Proposition 1.13 - If $\mu_0$ is a premeasure on $\mathcal{A}$ and $\mu^*$ is defined by (1.12) then
a.) $\mu^*|\mathcal{A} = \mu_0$
b.) every set in $\mathcal{A}$ is $\mu^*$-measurable
Proof a.) - Suppose $E\in\mathcal{A}$ then $\mu^*(E)\leq \mu_0(E)$ (since $E$,$\emptyset$ cover $E$). Now let $E\subset \bigcup_{1}^{\infty}A_j$ and $\{A_j\}_{1}^{\infty}\subset \mathcal{A}$, set $$E_j = E\cap \left(A_j \setminus \bigcup_{j' < j}A_j\right)\in\mathcal{A}$$ Then, $$\mu_0(E) = \sum_{1}^{\infty}\mu_0(E_j) \leq \sum_{1}^{\infty}\mu_0(A_j)$$ so $\mu_0(E)$ is a lowerbound of $\mu^*(E)$ in 1.12.
Proof b.) - Suppose $A\in\mathcal{A}$ and $E\subset X$. Let $\epsilon > 0$, choose $\{B_j\}_{1}^{\infty}\subset \mathcal{A}$ with $E\subset \bigcup_{1}^{\infty}B_j$ and $$\sum_{1}^{\infty}\mu_0(B_j) < \mu^*(E) + \epsilon$$ Then, $$\mu^*(E\cap A) + \mu^*(E\cap A^c) \leq \sum_{1}^{\infty}\mu_0(B_j\cap A) + \sum_{1}^{\infty}\mu_0(B_j\cap A^c) = \sum_{1}^{\infty}\mu_0(B_j) < \mu^*(E) + \epsilon$$
I am not sure if this is correct. Any suggestions is greatly appreciated.