Question about the Riesz representation theorem(s)

Question

I am looking at two seemingly same (but not quite) Riesz representation theorems:

(Wikipedia) Let $X$ be a locally compact Hausdorff space. Let $C_c(X)$ be the space of compactly supported continuous functions. Then a positive continuous linear functional $\Lambda : C_c(X) \to \mathbb R$ can be represented as an integral that is, $\Lambda$ corresponds to a (unique) measure $\mu$ (regular and Borel) such that $\Lambda (f) = \int_X f d \mu$ for all $f \in C_c(X)$.

(My lecture notes) Let $X$ be a locally compact, $\sigma$-compact metric space. Then a positive linear functional $\Lambda : C_c(X) \to \mathbb R$ can be represented by a (unique) measure $\mu$ (locally finite and positive) such that $\Lambda (f) = \int_X f d \mu$ for all $f \in C_c(X)$.

I can see that metric implies Hausdorff. But I find it impossible to remember all the conditions, namely, those about $X$ (just locally compact or $\sigma$-compact as well?) and those about the measure (surely every measure is positive by definition but requiring it to be Borel and regular seems to be stronger than locally finite).

My question(s): Which of these two versions is more general? Or are they the same? And how can I tell them apart, that is, remember which version comes with which assumptions on the space and the measure?

Thanks for your help.

Answer 1

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:

1. Regular and Borel.
2. 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:

1. The first theorem is more general, because it applies to more cases.
2. With the assumptions of the second theorem, they are equally strong.

Answer 2

To my taste, adjectives in a situation like this should be re-read in terms of what they accomplish (or don't) in terms of the goal at hand.

That is, locally compact, Hausdorff assures a good supply of continuous, compactly-supported functions, by Urysohn's lemma. (Otherwise, there might be too few continuous functions to approximate characteristic functions of compact sets.)

The sigma-compactness is (in my world) completely reasonable, and avoids various "pathologies" both in terms of outer regularity and measures on product spaces. It is not obvious in advance that it is needed, but... it is. (The interaction of these adjectives in metric spaces is a source of many homework exercises.)

And the outer and inner regularity, and "Borel"-ness, are certainly desirable, and unsurprising in a context that the purported measure is strongly related to continuous functions (assuming that the space is nice enough so there aren't pathologies!)