Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / D / 2 / b

The function is measurable and $0\le {\cos }^2(1/x)\le 1$. Since $(0,1)$ has finite Lebesgue measure, comparison gives ${\int }_0^{1}g\le 1$. Therefore $g$ is integrable on $(0,1)$, despite oscillating infinitely often near zero.

For $x>1$, $1/x\in (0,1)$ and ${\cos }^2(1/x)\ge {\cos }^2(1)>0$. Its integral over the infinite interval is bounded below by the integral of a positive constant. Therefore $g$ is not integrable on $(1,\infty )$. In fact it tends to one at infinity.