Duals of $L^\infty$ and $C_c$ (finitely additive vs. Radon measures) on Euclidean space

Question

According to Can we identify the duals of $\ell^\infty$ and $L^{\infty}$ with another "natural space"?, $$(\text{faM}_b^{\pm, \ll}(\mathbb R^d), \|\bullet \|_{TV}) \cong (L^\infty(\mathbb R^d), \|\bullet \|_\infty)^*, \qquad \nu\mapsto \left[f \mapsto \int f \,d\nu\right]$$ where the notation is read as "finitely additive signed measures of bounded total mass (i.e. "finite measure"), that are futhermore absolutely continuous w.r.t. Lebesgue measure on $\mathbb R^d$).

It is well known that $$(\text{RdM}_b^\pm(\mathbb R^d), \|\bullet \|_{TV}) \cong (C_c(\mathbb R^d), \|\bullet\|_\infty)^*, \qquad \rho \mapsto \left[f \mapsto \int f \,d\rho\right],$$ where the notation is read as "finite/bounded signed Radon measures on $\mathbb R^d$".

Since $C_c\subseteq L^\infty$, every continuous linear functional on $L^\infty$ should restrict to a continuous linear functional on $C_c$. So the above results say that for any $\nu \in \text{faM}_b^{\pm, \ll}(\mathbb R^d)$, there exists $\rho \in \text{RdM}_b^\pm(\mathbb R^d)$ s.t. for all $f\in C_c(\mathbb R^d)$, $$\int f \, d \nu = \int f \, d \rho.$$

Question: Is this a correct interpretation? If so, how can it be intuitively that somehow we can upgrade from finitely additive to $\sigma$-additive?

Answer 1

I think a better way to think about the relation is that compactly supported continuous functions are too coarse a tool to differentiate between finitely and countably additive measures. And finitely additive measures (on $\sigma$-algebras) are sufficiently wild that no concrete example exists.

Take $x\in\mathbb R^d$ and let $\delta_x$ be the point mass at $x$, a perfectly fine countably additive Radon measure. However, $\delta_x$ is not absolutely continuous with respect to the Lebesgue measure. We have to find a finitely additive measure absolutely continuous with respect to the Lebesgue measure that cannot be distinguished from $\delta_x$ using compactly supported continuous functions.

For each open neighborhood $O$ of $x$, let $\mathcal{F}_O$ be the family of measurable sets whose symmetric difference to $O$ has Lebesgue measure zero. The union of all such families is a free filter base that contains every neighborhood of $x$. Extend it to an ultrafilter and let $m$ be the finitely additive measure that assigns measure $1$ to every measurable set in the ultrafilter and measure $0$ to every measurable set outside the ultrafilter. Then, we have

$$f(x)=\int f~\mathrm d\delta_x=\int f~\mathrm d m$$

for all $f\in C_c(\mathbb R^d)$. The extension of the filter base is, in general, not unique, even when restricted to measurable sets. So, compactly supported continuous functions cannot differentiate between all finitly additive measures, either.