Let $X$ be a compact Hausdorff space, $C(X)$ the set of all real continuous functions on $X$, and $\mathcal{B}$ be the Baire $\sigma$-algebra of $X$, which is the $\sigma$-algebra generated by the functions in $C(X)$. Furthermore $M(X)$ is the set of all finite signed measures on $(X,\mathcal{B})$ with the norm $||\mu||=|\mu|(X)$, $|\mu|$ being the total variation of $\mu$. Finally, for all $\mu \in M(X)$ define $\phi_{\mu}:C(X)\rightarrow \mathbb{R}$ by $\phi_{\mu}(f)=\int f d\mu$.
It is possible to formulate the Riesz-Markov-Kakutani theorem as follows:
The application $\mu\mapsto \phi_{\mu}$ is a surjective isometry from $M(X),||.||$ onto $C(X)^{*},||.||_{*}$ the dual space of $C(X),||.||_{\infty}$.
In other words the topological dual of the Banach space $(C(X),||.||_{\infty})$ (with as usual $||f||_{\infty}=\sup_{x\in X} f(x))$) can be identified with $M(X)$.
My question is: since there is a functional analysis formulation of the theorem, is there a functional-analysis-flavoured proof of it ?
Note that this version of the Riesz-Markov-Kakutani theorem is much stronger than the usually stated one, which is concerned positive functionals on $\mathbb{R}$. The fact that the dual norm is the total variation one is equivalent to the fact that Baire measures are necessarily regular, a not so trivial fact proved in Halmos's Measure Theory.
I find proofs like the one in Rudin's Real and complex analysis disturbingly artificial and complex for such what seems such a natural and important result: an integral is nothing more or less than a bounded linear functional on continuous functions (when integrating over a compact space).
I had a professor who used Daniell's integral with which the Riesz-Markov-Kakutani theorem follows almost instantly. That is a very good proof, the best I've seen so far, but I'm still under the impression that there might be other ways to look at the problem.