Suppose that there are two continuous functions $\phi_1,\phi_2:A\subseteq\mathbb{R}^n\to\mathbb{R}$ and the set $$ D=\left\lbrace x\in A:\phi_1(x)\leq\phi_2(x)\right\rbrace $$ is compact and Jordan measurable, i.e. $\partial D$ is of measure zero. How can we prove that the set $$ S=\left\lbrace (x,y)\in\mathbb{R}^{n+1}: x\in D\mathrm{\ and\ } y\in[\phi_1(x),\phi_2(x)]\right\rbrace $$ is also compact and Jordan measurable?
For Jordan measurable, it is straightforward since $$ \partial S=\bigcup_{i=1,2}\left\lbrace (x,\phi_i(x))\in\mathbb{R}^{n+1}:x\in D\right\rbrace $$ and each of them is of measure zero obviously.
But for compactness, I only know that $\phi_1,\phi_2$ attain the maximum and minimum over the compact set $D$ and so there exist $M,m>0$ such that $$ S\subseteq D\times \left[m,M\right] $$ so it suffices to show that $S$ is closed since any closed subset of a compact set is compact. However, I have no idea how to construct a proof to show that $S$ is closed.
Note that $S=\left\{(x,y)\in D\times[m,M]\,\middle|\,y-\phi_1(x)\phi_2(x)-y\geqslant0\right\}$. So, if you define$$\begin{array}{rccc}\Psi\colon&D\times[m,M]&\longrightarrow&\mathbb R^2\\&(x,y)&\mapsto&\bigl(y-\phi_1(x),\phi_2(x)-y\bigr),\end{array}$$then $S=\Psi^{-1}\bigl([0,\infty)^2\bigr)$. Since $\Psi$ is continuous and $[0,\infty)^2$ is closed, this proves that $S$ is a closed set.