TLDR: How to prove $rba(\Omega) \cong L^1(\Omega,\mu)$ for some measure $\mu$ when $\emptyset\neq\Omega\subseteq\mathbb{R}^d$?
I study $rba(\Omega)$, the space of regular, finite, finite additive measures. Here it is shown that $rba(\Omega)$ for compact Hausdorff spaces is isomorphic to a direct sum of $L^1(\mu)$ spaces. Here it is however claimed that you do not need a direct sum and only one measure $\mu$ is sufficient. I would like to know how to prove that this is true. Based on the proof here I would think that it is possible to choose a probability measure with full support on $\Omega$. Then likely every $\nu\in rba(\Omega)$ satisfies either $\mu\ll\nu$ or $\mu_k\rightharpoonup\mu$ with each $\mu_k\ll\nu$. The map
$$
\phi:rba(\Omega)\to L^1(\Omega,\mu), \; \nu\mapsto \frac{d\mu}{d\nu}
$$
then defines the isometry. How do I prove this rigorously?
