Is the Carathéodory measurability criterion optimal in some sense? If $m$ is an outer measure defined on a set $X$, we say that a subset $E$ of $X$ is Carathéodory-measurable with respect to $m$ if for all subsets $A$ of $X$, we have $m(A)=m(A\cap E) + m(A\cap E^c)$.  And if $M$ is the set of all Carathéodory-measurable sets with respect to $m$, then $M$ is a complete sigma algebra on $X$
and $m$ restricted to $M$ is a complete measure on $X$.
My question is, is $M$ "optimal" in some way?  Is it the biggest or smallest subset of $P(X)$ such that $m$ restricted to that subset is a measure?  Is it the biggest or smallest subset of $P(X)$ such that $m$ restricted to that subset is a complete measure?
To put it another way, what is it that makes the Carathéodory measurability criterion the "best" criterion for measurability?  Or is it not the best, but just an arbitrary choice out of a sea of infinitely many equally good stronger and weaker measurability criteria?
 A: The answer is no, the Lebesgue measure can be extended to a translation invariant measure in a $\sigma$-algebra that properly contain the Lebesgue sigma álgebra. However this example is not trival. In fact it is  content of  a paper of Annals of Mathematics.
See: https://www.jstor.org/stable/1969435
A: Edit: Yes, the Caratheodory condition defines the largest $\sigma$-algebra $\sum$ such that the restriction of $m$ to $\sum$ is a measure and such that we also have $$m(A)=\inf_{A\subset E\in\sum}m(E)$$for all $A\subset X$. That follows from this:


Motivation for C's definition of measurability: Suppose $(X,\sum,\mu)$ is a measure space. Define the outer measure of $A\subset X$ by $\mu^*(A)=\inf_{A\subset E\in\sum}\mu(E)$. If $A\subset X$ and $E\in\sum$ then $\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E)$.


Proof. Suppose $A\subset X$ and $E\in\sum$. Choose $F$ with $A\subset F\in\sum$. Since $\mu$ is a measure, $A\cap E\subset F\cap E\in\sum$ and $A\setminus E\subset F\setminus E\in\sum$ we have $$\mu(F)=\mu(F\cap E)+\mu(F\setminus E)\ge\mu^*(A\cap E)+\mu^*(A\setminus E).$$Taking the inf over $F$ shows that $$\mu^*(A)\ge\mu^*(A\cap E)+\mu^*(A\setminus E).$$
The other inequality is clear, since it's easy to see that $\mu^*$  is subadditive: $\mu^*(B\cup  C)\le\mu^*(B)+\mu^*(C)$ for every $B$ and $C$. (If $B\subset E_a\in\sum$ and $C\subset E_2\in\sum$ then $$\mu^*(B\cup C)\le\mu(E_a\cup E_2)\le\mu(E_1)+\mu(E_2);$$now take the inf over $E_1$ and $E_2$ on the right side.)
