Let $(X, \mathcal{A}, \mu)$ be a measurable space with a $\sigma-$finite measure $\mu$, while $(X, \mathcal{A^{*}}, \mu^{*})$ is the completion of the underlying space.
Show:
For $B \in \mathcal{A^{*}}$ with $\mu^{*}(B) < \infty $, there exists $A \in \mathcal{A}$ so that $B \subseteq A$ and $\mu^{*}(B)=\mu(A)$
My ideas:
Since $\mu^{*}(B) < \infty$
and $\mu^{*}(B)=\inf_{(A_{n})_{n} \in \mathcal{U}(B)}\sum_{n\in \mathbb N}\mu(A_{n})$, while $\mathcal{U}(B):=\{(A_{n})_{n \in \mathbb N} \subseteq \mathcal{A}: B \subseteq \bigcup_{n \in \mathbb N}A_{n}\}$
Let's say $(B_{n})_{n} \subseteq \mathcal{U}(B)$, such that:
$\inf_{(A_{n})_{n} \in \mathcal{U}(B)}\sum_{n\in \mathbb N}\mu(A_{n}) = \sum_{n \in \mathbb N}\mu(B_{n})$
Since $B_{n} \in \mathcal{A}, \forall n \in \mathbb N$ and $\mathcal{A}$ a $\sigma-$Algebra since $\mu$ is measurable.
As $(B_{n})_{n} \in \mathcal{U}(B) \Rightarrow B \subseteq \bigcup_{n \in \mathbb N}B_{n}$
Firstly, $\bigcup_{n}B_{n}$ is simply a countable union of $B_{n}\in \sigma-$Algebra $\mathcal{A}$, so it is also $\in \mathcal{A}$
Secondly, We can create disjoint sets $(C_{n})_{n}$ such that $\bigcup_{n \in \mathbb N} C_{n} = \bigcup_{n \in \mathbb N} B_{n}$, this means that $\mu( \sqcup_{n \in \mathbb N}C_{n})=\sum_{n \in \mathbb N}\mu(C_{n})=\mu^{*}(B)$
Which proves what was asked. Am I correct?
Additional question: How could I go about showing the same $A-B \in \mathcal{A^{*}}$ rather than $B \in \mathcal{A^{*}}$?
A set $B \subset X$ belongs to ${\cal A}^*$ if and only if there exist sets $E,F,G \subset X$ with the property that
In this case $\mu^*(B) = \mu(E)$.
Let $A = E \cup F$. Then $A \in \cal A$, $B \subset A$, and $\mu(A) \le \mu(E) + \mu(F) = \mu(E) \le \mu(A)$. Thus $$\mu(A) = \mu(E) = \mu^*(B).$$
Next note that $$A \setminus B \subset F \setminus G.$$ Since $F \setminus G \subset F$ and $\mu(F) = 0$ you have $A \setminus B \in {\cal A}^*$ too.