# Is the $\sigma$-algebra generated by the Carathéodory condition maximal?

I want to approach the (Carathéodory) definition of a measurable set from a different perspective, one that I haven't yet managed to find anyone addressing.

Say we have some outer measure $$\mu^*$$ on the set $$X$$. Let $$M \subseteq \mathcal{P}(X)$$ be some set of sets such that countable additivity holds between all the (disjoint) members of this set - such that $$\forall A_n \in M, n\in \mathbb{N}, \mu^*(\bigcup_{n=1}^{\infty} A_n) = \sum_{n=1}^{\infty} \mu^*(A_n)$$. Say that such a set $$M$$ is maximal with respect to $$\mu^*$$ if it has the property that for any set $$A \subseteq X$$ such that $$A$$ is countably additive with respect to all the subsets of $$X$$ from which it is disjoint, then it follows that $$A$$ belongs to $$M$$ itself.

Question: is the $$\sigma$$-algebra $$M$$ defined by using the Carathéodory condition on the outer measure $$\mu^*$$ maximal? That is, are there no provably no extensions that could be made to this set $$M$$ that preserve countable additivity among all its members?

This seems to me like a reasonable question to ask in order to justify the Carathéodory condition as a definition of measurability in full generality. It's important to know that the measure's countably additive on that set, but to me, that doesn't actually justify the Carathéodory condition directly: it also seems to me to be very important to know that the measurable sets you get from Carathéodory constitute the biggest possible such set you can have. Regardless, I haven't managed to find anyone asking precisely this question. Perhaps treatments of measurability that introduce Carathéodory first, rather than justifying it in terms of something else, are uncommon; I don't know.