# Two continuous functions over a closed, bounded and Jordan-Measurable set: Show that its graph Jordan-Measurable

Let f,g: $$A\subset\mathbb{R}^{n}\mapsto\mathbb{R}$$ continuous functions over the Jordan-measurable, bounded and closed set A, such that $$f(x)\leq g(x) \forall x\in A$$ Show that $$B=\{(x,y)\in\mathbb{R}^{n+1}|f(x)\leq y \leq g(x)\}$$ is Jordan-measurable in $$\mathbb{R}^{n+1}$$ I have problems with the proof. I don't know how to complete this but we know that if the function is continuous, then f is uniformly continuous, then for every $$\epsilon>0$$ there is a $$\delta>0$$ such that if $$|x-y|<\delta \implies |f(x)-f(y)<\epsilon|$$.Then maybe we can cover the graph with a finite number of rectangles but, to use this argument I must prove that the graph is continuous I guess... I will appreciate any help to end this proof because I'm a little lost.

It is enough to show that the Jordan measure of the boundary of this set is $$0$$. Now the boundary of the set is included in $$(\partial A\times [m,M]) \cup \Gamma_f\cup \Gamma_g$$.
Let's show that $$m(\Gamma_f)$$ ( the Jordan measure of the graph of $$f$$) is $$0$$. Indeed, $$A\subset C$$ where $$C$$ is a fixed cube. Divide $$C$$ into small cubes so that the variation of $$f$$ over any $$c\cap A$$ is smaller that $$\epsilon$$. We get that $$\Gamma_f\subset$$ a finite cover of $$n+1$$-dim cubes of total measure $$<\epsilon \cdot m(C)$$.
Added:Take $$\delta$$ so that $$|x-y|< \delta \implies |f(x)-f(y)|< \epsilon$$. Now divide $$D$$ into small cubes of diameter $$<\delta$$. For any $$x,y \in c\cap A$$ we have $$f(x)-f(y)< \epsilon$$. It follows that $$0\le \sup f(c\cap A)- \inf f(c\cap A)\le \epsilon$$. Therefore, $$f(c\cap A)$$ is contained in a closed interval $$I_c$$ of length $$\epsilon$$. So the piece of the graph of $$f$$ over $$c\cap A$$ is contained in $$c\times I_c$$ which has a measure $$\le m(c)\times \epsilon$$.
• Could you please explain the part of "... so that the variation of $f$ over any $c\bigcap A$ is smaller than \epsilon" Then that implies that the graph is contained in a finite cover? I don't understand very well that, especially the part of the variation of $f$, I will really appreciate it Feb 26 '20 at 1:27
• @Eduardo Cuéllar: I added some details. It is worth drawing a picture with $A$ a segment divided into smaller segments. Feb 26 '20 at 1:36