= 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 $|\sin u|\leq|u|$ gives $|\sin(1/x^3)|\leq x^{-3}$. By comparison for the <Lebesgue integral>,
$$
\boxed{\int_0^\infty|\sin(1/x^3)|\,dx\leq1+\int_1^\infty x^{-3}\,dx=\frac32<\infty.}
$$
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.
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