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I want to prove that $f$ is integrable over a jordan measurable set $A$ iff the set of discontinuities of $f$ has measure zero, where $f:A\subset \mathbb R^n\to \mathbb R$ is a bounded function using the lebesgue theorem:

Theorem: Let $f:R\subset \mathbb R^n\to \mathbb R$ where $R$ is a rectangle. $f$ is integrable on $R$ iff the set of discontinuituies of $f$ has measure zero

I also have the following definition:

$f$ is integrable over a jordan measurable set $A$ iff $f_A$ is integrable over some rectangle $R$ such that $A\subseteq R$

My attempt:

$\Leftarrow$)Let $D_{f,A}$ be the set of discontinuities of $f$ over $A$, $D_{f_A,R}$ the set of discontinuities of $f_A$ (where $f_A=f(x)$, if $x\in A$ and $f(x)=0$, if $x\notin A$) over some rectangle $R$ such that $A\subseteq R$; I know that $D_{f_A,R}\subseteq \partial A\cup D_{f,A}$ where $\partial A$ is the boundary of $A$.

Then I have that $\partial A$ is jordan measurable with jordan measure zero (becuase by hypothesis $A$ is jordan measurable) hence $\partial A$ has measure zero; also $D_{f,A}$ has measure zero, therefore $\partial A\cup D_{f,A}$ has measure zero and this implies that $D_{f_A,R}$ has measure zero;

By Lebesgue theorem I have that $f_A$ is integrable on R and by definition $f$ is integrable on $A$

$\Rightarrow$)If $f$ is integrable over a jordan measurable set $A$ by definition $f_A$ is integrable over some rectangle $R$ such that $A\subseteq R$ iff $D_{f_A,R}$ has measure zero(by Lebesgue theorem)

But I don´t know how to prove that $D_{f,A}$ has measure zero from here.

I would really appreciate if you cand lend me a hand

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  • $\begingroup$ Note that $\,D_{f_A,R}=D_{f,A} \cup \partial A$. $\endgroup$ – Tony Piccolo Apr 3 '15 at 20:56
  • $\begingroup$ @TonyPiccolo can you give me a hint to prove $D_{f,A}\cup \partial (A)\subseteq D_{f_A,R}$: I have that if $x\in D_{f,A}$ this implies that $x\in D_{f_A,R}$ but if $x\in \partial(A)$ how can I guarantee that it´s in $D_{f_A,R}$? $\endgroup$ – user128422 Apr 4 '15 at 3:35
  • $\begingroup$ Sorry, I was wrong. See here the theorem 4.1.42 on p. 250. What do you think of the proof ? $\endgroup$ – Tony Piccolo Apr 4 '15 at 17:00
  • $\begingroup$ I don´t think we can guarantee that $D_{f,A}$ has measure zero: we have that: $f$ is integrable on $A$ iff $f_{\chi_A}$ is integrable on some rectangle $R$ such that $A\subset R$ iff $D_{f_A,R}$ has measure zero, and we also know that $\partial(A)$ has measure zero (because it´s jordan measurable) but does that implies that $D_{f,A}$ has measure zero?. In the proof of the link they only did one part of the proof $\endgroup$ – user128422 Apr 4 '15 at 18:04
  • $\begingroup$ I agree. Unfortunately till now I didn't find another iff about the matter. $\endgroup$ – Tony Piccolo Apr 4 '15 at 18:52

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