Finding the Boundary of a Jordan Region Q: Consider the set $D = \{(x, y) \in [0,2] \times [0,2]: y > x\}$, determine $\delta D$. Show that $D$ is a Jordan Region.
Theorem: A bounded set $D$ is a Jordan Region if and only if its boundary $\delta D$ is a set of volume $0$. 
I was wondering how we go about showing that $\delta D$ has volume $0$, I understand the premise for the theorem, but I am unsure of how to show that $\delta D$ has volume $0$. Any help would be much appreciated! 
 A: The boundary $\partial D$ is a union of three line segments:
$$\partial D = S_1 \cup S_2 \cup S_3 \\ =\{(x,y): x=0,\, 0\leqslant y \leqslant 2\}\cup\{(x,y): 0\leqslant x \leqslant 2,\, y= 2\} \cup \{(x,y): 0\leqslant x \leqslant 2,\, y= x\} $$
A set in $S \subset\mathbb{R}^2$ has volume (content) $0$ if for any $\epsilon > 0$ there exists a finite collection of rectangles $R_1, \ldots R_n$ such that $S \subset \bigcup_{j} R_j$ and $\sum_{j} vol(R_j) < \epsilon$.
In this case, the segments $S_1$ and $S_2$ each have length $2$ and for any $\epsilon > 0$, they are contained in rectangles $R_1$ and $R_2$, respectively, of width $2$ and height $ \epsilon/12$. The segment $S_3$ has length $2\sqrt{2}$ and is contained in a rectangle $R_3$  of width $2\sqrt{2}$ and height $ \epsilon/12\sqrt{2}$.
Thus, $\partial D \subset R_1 \cup R_2 \cup R_3$ and
$$vol(R_1) + vol(R_2) + vol(R_3) = 2 \cdot \frac{\epsilon}{12} + 2 \cdot \frac{\epsilon}{12} + 2\sqrt{2} \cdot \frac{\epsilon}{12\sqrt{2}} = \frac{\epsilon}{2} < \epsilon$$
