Motivation

I saw a minor difference the definition of Jordan content 0 sets between two books.

Some basic definitions

I make this list for my reference only. If I’m asked the true meaning of some geometric concepts, such as boundary, by someone seriously, then I can quickly refer to this list in order to remind myself what ideas should be introduced.

Unless otherwise specified, assume that $A \in \R^n,\vect{x}, \vect{x}_0,\vect{u},\vect{v} \in \R^n,\vect{u}_i \in \R^n\, \forall\,i \in \N$.

Norm

The norm of $\vect{x}$ is defined as $\norm{\vect{x}} := \sqrt{\sum\limits_{i = 1}^n x_i^2}$.

Distance

The distance between $\vect{u}$ and $\vect{v}$ is $\norm{\vect{u} - \vect{v}}$.

Open ball

$\mathcal{B}_r (\vect{x}_0) := \{\vect{x} \mid \norm{\vect{x} - \vect{x}_0} < r\}$

Interior point

$\vect{u}$ is an interior point in $A$ if $\,\exists\,r > 0\text{ s.t. } \mathcal{B}_r(\vect{u})\subseteq A$. ¹

Interior of a set

$\interior A := \{\vect{u} \mid \exists\,r > 0 \text{ s.t. } \mathcal{B}_r(\vect{u}) \subseteq A\}$

Open set

$A$ is open if $A = \interior A$.

Limit of $\{\vect{u}_k\}$

$\forall\,\varepsilon > 0,\exists\,K \in \N \text{ s.t. } \forall\,k \ge K, \norm{\vect{u}_k - \vect{u}} < \varepsilon$
We denote it as $\lim\limits_{k \to \infty} \vect{u}_k = \vect{u}$.

Closed set

$\vect{u}_k \in A\,\forall\,k \in \N, \lim\limits_{k \to \infty} \vect{u}_k = \vect{u} \implies \vect{u}\in A$

Boundary

$\partial A := \{\vect{x}\mid\forall\,r > 0,\exists\,\vect{u}\in A, \vect{v} \notin A \text{ s.t. } \vect{u},\vect{v} \in \mathcal{B}_r(\vect{x})\}$

Exterior point

$\vect{u}$ is an exterior point in $A$ if $\,\exists\,r > 0 \text{ s.t. } \mathcal{B}_r(\vect{u}) \subseteq \R^n \setminus A$.

Exterior of a set

$\exterior A := \{\vect{u} \mid \exists\,r > 0 \text{ s.t. } \mathcal{B}_r(\vect{u}) \subseteq \R^n \setminus A\}$

Some claims

If a subset $A$ of $\R^n$ is contained in a closed subset $B$ of $\R^n$, $\partial A \subseteq B$.

Proof. Let $\vect{u} \in \partial A. \forall\,k \in \N, \exists\,\vect{u}_k \in \mathcal{B}_{1/k}(\vect{u}) \cap A$. Then $\lim\limits_{k \to \infty} \vect{u}_k = \vect{u}$ because $\forall\,\varepsilon > 0,K > 1/\varepsilon, (k \ge K \implies \norm{\vect{u} - \vect{u}_k} < 1/k \le \varepsilon)$.
$\because \vect{u}_k \in A\,\forall\,k \in \N, {\displaystyle \lim_{k \to \infty} \vect{u}_k = \vect{u}}$, and $B$ is closed.
$\therefore\vect{u}\in B$.
Q.E.D.

$\R^n = \interior A \sqcup \partial A \sqcup \exterior A$

Proof. Let $\vect{x} \notin \interior A$. Then $\forall\,r > 0,\exists\,\vect{u} \in \mathcal{B}_r(\vect{x}) \setminus A$.

Case 1: $\forall\,r > 0,\mbox{there also exists}\, \vect{v} \in \mathcal{B}_r(\vect{x}) \cap A$.
Putting the two statements about $\vect{u}$ and $\vect{v}$ together, we get $\forall\,r > 0, \exists\,\vect{u} \notin A, \vect{v} \in A \text{ s.t. } \vect{u},\vect{v} \in \mathcal{B}_r(\vect{x})$.
$\therefore \vect{x} \in \partial A$

Case 2: $\neg\,(\forall\,r > 0,\exists\,\vect{v} \in \mathcal{B}_r(\vect{x}) \cap A)$
i.e. $\exists\,r > 0 \text{ s.t. } \nexists\,\vect{v} \in \mathcal{B}_r(\vect{x}) \cap A$.
$\therefore \exists\,r > 0 \text{ s.t. } \vect{v} \in A \uparrow \vect{v} \in \mathcal{B}_r(\vect{x})$²
i.e. $\exists\,r > 0 \text{ s.t. } \vect{v} \in \mathcal{B}_r(\vect{x}) \implies \vect{v} \notin A$
$\therefore \exists\,r > 0 \text{ s.t. } \mathcal{B}_r(\vect{x}) \subseteq \R^n \setminus A$
Hence $\vect{x}\in\exterior A$.
Q.E.D.

From the definition of the exterior of $A$, it’s clear that $\exterior A \subseteq \R^n \setminus A$. The reason is that $\vect{u} \in \mathcal{B}_r(\vect{u})$. If $\vect{u} \in \exterior A$, then $\exists\,r > 0 \text{ s.t. } \mathcal{B}_r(\vect{u}) \subseteq \R^n \setminus A$. Thus, $\vect{u} \in \mathcal{B}_r(\vect{u}) \subseteq \R^n \setminus A \implies \vect{u} \notin A$.

Another trivial fact is from the definition of the interior of $A$. $\interior A \subseteq A$ due to the same reason for $\exterior A \subseteq \R^n \setminus A$, that is, $\vect{u} \in \mathcal{B}_r(\vect{u})$. $\vect{u} \in \interior A \implies \exists\,r > 0 \text{ s.t. } \mathcal{B}_r(\vect{u}) \subseteq A$. Thus, $\vect{u} \in \mathcal{B}_r(\vect{u}) \subseteq A \implies \vect{u} \in A$.

More definitions

Unless otherwise specified, assume that $a_i,b_i \in \R\,\forall\,i \in \N$.

Closure

$\overline{A} := \interior A \cup \partial A$

Bounded set

$\exists\,M \ge 0 \text{ s.t. } \norm{\vect{x}} \ge M\, \forall\,\vect{x} \in A$

Generalized rectangle

$\vect{I} := \vect{I_1} \times \dots \times \vect{I_n}$, where $\vect{I_i} := [a_i,b_i]\,\forall\,i \in \N$.

Volume of generalized rectangle

$\volume \vect{I} := \prod\limits_{i = 1}^n (b_i - a_i)$

From the definition of the closure of a set, we can conclude that $A \subseteq \overline{A}$ since $\R^n = \interior A \sqcup \partial A \sqcup \exterior A$ and $A \cap \exterior A = \varnothing$. (proved) Thus, $A \subseteq \interior A \sqcup \partial A = \overline{A}$.

We can extend the first claim like this.

If a subset $A$ of $\R^n$ is contained in a closed subset $B$ of $\R^n$, $\overline{A} \subseteq B$.

Proof. $\overline{A} := \interior A \cup \partial A$. By the first claim, it’s already known that $\partial A \subseteq B$, so it remains to show that $\interior A \subseteq B$. Since $\interior A \subseteq A$ (proved) and it’s given that $A \subseteq B$, we’re done!
Q.E.D.

Comparing the definitions

That’s the definition of Jordan content 0 sets.

Suppose $D$ is a bounded subset of $\R^n$.
$\forall\,\varepsilon > 0,\exists$ generalized rectangle $\vect{I}_1, \dots,\vect{I}_m \subseteq \R^n$ s.t. $D \subseteq \bigcup\limits_{j = 1}^m \vect{I}_j$ and $\sum\limits_{j = 1}^m \volume \vect{I}_j < \varepsilon$.

I have read another version of the definition, and the only difference between those two versions is that $D$ is replaced by $\overline{D}$. It took me a short while to understand why this is OK.

First, $\vect{I}_j$ is closed for each $j$ since it’s the Cartesian product of closed intervals.
$\bigcup\limits_{j = 1}^m \vect{I}_j$ is also closed. (DeMorgan’s Law)
Hence, by the extended claim, one can also write $\overline{D} \subseteq \bigcup\limits_{j=1}^m \vect{I}_j$ instead of $D$ in the above definition.

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