
\vspace*{.3in}
\noindent{\bf Beals page 5, \#4:}
The inequality
$$
m^* (A_1\cup A_2) \leq m^* A_1 + m^* A_2
$$
holds for \emph{all} $A_1,A_2\subset\R$, so it remains for me to use
the hypothesis to prove the reverse inequality.

Given $\epsilon > 0$, let $\mathcal I$ be a finite or countable 
collections of intervals covering $A_1\cup A_2$ and satisfying
$$
|\mathcal I| \leq m^* (A_1\cup A_2) + \epsilon.
$$
For each interval $I\in\mathcal I$, let $I' = I\cap I_1$ and $I'' = I\cap
I_2$.  Note that 
\begin{itemize}
\item For any $I\in\mathcal I$, the corresponding sets $I'$ and $I''$ are 
open intervals satisfying
$$
|I'| + |I''| \leq |I|.
$$.
\item $\mathcal I' \eqdef \{I':I\in\mathcal I\}$ covers $A_1$
\item $\mathcal I'' \eqdef \{I'':I\in\mathcal I\}$ covers $A_2$.
\end{itemize}
Therefore,
$$
m^* A_1 + m^* A_2 \leq |\mathcal I'| + |\mathcal I''| \leq |\mathcal I|
\leq m^*(A_1\cup A_2) + \epsilon.
$$
But $\epsilon > 0$ was arbitrary, so I deduce that
$$
m^* A_1 + m^* A_2 \leq m^*(A_1\cup A_2)
$$


