Application of Caratheodory’s theorem on outer measure I have been going through Caratheodory’s theorem and its applications in measure theory and stumbled upon the following theorem,

Let $\mathscr{C}$ be a finitely additive class and $\lambda$ be the measure on class $\mathscr{C}$. Then,

*

*$\mathscr{C}^*$ is a $\sigma$-algebra. Besides, if we define $\mu(E)=\lambda^*(E)$ for $E\in\mathscr{C}^*$, then $(X, \mathscr{C}^*, \mu)$ is a complete measure space where $\mu$ is a measure defined on $\sigma$-algebra and $\lambda^*$ is the outer-measure.

*It holds $\mathscr{C}\subset\mathscr{C}^*$. Moreover, if $C\in\mathscr{C}$, it holds $\lambda(C)=\lambda^*(C)$.


I have been trying to prove the above. I think the first statement could be the direct consequence of Caratheodory but I have been stuck with the second.
First, I thought about proving the equality $\lambda(C)=\lambda^*(C)$, it would be a good idea to proceed by showing both $\lambda(C)\geq\lambda^*(C)$ and $\lambda(C)\leq\lambda^*(C)$. In here, I believe it is fairly easy to show the former through an implication of $C\in\mathscr{C}$ and in order to try to prove $\lambda(C)\leq\lambda^*(C)$, I rewrote it as $\lambda(C)\leq\sum_{j=1}^{\infty}\lambda(C_j)$ where the infimum implies the original inequality.
However, I am not sure as to how to proceed with showing $\mathscr{C}\subset\mathscr{C}^*$ here? So far I tried to rewrite this as trying to prove $$\lambda^*(E) \geq \lambda^*(E\cap C) + \lambda^*(E\cap C^c)$$
Not quite sure if it is the right direction or how to proceed from here.
 A: Fix $\epsilon > 0$. Let $A \in $ and let $E \subseteq X$. There exists a sequence $(U_{n})_{n = 1}^{\infty}$ in  such that $E \subseteq \bigcup_{n = 1}^{\infty} U_{n}$ and
$$\sum_{n = 1}^{\infty} \: \lambda(U_{n}) \: \leq \: \lambda^{\star}(E) + \epsilon$$
The sequences $(U_{n} \cap A)_{n = 1}^{\infty}$ and $(U_{n} \setminus A)_{n = 1}^{\infty}$ belong to  and cover $E \cap A$ and $E \setminus A$, respectively. By additivity of $\lambda$,
\begin{equation*}
    \begin{split}
        \lambda^{\star}(E \cap A) + \lambda^{\star}(E \setminus A) \: & \leq \: \sum_{n = 1}^{\infty} \: \lambda(U_{n} \cap A) + \sum_{i = 1}^{\infty} \: \lambda(U_{n} \setminus A) \\
        & = \: \sum_{n = 1}^{\infty} \: \lambda(U_{n}) \\ 
        & \leq \: \lambda^{\star}(E) + \epsilon\\
    \end{split}
\end{equation*}
This shows $ \subseteq ^{\star}$. 
Finally, let $C \in $. The sequence $(C, \emptyset, \emptyset...)$ trivially covers $C$, meaning $$\lambda^{\star}(C) \: \leq \: \lambda(C) + \sum_{n = 2}^{\infty} \lambda(\emptyset) = \lambda(C)$$
For the reverse inequality, let $(V_{n})_{n = 1}^{\infty}$ be a sequence in  such that $C \subseteq \bigcup_{n = 1}^{\infty} V_{n}$ and
$$ \sum_{n = 1}^{\infty} \: \lambda(V_{n}) \: \leq \: \lambda^{\star}(C) + \epsilon$$
By countable subadditivity of $\lambda$,
\begin{equation*}
    \lambda(C) \: \leq \: \sum_{n = 1}^{\infty} \: \lambda(V_{n}) \: \leq \: \lambda^{\star}(C) + \epsilon
\end{equation*}
