Ex3-10

Chapter 3 Question 10
We suppose |E_1|<\infty, |E_2|<\infty, otherwise both sides of the equality are unbounded. Now let A=E_1-E_2,B=E_1\cap E_2,C=E_2-E_1, so
\begin{align}
\left(E_1\cup E_2\right)=A\cup C\cup B 
\end{align}
then since A,B,C are measurable and disjoint, Theorem 3.23 gives
\begin{align}
|E_1\cup E_2|=|A\cup C\cup B |=|A|+ |C|+ |B| 
\end{align}
Since A=E_1-B, C=E_2-B and from Corollary 3.25, B\subset E_1 and B\subset E_2, we get
\begin{align}
|A|+ |C|+ |B| &=|E_1-B|+|E_2-B|+|B|\\
&=|E_1|-|B|+|E_2|-|B|+|B|=|E_1|+|E_2|-|B|
\end{align}
Thus
\begin{align}
|E_1\cup E_2|=|E_1|+|E_2|-|E_1\cap E_2|
\end{align}
This completes the proof.