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In my calculus course we extended the definition of riemann integrable to functions whose domain are jordan-measurable sets, but can we extend the definition if we just ask for the domain to be bounded? like this :

Let $f:D\subset \mathbb R^n\to \mathbb R$ be a bounded function over $D$ (which is a bounded set). $f$ is said to be riemann integrable over $D$ iff the function $f_D=\begin{cases} f(x), & \text{if x$\in$ D} \\[2ex] 0, & \text{if x$\notin$ D} \end{cases}$ is riemann integrable over some rectangle $R$ such that $D\subseteq R$. In that case we define the integral of $f$ over $D$ like $\int_{D}f=\int_Rf_D\,.$

Or are there an problems if we ask $D$ just to be bounded?

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Yes, you can define it like this. $D$ may not be Jordan measurable, but $f_D$ may be Riemann integrable. One such example would be $f(x)=0$ for all $x\in D$.

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  • $\begingroup$ I was checking on my notes and I have the following theorem: $D$ is jordan measurable iff $\chi_D$(the characteristic function of $D$) is integrable on some rectangle $R$ such that $D\subseteq R$; using this result for $f(x)=1$ if $x\in D$ if $f_D(x)$ is integrable on $R$ then that means that $D$ is jordan measurable $\endgroup$ – user128422 Apr 1 '15 at 21:44
  • $\begingroup$ Yes, but there might be an $f$ sucht that $f_D$ is integrable even if $D$ is not Jordán measurable. $\endgroup$ – Julián Aguirre Apr 1 '15 at 23:05

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