I will concentrate on comparing (3) and (4). The definition (1) is meant for finite signed measures, whereas all the other definitions are meant for arbitrary positive measures; (1) is equivalent to (4) in the case of finite positive measures. (2) appears to be equivalent to (4) ["locally finite" can mean "finite on compact sets", although it is sometimes taken to mean "finite on the elements of some topological basis"; these are equivalent in the LCH (locally compact Hausdorff) case]. Finally, (5) does not appear to be a definition at all, but rather a description of a definition.
Now then,
i) In the case of a second countable LCH space, every locally finite measure satisfies both (3) and (4) (Theorem 7.8 of [1]). This is the most commonly considered scenario in applications, which is why almost no one bothers to carefully sort out the differences between the different definitions.
ii) In the case of a sigma-compact LCH space, (3) and (4) are equivalent. The forward direction is Corollary 7.6 of [1]; the backwards direction follows from the forward direction together with (iv) below (but I'm sure there is an easier proof).
iii) (3) and (4) are not equivalent in general, even for LCH metrizable spaces (Exercise 7.12 of [1]).
iv) In an LCH space, there is a bijection between
A) measures satisfying (3),
B) measures satisfying (4), and
C) positive linear functionals on the space of continuous functions with compact support.
(The Riesz representation theorem gives either (A)<->(C) or (B)<->(C), depending on where you look; (A)<->(B) is in the Schwarz book mentioned by Joe Lucke; see also Exercise 7.14 of [1])
[1] G. B. Folland, Real Analysis: Modern Techniques and Their Applications
Note: In [1], "Radon" refers to measures satisfying (3).
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.
Best Answer
If $X$ is not $\sigma$-finite, the answer is NO. Here is a simple example.
Let $X=[0,1]$ with the usual topology. Let $\mu$ be defined on the Borel $\sigma$-algebra, by $\mu(\emptyset)=0$ and $\mu(E)=+\infty$ if $E\neq \emptyset$.
It is easy to see that$\mu$ is a regular measure. Let $g$ be the constant function, $g=1$.
We have $\int_E g = 0$ for all Borel sets $E$ of finite measure (because the only Borel set of finite measure is the empty set!) and, it is clear that we can not conclude $g=0$ a.e.