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Consider $f:\mathbb{R^2}\to\mathbb{R}$ defined by:

$$f(x,y):=\arctan\frac{1}{x-y}\quad\forall x\neq y$$ $$f(x,x):=0$$

Is the function Riemann integrable in the square $[0,1]\times[0,1]$?

I just have no clue how to attack this, surely basic, exercise.

F.Webber
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1 Answers1

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Well, this function has an uncountable number of discontinuities, so it should not be Riemann Integrable (assuming that same result holds for functions from $\mathbb R ^2$ as well as it does for $\mathbb R$

It is discontinuous on the entire diagonal of $[0,1]^2$, since as you approach the diagonal when $x<y$, you get the $\arctan (-\infty)=-\pi /2$, whereas when you approach it when $x>y$, you get $\arctan(\infty )=\pi /2 $, neither of which is 0., and for a function to be Riemann integrable you must have at most a countable number of discontinuties.

Alan
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  • I believe that, for a function to be Riemann integrable in a multidimensional space (in particular, in dimension 2), its discontinuities set must have measure zero. In this case, a line in $\mathbb{R^2}$ (like the diagonal $y=x$) does have measure zero, hence it shouldn't be a problem for the integrability. However, thanks for the answer, as I've thought that maybe by proving those are the only discontinuities we can prove the function is Riemann integrable. – F.Webber Dec 09 '14 at 05:46