Outer Jordan Content of Dilation

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Theorem

Let $M \subseteq \R^n$ be a bounded subspace of Euclidean $n$-space.

Let $c_1, c_2, \dotsc, c_n \in \R_{\ge 0}$ be non-negative real numbers.

Let $M' \subseteq \R^n$ be defined as:

$M' = \set {\tuple {c_1 x_1, c_2 x_2, \dotsc, c_n x_n} : \tuple {x_1, x_2, \dotsc, x_n} \in \R^n}$


Then:

$\map {m^*} {M'} = c_1 c_2 \dotsm c_n \map {m^*} M$

where $m^*$ denotes the outer Jordan content.


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Proof

Let $\epsilon > 0$ be arbitrary.

By Characterizing Property of Infimum of Subset of Real Numbers, let $C$ be a finite covering of $M$ by closed $n$-rectangles such that:

$\ds \sum_{R \mathop \in C} \map V C < \map {m^*} M + \frac \epsilon {c_1 c_2 \dotsm c_n + 1}$

Let $C'$ be defined as:

$\ds C' = \set {\closedint {c_1 a_1} {c_1 b_1} \times \dotso \times \closedint {c_n a_n} {c_n b_n} : \closedint {a_1} {b_1} \times \dotso \times \closedint {a_n} {b_n} \in C}$


For any $\tuple {c_1 x_1, \dotsc, c_n x_n} \in M'$, there must be some:

$R = \closedint {a_1} {b_1} \times \dotso \times \closedint {a_n} {b_n} \in C$

such that:

$\tuple {x_1, \dotsc, x_n} \in R$

by definition of a covering.

That is:

\(\ds a_1\) \(\le\) \(\, \ds x_1 \, \) \(\, \ds \le \, \) \(\ds b_1\)
\(\ds a_2\) \(\le\) \(\, \ds x_2 \, \) \(\, \ds \le \, \) \(\ds b_2\)
\(\ds \) \(\) \(\, \ds \vdots \, \) \(\ds \)
\(\ds a_n\) \(\le\) \(\, \ds x_n \, \) \(\, \ds \le \, \) \(\ds b_n\)

Therefore, by Real Number Ordering is Compatible with Multiplication:

\(\ds c_1 a_1\) \(\le\) \(\, \ds c_1 x_1 \, \) \(\, \ds \le \, \) \(\ds c_1 b_1\)
\(\ds c_2 a_2\) \(\le\) \(\, \ds c_2 x_2 \, \) \(\, \ds \le \, \) \(\ds c_2 b_2\)
\(\ds \) \(\) \(\, \ds \vdots \, \) \(\ds \)
\(\ds c_n a_n\) \(\le\) \(\, \ds c_n x_n \, \) \(\, \ds \le \, \) \(\ds c_n b_n\)

Or, in other words:

$\tuple {c_1 x_1, \dotsc, c_n x_n} \in \closedint {c_1 a_1} {c_1 b_1} \times \dotso \times \closedint {c_n a_n} {c_n b_n} \in C'$

Hence:

$C'$ is a finite covering of $M'$ by closed $n$-rectangles.

Therefore:

\(\ds \map {m^*} {M'}\) \(\le\) \(\ds \sum_{R \mathop \in C'} \map V R\) Definition of Outer Jordan Content
\(\ds \) \(=\) \(\ds \sum_{\sqbrk {\closedint {\mathbf a} {\mathbf b} } \mathop \in C} \prod_{i \mathop = 1}^n \paren {c_i b_i - c_i a_i}\) $\map V R$ as in Definition of Outer Jordan Content
\(\ds \) \(=\) \(\ds \sum_{\sqbrk {\closedint {\mathbf a} {\mathbf b} } \mathop \in C} \prod_{i \mathop = 1}^n c_i \paren {b_i - a_i}\)
\(\ds \) \(=\) \(\ds \sum_{\sqbrk {\closedint {\mathbf a} {\mathbf b} } \mathop \in C} c_1 c_2 \dotsm c_n \prod_{i \mathop = 1}^n \paren {b_i - a_i}\)
\(\ds \) \(=\) \(\ds c_1 c_2 \dotsm c_n \sum_{R \mathop \in C} \map V C\) $\map V R$ as in Definition of Outer Jordan Content
\(\ds \) \(\le\) \(\ds c_1 c_2 \dotsm c_n \paren {\map {m^*} M + \frac \epsilon {c_1 c_2 \dotsm c_n + 1} }\) By construction of $C$
\(\ds \) \(=\) \(\ds c_1 c_2 \dotsm c_n \map {m^*} M + \frac {c_1 c_2 \dotsm c_n} {c_1 c_2 \dotsm c_n + 1} \epsilon\)
\(\ds \) \(<\) \(\ds c_1 c_2 \dotsm c_n \map {m^*} M + \epsilon\)

As $\epsilon > 0$ was arbitrary, it follows from Real Plus Epsilon that:

$\map {m^*} {M'} \le c_1 c_2 \dotsm c_n \map {m^*} M$

$\Box$


Now, if every $c_i > 0$, we have that:

$M = \set {\tuple {\frac 1 {c_1} x_1, \dotsc, \frac 1 {c_n} x_n} : \tuple {x_1, \dotsc, x_n} \in M'}$

Therefore, the above argument can be applied with the roles of $M$ and $M'$ interchanged to prove that:

$\map {m^*} M \le \frac 1 {c_1 c_2 \dotsm c_n} \map {m^*} {M'}$

which is to say:

$c_1 c_2 \dotsm c_n \map {m^*} M \le \map {m^*} {M'}$


If the above does not hold, then $c_i = 0$ for some $i \in \set {1, 2, \dotsc, n}$.

Then:

$c_1 c_2 \dotsm c_n \map {m^*} M = 0 \le \map {m^*} {M'}$


In either case, we have:

$c_1 c_2 \dotsm c_n \map {m^*} M \le \map {m^*} {M'}$

The result follows from the two inequalities.

$\blacksquare$

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