# Does integrability on set implies integrability on (opportune) subset?

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!

• Look at the Lebesgue criteria for Riemann integrability 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
• I believe they are equivalent. – copper.hat Nov 18 '15 at 20:01