I am trying to prove the following:
Given a pre-measure $p : R \rightarrow [0,+\infty]$ where $R \subset 2^X$ is a semi-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?