Show that $|A\cup B|+|A\cap B| =|A|+| B|$ I want to show that $|A\cup B|+|A\cap B| =|A|+| B|$. I know that $|A\cup B| = A\cup(B-A)$ and that $B-A = B-(A\cap B)$. We also know that $A$ and $B$ and Lebesgue measurable. 
Here is my approach:
$$|A\cup B|+|A\cap B| = |A\cup(B-A)|+|B-(A\cap B)|$$
Since $A$ and $(B-A)$ are disjoint sets I can say
$$= |A|+|(B-A)|+|B-(A\cap B)|$$
and since $(A\cap B) \subset B$ I can say
$$= |A|+|(B-A)|+|B|-|(A\cap B)|$$
Rearranging I get
$$= |A|+|B| +|(B-A)|-|(A\cap B)|$$
So if I can show $|(B-A)|=|(A\cap B)|$ then I have what I need but I don't think it is possible. The closest I can get is $|(B-A)|=|(A\cap B)| = |(A - CB)|$. 
Is there something I am not seeing? 
 A: Let $E$ be a large set containing both $A,B$ and for each $C \subset E$ we define $\chi_C: E \to \{0 ,1 \} $ to be the characteristic function of the set, that is
$$\chi_C(x)=1 \mbox{ if } x \in C \mbox{ and 0 otherwise}$$
Then, the following are well known/easy to prove
$$ |C|= \sum_{x \in E} \chi(x)$$
$$\chi_{A \cup B}= \chi_A+\chi_B-\chi_A\chi_B$$
$$\chi_{A \cap B}= \chi_A\chi_B \,.$$
Then
$$|A\cup B|+|A\cap B|= \sum_{x \in E} \left( \chi_A(x)+\chi_B(x)-\chi_A(x)\chi_B(x)\right)  +\sum_{x \in E} \left( \chi_A(x)\chi_B(x)\right)$$
$$ =\sum_{x \in E} \left( \chi_A(x)\right) +\sum_{x \in E} \left( \chi_B(x)\right) =|A|+|B|$$
P.S. Characteristic functions are in my opinion the simplest way to prove relations with sets. The other useful relations are:
$$A=B \Leftrightarrow \chi_A=\chi_B$$
$$A \subset B \Leftrightarrow \chi_A \leq\chi_B$$
$$\chi_{A \backslash B}= \chi_A - \chi_A \chi_B$$
$$\chi_{A \Delta B}= \chi_A+\chi_B - 2\chi_A \chi_B=\left(\chi_A-\chi_B \right)^2$$
A: Although I'm pretty late to the party here's how I would have solve this problem. N.S. already gave a neat answer, but a second approach harms no one so here I go:
First decompose $A$, $B$ and $A\cap B$ into disjoint sets:
$(1)\;A = (A\setminus B)\cup(A\cap B)$
$(2)\;B = (B\setminus A)\cup (A\cap B)$
$(3)\;A\cup B = (A\setminus B)\cup (B\setminus A)\cup(A\cap B)$
This gives us
$(1^{\prime})\;|A| = |A\setminus B|+|A\cap B|$
$(2^{\prime})\;|B| = |B\setminus A|+|A\cap B|$
$(3^{\prime})\;|A\cup B| = |A\setminus B|+|B\setminus A|+|A\cap B|$
All that is left now is to bring together the equations above:
$$
\begin{align}
|A\cup B| &=|A\setminus B|+|B\setminus A|+|A\cap B|\\
&=|A|-|A\cap B|+|B|-|A\cap B|+|A\cap B|\\
&=|A|+|B|-|A\cap B|
\end{align}
$$
which implies $|A|+|B|=|A\cup B|+|A\cap B|$.
