I have come across a proof of Theorem 1.9 in Folland's Real Analysis from this thread.
Theorem 1.9 Let $(X, \mathcal X, \mu)$ be a measure space. Let $\overline{\mathcal X}$ be the completion of $\mathcal X$ with respect to $\mu$. There is a unique measure $\overline \mu$ on $(X, \overline{\mathcal X})$ such that $\mu = \overline \mu|_{\mathcal X}$. We also have $\overline \mu$ is complete.
My below attempt seems more concise than that from the other thread. Could you check if I made some subtle mistakes?
Proof Let $\mu^*$ be the outer measure induced from $\mu$. Let $\mathcal{M}$ be the collection of all $\mu^*$-measurable sets, i.e., those sets $A \subseteq X$ such that $$ \mu^*(E)=\mu^*(E\cap A) +\mu^*(E\cap A^c) \quad \forall E\subset X. $$
I proved previously that $\mu = \mu^*|_{\mathcal X}$, $\overline{\mathcal X} \subset \mathcal M$, and $\mu^*|_{\overline{\mathcal X}}$ is a complete measure. This proves the existence part. Let's prove the uniqueness part. Let $\mathcal N$ be the collection of all subsets of $\mu$-null subsets of $X$, i.e., $$ \mathcal N := \{A \subset X :\exists N \in \mathcal X \text{ such that } A \subset N \text{ and } \mu (N)=0\}. $$
Let $\nu$ be a measure on $(X, \overline{\mathcal X})$ such that $\mu = \nu|_{\mathcal X}$. Fix $A \in \overline{\mathcal X}$. Then $A = B \cup C$ for some $B \in \mathcal X$ and $C \in \mathcal N$. We have $$ \mu(B)=\nu(B) \le \nu(A) \le \nu (B)+\nu(C) = \nu(B)+0 = \mu(B). $$
It follows that $\nu(A) = \mu(B)$. This completes the proof.
