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\vspace*{.3in}
\noindent{\bf Beals page 5, \#7:}
The trick here is to remove smaller and smaller fractions of the remaining
intervals as the construction progresses.  Specifically, I define a 
decreasing sequence $\{C_k\}$ of closed sets $C_k\subset [0,1]$ as follows.
First I choose a sequence $\{a_j\}_{j\in\N}$ of positive real numbers such
that 
$$
s := \sum_{j=1}^\infty a_j < 1.
$$
I let $C_0 = [0,1]$ and $D_0 = [0,1] - C_0 = \emptyset$.  I divide $C_0$ into 
two closed intervals of equal length by removing an open interval of length 
$a_1$ centered at $1/2$.  I call the union of the remaining closed intervals 
$C_1$ and set $D_1 = [0,1] - C_1$.  

Likewise, given a closed set 
$C_k \subset [0,1]$ consisting of $2^k$ closed, pairwise disjoint 
intervals $I$ of equal length, I create the set 
$C_{k+1}\subset C_k$ by removing from each interval $I\subset C_k$
an open interval $J$ centered on the midpoint of $I$ such that 
$|J|\leq a_{k+1}|I|$.  Thus, $C_{k+1}$ consists of $2^{k+1}$ closed, 
pairwise disjoint intervals.  Moreover, since the sum of the lengths
of the closed intervals comprising $C_k$ is no greater than one, it follows
that the sum of the lengths of the intervals removed from $C_k$ to create
$C_{k+1}$ is no larger than $a_{k+1}$.  Stated in terms of the complements
$D_k$ and $D_{k+1}$ of $C_k$ and $C_{k+1}$ in $[0,1]$, I have
$$
m^* D_{k+1} \leq m^* D_k + a_{k+1}.
$$
Now if I let $C = \bigcap_{k\in\N} C_k$ and $D = \bigcup_{k\in\N} D_k$, I 
have $C \cup D = [0,1]$.  Hence
$$
m^* C \geq 1 - m^* D \geq 1 - \sum_{k=0}^\infty m^*(D_k - D_{k-1})
\geq 1 - \sum_{k=0}^\infty a_k = 1 -s > 0.
$$ 

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