Below, I tried to understand how to construct (from outer measure) the completion of a measure space, and how it is indeed complete. Could you verify if my understanding is fine?
Let $(X, \mathcal X, \mu)$ be a measure space. Let $\mu^*$ be the outer measure induced from $\mu$, i.e., $$ \mu^*(A):=\inf \left\{ \sum_{n=1}^{\infty} \mu (B_n) : (B_n) \subset \mathcal X , A \subset \bigcup_n B_n \right\} \quad \forall A \subset X. $$
Let $\mathcal{M}$ be the collection of all $\mu^*$-measurable sets, i.e., those sets $A \subseteq X$ such that $$ \mu^*(E)=\mu^*(E\cap A) +\mu^*(E\cap A^c) \quad \forall E\subset X. $$
By Theorem 4.2 in Amann/Escher's Analysis III, $\mathcal M$ is a $\sigma$-algebra on $X$ and the restriction of $\mu^*$ to $\mathcal M$ is a complete measure.
Theorem 1 $\mathcal X \subset \mathcal M$ and $\mu$ coincides with the restriction of $\mu^*$ to $\mathcal X$.
Proof Fix $A \in \mathcal X$ and $E \subset X$. Let's prove that $\mu^*(E) \ge \mu^*(E\cap A) +\mu^*(E\cap A^c)$. If $\mu^* (E) = \infty$, we are done. Let $\mu^* (E) < \infty$. There is a sequence $(B_n) \subset \mathcal X$ such that $E \subset \bigcup_n B_n$ and $\mu^* (E) > \sum_{n=1}^{\infty} \mu (B_n) - \varepsilon$. Clearly, $(B_n \cap A)_n \subset \mathcal X$ and $(E \cap A) \subset \bigcup_n (B_n \cap A)$. Similarly, $(B_n \cap A^c)_n \subset \mathcal X$ such that $(E \cap A^c) \subset \bigcup_n (B_n \cap A^c)$. Hence $\mu^* (E \cap A) \le \sum_{n=1}^{\infty} \mu (B_n \cap A)$ and $\mu^* (E \cap A^c) \le \sum_{n=1}^{\infty} \mu (B_n \cap A^c)$. As such, $\mu^* (E) > \mu^* (E \cap A) +\mu^* (E \cap A^c) - \varepsilon$. The claim then follows by taking the limit $\varepsilon \downarrow 0$. It is obvious that $\mu = \mu^* |_{\mathcal X}$. $\tag*{$\blacksquare$}$
Let $\mathcal N$ be the collection of all subsets of $\mu$-null subsets of $X$, i.e., $$ \mathcal N := \{A \subset X :\exists N \in \mathcal X \text{ such that } A \subset N \text{ and } \mu (N)=0\}. $$
Because $\mathcal X \subset \mathcal M$ and $(X, \mathcal M, \mu^* |_{\mathcal M})$ is a complete measure space, we get $\mathcal N \subset \mathcal M$. Then $\overline{\mathcal X} := \sigma(\mathcal X \cup \mathcal N) \subset \mathcal M$.
Theorem 2 $A \in \overline{\mathcal X}$ if and only if $A = B \cup C$ for some $B \in \mathcal X$ and $C \in \mathcal N$.
Proof It suffices to prove that $\mathcal A := \{B \cup C : B \in \mathcal X, C \in \mathcal N\}$ is a $\sigma$-algebra on $X$.
Let $A \in \mathcal A$, i.e., $A = B \cup C$ for some $B \in \mathcal X$ and $C \in \mathcal N$. There is $D \in \mathcal X$ such that $\mu(D)=0$ and $C \subset D$. Let's prove that $A^c \in \mathcal A$. Indeed, $$ \begin{align} A^c &= X \setminus (B \cup C) = (X \setminus B) \cap (X \setminus C) \\ &= (X \setminus B) \cap [(X \setminus D) \cup (D \setminus C)] \\ &= [(X \setminus B) \cap (X \setminus D)] \cup [(X \setminus B) \cap (D \setminus C)]. \end{align} $$ Because $[(X \setminus B) \cap (D \setminus C)] \subset D$, we get $[(X \setminus B) \cap (D \setminus C)] \in \mathcal N$. On the other hand, $[(X \setminus B) \cap (X \setminus D)] \in \mathcal X$. So $A^c \in \mathcal A$. Let $(A_n) \subset \mathcal A$. Then $A_n = B_n \cup C_n$ for some $B_n \in \mathcal X$ and $C_n \in \mathcal N$. There is $D_n \in \mathcal X$ such that $\mu(D_n)=0$ and $C_n \subset D_n$. Let's prove that $\bigcup_n A_n \in \mathcal A$. Indeed, $$ \begin{align} \bigcup_n A_n &= \bigcup_n (B_n \cup C_n) = \left (\bigcup_n B_n \right ) \cup \left (\bigcup_n C_n \right ). \end{align} $$ We have $\left (\bigcup_n B_n \right ) \in \mathcal X$ and $\left (\bigcup_n C_n \right ) \subset \left (\bigcup_n D_n \right )$. On the other hand, $\bigcup_n D_n \in \mathcal X$ is a $\mu$-null set. $\tag*{$\blacksquare$}$
Let $\overline \mu$ be the restriction of $\mu^*$ to $\overline X$.
Theorem 3 $(X, \overline{\mathcal X}, \overline \mu)$ is a complete measure space.
Proof Let $A \in \overline{\mathcal X}$ such that $\overline \mu(A)=0$. Fix $A' \subset A$. We will prove $A' \in \overline{\mathcal X}$. By Theorem 2, $A = B \cup C$ for some $B \in \mathcal X$ and $C \in \mathcal N$. It follows that $\mu^* (B) = \mu^* (C)=0$. Then $B \in \mathcal N$ and thus $A \in \mathcal N$. Clearly, $A' \in \mathcal N$. $\tag*{$\blacksquare$}$
Then $(X, \overline{\mathcal X}, \overline \mu)$ is called the completion of $(X, \mathcal X, \mu)$.
