I am trying to prove the following:
Given a pre-measure $p : R \rightarrow [0,+\infty]$ where $R \subset 2^X$ is a set-ring, define the outer measure $u^{\ast}(E) = \inf\{\sum_{j=1}^{\infty} p(A_j) : E \subset \bigcup_{j=1}^{\infty} A_j \}$ where $\inf\emptyset = +\infty$. Define the Caratheodory measurable sets as usual: $C = \{A \in 2^X : \forall B \in 2^X : u^{\ast}(B) = u^{\ast} ( B\cap A) + u^{\ast}(B \setminus A) \}$. Then $(X,C,u^{\ast})$ is a measure space, $R \subset C$, and $u^{\ast}$ restricts to $p$ on $R$. Show that if $\lambda$ is a measure on $C$ restricting to $p$ on $R$, and there exist an increasing sequence of sets in $R$, $\{X_n\}$ such that $u^{\ast}(X_n) = \lambda(X_n) < + \infty$, with $X_n \uparrow X$, and we already know that $\lambda(E) \leq u^{\ast}(E)$ for any $E \in C$ then $u^{\ast}(E) = \lambda(E)$ for any $E \in C$.
My attempt:
It suffices to show that $u^{\ast}(E \cap X_n) = \lambda(E \cap X_n)$ for any $X_n$ in the increasing sequence. Since both $u^{\ast}(E)$ and $\lambda(E)$ are the limit as $n \rightarrow \infty$ of these terms.
Since $E \in C$, by definition we have $u^{\ast}(X_n) = u^{\ast}(X_n \cap E) + u^{\ast}(X_n \setminus E) $. Moreover we have that $u^{\ast}(X_n) = \lambda(X_n)$ since $X_n \in R$, and since $\lambda$ is a measure and thus finitely additive, $\lambda(X_n) = \lambda(X_n \cap E) + \lambda(X_n \setminus E)$, which gives $$u^{\ast}(X_n \cap E) + u^{\ast}(X_n \setminus E) = \lambda(X_n \cap E) + \lambda(X_n \setminus E)$$ We have that $\lambda(X_n \cap E) \leq u^{\ast}(X_n \cap E)$ and $\lambda(X_n \setminus E) \leq u^{\ast}(X_n \setminus E)$ since we know that $\lambda \leq u^{\ast}$ on $C$. Since each term in the previous two inequalities are necessarily finite, this implies that $\lambda(X_n \cap E) = u^{\ast}(X_n \cap E)$ and $\lambda(X_n \setminus E) = u^{\ast}(X_n \setminus E)$ , giving the desired result.
Is this proof sound?
If $\mathscr{E}$ is a semiring and $\mu$ is a nonnegative $\sigma$-additive and $\sigma$-subadditive function on $\mathscr{E}$, i.e.
then the extension $\mu$ to all of $\sigma(\mathscr{E})$ may not be unique. However, the Caratheodory construction provides a maximal extension, that is, if $\mu$ is the measure built using the procedure of Caratheodory, and $\nu$ is any other extension of $\mu$ to $\sigma(\mathscr{E}$, then $$ \nu(E)\leq\mu(E),\quad\forall E\in\sigma(\mathscr{E})$$
When $\mathscr{E}$ is a ring, then the extension is unique.
Here is a sketch of the proof:
Let $\eta$ be another extension of $\mu$ to $(\Omega,\sigma(\mathscr{E}))$. For any $E\in\sigma(\mathscr{E})$, choose $B_n\in\mathscr{E}$ so that $E\subset\bigcup_nB_n=B$. Then $\eta(E)\leq\eta(B)\leq\sum_n\eta(B_n)=\sum_n\mu(B_n)$. Taking infima over all possible covers gives $\eta(E)\leq\mu^*(E)=\mu(E)$.
Assume that $\mathscr{E}$ is a ring and suppose that $E\in\sigma(\mathscr{E})$ with $\mu^*(E)=\mu(E)<\infty$. For any $\varepsilon>0$ choose $B_n\in\mathscr{E}$ so that $E\subset\bigcup_nB_n=B$ and $\mu(B)<\mu(E)+\tfrac{\varepsilon}{2}$. Hence, $\mu(B\setminus E)<\tfrac{\varepsilon}{2}$. Since $A_k:=\bigcup^k_{j=1}B_j\nearrow B$, we can choose $k$ so that $\mu(B)-\mu(A_k)=\mu(B\setminus A_k)<\tfrac{\varepsilon}{2}$. Since $\eta(B)=\eta(E)+\eta(B\setminus E)$, $\eta\leq \mu$ and $\eta=\mu$ on $\mathscr{E}$, it follows that $$ \begin{align} \eta(E)&=\eta(B)-\eta(B\setminus E)\geq\eta(A_k)-\eta(B\setminus E)\\ &\geq\mu(A_k)-\mu(B\setminus E)> \mu(B)-\varepsilon\geq\mu(E)-\varepsilon \end{align} $$ Letting $\varepsilon\searrow0$ we obtain that $\eta(E)=\mu(E)$.
This is also related to the maximality of the Daniell mean of the elementary integral $(\mathcal{E},I)$ where $\mathcal{E}$ is the space of bounded simple functions $\phi=\sum^n_{k=1}\mathbb{1}_{A_k}$ such that $|I(\phi)|=|\sum^n_{k=1}a_k\mu(A_n)|<\infty$.
Some references:
Kallenberg's Foundations of probability chapter 1 has an excellent exposition of the Caratheodory construction starting from semi rings.
Klaus Bichteler's Integration Theory: A functional approach, has an excellent presentation of the Daniell integral and is equivalence to the Lebesgue-Caratheodory construction. The book discusses the maximality that I mention above.