I am following the book $\\$ An Introduction to Measure and Integration written by Prof. Inder K Rana for my measure theory course. While going through this book I came across the aforesaid theorem which roughly states that every measure space has a completion. The proof that he has presented is also very rigorous and easy to read. But in the entire proof he didn't show anywhere that the measure space $(X, \mathcal S \cup \mathcal N, \overline {\mu})$ is complete. How to prove it?
Well we need to only to show that if $E \subseteq A \in \mathcal S \cup \mathcal N$ with $\overline {\mu} (A) = 0$ then $E \in \mathcal S \cup \mathcal N.$ Now the author has also shown that for any $S \cup N \in \mathcal S \cup \mathcal N$ with $S \in \mathcal S, N \in \mathcal N,\ \ $ $\overline {\mu} (S \cup N) = \mu (S).$ So $\mu (A) = 0 \implies \exists\ S \in \mathcal S$ such that $\mu (S) = 0,$ where $A = S \cup N,$ for some $S \in \mathcal S, N \in \mathcal N.$ This shows that $S \in \mathcal N.$ But then $A \in \mathcal N,$ since $\mathcal N$ is closed under countable unions. How does it imply that $E \in \mathcal S \cup \mathcal N$? Any help in this regard will be highly appreciated.
Thank you very much for your valuable time for reading.
Since $\mu(S) = 0$ we know that $S \cap E \in \mathcal N$. Since $N \in \mathcal N$ there exists $B \in \mathcal S$ with $\mu(B) = 0$ and $N \subset B$, so $(N \cap E) \subset N \subset B$ and hence $N \cap E \in \mathcal N$. Since $E \subset S \cup N$, $E = (S \cap E) \cup (N \cap E) \in \mathcal N$ so in particular $E \in \mathcal S \cup \mathcal N$.