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{\bf Question}

Let $F$ be a closed set.  Let $A_0$ be a set disjoint from $F$.

Let $\ds A_n=\{x|x\epsilon A_0, \, d(x,F)\geq\frac{1}{n}\}$.

Show that $A_1\subseteq A_2\subseteq \cdots \subseteq A_0$, and
deduce the existence of $\ds \lim_{n\to\infty}m^*(A_n)$ (it may be
$+\infty$).

Prove that $\ds A_0=\bigcup_{n=1}^\infty \hspace{0.3in} (\dagger)$

Write $D_n=A_{n+1}-A_n$.  Show that, provided $m\geq n+2,$

$\ds d(D_m,D_n)\geq\frac{1}{m(n+1)}>0$.

Consider the sums $\ds \sum_{k=1}^\infty m^*(D_{2k}), \,\,\,
\sum_{k=0}^\infty m^*(D_{2k+1})$.  If the first is infinite, prove
that $m^*(A_{2n})\to\infty$, and if the second is infinite, prove
that

$m^*(A_{2n+1})=+\infty$.  Deduce that if either sum is infinite
then

$\lim m^*(A_n)=+\infty\geq m^*(A_0)$.  If both sums are finite,
use $(\dagger)$ and

$\ds A_0=A_{2n}\cup\bigcup_{k=n}^\infty
D_{2k}\cup\bigcup_{k=n}^\infty D_{2k+1}$ to show that
$m^*(A_0)\leq \lim m^*(A_{2n})$.

Deduce finally that $m^*(A_n)\to m^*(A_0)$ as $n\to\infty$.  (None
of the $A$'s need be measurable).  Use this result to prove that a
closed set $S$ is measurable.

(Hint: in the definition of measurability, let $E-S=A_0$).



\vspace{0.25in}

{\bf Answer}

$\ds A_n=\{x|x\epsilon A_0, d(x,F)\geq\frac{1}{n}\}\subseteq A_0$

If $x\epsilon A_n$ then $x\epsilon A_0$ and $\ds
d(x,F)\geq\frac{1}{n}$

$\Rightarrow x\epsilon A_0$ and $\ds d(x,F)\geq\frac{1}{n+1}
\Rightarrow x\epsilon A_{n+1}$

Therefore $A_1\subseteq A_2\subseteq \cdots \subseteq A_0$

Hence $m^*(A_n)$ is an increasing sequence.  Thus $\lim
m^*(A_n)\leq m^*(A_0)$ (may be $+\infty$)

Suppose $x\epsilon D_m, \,\,\, y\epsilon D_n$.  Let $f\epsilon F$

$\ds d(y,x)\geq d(y,f)-d(x,f)$

$\ds \hspace{0.5in} \geq \frac{1}{n+1}-d(x,f)$ since $y\epsilon
D_n\subseteq A_{n+1}$

Now $x\epsilon D_m$ so $x\epsilon A_{m+1}-A_m$

Hence $\ds \frac{1}{m+1}\leq d(x,f)<\frac{1}{m}$

Thus $\ds d(y,x)\geq
\frac{1}{n+1}-\frac{1}{m}=\frac{m-n-1}{m(n+1)}\geq
\frac{1}{m(n+1)} \hspace{0.2in} (1)$

$\hspace{2.5in}$ provided $m\geq n+2$

Consider the sums $\ds \sum_{k=1}^\infty m^*(D_{2k})$, and $\ds
\sum_{k=1}^\infty m^*(D_{2k+1})$

Suppose $\ds \sum_{k=1}^\infty m^*(D_{2k})=+\infty$

Then $\ds
A_{2n}=A_1\cup\bigcup_{k=1}^{n-1}D_{2k}\cup\bigcup_{k=0}^{n-1}
D_{2k+1}\supseteq \bigcup_{k=1}^{n-1}D_{2k}$

Thus $\ds m^*(A_{2n})\geq
m^*\left(\bigcup_{k=1}^{n-1}D_{2k}\right)=
\sum_{k=1}^{n-1}m^*(D_{2k})$ by (1) and

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

Therefore $m^*(A_{2n}) \rightarrow +\infty$ as $n\rightarrow
\infty$

Similarly if the other sum is $+\infty$ then $m^*(A_{2k+1})\to
+\infty$.

Hence if either sum is $+\infty$, $m^*(A_n)\rightarrow +\infty$ as
$n\to\infty$

and so $m^*(A_n)\rightarrow m^*(A_0)=+\infty$ as $n\to\infty$.

If both sums are convergent then $\ds
A_0=A_{2n}\cup\bigcup_{k=n}^\infty D_{2k}\cup\bigcup_{k=n}^\infty
D_{2k+1}$ and so

$\ds m^*(A_0)\leq m^*(A_{2n})+m^*\left(\bigcup_{k=n}^\infty
D_{2k}\right)+m^*\left(\bigcup_{k=n}^\infty D_{2k+1}\right)$

$\ds\hspace{0.5in} \leq m^*(A_{2n})+\left(\sum_{k=n}^\infty
m^*(D_{2k})\right) +\sum_{k=n}^\infty m^*(D_{2k+1})$

Since both sums converge, both sums from $n$ to $\infty$ tend to
zero as $n\to\infty$.

Hence $m^*(A_0)\leq \lim m^*(A_{2n})=\lim m^*(A_n)$ by
monotonicity.

Hence $m^*(A_0)=\lim m^*(A_n)$

${}$

Let $F$ be a closed set.  Let $T$ be any set, let $A_0=T-F$

Let $\ds A_n=\{x|x\epsilon A_0,d(x,F)\geq\frac{1}{n}\}$

Then $d(A_n,T\cap F)>0$ and so by theorem 2.8

$m^*(T)=m^*((T\cap F)\cup(T-F))$

$\hspace{0.5in} \geq m^*((T\cap F)\cup A_n) \hspace{0.5in}$ (since
$A_n\subseteq T-F$)

$\hspace{0.5in}=m^*(T\cap F)+m^*(A_n) \hspace{0.5in}$  for all
$n$.

Therefore $m^*(T)\geq m^*(T\cap F)+\lim m^*(A_n)$

$\hspace{1.2in}=m^*(T\cap F)+m^*(T-F)$

Hence the result.




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