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Let $f:A\subset\mathbb{R}^n\to\mathbb{R}$ be Riemann integrable on $A$. If $B\subset A$, $f$ is not in general Riemann integrable on $B$.

Nevertheless, it would appear to me quite intuitive that, if we assume opportune restrictive conditions for $B$, integrability on $A$ would imply integrability on $B$.

As i said in the linked post, I think a hint may come from a particular case that I find here where it is said that, at least in the particular case where $A=[a,b]\subset\mathbb{R}$ and $B=[a',b']\subset [a,b]$, since $\int_A f(x_1,\ldots,x_n)dx_1\ldots dx_n$ and $\int_A \chi_B(x_1,\ldots,x_n)dx_1\ldots dx_n$ exist, then $\int_A \chi_B(x_1,\ldots,x_n)f(x_1,\ldots,x_n)dx_1\ldots dx_n$ also does and, by definition, is identical to $\int_B f(x_1,\ldots,x_n)dx_1\ldots dx_n$. Nevertheless, I am not how to use these considerations for the case where $B\subset A\subset\mathbb{R}^n$ with $n\ge 2$, once assumed opportune constraints on $B$. As a side note, I stress that I would not be able to understand a proof based on Lebesgue integration because I have no knowledge of how Riemann and Lebesgue integrals are related if $n>1$. I $\infty$-ly thank any answerer!

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Self-teaching worker
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    Look at the Lebesgue criteria for Riemann integrability https://en.wikipedia.org/wiki/Riemann_integral#Integrability. This does not use the Lebesgue integral, but does use the concept of measure zero. – copper.hat Nov 18 '15 at 18:25
  • @copper.hat Therefore the Lebesgue condition is valid for the case of $A\subset\mathbb{R}^n$, $n>1$, have I correctly understood? I fear that the proof of this fact would be too long to be an answer here...

    I only know the definition of Riemann integrability of $f$ on $A$ when $f$ is such that the limit $$\lim_{\delta\to 0}\sum_i\bar{f}(\xi_{1,i},\ldots,\xi_{n,i})\Delta V_i$$ where $\xi_{k,i}$, $k=1,\ldots, n$ are arbitrarily chosen points...

     – Self-teaching worker Nov 18 '15 at 19:47
  • ...in the $i$-th $n$-parallelepiped, whose volume is $\Delta V_i$, of the partition, whose mesh size is $\delta$, of a larger $n$-parallelepiped $P$ such that $D\subset P$, exists finite, independent from the partition and choice of $(\xi_{1,i},\ldots,\xi_{n,i})$; here $\bar{f}:P\to\mathbb{R}$ is the extension of $f$ such that $\forall\boldsymbol{x}\in D\quad\bar{f}(\boldsymbol{x})=f(\boldsymbol{x})$ and $\forall\boldsymbol{x}\in P\setminus D\quad\bar{f}(\boldsymbol{x})=0$. – Self-teaching worker Nov 18 '15 at 19:47
  • Is it equivalent to a Darboux sum based definition like in the $n=1$ case (in that case I will try to find a proof of the equivalence)? I heartily thank you! – Self-teaching worker Nov 18 '15 at 19:48
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    I believe they are equivalent. – copper.hat Nov 18 '15 at 20:01
  • @copper.hat I have stopped my studies and spent 3 days (while also working for my job) trying to understand it (and to prove the equivalence of the integrals defined by the Darboux and the Riemann sums to myself), which implies descending several rabbit holes which were unknown to me, but this proof of the fact you quote is amazingly interesting. $\infty$ thanks! – Self-teaching worker Nov 20 '15 at 20:51

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