Well-Defined Jordan Content Equals Content

Theorem
Let $M$ be a bounded subspace of Euclidean space.

Let the Jordan content of $M$ be $\map m M$.

Then the content $\map V M = \map m M$.

Proof
Let $C$ be a finite covering of $M$.

By, $\map V C \ge \map V M$.

Therefore, $\map V M$ is a lower bound of all $\map V C$.

So by the definition of greatest lower bound:
 * $\map V M \le \map {m^*} M$

Let $D$ be a finite covering of $S \setminus M$.

By the same reasoning:
 * $\map V {S \setminus M} \le \map {m^*} {S \setminus M}$

But:

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

But by hypothesis:
 * $\map V S - \map {m^*} {S \setminus M} = \map {m^*} M = \map m M$

So it follows that:
 * $\map V M = \map m M$