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?
 A: 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 $\mu^* = \nu$.
A: Suppose that there exists another extension $v$ of $\mu$ such that $(X,\Sigma^*,v)$ is complete. Let $Y \in \Sigma^*$. Then $Y=S \cup Q, S \in \Sigma$ for some $N \supseteq Q: N \in \Sigma, \mu(N)=0$. We have
$$v(Y)=v(S \cup Q)=v(S) \cup v(Q\setminus S)=v(S)=\mu(S)$$
since (to be more precise) (i) $S$ and $Q\setminus S$ are disjoint; (ii)  $v$ is an extension of $\mu$ to $\Sigma^*$ implying $\mu(N)=v(N)=0$ and $\mu(S)=v(S), \forall N,S\in \Sigma$; (iii)  $Q \setminus S \subseteq Q \subseteq N \in \Sigma \subseteq \Sigma^*$ and $(X,\Sigma^*,v)$ complete, implies by monotonicity, $v(Q \setminus S)=0$. 
Likewise, $\mu^*(Y)=\mu(S)$, and we are done.
