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\vspace*{.3in}
\noindent{\bf Beals page 11, \#1:}
If $E\subset\R$ is any set, then
$$
m^* E \leq m^*(E\cap A) + m^*(E\cap A^c)
$$
automatically, regardless of $m^* A$.  On the other hand, since $m^* A = 0$,
$$
m^*(E\cap A)  + m^*(E\cap A^c) 
\leq 
m^* A + m^* E = m^* E.
$$
Hence $m^* E = m^*(E\cap A) + m^*(E\cap A^c)$ for all $E\subset \R$.
That is, $A$ is measurable.
\qed
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