Suppose we have a $\gamma : 2^X \to \mathbb{R}_{\geqslant 0}$ that is monotonic ($A \subseteq B \implies \gamma(A) \leqslant \gamma(B)$), semi-$\sigma$-additive ($A \subseteq \bigcup_{i \in \mathbb{N}} A_i \implies \gamma(A) \leqslant \sum_{i \in \mathbb{N}} \gamma(A_i)$) and satisfies $\gamma(\varnothing) = 0$; in our class we called that pre-measure though it seems to be a slightly unconventional wording. Let's call $A \subseteq X$ $\gamma$-measurable iff the Carathéodory's criterion holds: $\forall B \subseteq X \quad \gamma(B) = \gamma(B \cap A) + \gamma(B \cap \bar{A})$, where $\bar{A}$ denotes complement $X \setminus A$. Denote with $\Sigma$ the totality of all $\gamma$-measurable $A \subseteq X$. In our class we proved that $\Sigma$ is a sigma-algebra and $\gamma|_{\Sigma}$ is a $\sigma$-additive measure on it (looks kind of similar to Carathéodory's extension theorem). Nothing special at the first glance.
However, it is unclear why should $\Sigma$ be maximal such $\sigma$-algebra that this restriction (or rather extension if $\gamma$ is an outer measure of some $\mu$ defined on a smaller domain) turns out to be possible. It could appear as that a contradiction to the Carathéodory's criterion 'breaks' additivity but that's not necessarily true: maybe the counterexample set $B$ doesn't belong to a $\sigma$-algebra generated by $\Sigma \cup \{ A \}$, where $A$ is the non-measurable set in question, thus making no need to check properties of the measure on it.
If it's true and $\Sigma$ is indeed not maximal, then there exist unmeasurable sets that paradoxically can me assigned a measure without causing any contradictions. That sounds... odd.
