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).
One standard example is the reals numbers times the reals with the discrete topology: $X = \mathbb{R} \times \mathbb{R}_d$.
This is a locally compact metrizable space. The compact subsets intersect only finitely many horizontal lines and each of those non-empty intersections must be compact. A Borel set $E\subset X$ intersects each horizontal slice $E_y$ in a Borel set.
Consider the following Borel measure where $\lambda$ is Lebesgue measure on $\mathbb{R}$:
$$
\mu(E) = \sum_{y} \lambda(E_y).
$$
This is easily checked to define an inner regular Borel measure and its null sets are precisely those Borel sets that intersect each horizontal line in a null set. In particular, the diagonal $\Delta = \{(x,x) : x \in \mathbb{R}\}$ is a null set. However, every open set containing $\Delta$ must intersect each horizontal line in a set of positive measure, so it must have infinite measure and hence $\mu$ is not outer regular.
Now define $\nu$ by the same formula as $\mu$ if $E$ intersects only countably many horizontal lines, and set $\nu(E) = \infty$ if $E$ intersects uncountably many horizontal lines. Now this measure $\nu$ is inner regular on open sets and outer regular on Borel sets.
Finally, you can check that $\mu$ and $\nu$ assign the same integral to compactly supported continuous functions in $X$.
Best Answer
I suppose $\nu_a$ stands for the absolutely continuous part and $\nu_b$ for then singular part.
$\|\mu+\nu\|=\|\mu\|+\|\nu\|$ if $\mu \perp \nu$. In Lebesgue decompostion we do have $\nu_a \perp \nu_b$. Hence, $\|\nu\|=\|\nu_a\|+\|\nu_b\|$.