This is from Folland's Real Analysis, Theorem 1.19. we are given $\mu$ as a complete Lebesgue-Stieljes measure. In the course of proving the following fact that:
if $E \subset \mathbb R$ and $E = H \cup N$ with $H$ a $F_\sigma$ set and $\mu(N) = 0$, then $E \in \mathcal A_{\mu},$
Folland states that the result is obvious since $\mu$ is complete. I GET HIS PROOF. However, I am wondering whether the above can be proven without using the fact that $\mu$ is complete, since the fact that it is given that $\mu(N) = 0$ IMPLIES that $N$ is in the domain of $\mu$ (also, please confer page 26 and 27 of Folland's Real Analysis), i.e. $N \in \mathcal A_{\mu}$. Since we know that $H \in \mathcal A_{\mu}$ and $N \in \mathcal A_{\mu}$, the fact that $A_{\mu}$ is closed under union could imply without using the completeness of $\mu$ that $E \in \mathcal A_{\mu}.$ I completely fail to see how my argument is false.
I have also searched several math stack exchange posts related to the exact same issue I was having (see here and the first comment on here . Again, I COMPLETELY GET THE PROOFS USING COMPLETENESS. My post is NOT the question of how I to get the proof using completeness, but whether my proof that does not use completeness of $\mu$ still works. If not, could you either point exactly where my argument fails? I am thinking that it was Folland's small mistake that he defined a null set that way.
I am now thinking that he should have stated the following instead:
if $E \subset \mathbb R$ and $E = H \cup N$ with $H$ a $F_\sigma$ set and $N \subset T$ where $T \in \mathcal A_{\mu}$ and $\mu(T) = 0$, then $E \in \mathcal A_{\mu}.$