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

I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.

Show that an open Euclidean ball $$B(x, r) := \{y \in \mathbb{R}^d : |y − x| < r\}$$ in $$\mathbb{R}^d$$ is Jordan measurable, with Jordan measure $$c_d r^d$$ for some constant $$c_d > 0$$ depending only on $$d$$.

Is there an elementary way to approach this problem?

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 $$\sum^M_{k=1}|R_k| Thus, $$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.

It's easy to incorrectly read this question and think we need to calculate an exact measure. Tao isn't asking for an exact answer, as you can infer by his remark after part (2), where he mentions the Gamma function and how he won't cover that. Actually, reading part (2) before starting part (1) makes the whole question a lot less mysterious.

Within part (1) of exercise 1.1.10, there are 3 sub-questions. Below I lay out an approach to these. I faced an issue with the last part which I mention below.

## Closed balls are Jordan measurable

You can prove this by splitting a ball into two pieces and using 1.1.7's result of measurability of the set "underneath" a function along with finite additivity to sum up the two measures.

Consider an arbitrary closed ball at the origin, $$\overline{B(0, r)} := \{y \in \mathbb{R}^d : |y| < r \}$$.
$$\overline{B(0, r)}$$ can be expressed as so: \begin{align} \overline{B(x, r)} &= \{y \in \mathbb{R}^d : y_1^2 + ... + y_d^2 \le r^2 \} \\ &= \{y \in \mathbb{R}^d : y_d^2 \le r^2 - (y_1^2 + ... + y_{d-1}^2) \} \\ &= \{y \in \mathbb{R}^d : 0 \le y_d \le (r^2 - y_1^2 - ... - y_{d-1}^2)^{\frac{1}{2}} \} &&\cup \{y \in \mathbb{R}^d : -(r^2 - y_1^2 - ... - y_{d-1}^2)^{\frac{1}{2}} \le y_d < 0) \} \\ &= \{(y', y_d) : y' \in \mathbb{R}^{d-1}; \, 0 \le y_d \le f(y') \} &&\cup \{(y', y_d) : y' \in \mathbb{R}^{d-1}; \, -f(y') \le y_d < 0 \} \\ &= \{(y', y_d) : y' \in M; \, 0 \le y_d \le f(y') \} &&\cup \{(y', y_d) : y' \in M; \, -f(y') \le y_d < 0 \} \\ \end{align} Where $$y'$$ is $$y$$ without the last dimension, $$M$$ is a box in $$\mathbb{R}^{d-1}$$ containing the projection of the ball, and $$f : \mathbb{R}^{d-1} \to \mathbb{R}$$ is defined as $$f(y') := (r^2 - y_1^2 - ... - y_{d-1}^2)^{\frac{1}{2}}$$. The first part of the union above is in the form $$\{(x,t) : x \in B; 0 \le t \le f(x) \}$$ that appeared in 1.1.7 (2). (The second part is almost in the same form—you need to derive a second version of 1.1.7 (2), or invert the function and translate).

$$\overline{B(0, r)}$$ is the union of two disjoint Jordan measurable sets, so it too is Jordan measurable. Furthermore, translation invariance of Jordan measurability means that any closed ball $$\overline{B(x, r)}$$ is Jordan measurable.

## Open balls are Jordan measurable

The set $$C = \overline{B(x, r)} \setminus B(x, r)$$ representing the 'outside cover' of the d-dimensional ball and is measurable with measure zero. This can be shown by a similar setup to the part above, then applying 1.1.7 (1).

Thus the open ball $$B(x, r) = \overline{B(x, r)} \setminus C$$ must also be measurable with \begin{align} m(B(x, r)) &= m(\overline{B(x, r)} \setminus C) \\ &= m(\overline{B(x, r)}) -m(C) \quad (\text{as } C \subset B(x, r)) \\ &= m(\overline{B(x, r)}) - 0 \\ &= m(\overline{B(x, r)}) \end{align}

So, open balls have the same measure as their closed counterpart.

## Measure is $$c_dr^d$$ for some $$c_d > 0$$

For a ball $$B(0,r)$$ in $$\mathbb{R}^d$$, consider the inscribed and circumscribed cubes. The circumscribed cube has side length $$l = 2r$$, so it has measure $$A = 2^dr^d$$. The inscribed cube has diagonal of length $$2r$$, so it has a side length $$l = \frac{2}{\sqrt{d}}$$ and then measure $$A = (\frac{2}{\sqrt{d}})^dr^d$$. Thus, we have the bounds: $$(\frac{2}{\sqrt{d}})^dr^d < m(B(0,r)) < 2^dr^d$$.

You can see that the true measure is some positive constant, and the constant has bounds depending only on the dimension. This answers part (2).

Issue: although the bounds depend only on $$d$$, I'm not sure how to prove that the exact measure also depends only on $$d$$.

HINT:

Cover the ball $$B(x, (1-\epsilon) r)$$ with finitely many cubes of diameter $$<\epsilon r$$. This Jordan cover $$C_1$$ will be contained in $$B(x,r)$$. Do a homothety of center $$x$$ and ratio $$\frac{1}{1-\epsilon}\$$. We get a Jordan cover $$C_2$$ of $$B(x,r)$$ contained in $$B(x,\frac{1}{1-\epsilon} r)$$, so contained in $$B(x,2r)$$ (assume $$\epsilon < \frac{1}{2})$$. Therefore, $$C_1\subset B(x,r) \subset C_2$$ and $$\mu(C_2)- \mu(C_1) = (1- (1-\epsilon)^n) \cdot \mu(C_2) \le \\ \le (1- (1-\epsilon)^n) \mu^{\star}( B(x,2r))\underset{\epsilon \to 0}{\rightarrow} 0$$

Note: remember how the Greeks approximated the circle with inscribed and circumscribed regular polygons.