$\newcommand{\scrF}{\mathscr{F}}$ Problem: Let $\mu^\ast$ be an outer measure defined on $\Omega$. Show that if $A \subseteq \Omega$ satisfies $\mu^\ast(A) = 0$ then $A$ is $\mu^\ast$-measurable. Deduce that the measure obtained by restricting $\mu^*$ to the collection of it's measurable sets is complete.
I think the first part is fine,
Attempt of Part 1: Let $T \subseteq \Omega$. Then note that $\mu^\ast(T \cap A) \leq \mu^*(A)$ and $\mu^\ast(T \cap A^c) \leq \mu^*(T)$ both by monotonicity. Then it follows that $$ \mu^\ast(T \cap A) + \mu^\ast(T \cap A^c)\leq \mu^\ast(T)+\mu^\ast(A) = \mu^\ast(T), $$ and so we conclude that $A$ is $\mu^\ast$-measurable.
The issue I have is with my attempt at showing the restriction is complete. To situate myself, here's what I think the definition of a complete measurable space is.
Definition: A measurable space $(\Omega, \mathscr{F}, \mu)$ is complete if the $\Sigma$-algebra contains all subsets of sets with measure $0$. Specifically if $A \subseteq B$ and $\mu(B) = 0$, then $A \in \mathscr{F}$.
So in light of this I had thought to write a solution as follows, where $\scrF_{\mu^\ast}$ is the collection of all $\mu^\ast$-measurable sets in $\Omega$.
Attempt:Now we can restrict $\mu^*|_{\scrF_{\mu^*}}$ and this forms a measurable space $(\Omega, \scrF_{\mu^*},\mu^*|_{\scrF_{\mu^*}})$. Now we want to show that this measurable space is complete; i.e it contains all subsets of sets with measure $0$, or even more specifically all subsets of sets with measure $0$ are $\mu^*|_{\scrF_{\mu^*}}$ measurable. Moving forward let $\phi = \mu^*|_{\scrF_{\mu^*}}$. Let $E \in \scrF_{\mu^*}$ such that $\phi(E) = 0$ and let $A$ be an arbitrary subset of $E$. It follows by monotonicity that $\phi(A) \leq \phi(E) = 0$ but since $\phi$ is an outer measure we know the range of $\phi$ is $[0,\infty]$ forcing $\phi(A) = 0$. Then from above it follows that $A \in \scrF_{\mu^*}$ and so the measurable space $(\Omega, \scrF_{\mu^*},\mu^*|_{\scrF_{\mu^*}})$ is in fact complete.
But is there a mistake here? Because by leveraging monotonicity I think I'm tacitly assuming that since $A \subseteq E$ and $E \in \scrF_{\mu^\ast}$ that $A \in \scrF_{\mu^\ast}$ but is this true? Are we guaranteed that the subset of the measurable set is itself measurable?
I don't know if this is a good question or I missed something and am turning my wheels. Thanks in advance for the clarification.
Edit 1: For reference what gave me pause was a section of Folland's book. On page 26 he writes
"If $\mu(E) = 0$ and $F \subset E$ then $\mu(F) = 0$ by monotonicity provided that $F \in \mathcal{M}$, but in general it need not be true that $F \in \mathcal{M}$."
So if I fell into this trap how do I fix the proof?
Edit 2: The more I think of it, what does the problem even mean. It seems like we're going in the opposite direction to me, if $\mu^\ast(A) = 0$ then $A$ is $\mu^\ast$ measurable, but if we didn't know it was $\mu^\ast$ measurable before how do we even assign it the value of $0$?
An outer measure is defined on all sets. since you verified that sets of outer measure zero are $\mu^*$ measurable, you obtain completeness by definition. the remark in Folland's book pertains to general measures, not necessarily obtained by restricting an outer measure to its measurable sets.