# Real Analysis Folland Proposition 1.13

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.

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.