The total variation norm $\mu$ of a complex measure can be defined as
$$\lVert \mu \rVert = \sup_{\lVert f \rVert_u \le 1} \left| \int f\ \mathrm{d}\mu\right|$$
Here $\lVert \cdot \rVert_u$ is the uniform norm. For complex Radon measures, I am wondering if the total variation can be approximated within $C_0$, the continuous functions that vanish at infinity:
$$\tag{1} \phantom{{\qquad (\mu \text{ Radon})}} \lVert \mu \rVert \overset{??}{=} \sup_{\substack{\lVert f \rVert_u \le 1 \\f \in C_0}} \left| \int f\ \mathrm{d}\mu\right| \qquad (\mu \text{ Radon})$$
$(1)$ is trivial for any positive measure, Radon or not. If $\mu$ is a (finite) signed measure, then the claim is $$ \tag{2} \mu^+ + \mu^- \overset{??}{=} \sup \left|\int_P f\ \mathrm{d}\mu^+ - \int_N f\ \mathrm{d}\mu^-\right| $$ where $\mu =\mu^+ - \mu^-$ is the decomposition into the positive and negative variations of $\mu$, and $P \sqcup N$ is a Hahn decomposition of the measure space. Then the Radon hypothesis allows us to apply Lusin's theorem: for every measurable $f$, we can approximate $f \chi_P - f \chi_N$ by a compactly supported continuous function $\phi$. We can modify $\phi$ to $$\widetilde{\phi} = \begin{cases}\phi & |\phi| < \lVert f\rVert_u \\ \operatorname{sgn}(\phi) \lVert f \rVert_u & |\phi| \ge \lVert f \rVert_u \end{cases}$$ $\widetilde{\phi}$ is still compactly supported and continuous, now with uniform norm $\le \lVert f \rVert_u$. Applying this construction to every $f$ with $\lVert f \rVert_u \le 1$ yields $(2)$.
I am not sure how to proceed to the general case, $\mu$ complex Radon. We do have $\mu = \mu_r + \mathrm{i} \mu_i$ for signed measures $\mu_r, \mu_i$, but I don't think their total variations have a nice relation to $\lVert \mu \rVert$; for example, $\mu_r$ and $\mu_i$ don't have to have disjoint support. Could you suggest how to prove (or disprove) $(1)$? TIA.