An open ball is a Jordan region. I am trying to prove that an open ball $B_r(a)$ in $R^n$ is a Jordan region.
This is a question from Wade's Introduction to Analysis, question 12.1.4 part (b).
To show this, I need to prove that there exists a rectangle R which contains $B_r(a)$ and the volume of the boundary of this ball is 0. 
I can see that I can always enclose an open ball by an n dimensional rectangle. But given a $r$, how do I precisely write down this rectangle? Also, I have no idea how to show that its volume is 0.
I also know that a projectable region is a Jordan region. Can I show that an open ball is projectable? Would it suffice as a proof? Any help is appreciated.
 A: To show that the volume of the boundary $S$ is zero, consider the region between two balls centered at $a$ given by  $$V_\epsilon = \{x: r - \epsilon \leqslant \|x-a\| \leqslant r + \epsilon\}$$  We have $S \subset V_\epsilon \subset R$ where $R$ is a rectangle enclosing the outer ball defining $V_\epsilon$.
Hence,
$$vol(S) = \int_R \chi_S \leqslant \int_R \chi_{V_\epsilon} = vol(V_\epsilon).$$
The volume of a ball of radius $r$ in $\mathbb{R}^n$ is $C_nr^n$ where $C_n$ is a constant . The volume of $V_\epsilon$ is then the difference between the volumes of two balls and is given by
$$vol(V_\epsilon) = C_n(r+ \epsilon)^n - C_n(r - \epsilon)^n = 2nC_n\epsilon + O(\epsilon^2).$$
Thus,
$$vol(S) \leqslant 2nC_n \epsilon + O(\epsilon^2).$$
Since $\epsilon > 0$ can be arbitrarily small it follows that $vol(S) = 0$.
Another approach is to show that the surface $S$ has measure $0$ by producing for any $\epsilon > 0$ a cover by rectangles with total volume less than $\epsilon.$  For general $n > 3$ this is quite complicated.  
See, for example, here.
