Jordan Regions II

1. Images of sets with Jordan content zero

Theorem (\(C^1\) Images of Sets of Jordan Content Zero are Jordan Content Zero)

If \(U \subset {\mathbb R}^n\) is open and \(\Phi : U \rightarrow {\mathbb R}^n\) is \(C^1\) on \(U\), then any bounded \(E \subset U\) such that \(\overline{E} \subset U\) which is Jordan content zero must have an image which is too, i.e., \(\Phi(E)\) is also Jordan content zero.

Proof

Meta (Main Ideas)

We know \(d(E, U^c) = \delta > 0\). Let \(K\) be the closure of the set of points distance at most \(\delta/2\) to the point \(E\). This will be a compact set.
We can cover \(E\) by nonoverlapping cubes of side length small enough so that their diameters are less than \(\delta/2\). The sum of the volumes of the cubes is without loss of generality less than \(\epsilon\) for any given \(\epsilon\).
There is a constant \(C\) such that \(||D_x \Phi|| \leq C\) for all points \(x \in K\).
This means that for any cube \(Q\) in our covering if \(d(Q)\) is its diameter and \(x_0\) is its center point, then \(\Phi(Q) \subset B_{C d(Q)/2}(\Phi(x_0))\).
This means that we can cover \(\Phi(Q)\) by a single cube of side length \(C d(Q)\). Its volume is then \(C^n (d(Q))^n = C^n |Q|\).

2. Jordan regions defined

Definition

A set \(E\) contained in a box \(R \subset {\mathbb R}^n\) is called a Jordan region when its boundary \(\partial E\) has Jordan content zero. We define the Jordan content of a Jordan region \(E\) to be

\[ \text{Vol}(E) := \inf_{\mathcal P} V(E, \mathcal P), \]

that is, the infimum over all partitions (of boxes \(R\) containing \(E\)) of the sum of volumes of the boxes in the partition which intersect \(E\).

Let us also define:

Outer/Upper Volume Sum
\[ \overline{V}(E,{\mathcal P}) = V(E,{\mathcal P}) := \sum_{\substack{R' \in {\mathcal P} \\ R' \cap E \neq \emptyset}} |R'| \]
Outer/Upper Volume
\[ \overline{V}(E) := \inf_{{\mathcal P}} \overline{V}(E,{\mathcal P}) \]
Inner/Lower Volume Sum
\[ \underline{V}(E,{\mathcal P}) := \sum_{\substack{R' \in {\mathcal P} \\ R' \subset E}} |R'| \]
Inner/Lower Volume
\[ \underline{V}(E) := \sup_{{\mathcal P}} \underline{V}(E,{\mathcal P}) \]

Important

It is easy to miss a subtle but technically nontrivial point, which is that having boundary \(\partial E\) with Jordan content zero is not literally the same as saying that the upper and lower volumes of \(E\) are equal. However, we can prove that these two criteria are in fact the same.

Proposition

A set \(E\) contained in the box \(R\) is a Jordan region if and only if its outer and inner volumes are equal, i.e.,

\[ \sup_{{\mathcal P}} \sum_{\substack{R' \in {\mathcal P}\\R' \subset E}} |R'| = \inf_{{\mathcal P}} \sum_{\substack{R' \in {\mathcal P}\\R' \cap E \neq \emptyset}} |R'| \]

where \({\mathcal P}\) ranges over all partitions of \(R\).

Proof

Every box which intersects \(E\) and is not contained in \(E\) must intersect \(\partial E\) because \(E \setminus \partial E\) and \(E^c \setminus \partial E\) are both open sets and boxes are connected. Therefore

\[ \sum_{\substack{R' \in {\mathcal P}\\R' \cap E \neq \emptyset}} |R'| - \sum_{\substack{R' \in {\mathcal P}\\ R' \subset E}} |R'| \leq V(\partial E,{\mathcal P}) \]

Letting \(V(\partial E,{\mathcal P})\) tend to zero shows that the asserted equality must hold when \(E\) is a Jordan region.

On the other hand, given any partition of \(R\), every point \(x \in \partial E\) must belong either to a box \(R' \in {\mathcal P}\) intersecting \(E\) and \(E^c\) or intersecting \(\partial R\). By hypothesis, there exists a partition \({\mathcal P}\) such that the sum of \(|R'|\) over \(R' \in {\mathcal P}\) such that \(R' \cap E \neq \emptyset\) and \(R' \cap E^c \neq \emptyset\) is less than \(\epsilon/4\). Further refining \({\mathcal P}\) if necessary (by taking grids which in each direction begin and end with extremely short intervals), it may be assumed that the sum of \(|R'|\) over those \(R' \in {\mathcal P}\) intersecting \(\partial R\) is also less than \(\epsilon/4\). Let \(R' \in {\mathcal P}\) be called bad when it is one of these two types. By the Union Touching Lemma, there is a partition \({\mathcal P}'\) of \(R\) such that the sum of volumes of boxes \(R'' \in {\mathcal P}'\) touching a bad box \(R' \in {\mathcal P}\) is small, i.e.,

\[ \sum_{\substack{R'' \in {\mathcal P}' \\ R'' \cap \bigcup_{R' \text{ bad}} R'} \neq \emptyset} |R''| \leq 2 \sum_{R' \text{ bad}} |R'| < \epsilon. \]

Every \(x \in \partial E\) must be contained in a bad box \(R' \in {\mathcal P}\), so consequently \(\partial E\) must be contained in the union of bad boxes \(R'\) and so any \(R''\) which contains a point \(x \in \partial E\) must also intersect a bad box \(R'\). Thus \(\overline{V}(\partial E,{\mathcal P}')\) is dominated by the sum over \(R''\) intersecting \(\bigcup_{R' \text{ bad}} R'\) and is thus smaller than \(\epsilon\).

To prove that a box \(E\) is a Jordan region, we must fix an \(\epsilon\) and then find a partition of some box \(R\) containing \(E\) such that the sum of the volumes of the boxes intersecting the boundary is at most \(\epsilon\).

One idea that works: subdivide \(E\) along each coordinate direction into \(N\) equal pieces, then give an approximate count of the number of boxes in the subdivision which intersect the boundary.

Theorem

Suppose \(E\) is some box in \({\mathbb R}^n\). Then \(E\) is a Jordan region and its volume is exactly \(|E|\).

Proof

Suppose \({\mathcal P}\) is a partition of some box \(R\) containing box \(E\). The partition \({\mathcal P}\) may be refined so that boxes \(R' \in \mathcal P\) which overlap with \(E\) are contained within \(E\) (because refining \({\mathcal P}\) decreases \(V(E,{\mathcal P})\)). In this case, the \(R'\) overlapping with \(E\) will themselves be a partition \(\mathcal P'\) of \(E\). Furthermore \(|R' \cap E| = |R'|\) (recall \(|\cdot|\) is product of side lengths). This means

\[{}\sum_{R' \in {\mathcal P}, \text{int}(R' \cap E) \neq \emptyset} |R' \cap E|{}\]

\[{}= \sum_{R' \in {\mathcal P}, \text{int}(R' \cap E) \neq \emptyset} |R'|.{}\]

The left-hand side is the sum of volumes of every box in \({\mathcal P}'\), so it is exactly \(|E|\). Thus \(V(E,{\mathcal P}) \geq |E|\) for any \({\mathcal P}\). We can also construct a partition \({\mathcal P}\) of \(E\) where there is equality (let \({\mathcal P}\) have \(E\) in it and no other boxes). Therefore \(\operatorname{Vol}(E) = |E|\).