98 lines
2.3 KiB
TeX
98 lines
2.3 KiB
TeX
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\documentclass{article}
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\input{../../preamble}
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\newcommand{\link}[2]{\lean{../..}
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{Common/Real/Geometry/Area} % Location
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{#1} % Fragment
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{#2} % Presentation
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}
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\begin{document}
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\header{Axiomatic Framework of Area}{Tom M. Apostol}
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We assume there exists a class $\mathscr{M}$ of measurable sets in the plane and
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a set function $a$, whose domain is $\mathscr{M}$, with the following
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properties:
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\section*{\verified{Nonnegative Property}}%
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\hyperlabel{sec:nonnegative-property}%
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For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$.
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\begin{axiom}
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\link{Nonnegative-Property}{Nonnegative Property}
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\end{axiom}
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\section*{\verified{Additive Property}}%
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\hyperlabel{sec:additive-property}%
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If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in
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$\mathscr{M}$, and we have $a(S \cup T) = a(S) + a(T) - a(S \cap T)$.
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\begin{axiom}
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\link{Additive-Property}{Additive Property}
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\end{axiom}
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\section*{\verified{Difference Property}}%
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\hyperlabel{sec:difference-property}%
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If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in
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$\mathscr{M}$, and we have $a(T - S) = a(T) - a(S)$.
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\begin{axiom}
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\link{Difference-Property}{Difference Property}
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\end{axiom}
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\section*{\verified{Invariance Under Congruence}}%
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\hyperlabel{sec:invariance-under-congruence}%
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If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is
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also in $\mathscr{M}$ and we have $a(S) = a(T)$.
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\begin{axiom}
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\link{Invariant-Under-Congruence}{Invariance Under Congruence}
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\end{axiom}
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\section*{\verified{Choice of Scale}}%
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\label{sec:choice-scale}
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Every rectangle $R$ is in $\mathscr{M}$.
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If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$.
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\begin{axiom}
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\link{Choice-of-Scale}{Choice of Scale}
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\end{axiom}
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\section*{\proceeding{Exhaustion Property}}%
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\hyperlabel{sec:exhaustion-property}%
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Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so
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that
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\begin{equation}
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\label{sec:exhaustion-property-eq1}
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S \subseteq Q \subseteq T.
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\end{equation}
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If there is one and only one number $c$ which satisfies the inequalities
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$$a(S) \leq c \leq a(T)$$ for all step regions $S$ and $T$ satisfying (1.1),
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then $Q$ is measurable and $a(Q) = c$.
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\begin{axiom}
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\link{Exhaustion-Property}{Exhaustion Property}
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\end{axiom}
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\end{document}
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