Exercise 1.1.13 (Jordan measure as limit of cardinality)

By definition of Jordan measurable set, for each $𝜖>0$ we can find an outer elementary cover $B\supset E$ and an inner elemetary cover $A\subset E$ such that $m(B)-m_J(E),m_J(E)-m(A)\le 𝜖$. By axiom of choice, construct an infinite sequence ${(B_n)}_{n\in \mathbb{N}}$ such that $m(B)-m_J(E)<\frac{1}{n}$ and similarly ${(A_n)}_{n\in \mathbb{N}}$ such that $m_J(E)-m(A)<\frac{1}{n}$. In Lemma 1.1.2 we have proven that the theorem assertion is true for elementary sets, i.e., $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:B_n=\frac{1}{N^d}\cdot \#\left(B_n\cap \frac{1}{N}{\mathbb{Z}}^d\right)$ and $A_n=\frac{1}{N^d}\cdot \#\left(A_n\cap \frac{1}{N}{\mathbb{Z}}^d\right)$. Notice that from

follows for all $N\in \mathbb{N}$:

by monotonicity of cardinality we thus obtain

or, in a more familiar form

Since this is true for any $N\in \mathbb{N}$, the inequality should be preserved for limits

In other words,

Since this is true for any $n\in \mathbb{N}$, we can again take limits

By the squeeze theorem (Analysis I, Theorem 6.4.14) we finally obtain

as desired.

We know the discretisation formula holds for all elementary sets. For simplicity of the presentation, let us denote $I_N(A)=\frac{1}{N^d}\#\left(A\cap \frac{1}{N}{\mathbb{Z}}^d\right)$.

Let $A$ be Jordan measurable, and let $𝜀>0.$ Let $E\subset A\subset F$ be elementary sets such that $m(F)<m(A)+𝜀$ and $m(E)>m(A)-𝜀$. Note that $E\cap \frac{1}{N}{\mathbb{Z}}^d\subset A\cap \frac{1}{N}{\mathbb{Z}}^d\subset F\cap \frac{1}{N}{\mathbb{Z}}^d$ so $I_N(E)\le I_N(A)\le I_N(F)$. Taking $N\to \infty$,

Taking $𝜀\to 0$ we get the formula.