= Solution
Both functions are continuous on the positive domains in question, so they are measurable. For the first function, the decreasing substitution $t=1/x$ and $|dx/dt|=t^{-2}$ give
$$
\int_0^1\left|\frac{\sin(1/x)}x\right|\,dx
=\int_1^\infty\frac{|\sin t|}{t}\,dt.
$$
For every integer $k\geq1$,
$$
\int_{k\pi}^{(k+1)\pi}\frac{|\sin t|}{t}\,dt
\geq\frac1{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}|\sin t|\,dt
=\frac{2}{(k+1)\pi}.
$$
Summing this lower bound gives a divergent <harmonic series>. Thus \b[$f$ is not <Lebesgue integrable> on $(0,1)$].
For the second function, $|\sin x|\leq x$ on $(0,1)$ and $|\sin x|\leq1$ everywhere give
$$
\int_0^\infty\frac{|\sin x|}{x^{5/4}}\,dx
\leq\int_0^1x^{-1/4}\,dx+\int_1^\infty x^{-5/4}\,dx
=\frac43+4<\infty.
$$
Therefore \b[$g$ is <Lebesgue integrable> on $(0,\infty)$]. The <sine power kernel integrability> threshold is $1<p<2$: one endpoint requires $p<2$, and the other requires $p>1$. The signed improper integral for $f$ does converge by the <Dirichlet test for improper integrals>, but conditional improper convergence does not give a finite integral of $|f|$.
\Image[/past-exam-of-the-integration-course-of-the-university-of-oxford/2017-sine-power-integrability.png]
{title=The integrability threshold for a sine power kernel and the divergent absolute tail at $p=1$}
{description=The right panel uses $t=1/x$ for the first function. Its signed truncated integrals settle to a limit while the absolute integrals grow without bound, above a harmonic lower bound.}
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