I appologize if the question in the title of the post is not clear. What I mean is this:
Can we prove that starting with the semi-algebra of intervals and the lenght function defined on it and proceeding by Caratheodory's method the $\sigma$-algebra $\mathcal{M}$ obtained of measurables sets is maximal in the sense that if we enlarge this family by adding a non-measurable set then the outer measure restricted to this new family $\mathcal{M}'$ loses the property of being $\sigma$-additive or the propery of being invariant under translations?
Let $\mathcal{M}'$ be a family containing $\mathcal{M}$ and such that there exist $M \in \mathcal{M}'$ and $X \subset \mathbb{R}$ satisfying $$m^*(X)<m^*(X\cap M) + m^*(X\cap M^c)$$ where $m^*$ is Lebesgue' outer measure.
I don't see how this can contradict $m^*$ being traslational invariant and $\sigma$-additive on $\mathcal{M}'$
You don't lose translation invariance, since $m^*$ is translation invariant.
But you immediately lose the $\sigma$-additivity: for your $M$ and $X$, $$ m^*((M\cap X)\cup(M^c\cap X))<m^*(M\cap X)+m^*(M^c\cap X), $$ so $m^*$ is not even additive in your enlarged $\sigma$-algebra.
In summary, $\mathcal M(\mathbb R)$ is maximal with respect to $\sigma$-additivity. It isn't with respect to translation invariance.