Outer Jordan Content Never Smaller than Inner Jordan Content

Theorem
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 $\map V {Q'} \le \map V Q$.

, 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:
 * $\map V S - \map {m^*} {S \setminus M} \le \map {m^*} M$