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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.

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Yes, the proof is correct. I copied your proof here just to add small details which may help to make it clearer.

Proof a.) - Suppose $E\in\mathcal{A}$ then $\mu^*(E)\leq \mu_0(E)$, (since $\{E,\emptyset,\emptyset,\cdots \}$ covers $E$). 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, $\{E_j\}_{1}^{\infty}$ is a family of disjoint sets in $\mathcal{A}$ and $E= \bigcup_{1}^{\infty}E_j$. Since $\mu_0$ is a premeasure, we have $$\mu_0(E) = \sum_{1}^{\infty}\mu_0(E_j) \leq \sum_{1}^{\infty}\mu_0(A_j)$$ so $\mu_0(E)$ is a lowerbound of $\{\sum_{1}^{\infty}\mu_0(A_j) : A_j \in \mathcal{A},\;E\subset \bigcup_{1}^{\infty}A_j\}$. So, from the definition of $\mu^*(E)$ (in 1.12), we get $$ \mu_0(E) \leq \mu^*(E)$$ So we can conclude that $$ \mu^*(E) = \mu_0(E) $$

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 $$\sum_{1}^{\infty}\mu_0(B_j) \leq \mu^*(E) + \epsilon$$ Note that, for each $j$, $(B_j\cap A), (B_j\cap A^c) \in \mathcal{A}$, $(B_j\cap A)$ and $(B_j\cap A^c)$ are disjoint and $B_j = (B_j\cap A)\cup (B_j\cap A^c)$. So, for each $j$, $$ \mu_0(B_j)= \mu_0(B_j\cap A)+ \mu_0(B_j\cap A^c)$$

Then, $$\mu^*(E\cap A) + \mu^*(E\cap A^c) \leq \sum_{1}^{\infty}\mu_0(B_j\cap A) + \sum_{1}^{\infty}\mu_0(B_j\cap A^c) = \sum_{1}^{\infty}\mu_0(B_j) \leq \mu^*(E) + \epsilon$$ Since $\epsilon$ is arbitrary, we have $$\mu^*(E\cap A) + \mu^*(E\cap A^c) \leq \mu^*(E) $$ Since $\mu^*$ is subadditive, we also have $$\mu^*(E\cap A) + \mu^*(E\cap A^c) \geq \mu^*(E) $$ So $$\mu^*(E\cap A) + \mu^*(E\cap A^c) = \mu^*(E) $$ So $A$ is $\mu^*$-measurable.

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