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{\bf Question}

Suppose that $\{S_1,\cdots S_n\}$ is a collection of sets, not
necessarily measurable, satisfying $d(S_i,S_j)&gt;0$ for $i\not=j$.
Extend theorem 2.8 to show that

$\ds m^*\left(\bigcup_{i=1}^n S_i\right)=\sum_{i=1}^n m^*(S_i)$


\vspace{0.25in}

{\bf Answer}

The result is true for $n=2$ by theorem 2.8.  Suppose true for
$n=k$, now consider $S_{k+1}$.  Let $d_i=d(S_i,S_{k+1})&gt;0$.  Let
$\ds d=\min_{1\leq i\leq k}d_i&gt;0$.  Then $\ds
d(S_{k+1},\bigcup_{i=1}^k)=d&gt;0$.  Hence by theorem 2.8

$\ds m^*(S_{k+1}\cup\bigcup_{i=1}^k S_i)=m^*(S_{k+1})+\sum_{i=1}^k
S_i$.  Hence the result.




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