Outer Jordan Content Never Smaller than Inner Jordan Content
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Theorem
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Let $M$ be a bounded subspace of Euclidean space.
Let $S$ be an orthotope enclosing $M$.
Let $\map {m^*} A$ denote the outer Jordan content of $A$.
Let $\map V A$ denote the content of $A$.
Then:
- $\map {m^*} M \ge \map V S - \map {m^*} {S \setminus M}$
Proof
Let $P$ be a finite covering of $M$, and $Q$ a finite covering of $S \setminus M$.
If $Q \supsetneq S$, we can find another finite covering $Q' = \set {Q_i \cap S : Q_i \in Q} \subseteq S$, where $\map V {Q'} \le \map V Q$.
Without loss of generality, assume that $Q \subseteq S$.
Then:
- $S \setminus Q \subseteq M \subseteq P$
It follows that:
- $\map V {S \setminus Q} \le \map V P$.
But because $Q \subseteq S$:
- $\map V {S \setminus Q} = \map V S - \map V Q$
Therefore:
- $\map V S - \map V Q \le \map V P$
Thus, by the definition of Outer Jordan Content:
- $\map V S - \map {m^*} {S \setminus M} \le \map {m^*} M$
$\blacksquare$