symmetric difference of finite union approximation of a set Ask a fundamental problem.
In the course of real analysis, my professor told us the following:  


*

*$A,B$ are measurable.  

*Approximate $A,B$ by finite union of intervals $\bar A=\bigcup_{i=1}^N(a_i,b_i),\bar B=\bigcup_{i=1}^N(c_i,d_i)$ such that $\mu^*(A\triangle \bar A)<\epsilon/2$ and $\mu^*(B\triangle \bar B)<\epsilon/2$.  

*By $(A\cup B)\triangle (\bar A \cup \bar B)\subset (A\triangle \bar A)\cup (B\triangle \bar B)$, we can show ....


My question is how to show "$(A\cup B)\triangle (\bar A \cup \bar B)\subset (A\triangle \bar A)\cup (B\triangle \bar B)$". 
There is another one: $\bar A \cap \bar B\subset (A\triangle \bar A)\cup (B\triangle \bar B)$  
How to show both?  Are they just from the basic properties of symmetric difference? 
 A: *

*We prove that 
$$
(A \cup B ) \bigtriangleup (C \cup D) \subset (A \bigtriangleup C) \cup (B \bigtriangleup D)
$$
\begin{align}
(A \cup B) -(C \cup D)&=(A  -C \cup D)\cup (B -C \cup D)
\\
&\subset(A -C) \cup (B -D)
\end{align}
Likewise
$$
(C \cup D)-(A \cup B)\subset (C-A) \cup (D -B)
$$
Thus
\begin{align}
(A \cup B ) \bigtriangleup (C \cup D)&=((A \cup B) -(C \cup D))\cup((C \cup D)-(A \cup B))
\\
&\subset ((A -C) \cup (B -D))\cup ((C-A) \cup (D -B))
\\
&=((A -C) \cup (C-A))\cup ((B -D)\cup (D -B))
\\
&=(A \bigtriangleup C) \cup (B \bigtriangleup D)
\end{align}

*We prove that 
$$
(A \cap B ) \bigtriangleup (C \cap D) \subset (A \bigtriangleup C) \cup (B \bigtriangleup D)
$$
\begin{align}
(A \cap B) -(C \cap D)&=(A \cap B)\cap (C \cap D)^c
\\
&=(A \cap B)\cap (C^c \cup D^c)
\\
&=((A \cap B)\cap C^c)\cup((A\cap B) \cap D^c)
\\
&\subset (A \cap C^{c}) \cup (B \cap D^{c}) 
\\
&=(A -C) \cup (B -D)
\end{align}
Likewise
$$
(C \cap D)-(A \cap B)\subset (C-A) \cup (D -B)
$$
Thus
\begin{align}
(A \cap B ) \bigtriangleup (C \cap D)&=((A \cap B) -(C \cap D))\cup((C \cap D)-(A \cap B))
\\
&\subset ((A -C) \cup (B -D))\cup ((C-A) \cup (D -B))
\\
&=((A -C) \cup (C-A))\cup ((B -D)\cup (D -B))
\\
&=(A \bigtriangleup C) \cup (B \bigtriangleup D)
\end{align}

