Is $g(x) = sin(1/x) \forall x \in \mathbb{R} \setminus \{0\}$ and $g(0) = 0$ Lebesgue integrable?

Question

Is $g(x) = sin(1/x) \forall x \in \mathbb{R} \setminus \{0\}$ and $g(0)=0$ Lebesgue integrable?

Answer 1

We have $\lim_{x\to \pm \infty} \frac{g(x)}{\frac{1}{x}}= \lim_{t\to 0}\frac{\sin t}{t} = 1$, and $\frac{1}{x}$ is not integrable.

Answer 2

No. Since $\sin(y)\geq \frac{2}{\pi}|y|$ for all $y\in [0,\frac{\pi}{2}]$, we have $g(x)\geq \frac{2}{\pi}\frac{1}{x}$ for all $x\in I:=[\frac{2}{\pi}, \infty)$. As $x\mapsto\frac{1}{x}$ is not integrable on $I$, it follows that $g$ is also not integrable on $I$. In particular, $g$ is not integrable on $\mathbb{R}\supset I$.