Exercise 1.1.15 (Uniqueness of Jordan measure)

Note that from these three properties we immediately deduce

(i)

[5.] monotonicity $\forall <!--\; FUNCTION\; APPLICATION\; -->E,F\in J({ℝ}^d):E\subset F⟹m^{\prime }(E)\le m^{\prime }(F)$

(ii)

[6.] finite subadditivity $\forall <!--\; FUNCTION\; APPLICATION\; -->E,F\in J({ℝ}^d):m^{\prime }(E\cup F)\le m^{\prime }(E)+m^{\prime }(F)$

In Exercise 1.1.3 we have have demonstrated that $m^{\prime }$ and $m$ must agree on the subset $E\left({ℝ}^d\right)\subset J\left({ℝ}^d\right)$ up to a constant $c=m^{\prime }\left({\left[0,1\right)}^d\right)$. Thus, we try to use elementary sets to estimate the alternative measure of $E$. Obviously, we will need the monotonicity property.

Pick an arbitrary $𝜖>0$, and take an elementary outer cover $\overline{E}$ and an elementary inner cover $\underline{E}$ of $E$ such that $m(\overline{E})$ and $m(\underline{E})$ do not differ from the Jordan measure $m(E)$ by more than $𝜖∕c$. We have, by the monotonicity of $m^{\prime }$:

Subtracting $\mathit{cm}(E)$ from both sides we obtain

Since $m(\overline{E})-m(E),m(E)-m(\underline{E})<𝜖∕c$ we obtain

In other words, we have shown that $|m^{\prime }(E)-\mathit{cm}(E)|\le 𝜖$ for any arbitrary $𝜖>0$. Thus, they must be equal, as desired.