It is known that the dual of $L^\infty (X, m)$ is the space of complex, finite, finitely additive measures on $X$ that are absolutely continuous with respect to $m$.
I would like to know whether one can get better properties of the measures in the dual if $X$ is a topological space and $m$ is Borel, regular, $\sigma$-finite. Most importantly, I would like to know whether these measures become countably-additive.
If $\omega \in L^\infty (X, m) ^*$ then there exist $\mu$, a complex, finite, finitely additive measure on $X$ that are absolutely continuous with respect to $m$, such that $\omega (f) = \int _X f \ \Bbb d \mu$.
Since the space $C_0 (X)$ of continuous functions that vanish at infinity embeds continuously into $L^\infty (X, m)$, it follows that $\omega$ restricts to a continuous linear functional on $C_0 (X)$, and can be therefore represented by some finite, regular measure $\nu$. A consequence of this is that $\int _X f \ \Bbb d \mu = \int _X f \ \Bbb d \nu \quad \forall f \in C_0 (X)$. Does this allow us to conclude that $\mu$ is countably additive?
(Quite frankly, I'd say that it doesn't, otherwise we could use the same technique to show that if $X$ is normal then the measures that form the dual of $C_b (X)$, which a priori are only finitely additive, should be countably additive. Nevertheless, I'm still hoping that $L^\infty (X, m)$ gets some nice properties from the topological properties of $m$.)
Alternatively, could one define a $L^\infty$ space of functions vanishing at infinity, call it $L_0 ^\infty$, and hope that its dual be some happy mix of the spaces $L^\infty (X, m)$ and $rca(X)$?
Even if we have a nice topological measure space, there is no way for us to expect any "regularity" of the measure from the Riesz-Representation theorem. Any possible nice properties of the resulting measure will come from the behavior of the linear functional in question. Intuitively, I think of the measures that are not countably additive as coming from linear functionals that do not arise as the restriction of a functional from $L^1$ (not exactly sure how correct this intuition is though).
Here's an example that should work. Let $X=[0,1]$ endowed with the Borel $\sigma$-algebra, $\mathfrak{M}=\mathcal{B}([0,1])$, and the Lebesgue measure, $\mu$. Clearly, $X$ is a topological space and $\mu$ is a Borel regular $\sigma$-finite measure.
Then consider the subspace of $L^\infty$, denoted by $S$, consisting of equivalence classes $[f]$ such that there is a measurable representative with the limit existing as $x\to 0$. Define the linear functional $L:S\to\mathbb{R}$ via $$L([f])=\lim_{x\to 0}\tilde{f}(x),$$ where $\tilde{f}$ is the associated representative. Now we show that $L$ is well defined, suppose that $\tilde{f}=\tilde{g}$ for $\mu$-a.e. $x\in X$, and that both $f_0:=\lim_{x\to 0}\tilde{f}(x)$ and $g_0:=\lim_{x\to 0}\tilde{g}(x)$ exist. Let $E:=\{x\in X~:~\tilde{f}(x)\neq\tilde{g}(x)\}$, which satisfies $\mu(E)=0$. Now fix $\varepsilon > 0$ and find $\delta > 0$ such that $|x|<\delta$ implies $|\tilde{f}(x)-f_0|<\varepsilon/2$ and $|\tilde{g}(x)-g_0|<\varepsilon/2$. Since $E$ contains no intervals, we see that there exists some point $x_0\in X$ with $|x_0|<\delta$ with $\tilde{f}(x_0)=\tilde{g}(x_0)$. So we deduce $$|f_0 - g_0|=|f_0-\tilde{f}(x_0)+\tilde{g}(x_0)-g_0|\leq|f_0-\tilde{f}(x_0)|+|g_0-\tilde{g}(x_0)|<\varepsilon.$$ Now by taking $\varepsilon\to 0$ we see that $f_0=g_0$, and so $L$ is well-defined. Similarly, we can show that $|L([f])|\leq\|[f]\|_{L^\infty}$.
Now use the Hahn-Banach theorem to extend $L$ to a linear functional $\Lambda\in(L^\infty(X,\mathfrak{M},\mu))^*$. Let $\lambda$ be the measure from Riesz-Representation theorem representing $\Lambda$, i.e. $$\Lambda(f)=\int_{[0,1]}f~d\lambda,\qquad f\in L^\infty(X,\mathfrak{M},\mu).$$ Now we have that $\lambda((0,1/n))=1$ for all $n\in\mathbb{N}$ (this follows from how we define the measure $\lambda$ in the proof of the representation theorem as $\lambda(E)=\Lambda(\chi_E)$, as well as the uniqueness). Now we obtain a contradiction if $\lambda$ were countably additive: if so, then we would have $\lambda(\emptyset)=1$, which clearly does not make sense.
To expand a bit on why your remark on the embedding of $C_0(X)$ into $L^\infty$ does not imply that the measure is countably additive, we can use the same example above. At first, you might think that the measure above is a Dirac delta at $0$, but in fact, we have that $\lambda(\{0\})=0$, but $\lambda((0,1])=1$. On the other hand, if we restrict $L$ to $C_0(X)=C(X)$ we obtain a linear functional $L:C(X)\to\mathbb{R}$ given by $L(f)=f(0)$. As a result, we deduce that the resulting measure from the Riesz-Representation theorem for $C(X)$ will be the Dirac delta measure at zero, $\nu =\delta_0$. So we see that the domain of the linear functional makes a very big difference on the measure which will represent it.
Regarding your last question about the dual of $L^\infty_0(X,\mathfrak{M},\mu)$, I'm not really sure what the dual space would look like right now.