Let $\mu^*$ be an outer measure and $\mu$ be a measure on $(\Omega,\mathcal F)$. $\mu^*$ can measure any subset of $\Omega$ while $\mu$ can only measure the sets in $\mathcal F$.
Let's take some set $A\subseteq\Omega$ that is not measurable. $\mu^*$ can assign a real value to that set but $\mu$ can't. One can complete a measure space though by "adding" all sets $A\subseteq B\in\mathcal F$ with $\mu(B) = 0$ to $\mathcal F$.
It is clear that a measure constructed from an outer measure produces a complete $\sigma$-algebra $\mathcal F^*$ because for $B\in\mathcal F^*$ with $\mu^*(B) = 0$ it follows that $\mu^*(A) = 0$ for $A\subseteq B$ due to monotony. Moreover, $A\in\mathcal F^*$ since $$\begin{align*} 0&\leq\mu^*(A\cap Q) + \mu(A\cap Q^\complement)\\&\leq\mu^*(B\cap Q) + \mu^*(B\cap Q^\complement)\\ &= \mu^*(B)\\ &= 0,\end{align*}$$ so $\mu^*(A) = \mu^*(A\cap Q) + \mu^*(A\cap Q^\complement)$ for any $Q\subseteq\Omega$. Restricting $\mu^*$ to $\mathcal F^*$ gives then a measure.
My questions:
- Completing the measure space was only concerned with null sets. What is with sets like $A\subseteq B\in\mathcal F$ and $\mu(B)>0$? Are they included in $\mathcal F^*$? Or does the term "completing" stems in fact from the fact that both approaches yield the same $\sigma$-algebras? (i.e. $\{F\cup N : \text{$F\in\mathcal F$ and $N$ is a null set}\} = \mathcal F^*$)
- Given the complete measure space, can we resurrect the incomplete measure space? That is, given $\mathcal F^*$, can we construct $\mathcal F$?