Open and closed unit balls are Jordan measurable It has earlier been proved that the image of a function $f:\mathbb{R}^2\to \mathbb{R}^3$ has zero content on $X:=[0,1]\times [0,1]$. Now I would like to prove that the open and closed unit balls in $\mathbb{R}^3$ are Jordan measurable. For a set to be Jordan measurable one has to ensure that the set is bounded and that its boundary has zero content.
My approach is to devise a function $f:X\to S^2\subset \mathbb{R}^3$, where $S^2$ is a unit sphere (either open or closed), and prove that it is Lipschitz. Then it will follow that the boundary of a unit ball in $\mathbb{R}^3$ has zero-content, and thus a unit ball, being a bounded set with a zero-content boundary, is Jordan measurable.

So let $f$ be defined as $f((x,y))=(x,y,\pm\sqrt{1-x^2-y^2})$, then $\|f(x,y)-f(s,t)\|\le\sqrt{(x-s)^2+(y-t)^2+1}\le \sqrt3$, and $\|(x,y)-(s,t)\|\le \sqrt2$. Let $c:=2$ be a constant, then it follows that $\|f(x,y)-f(s,t)\|\le c\|(x,y)-(s,t)\|$. Hence $f$ is Lipschitz on $X$. It follows that $S^2$ has zero content in $\mathbb{R}^3$. Since a unit ball is a bounded set, it follows that it is Jordan measurable.

Please let me know if my approach is correct. I'm a little bit concerned about my proof of the Lipschitz condition.
 A: There are some problems with this. Some more glaring than others.


*

*What do you mean $f((x,y)) = (x,y,\color{red}{\pm}\sqrt{1-x^2-y^2})$? This is not a well-defined function.

*Even if it were well-defined, this function is not defined on all of $X$. For instance at $x=y=1$, your function's $z$-coordinate is not real.

*I'm not sure you are taking the correct approach proving that $f$ is Lipschitz. You need to demonstrate that there exists a $C>0$ such that given $x,y\in X$, $|f(x)-f(y)|\le C|x-y|$, as you know. However, you are computing $\sup_{x,y}|f(x)-f(y)|$ and $\sup_{x,y}|x-y|$, and comparing them, which does not convince me that the function is Lipschitz.
You could do better by defining two separate functions $f_+$ and $f_-$ from $X\to \Bbb R^3$ as such:
$$
f_{\pm}(x,y) = \begin{cases} (x,y,\pm\sqrt{1-x^2-y^2}) & (x,y)\in B^2, \\ (x,y,0) & (x,y)\in X\setminus B^2,\end{cases}
$$
where $B^2\subset\Bbb R^2$ is the open unit ball. Showing these functions are Lipschitz on $B^2$ would suffice to show they are Lipschitz on $X$. For Lipschitz on $B^2$, note that $f_\pm$ are smooth on $B^2$ and that $B^2$ is convex. Then use the fact that the image of each of $f_\pm$ is measure $0$ by the theorem you cite and that the union of their images contains $S^2$ to conclude that $S^2$ has measure $0$.
