Let $\mathcal A_0$ be an algebra of subsets of a set $X$ and let $\mu$ be a probability measure on $\mathcal A.$ Let $\mu^*$ be the outer measure induced by $\mu$ such that for all $A \subseteq X$ we have $$\mu^*(A) : = \inf \left \{\mu^*(G)\ \bigg |\ G \supseteq A, G \in \widetilde {\mathcal A} \right \}$$ where $\widetilde {\mathcal A}$ is the collection of countable unions of elements of $\mathcal A_0.$ Consider the collection $\mathcal A : = \left \{A \subseteq X\ \bigg |\ \mu^*(A) + \mu^*(A^c) = 1 \right \}.$ Then we can prove that $\mathcal A$ is a $\sigma$-algebra of subsets of $X$ containing $\mathcal A_0$ and $\mu^*$ is a measure on $\mathcal A.$ Let $\sigma (\mathcal A_0)$ denote the $\sigma$-algebra generated by $\mathcal A_0.$ Then clearly $\sigma (\mathcal A_0) \subseteq \mathcal A.$
What our instructor claims that the measure space $(X, \mathcal A, \mu^*)$ is the completion of $(X,\sigma (\mathcal A_0), \mu^*).$ I am able to prove that $(X, \mathcal A, \mu^*)$ is a complete measure space. It is clear that $\mu^*(A \cup N) = \mu^*(A),$ for all $A \in \sigma (\mathcal A_0)$ and for all $\mu^*$-null sets $N.$ So in order to prove the proposed argument we need only to show that for any $A \in \mathcal A$ we have $A = B \cup N,$ for some $B \in \sigma (\mathcal A_0)$ and $N \subseteq M,$ for some $M \in \sigma (\mathcal A_0)$ with $\mu^*(M) = 0.$ What I found is that for any given $A \in \mathcal A$ there exists $B \in \sigma (\mathcal A_0)$ with $A \subseteq B$ such that $\mu^*(A) = \mu^*(B).$ Hence $\mu^*(B \setminus A) = 0.$ So from the definition of outer measure we can construct $C \in \sigma (\mathcal A_0)$ such that $B \setminus A \subseteq C$ and $\mu^*(C) = 0.$ This shows that $B \setminus A$ is a $\mu^*$-null set. Will it help anyway? I got stuck at this stage and couldn't proceed further. Any help in this regard will be appreciated.
Thanks in advance.