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.
My intuition for (1) is to approximate the smooth, compactly-supported functions with compactly-supported step functions.
This is backwards. You should approximate things with smooth compactly supported functions, not the other way around.
I think there is no way around using mollifiers; how else will you construct smooth functions? Here is my approach:
Let $\Omega$ be any open subset of $\mathbb R^n$. For $x\in\Omega$, let $d(x)$ be the distance from $x$ to $\partial\Omega$. Given $f\in L^2(\Omega)$, define
$$f_n(x) = \begin{cases} f(x)\quad &\text{ if }\ d(x)>1/n \\ 0\quad &\text{ if }\ d(x)\le 1/n \end{cases}$$
Note that $(f_n-f)^2$ converges to $0$ pointwise, and is dominated by $f^2$. By the dominated convergence theorem, $\|f_n-f\|_{L^2}\to 0$.
Next, approximate $f_n$ by $f_n*\phi_\epsilon$, where the mollifier $\phi_\epsilon$ is supported in a ball of radius $\epsilon<1/n$. This will ensure that the convolution is both smooth and compactly supported in $\Omega$. The $L^2$ convergence of mollified functions is a standard fact.
Best Answer
Try $X = \mathbb R$ with the usual topology, and counting measure on the rationals. The only continuous function in $L^1$ is $0$.