Basic Idea Behind Riesz-Markov-Kakutani Representation Theorem

Sorry if this seems like an overly elementary question but I've been rolling this around in my mind all day. Am I correct that the basic idea behind the Riesz-Markov-Kakutani represention theorem that I can take any (positive) linear functional $J:C_{c}(X) \to \mathbb{R}_{+}$ where $X$ is a locally compact Hausdorff space and re-write it/represent it as an integral such that (for some compact set $E \subset X$) as $$J[f]=\int_{E}f \, d\lambda$$ where $\lambda$ is a (unique) Borel measure? One example I am very tempted to conjecture right off the bat is the "evaluation-at-a-point functional" which for some $a \in E$ is given by $$J_{a}[f]=\int_{E} f \, d\delta_{a}=f(a)$$ where $\delta_{a}$ is the Dirac measure. Obviously the Dirac measure is locally finite and both inner and outer regular so it is a Borel measure. So am I correct in my thinking here? I also assume that there is some unique Borel measure $\nu$ that one could "represent" for example the total variation functional $V_{a}^{b}: BV([a,b]) \to \mathbb{R}_{+}$ as $$V^{b}_{a}[f]=\int_{[a,b]} f \, d\nu$$ and that this would hold for any positive linear functional defined on the space of compactly supported continuous functions?

Also, I read somewhere that measures are special cases of distributions and I've been trying to wrap my head around that idea and it seems to me that this theorem is pointing me in that direction. Can someone help me connect the dots here? I feel like I'm so close to seeing the big picture of how all of this fits together but I'm missing that key insight.

Edit: Another flash of insight I just had: for $\varphi \in \mathcal{D}(\mathbb{R})$ where $\mathcal{D}(\mathbb{R})$ is the space of test functions and $f:\mathbb{R} \to \mathbb{R}$ say we have a distribution acting on a test function $\varphi$ given by $$\langle T_{f},\varphi\rangle =\int_{\mathbb{R}} f(x) \varphi(x) \, d\mu(x)$$ and for a Borel measure $\mu$ we have a distribution acting on $\varphi$ given by $$\langle T_{\mu},\varphi \rangle=\int_{\mathbb{R}} \varphi(x) \, d\mu(x)$$ so perhaps $\varphi$ is a "smoothed out" version of the indicator function?

• Variation is not a linear functional. So you're not going to be able to represent variation in the way that you have stated. Distributions are more general than continuous linear functionals on $C_{c}(\mathbb{R})$. – DisintegratingByParts Aug 23 '16 at 9:17
• There are continuous functions on $[a,b]$ that fail to have finite total variation. Also, there may not be a compact $E$ that works for all $f$: think of Lebesgue measure on $\mathbb{R}$. – fourierwho Dec 6 '17 at 19:47