Let $f:[1,\infty)\to\mathbb{R}$ be a function defined by $$f(x)=\left\{\begin{array}{rl} \frac{1}{x^{2}}, & x\in\mathbb{Q}\cap[1,\infty), \\ -\frac{1}{x^{2}}, & x\in\mathbb{Q}^{c}\cap[1,\infty). \end{array} \right.$$
Then, is it true that $f$ is improper integrable on $[1,\infty)$?
Is it enough to guarantee that $\int_{1}^{\infty}\vert{f}\vert<\infty$?.
Or could the above statement be true, depending on whether it is Riemann or Lebesgue sense?
Give some comments. Thank you!
It is not integrabe in Riemann sense because it is discontinuous everywhere. [Riemann integrable functions are continuous almost everywhere]. It is integrable in Lebesgue sense and the integral is $-1$. The function is equal to $-\frac 1 {x^{2}}$ almost everywhere and $\int_1^{\infty} -\frac 1{x^{2}}\, dx=-1$.