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.]

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 $N_1 \cup N_2$.
