Show that $(E\cup Z_1)\setminus Z_2$ has the form $E\cup Z$ I'm trying to do an exercise as follows:

Let $(X, {\mathbf X}, \mu)$ be a measure space and let ${\mathbf Z}=\{E\in {\mathbf X}:\mu(E)=0\}$. Let $\mathbf X'$ be the family of all subsets of $X$ of the form $(E\cup Z_1)\setminus Z_2, E\in \mathbf X$, where $Z_1$ and $Z_2$ are arbitrary subsets of sets belonging to $\mathbf Z$. Show that a set is in $\mathbf X'$ if and only if it has the form $E\cup Z$ where $E\in \mathbf X$ and $Z$ is a subset of a set in $\mathbf Z$.

My proposed answer was if $Q=E\cup Z$ where $E\in \mathbf X$ and $Z\subset P\in\mathbf Z$ then $Q=E\cup Z\setminus(P\setminus Z)$ since $Z$ and $P\setminus Z$ are both subsets of $P\in \mathbf Z$. This seems to be wrong since it assumes $P\cap E=\emptyset$. Also, I cannot seem to work out how to go the other way around. I tried defining the set $R=\{x\in X:f(x)>0\}$. By the definition of sigma algebra, $R\in \mathbf X$. Then I tried taking intersections and complements with $R$. However I just keep getting messy expressions which never resolve to the required $E\cup Z$. Is this method with $R$ a good idea or did I miss something obvious?
[This is part of exercise 3.L. of The Elements of Integration and Lebesgue Measure by R. G. Bartle.]
 A: Clearly $E \cup Z = (E \cup Z) \setminus \emptyset \in \mathbf X'$.
On the other hand, suppose that $E \in \mathbf X$, $N_1,N_2 \in \mathbf Z$, $Z_1 \subset N_1$, and $Z_2 \subset N_2$. 
Now work out that $$(E \cup Z_1) \setminus Z_2 = (E \setminus N_2) \cup \bigg[(E \cap (N_2 \setminus Z_2)) \cup (Z_1 \setminus Z_2)\bigg]$$ 
which belongs to $\mathbf X'$, since the set in brackets is a subset of $N_1 \cup N_2$.
A: The "if" part in my view is trivial, if you consider $Z_1=Z$ and $Z_2=\emptyset$.
The "only if" part is non-trivial. First Let $X=\left(E\cup Z_1\right)\backslash Z_2$. Pick an increasing sequence $\left(X_i\right)\in \bf{X}$ such that $X_i\subset X$ for all $i\geq1$ and $\mu\left(X_i\right)\rightarrow\mu\left(E\right)$ as $i\rightarrow\infty$. This is possible because $Z_1$ and $Z_2$ are null sets. (You can prove this claim.) Then, define $E^\prime=\bigcup_{i\geq1}X_i$ and $Z^\prime=X\backslash E^\prime$. We then have all the desired properties: $E^\prime\cup Z^\prime=E$ (since $E^\prime \subset X$), $E^\prime \in \bf{X}$, and $Z^\prime$ is a subset of a set in $\bf{Z}$ (for the last claim I give a proof below).
Claim: $Z^\prime$ is a subset of a set in $\bf{Z}$.
Note that $Z^\prime = X\backslash E^\prime = \left(E\cup Z_1\right)\backslash Z_2 \backslash E^\prime $. Let $P_1\in\bf{Z}$ such that $Z_1\subset P_1$. Then, $Z^\prime = \left(E\cup Z_1\right)\backslash Z_2 \backslash E^\prime \subset \left(E\cup P_1\right)\backslash E^\prime \in \bf{X}$. So it now only remains to prove that $\left(E\cup P_1\right)\backslash E^\prime \in \bf{Z}$. This is simple computation from now on! $\mu\left(\left(E\cup P_1\right)\backslash E^\prime\right)=\mu\left(E\cup P_1\right)-\mu\left(\left(E\cap E^\prime\right)\cup \left(P_1\cap E^\prime\right)\right)=\mu\left(E\right)-\mu\left(E^\prime\right)=0$. So the claim is true indeed.
