I think the first version is more general. It clearly has weaker assumptions, as you noted. As for the conclusions, they both yield measures that correspond to the functional in exactly the same way, and are unique, which are described as following:
- Regular and Borel.
- Locally finite and positive.
In the first case, I think we can safely assume that "positive" is implicit (because it almost always is, if not stated otherwise, and because otherwise the resulting functional would not be positive!). Similarly, if the measure was not locally finite, then I believe that the resulting „functional” would not be bounded (or even finite), so we can also assume that implicitly, so the first version does not provide anything more with the assumptions of the second one.
As for the second case, I believe that there is also implicit assumptions of regularity and Borelness, because otherwise the measure would be unlikely to be unique, because by the first version we already have a Borel measure, but a Borel measure can always (except for trivial cases, e.g. when every set is Borel) be extended to a larger $\sigma$-field, defying uniqueness (remember that the integral of a continuous functions depends only on the Borel part of the measure), and any finite Borel measure on a metric space is already regular, and this should extend to $\sigma$-finite in the obvious way. (Note that locally finite measure on a $\sigma$-compact space is $\sigma$-finite.)
Summing it all up, unless I made some large mistakes:
- The first theorem is more general, because it applies to more cases.
- With the assumptions of the second theorem, they are equally strong.
One standard example is the reals numbers times the reals with the discrete topology: $X = \mathbb{R} \times \mathbb{R}_d$.
This is a locally compact metrizable space. The compact subsets intersect only finitely many horizontal lines and each of those non-empty intersections must be compact. A Borel set $E\subset X$ intersects each horizontal slice $E_y$ in a Borel set.
Consider the following Borel measure where $\lambda$ is Lebesgue measure on $\mathbb{R}$:
$$
\mu(E) = \sum_{y} \lambda(E_y).
$$
This is easily checked to define an inner regular Borel measure and its null sets are precisely those Borel sets that intersect each horizontal line in a null set. In particular, the diagonal $\Delta = \{(x,x) : x \in \mathbb{R}\}$ is a null set. However, every open set containing $\Delta$ must intersect each horizontal line in a set of positive measure, so it must have infinite measure and hence $\mu$ is not outer regular.
Now define $\nu$ by the same formula as $\mu$ if $E$ intersects only countably many horizontal lines, and set $\nu(E) = \infty$ if $E$ intersects uncountably many horizontal lines. Now this measure $\nu$ is inner regular on open sets and outer regular on Borel sets.
Finally, you can check that $\mu$ and $\nu$ assign the same integral to compactly supported continuous functions in $X$.
Best Answer
These are three different theorems and there's no relation between 1) and the others except in the case when $X$ is finite, where all three theorems coincide (since $\ell^2(X)$, $C_0(X)$ and $C_c(X)$ then are the same topological vector space). Note however that $C_0(X)$ and $C_c(X)$ are never Hilbert spaces (unless the locally compact space $X$ is empty or a point), so 2) and 3) can't have a direct relation to 1).
Since positivity implies continuity 2) can be interpreted as characterizing continuous linear functionals on $C_c{(X)}$ as well, and I'm addressing this version below.
The results 2) and 3) are closely related and often 3) is proved as a corollary of 2).
Note that the space $C_c(X)$ is dense in $C_{0}(X)$ with respect to the sup-norm. So a continuous linear functional on $C_c(X)$ (= a signed Radon measure) extends (uniquely) to a continuous linear functional $C_{0}(X)$ (= a signed bounded Radon measure) if and only if it is of bounded variation. Moreover, 2) and 3) coincide if $X$ is compact (and there are also proofs of 3) reducing it to that case).