There are $E, F \in \mathcal{M}$ satisfying $E \subset A \subset F$ and $\mu (F \backslash E) = 0.$

Question

Let $(X, \mathcal{M}, \mu)$ be a complete measure space. Show that if $A \in \mathcal{M},$ then there are $E, F \in \mathcal{M}$ satisfying $E \subset A \subset F$ and $\mu (F \backslash E) = 0.$

My efforts: Since $A \in \mathcal{M}$ then $A = B \cup C,$ where $B \in \mathcal{M}$ and $C \subset N,$ with $\mu(N) = 0.$ Let $E = B$ and $F = B \cup N.$ Therefore, $E \subset A \subset F$ and $\mu (F \backslash E) \leq \mu(N) = 0.$

Is that correct?

Edit problem statement:

Let $(X, \mathcal{M}, \mu)$ be a measure space and $(X, \bar{\mathcal{M}}, \bar{\mu})$ its completion. Show that $A \in \bar{\mathcal{M}}$ then $\exists E, F \in \mathcal{M}$ satisfying $E \subset A \subset F$ and $\mu (F \backslash E) = 0.$

Answer 1

The OP says that the definition of the completion of a measure space in force here is the definition here.

If you look at that definition you see that what we need to prove here is this:

Exercise. If $A\in\overline{\mathcal M}$ then $A\in\overline{\mathcal M}$.

(Presumably in whatever context the question came from the completion was actually given a formally different definition?)