\documentclass{amsart}

\usepackage{enumerate}
\usepackage{graphics}

\renewcommand{\inf}{\operatorname{inf}}
\newtheorem*{theorem}{Theorem}
\newtheorem*{corollary}{Corollary}
\newtheorem*{lemma}{Lemma}
\newtheorem*{proposition}{Proposition}

\theoremstyle{definition}
\newtheorem*{definition}{Definition}

\title{Upper and Lower Integrals}

\author{Dennis Snow}

%\date{September 16, 1998}

\begin{document}

\maketitle

Upper and lower integrals are introduced in section 1.17 of the text.
We start with a bounded function $f$ on $[a,b]$ and consider step
functions $s$ and $t$ on $[a,b]$ that satisfy
  \[
    s(x) \le f(x) \le t(x)\quad\mbox{for all}\quad a \le x \le b.
  \]
Let us abbreviate this by writing $s \le f$ and $f \le t$.  The text then
defines the following subsets of real numbers,
  \[
    S = \bigg\{\int_a^b s(x)\,dx \; \big| \; s \le f \bigg\},\quad
    T = \bigg\{\int_a^b t(x)\,dx \; \big| \; f \le t \bigg\},
  \]
and defines the lower and upper integrals of $f$ to be, respectively,
  \[
    \underline{I}(f) = \sup S, \quad \overline{I}(f) = \inf T,
  \]

An interesting property of the sets $S$ and $T$ not mentioned in the
text is that they are intervals.

\begin{proposition}
\begin{enumerate}[a)]
\item If $u\in S$ then $y\in S$ for every $y\le u$.
\item If $v\in T$ then $y\in T$ for every $y\ge v$.
\end{enumerate}
\end{proposition}

\begin{proof} To prove a), let $s \le f$ be a step function whose
integral is $u = \int_a^b s(x)\,dx$ and let $y \le u$.  Let $P =
\{x_0,x_1,\ldots,x_n\}$ be a partition of $[a,b]$ such that $s(x) = s_k$
for $x_{k-1}<x < x_k$, $1 \le k \le n$.  We now alter the step function
$s$ by lowering its value on the first subinterval so that its integral
will achieve the given value~$y$.
\[
\resizebox{3in}{!}{\includegraphics{lower.eps}}
\]
Define the step function
  \[ s_y(x) =\left\{\begin{array}{ll}\displaystyle
          s(x) - \frac{u - y}{x_1 - x_0} & \textrm{for}\ a = x_0 \le x < x_1,\\
          s(x)                           & \textrm{for}\ x_1 \le x \le x_n = b.
          \end{array}\right.
  \]
Since $(u - y)/(x_1-x_0) \ge 0$, $s_y(x) \le s(x) \le f(x)$ for all $a
\le x \le b$. Moreover,
  \begin{eqnarray*} \int_a^b s_y(x)\,dx
                      &=& \big(s_1 - \frac{u-y}{x_1-x_0}\big)(x_1 - x_0) +
                          \sum_{k=2}^n s_k(x_k - x_{x-1})\\
                      &=& y - u + \sum_{k=1}^n s_k(x_k - x_{x-1})\\
                      &=& y - u + \int_a^b s(x)\,dx\quad = \quad y
  \end{eqnarray*}
Therefore, $y \in S$.

The proof of b) is similar:
Let $t\ge f$ be a step function whose integral
is $v = \int_a^bt(x)\,dx$ and let $y\ge v$.
Let $P = \{x_0,x_1,\ldots,x_n\}$ be a partition of $[a,b]$ such that
$t(x) = t_k$ for $x_{k-1}< x <x_k$, $1 \le k \le n$. Define the step
function
  \[
  t_y(x) =\left\{\begin{array}{ll}\displaystyle
    t(x) + \frac{y - v}{x_1 - x_0} & \textrm{for}\ a = x_0 \le x < x_1,\\
    t(x)                           & \textrm{for}\ x_1 \le x \le x_n = b.
    \end{array}\right.
  \]
Since $(y-v)/(x_1-x_0)\ge 0$, $t_y(x) \ge t(x) \ge f(x)$ for all $a \le
x \le b$. As above, this new step function has been arranged so that
its integral is the number~$y$:
  \begin{eqnarray*} \int_a^b t_y(x)\,dx
     &=& \big(t_1 + \frac{y-v}{x_1-x_0}\big)(x_1 - x_0) +
         \sum_{k=2}^n t_k(x_k - x_{x-1})\\
     &=& y - v + \sum_{k=1}^n t_k(x_k - x_{x-1})\\
     &=& y - v + \int_a^b t(x)\,dx\quad = \quad y
  \end{eqnarray*}
Therefore, $y \in T$.

\end{proof} 

This proposition shows that the lower integral, $\underline{I}(f)$,
is the right hand endpoint of the interval~$S$, and the upper integral,
$\overline{I}(f)$, is the left hand endpoint of the interval~$T$. These
endpoints are not usually in the intervals, but they can be, for example,
when $f$ itself is a step function. A bounded function is integrable when
$\underline{I}(f) = \overline{I}(f)$.  This means that there is no gap
between the  intervals $S$ and $T$, and they share a common endpoint,
namely, the number representing $\int_a^bf(x)\,dx$.

\end{document}
