Uniqueness of completion of a measure space Let $(X,\Sigma,\mu)$ be a measure space.
Define $P=\{S\subset X : \exists N\in \Sigma(\mu(N)=0 \land S\subset N)\}$.
And let $\Sigma^*$ be the $\sigma$-algebra generated by $P\cup \Sigma$.
I have proved there exists a complete measure $\mu^*$ on the measurable space $(X,\Sigma^*,\mu^*)$, which is an extension of $\mu$.
However, it is in wikipedia that such extension is unique for $\Sigma^*$.
How do i prove this?

Let $\nu$ be another such complete measure, then $\nu|_{\Sigma} = \mu^*|_{\Sigma} = \mu$.
Now suppose $A \in \Sigma^*$. Note that $\Sigma^* = \{S\cup Q \mid S \in \Sigma, Q \in P\}$, so let $A = S\cup P$ with $S \in \Sigma$ and $Q \in P$, then 
$$\nu(A) = \nu(S\cup Q) \leq \nu(S) + \nu(Q) = \nu(S)$$ 
where the last equality holds because $Q \subseteq N$ for some $N \in \Sigma$ with $\nu(N) = \mu(N) = 0$, and $\nu$ is complete. As $S \in \Sigma$, $\nu(S) = \mu^*(S)$, so
$$\nu(A) \leq \nu(S) = \mu^*(S) \leq \mu^*(S\cup Q) = \mu^*(A).$$
Likewise $\mu^*(A) \leq \nu(A)$, so $\mu^* = \nu$.
