= Solution
For a nonnegative <measurable function>, the <Lebesgue integral> is defined through nonnegative <simple functions>:
$$
\boxed{\int_{\mathbb R}f\,dx=\sup\left\{\int_{\mathbb R}s\,dx:s\text{ simple},\ 0\leq s\leq f\right\}.}
$$
The value may be infinite. A real measurable $g$ is <Lebesgue integrable> when $\int|g|dx<\infty$; then $\int g=\int g^+-\int g^-$, where $g^+=\max(g,0)$ and $g^-=\max(-g,0)$.
If $|g|\leq h$, every simple function below $|g|$ is also below $h$. Taking the defining suprema proves <monotonicity of the Lebesgue integral>:
$$
0\leq\int|g|\,dx\leq\int h\,dx<\infty.
$$
Hence \b[$g$ is integrable], and both its positive and negative parts have finite <integrals>.
The oscillatory function $k$ is measurable because it is continuous off the singleton $\{0\}$ and has a specified value there. The elementary bounds $|\sin u|\leq\min(1,|u|)$ give the integrable majorant
$$
|k(x)|\leq H(x)=\begin{cases}1,&|x|\leq1,\\|x|^{-3},&|x|>1.\end{cases}
$$
Indeed, $\int H=2+2\int_1^\infty x^{-3}dx=3$. The comparison just proved gives $\int|k|\leq3$, so \b[$k$ is Lebesgue integrable over the whole real line], despite its oscillations near zero. Its <odd> symmetry also gives $\int k=0$, now legitimately because absolute integrability has been established.
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