bookshelf/Common/Real/Geometry/Area.tex

\documentclass{article}

\input{../../../preamble}

\newcommand{\lean}[2]{\href{./Area.html\##1}{#2}}

\begin{document}

\header{Axiomatic Framework of Area}{Tom M. Apostol}

We assume there exists a class $\mathscr{M}$ of measurable sets in the plane and
  a set function $a$, whose domain is $\mathscr{M}$, with the following
  properties:

\section*{\verified{Nonnegative Property}}%
\hyperlabel{sec:nonnegative-property}%

For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$.

\begin{axiom}

  \lean{Nonnegative-Property}{Nonnegative Property}

\end{axiom}

\section*{\verified{Additive Property}}%
\hyperlabel{sec:additive-property}%

If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in
  $\mathscr{M}$, and we have $a(S \cup T) = a(S) + a(T) - a(S \cap T)$.

\begin{axiom}

  \lean{Additive-Property}{Additive Property}

\end{axiom}

\section*{\verified{Difference Property}}%
\hyperlabel{sec:difference-property}%

If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in
  $\mathscr{M}$, and we have $a(T - S) = a(T) - a(S)$.

\begin{axiom}

  \lean{Difference-Property}{Difference Property}

\end{axiom}

\section*{\verified{Invariance Under Congruence}}%
\hyperlabel{sec:invariance-under-congruence}%

If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is
  also in $\mathscr{M}$ and we have $a(S) = a(T)$.

\begin{axiom}

  \lean{Invariant-Under-Congruence}{Invariance Under Congruence}

\end{axiom}

\section*{\verified{Choice of Scale}}%
\label{sec:choice-scale}

Every rectangle $R$ is in $\mathscr{M}$.
If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$.

\begin{axiom}

  \lean{Choice-of-Scale}{Choice of Scale}

\end{axiom}

\section*{\proceeding{Exhaustion Property}}%
\hyperlabel{sec:exhaustion-property}%

Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so
  that
  \begin{equation}
    \label{sec:exhaustion-property-eq1}
    S \subseteq Q \subseteq T.
  \end{equation}
If there is one and only one number $c$ which satisfies the inequalities
  $$a(S) \leq c \leq a(T)$$ for all step regions $S$ and $T$ satisfying (1.1),
  then $Q$ is measurable and $a(Q) = c$.

\begin{axiom}

  \lean{Exhaustion-Property}{Exhaustion Property}

\end{axiom}

\end{document}