Measure extension via inner measure Is it possible to use "inner measure" in the proof of the Caratheodory extension theorem?
I'm trying to understand why we prefer outer one instead, although definitions of outer and inner measures are dual.
N. B.: I mean $\mu_*(A) = \sup \{\sum \mu(E_i) | E_i\mbox{ disjoint}, E_i \subset A\}$ is the "inner measure," generated by some pre-measure $\mu$, defined on some ring $\mathcal R$. It's not in general use.
So let me explain the problem:
We can proof superadditivity of $\mu_*$ defined above.
Then we consider $\mathcal M (\mathcal R) = \{A \;|\; \forall E  \in \mathcal R \;\;\mu_*(E\cup A) + \mu_*(E\setminus A) = \mu_*(E)\}.$  
Now it's not hard to show that $\mathcal M(\mathcal R)$ closed under finite union and completion. 
But I can't prove that $\mathcal M(\mathcal R)$ closed under countable union. Maybe it's not true in general? What is the difference in such symmetrical concepts of inner in outer measure?
 A: I think there is very simple counterexample to my construction (with disjoint sets):
Let $\mathcal R$ be the ring of intervals (with and without endpoints) with rational endpoints with usual $\mu$ a.k.a "length" of interval. 
Suppose that our construction give us a real good measure $\mu_*$ with some $\mathcal M(\mathcal R)$. Then $\mu_*(X) = 0$ for every countable $X$ and every such $X$ is clearly in $\mathcal M(\mathcal R).$ Also it's not difficult to show that $\mathcal R \subset \mathcal M (\mathcal R)$ and $\mu_*$ really extends $\mu.$
Then we conclude that $\mu_*([0,1]\setminus \mathbb Q) = 1,$ but it's not equal the value of $\mu_*([0,1])\setminus \mathbb Q),$ computed by definition of $\mu_*$ - then it will be 0, because every interval with non-zero length contains a rational point, so only $E \in \mathcal R$ contained in $[0,1]\setminus \mathbb Q$ is $\varnothing.$
Also it is counterexample for such construction of $\mu_*$: $\mu_*(A) =\sup \{\mu(E) | A \supset E \in \mathcal R\},$ which seems to be more dual than mine.
