Jordan Regions III
1. Properties of Jordan Regions
Theorem
If \(E_1,\ldots,E_N\) are Jordan regions, then \(E_1 \cup \cdots \cup E_N\) is a Jordan region and
\[{}\text{Vol}(E_1 \cup \cdots \cup E_N){}\]
\[{}\leq \text{Vol}(E_1) + \cdots + \text{Vol}(E_N). {}\]
Proof
Theorem
If \(E_1, E_2\) are Jordan regions, then \(E_1 \cap E_2\) is a Jordan region and \(E_1 \setminus E_2\) is a Jordan region. Moreover,
\[ \text{Vol}(E_1) = \text{Vol}(E_1 \cap E_2) + \text{Vol}(E_1 \setminus E_2). \]
Proof
2. Example: Bounded Convex Sets are Jordan Regions
Theorem
Every bounded convex \(E\) set in \({\mathbb R}^n\) is a Jordan region.
Proof
Meta (Idea of the Proof)
The rough idea is that every cube \(Q\) intersecting \(\partial E\) can be “seen” from the outside, meaning that there is a line from the cube's center along a coordinate direction out to one of the faces of the containing box \(R\) and that line doesn't pass through any other boxes which intersect \(\partial E\). This means you can count the boxes by counting the number of faces which can be “seen.” This is also really important because the number of faces which can be seen scales like \(2^{(n-1)\ell}\) when the side length is \(2^{-\ell}\), so the total volume of boxes which might be “seen” decays like a constant times \(2^{-\ell}\).
This is heuristic is not quite accurate as written—sometimes boxes \(Q\) that need to be counted are obscured and can't be “seen.” However, it is true that all such boxes are “close” to the surface. If you let each line of sight pierce the collection of boxes for a little while (if you allow it to pass through \(n+1\) boxes in dimension \(n\) before you stop counting), then you are guaranteed to reach all boxes that must be counted.
The idea is that if a box \(Q\) of interest is sufficiently far in all directions from boxes which may be “seen,” then the obscuring boxes contain points whose convex hull encloses a neighborhood of \(Q\), and that means that \(Q\) didn't intersect the boundary of \(E\) in the first place.
Let \({\mathcal P}\) be a partition of a cube \(R\) containing \(E\) into cubes of side length \(2^{-\ell}\). Let \(Q\) be a cube in \({\mathcal P}\) which intersects \(\partial E\). Let \(e_1,\ldots,e_n\) be the standard unit coordinate vectors. It must be the case that there is an index \(j \in \{1,\ldots,n\}\) and a choice of \(\pm\) such that the cube \(\pm k 2^{-\ell} e_j + Q\) is entirely outside of \(E\) when \(k\) is an integer greater than \(n\). If not, the convex hull of the \(2n\) points in each of these boxes would contain a neighborhood of \(Q\) and consequently \(Q\) would not intersect \(\partial E\). This fact is a consequence of the lemma below.
When the sign \(\pm\) and the index \(j\) are as above, we say that \(Q\) is above face \(+j\) or the face \(-j\) (depending on the sign of \(\pm\).) Along any line in the \(j\)-th coordinate direction, there can be no more than \(n+1\) cubes that are above the face \(+j\) and no more than \(n+1\) cubes that are above the face \(-j\), as otherwise one of the cubes would necessarily violate the defining condition of being above face \(\pm j\) for the other one. And since every cube intersecting \(\partial E\) is above one of these faces, there can be at most \(2n(n+1) (2^\ell/r)^{n-1}\), where \(r\) is the side length of \(R\). That means that the sums of their volumes is at most \(2n(n+1) r^{-n+1} 2^{-\ell}\), which tends to zero as \(\ell \rightarrow \infty\).
Lemma
Let \(e_1,\ldots,e_n\) be the standard unit coordinate vectors. If \(K\) is a convex set in \({\mathbb R}^n\) which contains vectors of the form \(\pm c^{\pm}_j e_j + \delta^{\pm}_j\) for each \(j\) and each choice of \(\pm\), where \(c^{\pm}_j \geq n\) and each \(\delta^{\pm}_j\) is a vector all of whose coordinates have magnitude at most \(1/2\), then the set \(K\) contains \([-1/2,1/2]^n\).
Proof
It suffices to show that all vertices of \([-1/2,1/2]^n\) are contained in \(K\). The proof is by induction on dimension. The case \(n=1\) is trivial: any point in \([-3/2,-1/2]\) and any point in \([1/2,3/2]\) have convex hull containing all of \([-1/2,1/2]\). So suppose \(n \geq 2\). Consider the value of \(\theta\) for which
\[ v:= (1-\theta) (\pm c_j^{\pm} e_j + \delta_j^{\pm}) + \theta ( c^+_n e_n + \delta_n^+) \]
has \(+1/2\) in its \(n\)-th coordinate position, where \(j < n\). This \(\theta\) satisfies
\[ \theta = \frac{\frac{1}{2} - (\delta_j^{\pm})_n}{c^+_n + (\delta_n^+)_n - (\delta_j^{\pm})_n}\]
where \((\delta_j^{\pm})_n\) is the \(n\)-th coordinate of \(\delta_j^{\pm}\) and so on. Because \(c_n^+ \geq 2\) and \(|(\delta_n^+)_n|, |(\delta_j^{\pm})_n| \leq 1/2\), the numerator and denominator can never be negative and the denominator in particular is at least \(1\). As a function of \((\delta_j^{\pm})_n\), \(\theta\) is decreasing when \((\delta_j^{\pm})_n \leq c^+_n + (\delta_n^+)_n - (1/2)\), which includes the entire interval \([-1/2,1/2]\). Thus
\[ 0 \leq \theta \leq \frac{\frac{1}{2} + \frac{1}{2}}{c^+_n + (\delta_n^+)_n + \frac{1}{2}}. \]
Now as a function of \((\delta_n^+)_n\), the numerator is minimized when it equals \(-1/2\), so
\[ 0 \leq \theta \leq \frac{1}{c_n^+} \leq \frac{1}{n}. \]
So the vector \(v\) must equal
\[ \pm \tilde c_j^{\pm} e_j + \tilde \delta_j^{\pm}\]
for \(\tilde c_j^{\pm} \geq (n-1) c_j^{\pm}/n \geq (n-1)\) and \(\tilde \theta_j^{\pm}\) again having coordinate entries bounded by \(1/2\) in magnitude (because it's a convex combination of two such vectors). This means that \([-1/2,1/2]^{n-1} \times \{1/2\}\) is contained in \(K\), and by symmetry and convexity, all of \([-1/2,1/2]^n\) belongs to \(K\) as well.