Examine the convergence of the improper integral $\int_1^{\infty} f(x) dx$

Examine the convergence of the improper integral $\int_1^{\infty} f(x) dx$ where
$f(x) = \frac{1}{x^3}$ if $x$ be rational $\geq 1$ and
$f(x) = \frac{-1}{x^3}$ if $x$ be irrational $>1$.

The solution given is straight forward $\int_{1}^{\infty}|f(x)|dx= \int_{1}^{\infty}\frac{1}{x^3}dx$ so its convergent, since is $f(x)$ absolutely convergent so is $\int_1^{\infty} f(x) dx$.

But I don't understand why the set of discontinuity points of $f(x)$ is infinite and has an infinite number of limit points. If so, it's not even Riemann integrable in the first place. How, then, can we comment on its convergence?