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I'm aware of the result in dimension 1 that given $f \in R([a,b])$ ($R$ for Riemann integrable) then $f$ is continuous almost everywhere. This result is given passing through Lebesgue Measure, that can be given also in $\mathbb{R}^n$. So my question is: The result stated above is true when we pass in $\mathbb{R}^n$, that is: is a function $f: A \subset \mathbb{R}^n \rightarrow\mathbb{R} $ Riemann-integrable necessarly continuous a.e.?

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Just for $A={\bf{R}}^{n}$, if $f$ were improper Riemann integrable, then for any block $I_{(N_{1},...,N_{n})}=[N_{1},N_{1}+1]\times\cdots\times[N_{n},N_{n}+1]$, where $(N_{1},...,N_{n})\in{\bf{Z}}^{n}$, $f$ is Riemann integrable on that block, so it is continuous a.e. on that block. Since the countable union of all such blocks is ${\bf{R}}^{n}$, then $f$ is continuous a.e. on the whole space.

For any other $A$, it is hard to tell because Riemann integral or even improper Riemann integral is defined with certain conditions regarding of $A$.

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