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 $&gt;1$.

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 what I don't understand is, the set of points of discontinuity of  $f(x)$ is infinite and has infinite number of limit points. If so its not even riemann integrable in the first place. How then can we comment on its convergence ?
We aren't integrating f(x), so we don't have to worry about f(x) being continuous. Rather, we are integrating g(x) = (1/x)^3, which is continuous. 

Since |f(x)| = g(x), we can say that
Since we know g(x) > f(x) and -g(x) &lt; f(x).
If g(x) approaches C, then f(x) is between C and -C. Therefore, it approaches a finite value. 

Basically, we avoid having to worry about integrating f(x) by integrating a different function, g(x), instead and comparing the two.

Edit because my answer was unsatisfactory: Basically squeeze theorem is what proves the correctness of this method. Because squeeze theorem does not require a function to be continuous, this method works.
We aren't integrating f(x), so we don't have to worry about f(x) being continuous. Rather, we are integrating g(x) = (1/x)^3, which is continuous. 

Since |f(x)| = g(x), we can say that
Since we know g(x) > f(x) and -g(x) &lt; f(x).
If g(x) approaches C, then f(x) is between C and -C. Therefore, it approaches a finite value. 

Basically, we avoid having to worry about integrating f(x) by integrating a different function, g(x), instead and comparing the two.

Edit because my answer was unsatisfactory: Basically squeeze theorem is what proves the correctness of this method. Because squeeze theorem does not require a function to be continuous, this method works.