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.