follows that $T$ must be identically $0$.
What I am most worried about in the above proof, is that the functions $v$ only take values in the non-negative reals. In short, I am asking: does $$ \int_\mathbb{R} v(t)\ d\mu(t) = 0 \quad \forall v \in C_c(\mathbb{R},[0,\infty)) $$ indeed imply that $\mu$ is the zero measure, given that $\mu$ is a complex-valued Radon measure on $\mathbb{R}$ with finite total variation?
I would appreciate any mention of other technicalities that I may be missing in the above argument. Thank you.