Past exam of the integration course of the University of Oxford / 2023 / 2 / a / i / Solution

The function is continuous on $(0,\infty )$, hence a measurable function. On $(0,1)$ its absolute value is at most one, while on $[1,\infty )$ the elementary estimate $|\sin u|\le |u|$ gives $|\sin (1/x^3)|\le x^{-3}$. By comparison for the Lebesgue integral,It is therefore a Lebesgue integrable function. The oscillations near zero do not obstruct absolute integrability because their amplitude stays bounded on an interval of finite measure.