Open balls in $\mathbb{R}^d$ are Jordan Measurable

I think it's a fairly involved calculation to prove this from scratch (using covers).

On the other hand, we have

$1).\ $ the content of a Jordan measurable set $S$ is $c(S)=\int 1_S$ (this is easy to prove),

$2).\ $ if $c(\partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),

$3).\ $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_n\pi{(n/2)}r^n,$

so it suffices to prove that $c(\partial B)=0$ because then we have that $B$ is Jordan measurable and

$c(B)=\int 1_B=C_n\pi{(n/2)}r^n.$

Since $\partial B=\partial \overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0,\ r=1$.

Now, the graph of the continuous function $f$, from the $n-1$-ball: $ x\mapsto \sqrt{1-\|x\|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.

So, to conclude the proof, we only need show that the graph of $f,\ $ Gr$(f)$, has Jordan content zero:

Let $\epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $\delta>0$ such that $\|x-y\|<\delta\Rightarrow \|f(x)-f(y)\|<\epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1\le k\le M$, choosing $M$ large enough so that $x,y\in Q_k\Rightarrow \|f(x)-f(y)\|<\epsilon.$ Choose $x_k\in Q_k$ for each $1\le k\le M.$ Finally, define $R_k=\{(x,y):x\in Q_k;\ |y-f(x_k)|<\epsilon\}.$ Then, by construction, Gr$(f)$ is contained in $\bigcup_k R_k$ and $\sum^M_{k=1}|R_k|<M|Q_k|(2\epsilon)=2\epsilon.$ Thus, $c^*($Gr$f)=0$.

A symmetry argument or the above analysis applied to the map $ x\mapsto -\sqrt{1-\|x\|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.

The result follows.