About uniqueness in the Caratheodory's Theorem Caratheodory's Theorem asserts that if $\mathcal{A}$ is an algebra of subsets of $X$ and $\mu_0:\mathcal{A}\to \mathbb{R}\cup\{\infty\}$ is $\sigma-$additive on $\mathcal{A}$, then there is a measure $\mu$ defined on $\sigma(\mathcal{A})$ (the sigma-algebra generated by $\mathcal{A}$). Furthermore, this extension is unique if $\mu_0$ is $\sigma-$finite.
Let me remind you that the idea of the proof is to define an outer measure $\mu_*$ and prove that restricted to the caratheodory measurable sets is a measure. Of course one also have to prove that this class of sets is a $\sigma-$algebra. The outer measure is defined for all subsets of $X$:
$$\mu_*(E) = \inf \left \{\sum_i \mu_0(A_i) \ \Bigl|\Bigr. \ E\subset \bigcup A_i, A_i \in \mathcal{A} \right \}$$
I know two proofs of the uniqueness. One of these use the so called monotone class and the fact that the smallest monotone class that contains $\mathcal{A}$ is $\sigma(\mathcal{A})$. My question is with the second proof. The same is as follows:
Let $\nu$ is another measure on $\sigma(\mathcal{A})$ which coincides with $\mu_0$ on $\mathcal{A}$. Put $F \in \sigma(\mathcal{A})$ of finite measure. If $F\subset \bigcup A_i$ with $A_i \in \mathcal{A}$ hence
$$\nu(F) \leq \sum_i\nu(A_i) = \sum_i\mu(A_i) $$
taking infimum $\nu(F)\leq\mu(F)$ (remember $\mu$ comes from the outer measure). Now consider $F\subset \bigcup_i A_i$ such that $\mu(\bigcup A_i)\leq \mu(F) + \epsilon$. Let me call $E:= \bigcup A_i$. As $F$ has finite measure then this condition implieas $\mu(E\backslash F) \leq \epsilon$. Furthermore, $\mu(E) = \nu(E)$ because $\mu$ y $\nu$ agree on $\bigcup_{i=1}^n A_i \ \forall n$ and this is an increasing sequence of sets. Now:
$\mu(F) \leq \mu(E) = \nu(E) = \nu(E\backslash F) + \nu(F) \leq \mu(E\backslash F) + \nu(F) \leq \epsilon + \nu(F)
$
since $\epsilon$ is arbitrary the proof is done for all $F \in \sigma(\mathcal{A})$ of finite measure. Using this and the condition that $\mu_0$ is $\sigma-$finite the result is extended to all $F \in \sigma(\mathcal{A})$.
What i can't see is where we are using the condition $F \in \sigma(\mathcal{A})$. I mean, we have $\mathcal{M}$ the $\sigma-$algebra of caratheodory measurable sets which is larger and maybe one can obtain uniqueness here. If a examine the previous proof I can't see why we can't prove the result for this $\sigma-$algebra. Indeed, we have the existence of the measure since, as I comment before, $\mu_*$ is a measure on $\mathcal{M}$. To prove the uniquess I repite the same proof, word by word. Of course in this case $\nu$ is a measure on $\mathcal{M}$.
What is wrong?
 A: Your proof is correct. If $\mu_0$ is $\sigma$-finite, your proof shows that, for any sigma-algebra $\mathcal{S}$, such that $\mathcal{A}\subseteq \mathcal{S} \subseteq \mathcal{M}$, there a unique extension of $\mu_0$ to $\mathcal{S}$.
Answering to your comment: your proof does not work for sigma-algebras larger than $\mathcal{M}$, because it uses the fact that $\mu_*$ when restricted to the sigma-algebra is a measure (and $\mathcal{M}$ is the largest sigma-algebra containing $\mathcal{A}$ such that  $\mu_*$ , restricted to it, is a measure -- see proof below). In fact, there are examples of sigma-finite $\mu_0$ having more than one extension to a sigma-algebra larger than $\mathcal{M}$. Of course such extensions don't come from the outer measure. Let us see one such example.
Example:  Let $X =\{a,b\}$, where $a \neq b$. Let $\mathcal{A}=\{\emptyset, X\}$. Clearly,  $\mathcal{A}$ is an algebra (in fact a sigma-algebra).  Let us define $\mu_0(\emptyset)=0$ and $\mu_0(X)=1$. Clearly $\mu_0$ $\sigma-$additive on $\mathcal{A}$.
Let $\mu_*$ be the outer measure induced by $\mu_0$. Then, we have $\mu_*(\emptyset)=0$ and, for all $E \subseteq X$, if $E \neq \emptyset$, $\mu_*(E)=1$.
It is easy to check that $E$ is $\mu_*$-measurable if and only if $E=\emptyset$ or $E=X$.
That means, $\mathcal{M}= \{\emptyset, X\}= \mathcal{A}$.
Now, let $\mathcal{S} = \{\emptyset, \{a\}, \{b\}, X\}=2^X$. Clearly, $\mathcal{S}$ is a
sigma-algebra.
Now, for each $r \in [0,1]$, consider $\mu_r$ defined on $\mathcal{S}$ by
$\mu_r(\emptyset)=0$, $\mu_r(\{a\})=r$,  $\mu_r(\{b\})=1-r$ and  $\mu_r(X)=1$.
For each $r \in [0,1]$, $\mu_r$ is different extension of $\mu_0$ to $\mathcal{S}$.
Now let us prove that

$\mathcal{M}$ is the largest sigma-algebra containing $\mathcal{A}$ and such that  $\mu_*$ , restricted to it, is a measure.

Proof: Let $\mathcal{S}$ be a sigma-algebra of subsets of $X$ such that $\mathcal{A} \subseteq \mathcal{S}$ and such that $\mu_*$ , restricted to it, is a measure.
Given $E\in \mathcal{S}$, we want to prove that, for all $A \subseteq X$,
$$\mu_*(A) = \mu_*(A\cap E) + \mu_*(A\cap E^c)$$
If $\mu_*(A)=\infty$, this an immediate consequence of the sub-additivity of the outer measure. Now suppose that  $\mu_*(A)<\infty$. Then there is $B\in \sigma(\mathcal{A})$ such that $A \subseteq B$ and $\mu_*(B)=\mu_*(A)$ (such $B$ is called a measurable cover of $A$).
Since $\mathcal{A} \subseteq \mathcal{S}$ and $\mathcal{S}$ is a sigma-algebra, we have that  $\sigma(\mathcal{A}) \subseteq \mathcal{S}$. So $B\in \mathcal{S}$. Since $E \in \mathcal{S}$ and  $\mu_*$ , restricted to $\mathcal{S}$, is a measure, we have, using the monotonicity of the outer measure:
$$\mu_*(A\cap E) + \mu_*(A\cap E^c) \leqslant \mu_*(B\cap E) + \mu_*(B\cap E^c)=\mu_*(B)= \mu_*(A)$$
and, by the sub-additivity of the outer measure, we have
$$ \mu_*(A) \leqslant \mu_*(A\cap E) + \mu_*(A\cap E^c)$$
So
$$ \mu_*(A) = \mu_*(A\cap E) + \mu_*(A\cap E^c)$$
Thus, we have proved that given any  $E\in \mathcal{S}$, then $E$ is $\mu_*$-measurable, that is, $E\in \mathcal{M}$. So $\mathcal{S} \subseteq \mathcal{M}$.
