# Jordan Content of Unit Sphere in $\mathbb{R}^3$

I'm supposed to show that the jordan content is $0$. The definition for a set $S$ having jordan content zero I have to work with is :$\forall\epsilon>0$ there is a finite collection of generalized rectangles in $\mathbb{R}^n$ that covers $S$ the sum of these rectangles volumes being less that $\epsilon$.

So I have the set $S = \{(x,y,z) \in \mathbb{R}^3\mid x^2 + y^2 + z^2 =1\}$. I can't just build $1 \cdot 1 \cdot \frac{1}{k}$ boxes (with $k > \frac{1}{\epsilon}$) and call it a day because the sum obviously isn't going to be less than epsilon for all epsilon.

I think another approach would be to try and build a sequence of generalized rectangles such that the first sequence has two boxes each covering half of the sphere and then sub divide and keep only the boxes that intersect the sphere. But I'm having trouble formalizing this idea.

Can somebody point me in the right direction / give any advice?

Thanks in advance.

-

## 2 Answers

Your rectangles need to get small in all three dimensions. Roughly speaking, you will have cubes of size $\frac 1k$, volume $\frac 1{k^3}$ you will need of order $4 \pi k^2$ of them, so the total volume is of order $\frac {4 \pi}k$. You should be able to find a collection of cubes that cover the sphere. Even if it is larger than that, as long as it doesn't grow too fast with $\frac 1k$ you are set.

-

The real trouble of this problem is to avoid messy calculations, rather than the construction of the sequence of covers. It may be easier to work with a ball cover instead of a retangular cover. Since the ratio between the volume of a ball and its circum-cube is a constant ($\frac43\pi/2^3$). If the total volume of the ball cover shrinks to zero, so does the total volume of its circum-cubic cover.

So, in polar coordinates, let $\theta$ denotes the inclination and $\phi$ denotes the azimuth. We consider $(2n-1)\times4n$ balls with common radius $\frac{2\pi}n$ and centres $$(1,\theta,\phi)\in\{1\}\times\left\{\frac{j\pi}{2n}:j=-(n-1),-(n-2),\ldots,n-1\right\}\times\left\{\frac{k\pi}{2n}:k=0,1,\ldots,4n-1\right\}.$$

Clearly, every point on the unit sphere belongs to, in polar coordinates, some patch of the form $$\{1\}\times\left[\frac{(j-1)\pi}{2n}, \frac{(j+1)\pi}{2n}\right]\times\left[\frac{(k-1)\pi}{2n}, \frac{(k+1)\pi}{2n}\right].$$ Now the slightly messy part is to show that this patch lies inside the interior of the ball with centre $\left(1,\frac{j\pi}{2n}, \frac{k\pi}{2n}\right)$ and radius $\frac{2\pi}{n}$. This is a bit tedious because to calculate distance to the ball centre, you need to convert coordinates back to the Euclidean ones. Once this is done, the rest is easy because the total volume of the ball cover is $(2n-1)(4n)\frac43(\frac{2\pi}n)^3=O(\frac1n)$.

-