Theorem: Outer Jordan Content of Separated Union

Let $M \subseteq \R^n$ be a bounded subset of Euclidean $n$-space.
Fix $k \in \set {1, 2, \dots, n}$ and $\gamma \in \R$.
For $\vec x \in \R^n$, let $\vec x = \tuple {x_1, x_2, \dots, x_n}$
Let:
:$A = \set {\vec x \in M : x_k < \gamma}$
:$B = \set {\vec x \in M : x_k > \gamma}$
Then:
:$\map {m^*} M = \map {m^*} A + \map {m^*} B$
where $m^*$ denotes the outer Jordan content.

Proof:
Let $C = \set {\vec x \in M : x_k = \gamma}$.
By Trichotomy Law for Real Numbers:
:$M = A \cup B \cup C$
and so by Outer Jordan Content is Subadditive:
:$\map {m^*} M \le \map {m^*} A + \map {m^*} B + \map {m^*} C$
But by Outer Jordan Content of Degenerate Set:
:$\map {m^*} C = 0$
and so:
:$\map {m^*} M \le \map {m^*} A + \map {m^*} B$
{{qed|lemma}}
Now, fix $\varepsilon > 0$.
By Characterizing Property of Infimum of Subset of Real Numbers, there is a finite covering $\CC$ of $M$ by closed $n$-rectangles such that:
:$\ds \sum_{R \mathop \in \CC} \map V R < \map {m^*} M + \varepsilon$
Let:
:$\CC_A = \set {R \in \CC : b_k \le \gamma}$
:$\CC_B = \set {R \in \CC : a_k \ge \gamma}$
:$\CC_C = \set {R \in \CC : a_k < \gamma < b_k}$
where $\ds R = \prod_{i \mathop = 1}^n \closedint {a_i} {b_i}$
By Trichotomy Law for Real Numbers, $\CC_A, \CC_B, \CC_C$ form a partition of $\CC$, so:
:$\ds \sum_{R \mathop \in \CC} \map V R = \sum_{R_A \mathop \in \CC_A} \map V {R_A} + \sum_{R_B \mathop \in \CC_B} \map V {R_B} + \sum_{R_C \mathop \in \CC_C} \map V {R_C}$
For each $R \in \CC_C$, define:
:$R^A = R \cap \set {\vec x : x_k \le \gamma}$
:$R^B = R \cap \set {\vec x : x_k \ge \gamma}$
Then $R^A$ and $R^B$ are closed $n$-rectangles and:
:$\ds \map V {R^A} + \map V {R^B} = \paren {\gamma - a_k} \prod_{i \mathop \ne k} \paren {b_i - a_i} + \paren {b_k - \gamma} \prod_{i \mathop \ne k} \paren {b_i - a_i} = \map V {R}$
Define:
:$\CC^A = \set {R^A : R \in \CC_C}$
:$\CC^B = \set {R^B : R \in \CC_C}$
Then:
{{begin-eqn}}
{{eqn | l = \sum_{R_A \mathop \in \CC^A} \map V {R_A} + \sum_{R_B \mathop \in \CC^B} \map V {R_B}
      | o = \le
      | r = \sum_{R \mathop \in \CC_C} \map V {R^A} + \sum_{R \mathop \in \CC_C} \map V {R^B}
      | c = Inequality is necessary, since distinct $R_C$ may have the same $R_C^A$, etc.
}}
{{eqn | r = \sum_{R \mathop \in \CC_C} \paren {\map V {R^A} + \map V {R^B} }
}}
{{eqn | r = \sum_{R \mathop \in \CC_C} \map V {\CC_C}
}}
{{end-eqn}}
Combining the above:
{{begin-eqn}}
{{eqn | l = \sum_{R_A \mathop \in \CC_A \cup \CC^A} \map V {R_A} + \sum_{R_B \mathop \in \CC_B \cup \CC^B} \map V {R_B}
      | o = \le
      | r = \sum_{R_A \mathop \in \CC_A} \map V {R_A} + \sum_{R_B \mathop \in \CC_B} \map V {R_B} + \sum_{R_A \mathop \in \CC^A} \map V {R_A} + \sum_{R_B \mathop \in \CC^B} \map V {R_B}
      | c = Similar to above, since $R_A$ and $R^A$ may share elements, etc.
}}
{{eqn | o = \le
      | r = \sum_{R_A \mathop \in \CC_A} \map V {R_A} + \sum_{R_B \mathop \in \CC_B} \map V {R_B} + \sum_{R_C \mathop \in \CC_C} \map V {R_C}
}}
{{eqn | r = \sum_{R \mathop \in \CC} \map V R
}}
{{eqn | o = <
      | r = \map {m^*} M + \varepsilon
}}
{{end-eqn}}
If it can be shown that $\CC_A \cup \CC^A$ is a covering of $A$, and $\CC_B \cup \CC^B$ is a covering of $B$, then:
:$\ds \map {m^*} A \le \sum_{R_A \mathop \in \CC_A \cup \CC^A} \map V {R_A}$
:$\ds \map {m^*} B \le \sum_{R_B \mathop \in \CC_B \cup \CC^B} \map V {R_B}$
and so:
:$\map {m^*} A + \map {m^*} B < \map {m^*} M + \varepsilon$
would follow.
The result would then be proven by Real Plus Epsilon.
{{qed|lemma}}
Consider some $\vec x = \tuple {x_1, \dots, x_n} \in A$.
By definition of $A$, $x_k < \gamma$.
Since $\CC$ is a covering of $M$, there is some $\ds R = \prod_{i \mathop = 1}^n \closedint {a_i} {b_i} \in \CC$ such that:
:$a_i \le x_i \le b_i$
for each $i \in \set {1, \dots, n}$.
In particular, $a_k \le x_k < \gamma$, so either $R \in \CC_A$ or $R \in \CC_C$.
If $R \in \CC_A$, then also $R \in \CC_A \cup \CC^A$.
If instead $R \in \CC_C$, then:
:$\vec x \in R^A \in \CC^A$
Thus, in either case, $\vec x$ is an element of some rectangle $R_{\vec x} \in \CC_A \cup \CC^A$.
Since $\vec x \in A$ was arbitrary, $\CC_A \cup \CC^A$ is a covering of $A$.
The proof that $\CC_B \cup \CC^B$ is a covering of $B$ is similar.
The result then follows from the remarks above.
{{qed}}
Category:Outer Jordan Content
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