= Solution
The <function> is continuous for $x>0$, hence <Lebesgue measurable>. Near zero, the bound $|\sin x|\leq x$ gives
$$
\int_0^1\left|\frac{\sin x}{x^q}\right|\,dx
\leq\int_0^1x^{1-q}\,dx=\frac{1}{2-q}<\infty\qquad(0<q<2).
$$
Thus \b[$f$ is <Lebesgue integrable> on $(0,1]$ for $q<2$]. The convergence is an instance of <power singularity integrability>.
For the tail, take the pairwise <disjoint> <intervals>
$$
I_k=\left[k\pi+\frac\pi6,\ k\pi+\frac{5\pi}6\right],\qquad k\geq1.
$$
On each $I_k$, $|\sin x|\geq1/2$, $x\leq(k+1)\pi$, and the interval length is $2\pi/3$. Therefore
$$
\int_{I_k}\frac{|\sin x|}{x^q}\,dx
\geq\frac{\pi}{3}\bigl((k+1)\pi\bigr)^{-q}
=\frac{\pi^{1-q}}{3(k+1)^q}.
$$
Summing the nonnegative <integrals> over these <disjoint> <intervals> yields
$$
\int_1^\infty\frac{|\sin x|}{x^q}\,dx
\geq\frac{\pi^{1-q}}3\sum_{k=1}^\infty\frac1{(k+1)^q}=\infty
\qquad(0<q\leq1),
$$
by divergence of the <p-series>. Hence \b[$f$ is not <Lebesgue integrable> on $[1,\infty)$ for $0<q\leq1$].
The test is <absolute integrability>: oscillatory sign cancellation does not make the absolute-value <integral> finite. A conditionally convergent <improper integral> is a different notion. The <sine power kernel integrability> criterion separates these endpoint thresholds.
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