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).
The total variation $|\mu|$ of $\mu$ is defined via
$$|\mu|(B) = \sup_{\pi} \sum_{A \in \pi} \|\mu(A)\|_{\mathbb{R}^n}$$
where the supremum ranges over all finite partitions of the Borel set $B$ into disjoint Borel sets.
In the scalar case, this coincides with the definition via the Hahn decomposition. In the vector-valued case, however, there is no relation between the Hahn decompositions of $\mu_i$ and $|\mu|$ (or, at least, I am not aware of any connection). Similarly, there is no relation between the total variation $|\mu|$ of the vector measure $\mu$ and the vector of total variations $(|\mu_1|, \ldots, |\mu_n|)$.
Best Answer
Ok, following @Martin hint, here's why $|\mu|$ is Radon.
Write $\mu=(\mu_1 - \mu_2) +i (\mu_3-\mu_4)$ Using Jordan's decomposition.
Show that there are positive constants $c_1,c_2$ such that
$c_1 (\mu_1+\mu_2+\mu_3+\mu_4)\leq |\mu|\leq c_2(\mu_1+\mu_2+\mu_3+\mu_4)$
Show that when $\mu=\mu_1-\mu_2$ is real, $|\mu|=\mu_1+\mu_2$
Show that for a finite positive borel measure $\mu$, $\mu$ is Radon iff for every borelian $E$, $\epsilon >0$ there are $V,K$, $V$ open, $K$ compact, $K\subseteq E\subseteq V$ such that $\mu(V\setminus K)<\epsilon$.
This actually shows that $|\mu|$ Radon implies $\mu$ Radon