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{\bf Question}

We have defined $m^*$ by $\ds
m^*(S)=\inf\left\{\sum_{i=1}^\infty|R_i|:\bigcup_{i=1}^\infty
R_i\supseteq S\right\}$

Let $n_{\delta}^*$ be defined by

$\ds
n_{\delta}^*(S)=\inf\left\{\sum_{i=1}^\infty|R_i|:\bigcup_{i=1}^\infty
R_i\supseteq S, \, |R_i|&lt;\delta, \,\, i=1,2,\cdots\right\}$

Show that $n_{\delta}^*$ is a monotonic function of $\delta$.

Define $n^*$ by $\ds n^*(S)=\lim_{\delta\to0+}n_{\delta}^*(S)$

Show that $n^*(S)=m^*(S)$



\vspace{0.25in}

{\bf Answer}

If $\delta&lt;\delta'$ then $n_\delta^*(S)$ is an infimum taken over
a smaller set than $n_{\delta'}^*(S)$

Hence $n_\delta^*(S)\geq n_{\delta'}^*(S)$

Thus $n_\delta^*(S)$ tends to a limit as $\delta\to0+$.

Now $n_\delta^*(S)$ is an infimum taken over a smaller set than
$m^*(S)$.  Therefore $m^*(S)\leq n_\delta^*(S)\leq n^*(S)$

Now for all $\epsilon$ there exists $\ds \{R_i\} \hspace{0.2in}
\sum_{i=1}^\infty|R_i|&lt;m^*(S)+\epsilon$

Let $\delta&gt;0$ and choose each rectangle $R_i$ into sub-rectangles
$\{R_{ij}\}_{j=1}^{m_j}$ so that $|R_{ij}|&lt;\delta$.

Then $\ds |R_i|=\sum_{j=1}^{m_i}|R_{ij}| \hspace{0.2in}
\bigcup_jR{ij}=R_i$

So $\ds \bigcup_{ij}R_{ij}\supseteq S \hspace{0.2in}
|R_{ij}|&lt;\delta$.

Hence $\ds n_\delta^*(S)\leq
\sum|R_{ij}|=\sum|R_i|&lt;m^*(S)+\epsilon$

Therefore $n_\delta^*(S)\leq m^*(S)+\epsilon \hspace{0.2in}$ for
all $\epsilon$, for all $\delta$

Therefore $n^*(S)\leq m^*(S)+\epsilon$

Therefore $n^*(S)\leq m^*(S)$

Therefore $n^*(S)= m^*(S)$

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