Assuming countable choice, every measure space has a unique complete extension to its completion

By [L1] and [L2], $\bar{\mu }$ is a function on the sigma-algebra $\bar{\mathcal{A}}$, and $\bar{\mu }(A)=\mu (A)$ for $A\in \mathcal{A}$ by the representation $A=A\cup \mathrm{\varnothing }$; in particular $\bar{\mu }(\mathrm{\varnothing })=0$.

Let $(E_k)$ be disjoint in $\bar{\mathcal{A}}$. By [L6] each family of completed representations is nonempty, so [L5] chooses $E_k=A_k\cup N_k$ with measurable $A_k\subseteq E_k$ and $N_k$ contained in a measurable null set $Z_k$. Then the $A_k$ are disjoint, ${\bigcup }_k{E}_k=({\bigcup }_k{A}_k)\cup N$ with $N\subseteq {\bigcup }_k{Z}_k$, and ${\bigcup }_k{Z}_k$ is null by [L3].

If $E\in \bar{\mathcal{A}}$ and $\bar{\mu }(E)=0$, choose a representation $E=A\cup N$ from [L6] with $N\subseteq Z$ null. Then $A\cup Z$ is measurable and null, and every $S\subseteq E$ is represented by the empty measurable core plus the sub-null set $S\subseteq A\cup Z$.

Let $\lambda$ be any complete measure on $\bar{\mathcal{A}}$ extending $\mu$. For a representation $E=A\cup N$ from [L6], with $N\subseteq Z$ and $\mu (Z)=0$, one has $\lambda (Z)=0$ and completeness gives $\lambda (N)=0$. Replacing $N$ by $N\setminus A$ makes the union disjoint without changing $E$, so finite additivity gives $\lambda (E)=\lambda (A)=\mu (A)=\bar{\mu }(E)$.

Countable additivity of $\mu$ applied to the measurable cores in step 1.2 gives $\bar{\mu }({\bigcup }_k{E}_k)=\mu ({\bigcup }_k{A}_k)={\sum }_k\mu (A_k)={\sum }_k\bar{\mu }(E_k)$; with step 1.1, this proves that $\bar{\mu }$ is a measure.

Step 1.3 shows that every subset of every $\bar{\mu }$-null set belongs to $\bar{\mathcal{A}}$ and has value $0$, so the completed measure space is complete.

Steps 2.1, 2.2 and 1.4 prove respectively that $\bar{\mu }$ is an extending measure, is complete, and is the unique complete extension on $\bar{\mathcal{A}}$.