Outer Jordan Content of Right Triangle

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Theorem

Let $T \subseteq \R^2$ be defined as:

$T = \set {\tuple {x, y} \in \R^2 : x \ge 0 \land y \ge 0 \land x + y \le 1}$

Then:

$\map {m^*} T = \dfrac 1 2$

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Proof

Let $\epsilon > 0$ be arbitrary.

By the Axiom of Archimedes, let $n \in \N$ such that:

$n > 2 \epsilon$

Define $C \subseteq \powerset {\R^2}$ as:

$C = \set {\closedint {\dfrac p n} {\dfrac {p + 1} n} \times \closedint {\dfrac q n} {\dfrac {q + 1} n} : p, q \in \set {0, 1, \dotsc, n - 1} \land p + q < n}$

By construction, $C$ is a finite set of closed $2$-rectangles.

Let $\tuple {x, y} \in T$ be arbitrary.

Let $p \in \Z_{\ge 0}$ be the smallest non-negative integer such that:

$\dfrac {p + 1} n \ge x$

Because:

$\dfrac {\paren {n - 1} + 1} n = 1 \ge x$

we have:

$p \le n - 1$

If $\dfrac {p'} n < x$ for any $p' \in \Z_{\ge 0}$, then:

$\dfrac p n < x$

by minimality.

Otherwise, $p = 0$ as the smallest non-negative integer overall, so:

$\dfrac p n = 0 \le x$

In either case:

$\dfrac p n \le x$

In like manner, let $q \in \Z_{\ge 0}$ be the smallest non-negative integer such that:

$\dfrac {q + 1} n \ge y$

By the same argument as before:

$\dfrac q n \le y$

We also have:

\(\ds \frac {\paren {n - 1 - p} + 1} n\)

\(=\)

\(\ds \frac {n - p} n\)

\(\ds \)

\(=\)

\(\ds 1 - \frac p n\)

\(\ds \)

\(\ge\)

\(\ds 1 - x\)

as $\dfrac p n < x$ by minimality of $p$; in the case where $\dfrac {p + 1} n \ge x$ for all $p$, then $p = 0$, and the step follows from $x \ge 0$ Although this article appears correct, it's inelegant. There has to be a better way of doing it. In particular: too much going on here; separate this out so as to implement this step (or however many steps this actually subsumses) into a more explicit structure, perhaps as a lemma; however it is best implemented; I rather lack the time and inclination to analyse this directly myself You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Improve}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

\(\ds \)

\(\ge\)

\(\ds y\)

as $x + y \le 1$ by hypothesis

\(\ds \leadsto \ \ \)

\(\ds q\)

\(\le\)

\(\ds n - 1 - p\)

by minimality of $q$

\(\ds \leadsto \ \ \)

\(\ds p + q\)

\(\le\)

\(\ds n - 1\)

\(\ds \)

\(<\)

\(\ds n\)

Therefore:

$\closedint {\dfrac p n} {\dfrac {p + 1} n} \times \closedint {\dfrac q n} {\dfrac {q + 1} n} \in C$

Since, by the previous inequalities:

$\dfrac p n \le x \le \dfrac {p + 1} n$

$\dfrac q n \le y \le \dfrac {q + 1} n$

it follows that:

$\tuple {x, y} \in \closedint {\dfrac p n} {\dfrac {p + 1} n} \times \closedint {\dfrac q n} {\dfrac {q + 1} n}$

As $\tuple {x, y} \in T$ was arbitrary, we have that $C$ is a covering of $T$.

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