\documentclass[11pt]{article}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}

\textwidth = 6.5 in
\textheight = 9 in
\oddsidemargin = 0.0 in
\evensidemargin = 0.0 in
\topmargin = 0.0 in
\headheight = 0.0 in
\headsep = 0.0 in
\parskip = 0.2in
\parindent = 0.0in

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}{Definition}

\begin{document}







\vspace*{.3in}
\noindent{\bf Beals page 5, \#9:}
Note that $m^*A$ is finite because $A$ is bounded.
Suppose, in order to obtain a contradiction, that $m^* A \neq 0$.  Then (by
setting $\epsilon = m^*A > 0$), I can find a countable collection 
$\mathcal{I} = \{I_k\}_{k\in\N}$ of open intervals covering $A$ such that
$|\mathcal I| < m^*A + \epsilon = 2m^* A$.  But then
$$
m^* A \leq \sum_{k=1}^\infty m^*(A\cap I_k) 
       \leq \frac12 \sum_{k=1}^\infty |I_k| 
       = \frac12|\mathcal I| < m^* A.
$$
This contradiction proves that $m^*A = 0$.
\q \end{document}
