Let $(X,\mathscr{A},\mu)$ be a measure space. Define $\mathscr{N}=\{N\in \mathscr{A}: \mu(N)=0\}$ and set $\overline{\mathscr{A}}=\{E\cup F: E\in \mathscr{A}, F\subset N , \text{ for some }N\in \mathscr{N}\}$, define $\overline{\mu}:\overline{\mathscr{A}}\to[0,\infty]$ as $\overline{\mu}(E\cup F)=\mu(E)$. Show $\overline{\mathscr{A}}$ is a sigma algebra and $\overline{\mu}$ is complete measure.
A measure whose domain includes all subsets of null sets is called complete.
I showed that the set is a sigma algebra, but I am not sure how to show the defined measure is complete, it seems trivial by definition of the set $\overline{\scr{A}}$.
Let $N \in \mathscr{N}$, $F\subset N$ and WTS $F\in \overline{\mathscr{A}}$. By definition of $\overline{\scr{A}}$, for any $E\in \scr{A}$, $E\cup F\in \overline{\scr{A}}$, where $F\subset N\in \scr{N}$. Consider $E=\emptyset$, then $F=E\cup F \in \overline{\scr{A}} $.
Note that $\bar{\mu}(\varnothing)=\bar{\mu}(\varnothing \cup \varnothing)=\mu(\varnothing)=0 $. Given a sequence $\left(A_{n}\right)_{n \in \mathbb{N}} $ of disjoint sets in $ \overline{\mathscr{A}} $, there exists a sequence $ \left(E_{n}\right)_{n \in \mathbb{N}} $ of sets in $ \mathscr{A} $ and a sequence $ \left(F_{n}\right)_{n \in \mathbb{N}} $ of subsets of measure zero sets from $ \mathscr{A} $ such that $ A_{n}=E_{n} \cup F_{n} $ for all $ n \in \mathbb{N} $. Note that $ \left(E_{n}\right)_{n \in \mathbb{N}} $ is pairwise disjoint, and $ \cup_{n \in \mathbb{N}} F_{n} $ is a subset of a measure zero set. Therefore $$\bar{\mu}\left(\cup_{n \in \mathbb{N}} A_{n}\right)=\bar{\mu}\left(\left(\cup_{n \in \mathbb{N}} E_{n}\right) \cup\left(\cup_{n \in \mathbb{N}} F_{n}\right)\right)=\mu\left(\cup_{n \in \mathbb{N}} E_{n}\right)=\sum_{n \in \mathbb{N}} \mu\left(E_{n}\right)=\sum_{n \in \mathbb{N}} \bar{\mu}\left(A_{n}\right) .$$ This shows that $ \bar{\mu} $ is a measure. Let $ A \subseteq X $, and suppose there exists $ B \in \overline{\mathscr{A}} $ such that $ A \subseteq B $ and $ \bar{\mu}(B)=0 $. Then $ B=E \cup F $ for some $ E \in \mathscr{A} $ and $ F $ a subset of a measure zero set $ N \in \mathscr{A} $. It follows that $ A \subseteq E \cup N $, where $ E \cup N \in \mathscr{A} $ and $ \mu(E \cup N) \leq \mu(E)+\mu(N)=\bar{\mu}(E)+0 \leq \bar{\mu}(B)=0 $. Therefore $ A=\varnothing \cup A \in \overline{\mathscr{A}} $, which shows that $ \bar{\mu} $ is complete.