Riesz-Markov-Kakutani representation theorem: qualitative or abstract proofs. I am referring to the famous result which states:
Let X be a locally compact Hausdorff space. For any positive linear functional $\psi$ on $C_c(X)$, there is a unique regular Borel measure μ on X such that
$$\psi(f)=\int_{X}fd\mu(x)\ \\ \forall f\in C_c(x)$$
The proofs I read of this result involve very tricky and seemingly articificial calculations: they are very quatitative and make a lot of use of $\epsilon-\delta$ arguments. I absolutely do not mean they are less truthful whatsoever, nonetheless I find it strange there is not a 'qualitative' proof of this result, as we have for a lot of important duality theorems. 
Do anyone know some enlightening proof?
 A: Folland's proof (Theorem 7.2 in 1st edition Real Analysis) is quite conceptual, though in total the proof takes about 2.5 pages. The idea suggesting that this theorem should be true is very simple, though; the bulk of the proof is checking that the idea doesn't just give you a measure, but a Radon measure whose associated linear functional on $C_c(X)$ is the one you started with. 
The idea is that $X$ being locally compact and Hausdorff implies (via Urysohn's lemma) that we can approximate the characteristic function of any open subset of $X$ from both above (when the closure is compact) and below (always) by continuous, compactly supported functions. This lets us define a measure for any continuous linear functional.
More precisely, for any open subset $U \subset X$, the characteristic function $\chi_U$ satisfies
$$
\chi_U(x) = \sup_{f: U \to [0, 1]} f(x),
$$
where of course I'm restricting to $f \in C_c(X)$. Then if $\psi \in C_c(X)^*$, we define the $\psi$-measure of $U$ to be
$$
\mu_\psi(U) = \sup_{f: U \to [0, 1]} \psi(f). 
$$
Since the Borel $\sigma$-algebra is generated by open subsets, this suffices to define $\mu_\psi(E)$ for any Borel set $E \subset X$ (via Caratheodory's theorem). 
As mentioned in the comments, checking that this definition gives a Radon measure, and $\int \, f \, d\mu_\psi = \psi(f)$ requires some work and estimates, but nothing quite as nitty-gritty as $\varepsilon-\delta$ arguments. I think by construction $\mu_\psi$ looks promisingly like it will be inner-regular, at least. It may even look promising that in the case $\overline{U}$ is compact, we could use an $\inf$ over $f \in C_c(X)$ with $f \ge \chi_U$ above when defining $\mu_\psi(U)$ and get the same value. I'll leave it to your imagination (or to Folland's) to check $\int\,f\,d\mu_\psi = \psi(f)$. 
