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.
The main difference is that in the first case, the functional is defined on the space of continuous functions (on a compact set) whereas in the second case the functional operates on the integrable functions.
For completeness' sake, I add that there also exists a version for continuous functions for which the integration is performed with respect to a measure. This result is much more general than the ones you mentioned, but in your case, the measure theoretic version and the one with a BV-function are essentially the same. The reason for this is that to each signed measure $\mu$ on an interval corresponds a function with bounded variation $g$ (and vice versa) such that the Riemann-Stieltjes-integral of a continuous function $f$ with respect to $g$ coincides with the Lebesgue-integral of $f$ with respect to $\mu$. See this post for some details.
As for the measure-theoretic versions, whether you have to use a real (i.e. signed) measure or a complex measure just depends on whether your functional is real- or complex-valued. This will most often coincide with whether your continuous/integrable functions are real- or complex-valued, since a linear functional is typically defined to be a linear mapping from a vector space into the underlying scalar field, which can only be $\mathbb{R}$ in the case of real-valued functions.
The Riesz theorem for Hilbert spaces is, although named the same, a completely different story. This theorem is about the interplay of continuous functionals and the inner product whereas the other theorems are concerned with the representation of functionals by means of an integral. The deeper similarity of all of them is that they offer characterisations of a dual space.
EDIT: As pointed out by Martin Argerami, the last paragraph is incorrect, I quote:
Any Hilbert space can be represented as $H=L^2(X, \mu)$ for an appropriate measure space $X$, and so the Riesz Representation Theorem says that you can write a bounded functional $\phi$ by$$\phi(f)=\langle f,g\rangle = \int_X f\,\bar g\,d\mu.$$
Therefore, there is a much more immediate similarity between the Riesz theorems.
Best Answer
Yes, you extend $l$ to $\lambda \in C(\mathbb{C}^n)^\ast$. Then, since $\lambda$ is continuous, there is a compact $K \subset \mathbb{C}^n$ with
$$\lvert \lambda(f)\rvert \leqslant M\cdot \max \left\{\lvert f(z)\rvert : z \in K\right\},$$
and you use the Riesz representation theorem for continuous linear functionals on $C(K)$ (by Tietze's extension theorem, the restriction $\rho \colon C(\mathbb{C}^n) \to C(K)$ is surjective, and the kernel of $\rho$ is contained in $\ker\lambda$, so the induced functional on $C(K)$ is well-defined and uniquely determined by $\lambda$), which gives you a representing Borel measure $\mu$ on $K$.