Let $(X,\cal{M},\mu)$ be a measure space, and for $E,F\in \cal{M}$ write $E \sim F$ iff $\mu(E \Delta F)=0$. Let $\widetilde{\cal{M}}$ be the set of equivalence classes in $\cal{M}$ for $\sim$; for $E\in \cal{M}$, write $E^{•}\in\widetilde{\cal{M}}$ for its equivalence class.there is a partial ordering $\subset$ on $\widetilde{\cal{M}}$ defined by saying that, for $E, F\in\cal{M}$ ,
$E^{•}\subset F^{•} \Longleftrightarrow \mu(E\setminusF)=0$
operations $\bigcap,\bigcup,\setminus$ on $\widetilde{\cal{M}}$ defined by saying that
$E^{•}\bigcap F^{•}=(E\bigcap F)^{•}$
$E^{•}\bigcup F^{•}=(E\bigcup F)^{•}$
$E^{•}\setminus F^{•}=(E\setminus F)^{•}$
What does it mean These operation on $\widetilde{\cal{M}}$ is well-defined?
Given $A,B,C,D\in \mathcal{M}$ with $A\sim B$ and $C\sim D$:
First we know $\mu(A\setminus B) = 0$ if $A\sim B$: $$ \mu (A\triangle B) = \mu(A\setminus B\cup B\setminus A) = \mu (A\setminus B)+\mu(B\setminus A) = 0 \tag{$\square$} $$
Now we also have following rules for sets $E,F,G$:
$$ \begin{align} E\setminus (F\cap G) &= (E\setminus F) \cap (E\setminus G)\tag{$\bullet$}\\ (E\cap F)\setminus G &= E\cap(F\setminus G) \tag{$\star$} \end{align} $$
You have to show it this way:
$$\begin{align} (A\cap C )\triangle (B\cap D) &= \big[(A\cap C)\setminus(B\cap D)\big]\cup\big[(B\cap D)\setminus (A\cap C)\big] \\ &\overset{\bullet}{=} \big[(( A\cap C)\setminus B)\cap ((A\cap C)\setminus D) \big ]\cup \big[((B\cap D)\setminus A)\cap((B\cap D)\setminus C)\big] \\ &\overset{\star}{=} \underbrace{\big[(C\cap (A\setminus B)) \cap (A\cap (C\setminus D)) \big ]}_{=:\ K} \cup \underbrace{[(D\cap (B\setminus A)) \cap (B\cap (D\setminus C))\big]}_{=:\ L}\\ \end{align} $$
So we know that: $$\mu(K) \le \mu(C\cap(A\setminus B))+\mu(A\cap(C\setminus D) \overset{\square}{=} 0$$ Same goes for $L$ so we now know: $$ A\cap C \sim B\cap D \quad\Leftrightarrow\quad A^\bullet\cap C^\bullet = (A\cap C)^\bullet$$