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?

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.
