I have the following function: $f:[-\pi,\pi]\rightarrow [-\pi,\pi]\ \ f(x,y) = \left\{\begin{matrix} \frac{xy}{x^4+y^2}, & (x,y) \neq (0,0)\\ 0,& (x,y) = (0,0) \end{matrix}\right.$
I need to decide if the function is or it is not Riemann Integrable.
I am trying to do this by proving that the function is continuous. If it is it means that is Riemann Integrable too. And it is not because I found two paths for which the approach is different. The two paths are: $x = 0$ and $x = y$. Because of this, the function is not continuous and therefore is not Riemann Integrable, right?
I am not sure if this is the correct answare so if anyone could point me in the right direction it would be great. Thank you!


1 Answer 1


Because of this, the function is not continuous and therefore is not Riemann Integrable, right?

Not right. A discontinuous function can still Riemann integrable. For example, consider the function $f:[-1, 1] \to \Bbb R$ defined as $$f(x) = \begin{cases}1 & x \neq 0\\0 & x = 0\end{cases}.$$ This function is discontinuous at $0$ but still Riemann integrable on $[-1, 1]$.

Even in your question, the function is only discontinuous at one point and so, that would not matter.

However, your function still is not Riemann integrable. This is because your function is not bounded. To see this, let $x = 1/n$ and $y = x^2$. Then, we have $$f(x, y) = \dfrac{x^3}{2x^4} = \dfrac{n}{2}.$$

  • $\begingroup$ I'm not sure how much you know about Riemann integration to conclude from the above. You may point out if any point needs more elaboration. $\endgroup$ Jun 5, 2020 at 9:30
  • $\begingroup$ Ok so $x=\frac{1}{n}$ is a sequence, right? I am sorry but I am not sure I fully understand that. $\endgroup$ Jun 5, 2020 at 10:05
  • $\begingroup$ Okay, let me elaborate. I wanted to show that $f$ isn't bounded. One way to do so is to show that $f$ can take arbitrarily large values. So, I've shown that $f(1/n, 1/n^2) = n/2$. Note that $(1/n, 1/n^2) \in [-\pi, \pi]^2$ and that $n/2$ can be made arbitrarily large; that is to say, given any $M\in\Bbb R$, there exists $n\in\Bbb N$ such that $n/2 > M$. This shows that $f$ can take arbitrarily large values in the domain. $\endgroup$ Jun 5, 2020 at 10:27
  • $\begingroup$ Thank you so much. I did a bit of reseasch about this property and now everything is clear. $\endgroup$ Jun 5, 2020 at 12:23
  • $\begingroup$ Great. One more thing I'd point out: Depending on your definition of Riemann integration, requiring $f$ to be bounded can either be an a priori requirement or something that is a consequence of your definition. If you've studied it via tagged partitions, then it is a result of the definition. However, if you've studied it via upper and lower sums, then the definition requires $f$ to be bounded, to begin with. $\endgroup$ Jun 5, 2020 at 12:26

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