This may be too vague to end up being useful, but:
Are there any (natural? reasonable?) conditions that can be imposed on an outer measure $\phi^*:{\mathcal P}(S)\to[0,\infty]$ to ensure that the $\sigma$-algebra of measurable sets (obtained through Caratheodory's construction) is non-trivial?
By non-trivial, I mean that at least it contains some sets other than $\emptyset,S$; even better if the condition ensures the $\sigma$-algebra is infinite.
The trivial necessary and sufficient condition is that the initially given external measure is the external measure of some sigma-measure. In other terms, a closed object for the Galois correspondence between positive sigma-measures (on sigma-algebras) and positive external measures (defined on all subsets). Is this trivial condition completely useless? Well, you learn that "regular" external measures are the essentially same as "comp0lete" measures. You can also obtain that when you start from a finitely additive measure on a semiring, or a lattice, of sets, then all the initially given measurable sets (and even the Peano - Jordan measurable ones) are still measurable for the sigma-measure that you obtain with the two-step Caratheodory process. Some books on measure theory also consider the case of a initially given finitely subadditive measure (on a ring of sets).
There is another process, again with a Galois correspondence, to "bi-complete" a measure: not only you want that a subset of a zero-measure set again has measure zero, but also that if a set has infinite measure then it has a subset with finite nonzero measure (or, equivalently, the measure is the sup of the measure of the integrable subsets).
I think that standard measure theory text should have all this (Fremlin should be freely available online). A more sophisticated question is the distinction between the above measures (complete and locally finite) and the measures that are direct sum of finite measures. A sigma-finite and complete measure is a direct sum of finite complete measures and a direct sum of finite measures is complete and locally finite, but neither of the implications can be reversed (the counterexamples for the first are very easy, for the second they are not). In standard books, you can check also the relation of the above conditions with the following possible properties of a complete (positive sigma-additive) measure: (a) the Boolean algebra of measurable sets modulo measure zero sets is complete; (a') an analogue for the vector lattice of measurable functions modulo almost everywhere zero functions; (b) the natural duality between integrable and bounded measurable functions gives the dual of the Banach space of integrable functions (modulo "almost everywhere". The dual of bounded measurable functions is never reduced to the integrable ones [i.e. absolutely continuous sigma-measures], except for trivial [finite dimensional] cases; the dual of the other usual spaces is the "expected one" for all complete measures); (c) there is a "lifting" for the bounded measurable functions modulo "almost everywhere" to the bounded measurable functions. [Hint: (c) is subtly different from the preceding properties, check in the standard books]
I added this last part about "locally finite" measures becouse it is related to the other question of yours, but note that it is not exactly the same as the Bourbaki distinction (some measure theory books write about the distinction between the tau extension and the sigma-extension for a Daniell integral). I hope that with all these keywords you can check in your measure theory books.
One commonly used condition, even found in the original paper of Caratheodory, is what we now call a metric outer measure. $(S,d)$ is a metric space. A metric outer measure is an outer measure $\mu$ such that: If two sets $A, B \subseteq S$ have positive distance (there is $\delta > 0$ so that for all $a \in A$ and $b \in B$ we have $d(a,b) \ge \delta$), then $\mu(A \cup B) = \mu(A) + \mu(B)$. Then: if $\mu$ is a metric outer measure, then at least all Borel sets in $S$ are $\mu$-measurable sets.
