Construction of the completion of a measure space Let $(X,M, \mu)$ be a measure space. Let $\overline{M}$ be collection of $E \cup Z$ such that $E \in M$ and $Z \subset F$, where $F \in M,$ and $\mu(F) = 0.$ We also know $\bar{\mu}(E) = \mu(E).$
a) Show $\overline{M}$ is smallest sigma algebra containing $M$ and all subsets of elements of $M$ that have measure $0$.
Attempt: Notice if $E_{1} \subset \overline{M}$ then $E_{1} = E \cup Z.$ Since $E, Z \in M$ then $E_{1}^c = (E \cup F)^c \cup (F-Z)$ is in $M$ since $(E \cup F),(F-Z) \in M.$ Hence $\overline{M}$ is sigma algebra. If we let $Z = \emptyset$,then we see $M \subset \overline{M}.$ And since $\emptyset \in M,$ then let $E = \emptyset$, and then all subsets of $M$ with measure $0$ is in $M.$ Does this look okay? I am not sure how to show $\overline{M}$ is the smallest sigma algebra containing $M$
b) Show $\bar{\mu}$ is a measure of $\overline{M},$ and this measure is a completion. 
Attempt: First check $\bar{\mu}$ is a measure of $\overline{M}.$ Notice $\bar{\mu}(\emptyset) = \mu(\emptyset) = 0.$ If $A_{1} \subset A_{2}$, and $A_{1} = B_{1} \cup C_{1}, A_{2} = B_{2} \cup C_{2},$ then $\bar{\mu}(A_1) = \mu(B_1) \le \mu(B_2) = \bar{\mu}(A_2).$ Lastly $\bar{\mu}(\cup_{i=1}^{n}A_{i}) = {\mu}(\cup_{i=1}^{n}B_{i}) \le \sum_{i=1}^{n}\mu(B_i).$ Next for completion, we need to show if $F \in \overline{M}$ and $\bar{\mu}(F) = 0,$ if $E\subset F$ then $E \in \overline{M}.$ I am not sure how to this. 

The argument "since $E, Z \in M$" is incorrect, since $Z$ needn't be in $M$. It is true, though, that $Z \subset F$ for some null measurable set $F$.
Here is a suggestion:
Without loss of generality, if $E \cup Z \in \overline{M}$, while $Z \subset F \in Null(M)$, one can assume that $E \cap F = \phi$. Indeed, otherwise one could write $E \cup Z = (E \setminus F) \cup [(F \cap E) \cup Z]$, and have $E \setminus F \in M, (F \cap E) \cup Z \subset F \in Null(M)$
So, $E^c = E^c \setminus F \cup F \in \overline{M}$, since $E^c \setminus F \in M, F \in Null(M)$
