Jordan measure of a ball

Question

Currently I am trying to prove the following sentence:

Let $B(0,r)$ be an open ball. Show that $B(0,r)$ is Jordan measurable, with the Jordan measure $$m_{J} (B) = c_d \cdot r^d$$ for some constant $c(d)$ depending only on the dimension $d$, such that $$\left(\frac{2}{\sqrt{d}}\right)^d \leq c_d \leq 2^d$$

For me a key observation was that the ball $B^{(d)}(0,r)$ is the same as the area under/above the graph of $$\pm f:B^{(d-1)}(0,r)\to\mathbb{R}, \quad x\mapsto \pm\sqrt{r^2-\|x\|^2}$$ I wrote a relatively lengthy proof that by induction shows that the ball/area under this graph is Jordan measurable (by building an outer cover and the inner cover that do not differ in measure by more than $\epsilon$).

However, when it comes to establishing $c_d$ and its quantitative bounds, I am completely stuck. All of the sources I found in internet (such as this) use integrals to establish some kind of quantitative bounds, but I am not allowed to use them at all. Please help!

Also, see https://math.stackexchange.com/q/3058670/27978 – copper.hat Mar 18 '20 at 21:23 — copper.hat, Mar 18 '20 at 21:23
I suspect you are missing a $\ ^d$ on the lower bound. – copper.hat Mar 19 '20 at 04:05 — copper.hat, Mar 19 '20 at 04:05

score 2 · Accepted Answer · edited May 07 '20 at 08:43

Note that $B(0,1) \subset (-1,1)^d= B_\infty(0,1)$ and $m (-1,1)^d = 2^d$.

Note that if $x \in (-L,L)^d$ then $\|x\| \le \sqrt{d}L$ and so $(-{1 \over \sqrt{d}}, {1 \over \sqrt{d}})^d \subset B(0,1)$ and $m (-{1 \over \sqrt{d}}, {1 \over \sqrt{d}})^d = ({ 2 \over \sqrt{d}})^d$.

Hence $({ 2 \over \sqrt{d}})^d \le c_d \le 2^d$.

Note:

If a set $A$ is Jordan measurable, then for $r>0$ the set $rA = \{ra | a \in A\}$ is also Jordan measurable. If $R$ is a rectangle, it is easy to see that $m (rR) = r^d \cdot m(R)$, so it follows that $m (rA) = r^d \cdot m(A)$.

In particular, $m(B(0,r)) = r^d \cdot m(B(0,1))$.

The measure of $B(0,1)$ only depends on the dimension $d$.

Addendum:

I struggled to come up with a short proof of Jordan measurability. My first attempt was circular. The following is reasonable:

It is sufficient to show that $S=\partial B(0,1)$ has content zero. Clearly it is sufficient to show that $S_+ = \{ x \in S | x_d \ge 0 \}$ has content zero.

Define $f:[-2,2]^{d-1} \to \mathbb{R}$ by $f(y) = \begin{cases} \sqrt{1-\|y\|^2}, & \|y\| \le 1 \\ 0, & \text{otherwise} \end{cases}$. Since $f$ is continuous, it is Riemann integrable. Note that $S_+ \subset \operatorname{graph} f$, where $\operatorname{graph} f = \{ (y,f(y)) | y \in [-2,2]^{d-1} \}$.

Since $f$ is Riemann integrable, for any $\epsilon>0$ there is some partition $P$ such that $U(f,P)-L(f,P) < \epsilon$. Note that $\operatorname{graph} f \subset \cup_{R \in P} R \times [\inf_R f, \sup_R f]$ and $m(\cup_{R \in P} R \times [\inf_R f, \sup_R f]) = U(f,P)-L(f,P) < \epsilon$.

Hence $S_+$ and consequently $S$ has Jordan content zero.

Thank you very much! But how do I show that the volume of a ball ONLY depends on $r$ and $d$ (the first part of the exercise)? Because I do not see how this follows by squeezing. — qarabala, Mar 19 '20 at 07:58
@CebişMellim: I added a note, I'm not sure if that addresses your comment? — copper.hat, Mar 19 '20 at 16:51

Jordan measure of a ball

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