A question on Riemann integration

Question

Consider the function $f(x)=x$ when $x$ is rational and $f(x)=-x$ when $x$ is irrational. I want to show that the function is not Riemann integrable on any interval $[a,b]$. Clearly the function is only continuous at $x=0$. So if zero is not an element of $[a,b]$ the function is discontinuous every where. So the function has infinitely many discontinuities. Is telling this sufficient to prove that this is not Riemann integrable?

Answer 1

It is a theorem that a function is Riemann integrable if and only if its points of discontinuity form a measure zero set; in particular, this means that the function is not Riemann integrable if it is discontinuous everywhere. (Infinitely many discontinuities is not enough; the function $f(\frac{m}{n}) = \frac{1}{n}$ when $\frac{m}{n}$ is rational and in lowest terms, and $f(x) = 0$ otherwise, is discontinuous on all rational numbers and is Riemann integrable.)

However, this theorem may be too advanced for your context; my guess is that you want to show that on some subinterval, the upper sums for any partition are bounded below by a positive number and the lower sums bounded above by a negative number. Use the fact that in any interval, there are both rational and irrational numbers.