completion of a $\sigma$- algebra Let $(X,\textbf{X}, \mu)$ be a measure space, $\textbf{Z}=\{E\in \textbf{X}: \mu(E)=0\}$ and let $\textbf{X}'$ be the family of all subsets of $X$ of the form
$(E\cup Z_1)-Z_2$, where $E\in \textbf{X}$, $Z_1$ and $Z_2$ are arbitrary subsets of sets belonging to $\textbf{Z}$. Show that a set is in $\textbf{X}'$ if and only if it has the form $E\cup Z$ where $E\in \textbf{X}$ and $Z$ is a subset of a set in $\textbf{Z}$.
proof:
Let $N=\{E\subseteq X: \exists M\in \textbf{Z}, E\subseteq M\}$.
Suppose that $F\in \textbf{X}'$  then $F=(E\cup Z_1)-Z_2$, where $E\in \textbf{X}$ and $Z_1,Z_2 \in N$. Thus
$F=(E-Z_2)\cup (Z_1-Z_2)$
I know that $Z_1-Z_2=Z_3\in N$ but I don't know why $E-Z_2\in \textbf{X}$.
 A: Instead of writing $F=(E−Z_2)∪(Z_1−Z_2)$, you would instead write $F=(E−M)∪((M+Z_1)−Z_2)$, where $M$ is the element of $\bf Z$ so $Z_2\subseteq M$ (which exists by definition of $N$).
Then since $\bf Z \subseteq \bf X$, we have $(E−M)\in \bf X$.  Also, $(M+Z_1)−Z_2\subseteq M+Z_1\subseteq M+M'$, where $M'$ is the element of $\bf Z$ so $Z_1\subseteq M$ (by definition of $N$).  Therefore, $M+M'\in\bf Z$, and $(M+Z_1)−Z_2\in\bf X$.
But seriously, do not use bold letters and non-bold letters together in future posts.  Even when proofreading my own post, they were pretty hard to tell apart (e.g. for the Z's, the main difference is that the slant of non-bold Z's has a slightly higher slope than the slant of bold Z's).
A: *

*Suppose that $F\in \textbf{X}'$ then $F=(E\cup Z_1)-Z_2$. Since $Z_1, Z_2\in N$ there are $M_1, M_2\in Z$ such that $z_1\subseteq M_1$ and $Z_2\subseteq M_2$. Thus


$F=(E\cup Z_1)\cap Z_2^c$
$ =(E\cap Z_2^c)\cup (Z_1\cap Z_2^c)$
$ =(E\cap(M_2^c\cup (M_2\cap Z_2^c)))\cup (Z_1\cap Z_2^c)$
$ =((E\cap M_2^c)\cup (E\cap (M_2\cap Z_2^c)))\cup (Z_1\cap Z_2^c)$
$ =(E\cap M_2^c)\cup ((E\cap (M_2\cap Z_2^c))\cup (Z_1\cap Z_2^c))$
Then  $E_1=E\cap M_2^c\in \textbf{X}$ and 
$Z=(E\cap (M_2\cap Z_2^c))\cup (Z_1\cap Z_2^c)\subseteq M_2\cup M_1 $. Hence $Z\in N$ and $F=E_1\cup Z$.  
