\documentclass{amsart}

\newtheorem*{theorem}{Theorem}

\title{Integrability on Subintervals}

\author{Math 166: Calculus II}

%\date{January 13, 1999}

\begin{document}

\maketitle

\thispagestyle{empty}

The purpose of this note is to fill in a gap in Apostol's Calculus
text. The following theorem is very useful in practice and simplifies
the statements of several theorems in the text.

\begin{theorem}

Let $f$ be a function defined on an interval $[a,b]$. If $f$ is
integrable on $[a,b]$, then $f$ is integrable on any subinterval
$[c,d]\subset[a,b]$.

\end{theorem}

\begin{proof}

We shall prove the logically equivalent statement: Suppose there exists a
subinterval $[c,d]\subset[a,b]$ on which $f$ is not integrable. Then $f$
is not integrable on the larger interval $[a,b]$.

To show $f$ is not integrable on $[a,b]$ we must show that there is a gap
between the upper and lower integrals of $f$. In other words, there is a
constant $M$ such that given any step functions $s$ and $t$ on $[a,b]$
satisfying $s(x) \le f(x) \le t(x)$ for $a\le x \le b$, we have
  \[
  \int_a^b s - \int_a^b t \ge M
  \]
Since we are assuming $f$ is not integrable on $[c,d]$, we know there is a
constant $M$ such that
  \[
  \int_c^d s - \int_c^d t \ge M
  \]
This inequality immediately implies the previous one by additivity of
intervals and linearity of integrals for step functions:
  \[
  \int_a^b s - \int_a^b t = \int_a^c (s - t)
      + \int_c^d (s - t) + \int_d^b (s - t) \ge 0 + M + 0
  \]

\end{proof}



\end{document}
