Let $f:A\subset\mathbb{R}^n\to\mathbb{R}$ be Riemann integrable on $A$. If $B\subset A$ does the Riemann integral$$\int_B f(x_1,\ldots,x_n)dx_1\ldots dx_n$$exist finite and, if it does, how can it be proved?

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 sure whether these considerations are true for $B\subset A\subset\mathbb{R}^n$ with $n\ge 2$ and, if they are, I would have to know how to prove that the product of two Riemann integrable functions is Riemann integrable. As a side note, 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 heartily thank any answerer!


Take $A=[0,1]$, $f(x)=1$ and $B=\mathbb{Q}\cap [0,1]$.

The Riemann integral on $A$ is equal to one while the Riemann integral on $B$ is undefined.

  • $\begingroup$ If we add opportune constraints on $B$, would the integrability orf $f$ on $B$ follow? I think that it is more correct to ask a separate question for that: math.stackexchange.com/questions/1535549/… $\endgroup$ – Self-teaching worker Nov 18 '15 at 17:52
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    $\begingroup$ @Self-teachingDavide check out some of the related questions. $\endgroup$ – JP McCarthy Nov 18 '15 at 18:05
  • $\begingroup$ 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 theorem is a very interesting related result. $\endgroup$ – Self-teaching worker Nov 20 '15 at 20:53

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