Well-Defined Jordan Content Equals Content

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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 Common Notion $5$, $\map V C \ge \map V M$.

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:

\(\ds \map V M + \map V {S \setminus M}\)

\(=\)

\(\ds \map V S\)

\(\ds \map V M + \map {m^*} {S \setminus M}\)

\(\ge\)

\(\ds \map V S\)

\(\ds \map V M\)

\(\ge\)

\(\ds \map V S - \map {m^*} {S \setminus M}\)

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:

$\map V M = \map m M$

$\blacksquare$