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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{The definition of the Riemann Integral}

\begin{thm}
If \( f \) is continuous on \( [a,b] \), then \[
\int _{a}^{b}f(x)dx\]
 exists.
\end{thm}
\textbf{\emph{Proof:}}

This proof is more sophisitcated than most in this course, but it's nice to
be able to include it, so that the claims of existence of the integral that
we have already made use of can be justified.

Since \( f \) is continuous on a bounded interval, including the endpoints,
given any \( \epsilon >0 \), there is a \( \delta >0 \) so that \( |f(x)-f(y)|<\epsilon /(b-a) \)
whenever \( |x-y|<\delta  \). Take a partition \( \mathcal{P} \) of mesh \( \left\Vert \mathcal{P}\right\Vert <\delta  \),
then on any subinterval in that partition, if \( M_{i} \) is the maximum of
\( f \) on that subinterval, and \( m_{i} \) is the minimum, \( M_{i}-m_{i}\leq \epsilon /(b-a) \),
and so any two Riemann sums, \( R_{1}(f) \) and \( R_{2}(f) \), differ by
no more than: \begin{eqnarray*}
|R_{1}(f)-R_{2}(f)| & \leq  & \sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\\
 & \leq  & \sum _{i=1}^{n}\frac{\epsilon \Delta x_{i}}{b-a}\\
 & = & \frac{\epsilon (b-a)}{b-a}\\
 & = & \epsilon .
\end{eqnarray*}
Thus, for any \( \epsilon >0 \), a small enough mesh can be found so that the
difference between any two Riemann sums with that mesh will be less than \( \epsilon  \),
so the limit as the mesh goes to 0 does exist.
\end{document}
