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 $\le m(c)\times \epsilon$.

  • $\begingroup$ 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 $\endgroup$
    – BlueRedem1
    Feb 26 '20 at 1:27
  • 1
    $\begingroup$ @Eduardo Cuéllar: I added some details. It is worth drawing a picture with $A$ a segment divided into smaller segments. $\endgroup$
    – orangeskid
    Feb 26 '20 at 1:36

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