Outer Jordan Content of Dilation
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.
$\blacksquare$