Is the function locally integrable

Question

How can we show that the function $$u(x)= \begin{cases}
\sin(1/x) & \text{if } x \neq 0 \\
0 & \text{if } x = 0
\end{cases}$$
is locally integrable? I know that continuous functions are locally integrable. So, it is locally integrable on $\Bbb R-\{0\}$. So, for any closed interval $[a,b]$ with $a<0<b$, should I investigate the following limits?
$$\int_a^bu(x)=\lim_{\epsilon\to0^{-}}\int_a^\epsilon u(x)dx+\lim_{\epsilon\to0^{+}}\int_\epsilon^b u(x)dx.$$ Or is there more elegant way to show it?
Thanks!

Answer 1

Note that $u$ is continuous on $\mathbb{R}\setminus\{0\}$, thus $u$ is measurable. Also, $|u|\le 1$, so $u$ is locally integrable.