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:
- $\map V M = \map m M$
$\blacksquare$