Background Information:
In the theorem here we fix a complete Lebesgue-Stiltjes measure $\mu$ on $\mathbb{R}$ associated to the increasing right continuous function $F$, and we denote by $M_{\mu}$ the domain of $\mu$. Thus, for any $E\in M_{\mu}$ $$\mu(E) = \inf\{\sum_{1}^{\infty}[F(b_j) - F(a_j)]:E\subset \bigcup_{1}^{\infty}(a_j,b_j]\} = \inf\{\sum_{1}^{\infty}\mu((a_j,b_j]):E\subset \bigcup_{1}^{\infty}(a_j,b_j]\}$$
Theorem 1.18 - If $E\in M_{\mu}$, then \begin{align*} \mu(E) &= \inf\{\mu(U): E\subset U, U \ \text{open}\}\\ &= \sup\{\mu(K):K\subset E, K \ \text{compact}\} \end{align*}
Theorem 1.19 - If $E\subset \mathbb{R}$, the following are equivalent
a.) $E\in M_{\mu}$
b.) $E = V\setminus N_1$ where $V$ is a $G_{\delta}$ set and $\mu(N_1) = 0$
c.) $E = H\cup N_2$ where $H$ is a $F_{\sigma}$ set and $\mu(N_2) = 0$
Lebesgue measure $m^n$ on $\mathbb{R}^n$ is the completion of the $n$-fold product of Lebesgue measure on $\mathbb{R}$ with itself, that is, the completion of $m\times \ldots \times m$ on $B_{\mathbb{R}}\otimes \ldots \otimes B_{\mathbb{R}} = B_{\mathbb{R}^n}$, or equivalently the completion of $m\times \ldots \times m$ on $L\otimes \ldots \otimes L$.
The domain $L^n$ of $m^n$ is the class of Lebesgue measurable sets in $\mathbb{R}^n$.
In what follows, if $E = \prod_{1}^{n}E_j$ is a rectangle in $\mathbb{R}^n$, we shall refer to the sets $E_j\subset \mathbb{R}$ as the sides of $E$.
Question:
Suppose $E\in L^n$
a.) $m(E) = \inf\{m(U):E\subset U, U \ \text{open}\} = \sup\{m(K):K\subset E, K \ \text{compact}\}$.
b.) $E = A_1\cup N_1 = A_2\setminus N_2$ where $A_1$ is a $F_{\sigma}$ set, $A_2$ is a $G_{\delta}$ set, and $m(N_1) = m(N_2) = 0$.
c.) If $m(E) < \infty$, for any $\epsilon > 0$ there is a finite collection $\{R_j\}_{1}^{n}$ of disjoint rectangles whose sides are intervals such that $m(E \ \triangle \ \bigcup_{1}^{n}R_j) < \epsilon$.
Proof a.) - Let $E\in L^n$ and $\epsilon > 0$ then by definition of product measures there exists a countable family $\{T_j\}$ of rectangles such that $E\subset \bigcup_{1}^{\infty}T_j$ and $\sum_{1}^{\infty}m(T_j) \leq m(E) + \epsilon$. For each $j$, applying theorem 1.18 to the sides of $R_j$ we can find a rectangle $T_j\subset U_j$ whose sides are open sets such that $m(U_j) < m(T_j) + \epsilon 2^{-j}$. Let $U = \bigcup_{1}^{\infty}$, then $U$ is open and $$m(U) \leq \sum_{1}^{\infty}m(U_j) \leq m(E) + 2\epsilon$$ thus we can define $m(E) = \inf\{m(U):E\subset U, U \ \text{open}\}$.
I am not sure how to show the second part.
Proof b.) Not sure
Proof c.) Let $m(E) < \infty$ and $\epsilon >0$. So we know that $m(U_j) < \infty$ for all $j$. Now since the sides of $U_j$ are countable unions of open intrvals, take a sufficient number of subunions to obtain rectangles $V_j\subset U_j$ whose sides are finite unions of intervals such that $m(U_j) \leq m(V_j) + \epsilon 2^{-j}$. Take $N$ to be sufficiently large,then $$m(E\setminus \bigcup_{1}^{N}V_j) \leq m(\bigcup_{1}^{N}U_j\setminus V_j) + m(\bigcup_{N+1}^{\infty}U_j) < 2\epsilon$$ and $$m(\bigcup_{1}^{N}V_j\setminus E) \leq m(\bigcup_{1}^{\infty}U_j\setminus E) < \epsilon$$ thus we have $$m(E\triangle \bigcup_{1}^{N}U_j) < 3\epsilon$$
I am not sure if more details needed to be added to make this more clear. Any suggestions is grealy appreciated.