The notion of exterior measure seems quite natural to me after having seen the definition of the Lebesgue exterior measure as the infimum of the measures associated to countable covers of the sets by intervals or rectangles.
Let us first assume that we are on a space $X$ of finite measure, say the unit cube. Let us try to single out a class of sets for which $\sigma$-additivity may holds. Let $A$ be a set in such a class. If we want to measure $A$, we also want to measure its complement in $X$ and the two sets are disjoints so that at least the following equality should be satisfied.
$$\mu^*(A)+\mu^*(X\backslash A) = \mu^*(X).$$
This property can be shown to be equivalent to Caratheodory definition of measurability if $X$ is of finite measure and the measure is regular (for all subsets $A$ of $X$, there exists a measurable set $B$ containing $A$ with the same outer measure, a fact that can be guaranted by replacing $\mu^*(A)$ by $inf\{\mu^*(B) \mid A \subset B, \ B \ \mu^*-\hbox{measurable}\}$), and it looks quite natural to me.
If $X$ is not of finite measure, this definition is not very restrictive because $everything + \infty = \infty$. In particular, it is satisfied for all bounded sets in ${\bf R}^d$ wrt to Lebesgue exterior measure. So we need to be slightly more restrictive and asks that the previous property holds in restriction, say, to balls $B_N$ of radius $N$, $N>0$ if they are of finite measure.
$$ \mu^*(B_N) = \mu^*(B_N \cap A) + \mu^*(B_N \cap A^c), \quad \forall N.$$
And here again, this can be shown to be equivalent to Caratheodory's definition.
If we are not on ${\bf R}^d$ but on some general abstract set $X$ with infinite measure, the only choice here is to restrict to all sets $E$ of finite measure.
$$ \mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c), \quad \forall E \subset X \hbox{ such that } \mu^*(E)<\infty.$$
Note that this definition is obviously satisfied if $E$ is of infinite measure and we get the general definition of Caratheodory.