Exercise 1.1.8 (Triangles are Jordan measurable)

Question

Let $A,B,C$ be three points in $\mathbb{R}^2$.
\begin{enumerate}[label=(\roman*)]
	\item Show that the solid triangle with vertices $A,B,C$ is Jordan measurable.
	\item Show that the Jordan measure of the solid triangle is equal to
	$$\frac12 \big|(B-A)\wedge (C-A)\big|$$
	where $|(a,b)\wedge (c,d)| = |ad-bc|$.
\end{enumerate}

Elu Thingol · Accepted Answer

\documentclass{article}
\usepackage[british]{babel}
\usepackage{amsmath}
\usepackage{float}
\usepackage{tikz}
\def\graphscale{0.7}
\def\graphcolour{blue!50!cyan}

\begin{document}
The previous solution had some parts that were not entirely correct. An updated solution will be posted this week.\par
2021-11-10

\begin{figure}[H] \centering
    \begin{tikzpicture}[scale=\graphscale]
      \draw[thick,->] (0,0) -- (9,0) node[anchor=north west] {$d_1$};
      \draw[thick,->] (0,0) -- (0,6) node[anchor=south east] {$d_2$};
      \draw (1,1) -- (5,5); 
ode at (3,4) {$f_{AB}$};
      \draw (5,5) -- (7,3); 
ode at (6.5,4.5) {$f_{BC}$};
      \draw (7,3) -- (1,1); 
ode at (4,1.5) {$f_{AC}$};
      \draw[dashed] (1,1) -- (1,0); 
ode[label=left:	extbf{A}] at (1,1){}; 
ode[label=below:$A_{1}$] at (1,0){};
      \draw[dashed] (5,5) -- (5,0); 
ode[label=above:	extbf{B}] at (5,5){}; 
ode[label=below:$B_{1}$] at (5,0){};
      \draw[dashed] (7,3) -- (7,0); 
ode[label=right:	extbf{C}] at (7,3){}; 
ode[label=below:$C_{1}$] at (7,0){};
  \end{tikzpicture}
    \caption{The edges $A,B,C$ of the triangle formed by the intersection of $f_1,f_2,f_3$}
  \end{figure}

\begin{figure}[h] \centering
      \def\A{(-3,3)} \def\B{(-1,1)} \def\C{(-2,-1)}
      \def\AA{(-1,4)} \def\BB{(1,2)} \def\CC{(0,0)}
\begin{tikzpicture}[scale=\graphscale]
\draw[thick,->] (-3,0) -- (1.5,0) node[anchor=north west] {$X_1$};
\draw[thick,->] (0,-1.5) -- (0,3) node[anchor=south east] {$X_2$};
\draw[fill=\graphcolour,fill opacity=0.3] \A node[anchor=north]{$A$}
  -- \B node[anchor=north]{$B$}
  -- \C node[anchor=east]{$C$}
  -- cycle;
\end{tikzpicture} \hspace{5mm}
%
\begin{tikzpicture}[scale=\graphscale]
\draw[thick,->] (-3,0) -- (1.5,0) node[anchor=north west] {$X_1$};
\draw[thick,->] (0,-1.5) -- (0,3) node[anchor=south east] {$X_2$};
\draw[fill=\graphcolour,fill opacity=0.33] \AA node[anchor=south east]{$A$}
  -- \BB node[anchor=west]{$B$}
  -- \CC node[anchor=north east]{$C$}
  -- cycle;
\end{tikzpicture} \hspace{5mm}
\caption{Moving the triangle to the origin.}
\end{figure}

\begin{figure}[H] \centering
    \begin{tikzpicture}[scale=\graphscale]
      \filldraw[fill=\graphcolour,fill opacity=0.33,draw=white] (2,0) -- (2,2) -- (6,6) -- (6,0) -- (2,0);
      \draw[thick,->] (0,0) -- (8,0) node[anchor=north west] {$X_1$};
      \draw[thick,->] (0,0) -- (0,7) node[anchor=south east] {$X_2$};
      \draw (2,2) -- (6,6);
      \draw[dashed] (2,2) -- (6,2);
      \draw[dashed] (2,2) -- (2,0);
      \draw[dashed] (6,6) -- (6,0);
      
ode at (6,-0.5) {b}; 
ode at (6,6.5) {$f(b)$}; 
      
ode at (2,-0.5) {a}; 
ode at (1.5,2.5) {$f(a)$};
      
ode at (8,1) {$(b-a)\cdot |f(a)|$};
      
ode at (9,4) {$\dfrac{1}{2} (b-a)\cdot |f(b) - f(a)|$};
    \end{tikzpicture}
    \caption{The intuition behind the formula for the area under the line-shaped functions.}
  \end{figure}

\begin{figure}[H] \centering
    \begin{tikzpicture}[scale=\graphscale]
      \draw[thick,->] (0,0) -- (6,0) node[anchor=north west] {$X_1$};
      \draw[thick,->] (0,0) -- (0,7) node[anchor=south east] {$X_2$};
      \draw (1,1) -- (5,5);
      
ode at (5,-0.5) {b}; 
ode at (5,5.5) {$f(b)$}; 
      
ode at (1,-0.5) {a}; 
ode at (0.5,1.5) {$f(a)$};
      \foreach \x in {1,2,...,4} {
        \draw[fill=\graphcolour,fill opacity=0.33] (\x, 0) rectangle (\x + 1, \x);
    }
  \end{tikzpicture}%
  \begin{tikzpicture}[scale=\graphscale]
      \draw[thick,->] (0,0) -- (6,0) node[anchor=north west] {$X_1$};
      \draw[thick,->] (0,0) -- (0,7) node[anchor=south east] {$X_2$};
      \draw (1,1) -- (5,5);
      
ode at (5,-0.5) {b}; 
ode at (5,5.5) {$f(b)$}; 
      
ode at (1,-0.5) {a}; 
ode at (0.5,1.5) {$f(a)$};
      \foreach \x in {1,1.5,...,4.5} {
        \draw[fill=\graphcolour,fill opacity=0.33] (\x, 0) rectangle (\x + 0.5, \x);
    }
  \end{tikzpicture}%
  \begin{tikzpicture}[scale=\graphscale]
      \draw[thick,->] (0,0) -- (6,0) node[anchor=north west] {$X_1$};
      \draw[thick,->] (0,0) -- (0,7) node[anchor=south east] {$X_2$};
      \draw (1,1) -- (5,5);
      
ode at (5,-0.5) {b}; 
ode at (5,5.5) {$f(b)$}; 
      
ode at (1,-0.5) {a}; 
ode at (0.5,1.5) {$f(a)$};
      \foreach \x in {1,1.2,...,4.8} {
        \draw[fill=\graphcolour,fill opacity=0.33] (\x, 0) rectangle (\x + 0.2, \x);
    }
    \end{tikzpicture}
    \caption{Strategy for inside covering the area under the graph of a linear-shaped function.}
  \end{figure}
\end{document}

Exercise 1.1.8 (Triangles are Jordan measurable)

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Exercise 1.1.8 (Triangles are Jordan measurable)

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