Caratheodory criterion, sufficient but is it necessary? When constructing the Lebesgue measure on $\mathbb{R}$, it is shown that the sets that satisfy the caratheodory criterion form a sigma-algebra (the Lebesgue sigma-algebra), and also that the countable additivity of the Lebesgue outer measure is satified on this sigma-algebra.
But is the caratheodory necessary? That is, could there be a bigger sigma-algebra, or another sigma-algebra not comparable with the Lebesgue sigma-algebra, where the caratheodory criterion is not satisfied by all the sets, but countable additivity still holds?
 A: Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$, let $\Lambda$ be the Lebesgue $\sigma$-algebra on $\mathbb{R}$ and let $m^*$ be the Lebesgue outer measure define on $P(\mathbb{R})$. We know that: $\mathcal{B}\subseteq \Lambda$ and $m^*$ restricted to $\Lambda$ is countably additive.
Claim: Let $\Sigma$ be a $\sigma$-algebra on $\mathbb{R}$. Suppose all intervals are in $\Sigma$ and $\Sigma \nsubseteq \Lambda$, then $m^*$ is not additive on $\Sigma$.
Proof: Since all intervals  are in $\Sigma$ and $\Sigma$ is a $\sigma$-algebra, we have that $\mathcal{B} \subseteq \Sigma$.
Since  $\Sigma \nsubseteq \Lambda$, there $E\in \Sigma$ such that $E\notin \Lambda$. Now consider such $E$. Since $E\notin \Lambda$, there is $A\in P(\mathbb{R})$ such that
$$ m^*(A) < m^*(A\cap E) + m^*(A\cap E^c)$$
In particular, $  m^*(A) < +\infty$. So there is a Borel measurable cover of $A$, that is:   $B\in \mathcal{B}$ such that $A\subseteq B$ and $m^*(B)=m^*(A)$. So, we have:
$$m^*(B)= m^*(A) < m^*(A\cap E) + m^*(A\cap E^c) \leqslant m^*(B\cap E) + m^*(B\cap E^c)  \tag{1}$$
(since $A\cap E \subseteq B \cap E$, by $m^*$ monotonicity, we have $m^*(A\cap E) \leqslant m^*(B\cap E)$. In a similar way, we have we have $m^*(A\cap E^c) \leqslant m^*(B\cap E^c)$).
Since $B\in \mathcal{B}\subseteq \Sigma$, $E\in \Sigma$, we have that $ B\cap E \in \Sigma$ and $B\cap E^c \in \Sigma$. Note that $(B\cap E) \cap (B\cap E^c)=\emptyset$  and $B=(B\cap E) \cup (B\cap E^c)$. But, from $(1)$, we have
$$m^*(B)< m^*(B\cap E) + m^*(B\cap E^c) $$
So $m^*$ is not additive on $\Sigma$.
Conclusion: If $\Sigma$ is a $\sigma$-algebra on $\mathbb{R}$ such that all intervals are in $\Sigma$ and $m^*$ is additive on $\Sigma$ then $\Sigma \subseteq \Lambda$.
Remark 1: The same proof above works if $\Sigma$ contains any family  of set that generates $\mathcal{B}$, the Borel $\sigma$-algebra. For instance,  the family of open intervals or the family of closed intervals, etc..
Remark 2: If $\mathcal{B} \nsubseteq \Sigma$ then it is easy to build an exemple of a $\sigma$-algebra $\Sigma$ such that  $ \Sigma \nsubseteq \Lambda$ and $m^*$ is countably additive on $\Sigma$. In fact, let $E \subseteq \mathbb{R}$ such that $E \notin \Lambda$ ($E$ is not Lebesgue measurable).  Let
$$\Sigma=\{\emptyset, E, E^c, \mathbb{R}\}$$
then $\Sigma$ is a $\sigma$-algebra, $ \Sigma \nsubseteq \Lambda$ and $m^*$ is countably additive on $\Sigma$.
Answering the questions:

*

*Is the Caratheodory criteria necessary?

If you are considering $\sigma$-algebras that contain the set of all intervals (or equivalently, contain $\mathcal{B}$), the answer is YES.  Otherwise, the answer is NO.


*Could there be a bigger sigma-algebra, (bigger than Lebesgue sigma-algebra) where the caratheodory criterion is not satisfied by all the sets, but countable additivity still holds?

NO. Any  $\sigma$-algebra $\Sigma$ strictly bigger than Lebesgue $\sigma$-algebra, contains the set of all intervals and according to Claim,  $m^*$  is not additive on  $\Sigma$.


*Could there be another sigma-algebra
not comparable with the lebesgue measurable sets, where the caratheodory
criterion is not satisfied by all the sets, but countable additivity still
holds?

If you are considering $\sigma$-algebras that contain the set of all intervals (or equivalently, contain $\mathcal{B}$), the answer is NO (see Claim above).
If you are considering any $\sigma$-algebra (not necessarily containing the set of all intervals) then the answer is YES, and the exemple is rather trivial (see Remark 2 above).
