Am trying to do exercise $7$-$(a)$ in Bauer's book Measure and Integration Theory:
(a) Show that every measure $\mu$ on a $\sigma$-algebra $\mathcal{A}$ in a set $\Omega$ can be completed. That is, $\mu$ can be extended to a complete measure $\mu_0$ on a $\sigma$-algebra $\mathcal{A}_0$ in $\Omega$, $\mathcal{A}\subset\mathcal{A}_0$, in such a way that every complete measure $\mu'$ on a $\sigma$-algebra $\mathcal{A}'$ in $\Omega$, $\mathcal{A}\subset\mathcal{A}'$, which extends $\mu$ is also an extension of $\mu_0$.
(c) Show that the $\mu$-completion $\mathcal{A}_0$ of a $\sigma$-algebra $\mathcal{A}$ in $\Omega$ consists of all sets $A\cup N$ with $A\in \mathcal{A}$ and $N$ a subset of a $\mu$-null set. For every such set $\mu_0(A \cup N)=\mu(A)$.
My difficulty is answering $(a)$ without doing $(c)$ first, which is the way completion is usually presented. Here is my work so far:
Let $\mathcal{N}$ be the set of all subsets of $\mu$-null sets and let $\mathcal{A}_0$ be the $\sigma$-algebra generated by $\mathcal{A} \cup \mathcal{N}$. Let $\mu^{*}$ be the outer measure defined on $\mathcal{P}(\Omega)$ by
$$\mu^{*}(A)=\inf \bigg\{ \sum_{n=1}^{\infty}\mu(A_n): (A_n)\subset \mathcal{A}, A\subset \bigcup_{n\in\mathbb{N}}A_n \bigg \}$$
and let $\mathcal{A}^*$ be the $\sigma$-algebra of $\mu^*$ measurable sets (Carathéodory's criterion). We have that $\mu^*$ agrees with $\mu$ on $\mathcal{A}$. By construction, each set in $\mathcal{A}$ is $\mu^*$ measurable. By Carathéodory's theorem, $\mu^*$ is a complete measure on $\mathcal{A}^*$, and so we see that $\mathcal{A}^*$ also contains $\mathcal{N}$. Hence $\mathcal{A}_0 \subset \mathcal{A}^*$, and $\mu_0=\mu^*|_{\mathcal{A}_0}$ is a measure on $\mathcal{A}_0$ which extends $\mu$.
But how can I show that this measure is complete and that it is extended by any complete extension $\mu'$ of $\mu$?
EDIT: I don't see a more efficient way than to prove that $\mathcal{A}_0$ consists of all sets of the form $A\cup N$ with $A\in \mathcal{A}$ and $N\in \mathcal{N}$. Then we check that $\mu_0(A\cup N)=\mu(A)$ for such set and we have an explicit characterization of $\mu_0$ on $\mathcal{A}_0$.
To show that $\mu_0$ is complete, let $A \subset B \in \mathcal{A}_0$ with $\mu_0(B)=0$. Write $B=C\cup N$ with $C\in \mathcal{A}$ and $N\in \mathcal{N}$. Then $B\setminus C\subset N$ so $B\setminus C \in \mathcal{N}$. Also,
$$0\leq \mu(C)=\mu_0(C)\leq \mu_0(C\cup N)=\mu_0(B)=0$$
which forces $\mu(C)=0$. Hence $C\in \mathcal{N}$ as well. Since the union of two $\mu$-null sets is again a null set we see that $B=C\cup B\setminus C \in \mathcal{N} $. We conclude $A\in\mathcal{N}\subset \mathcal{A}_0$.
Finally we show that any complete extension $\mu'$ of $\mu$ extends $\mu_0$. Let $\mu'$ be a complete measure on a $\sigma$-algebra $\mathcal{A}'$, $\mathcal{A}\subset \mathcal{A}'$, which extends $\mu$. For $\mu'$ to be complete we must have $\mathcal{N}\subset \mathcal{A}'$. Thus $\mathcal{A}_0\subset \mathcal{A}'$. Also, $N\subset M \in \mathcal{A}$ with $\mu(M)=0$ implies
$$ 0\leq \mu'(N)\leq \mu'(M)=\mu(M)=0$$
so $\mu'$ and $\mu$ agree on both $\mathcal{A}$ and $\mathcal{N}$. Hence if we write $A=C\cup N \in \mathcal{A}_0$ with $C\in \mathcal{A}$ and $N\in \mathcal{N}$ we get
$$\mu'(A)=\mu'(C)+\mu'(A\setminus C)=\mu'(C)=\mu_0(C)=\mu_0(C)+\mu_0(A \setminus C) = \mu_0(A)$$ where the second and fourth equalities are because $A\setminus C \in \mathcal{N}$. We conclude that $\mu'$ extends $\mu$.
We have shown that $\mu_0$ is the completion of $\mu$, i.e. the smallest complete extension of $\mu$.
As NateEldredge pointed out, if $\mu$ is $\sigma$-finite, then we can actually show the inlusion $\mathcal{A}^* \subset \mathcal{A}_0$, which means that $\mu_0=\mu^{*}|_{\mathcal{A}^{*}} $ is the completion of $\mu$.
Am I understanding this correctly?