Proving Lebesgue integrability of a bounded function

Question

Let $u:\mathbb{R}\to\mathbb{R}$ be a simple function such that \begin{equation} |u(x)|\le\frac{x^2}{1+|x|^3}\end{equation} for all $x\in\mathbb{R}$. Prove that $u$ is Lebesgue integrable.

My attempt: Note that $f(x):=\frac{x^2}{1+|x|^3}$ is measurable as the ratio of two measurable functions. Moreover, $|u(x)|\le f(x)$ and by the dominance principle $u(x)$ is integrable as it is simple and bounded above so its integral will be finite. This is sufficient for integrability. Is this correct?

Answer 1

Suppose $u=\sum\limits_{j=1}^{n}c_j\chi_{E_j}$ with $c _j \neq 0$ and $E_j$'s disjoint. Each $E_j$ is a bounded set: Otherwise, $0< |c_j|\leq \frac {x_i^{2}}{1+|x_i|^{3}} \to 0$ along a sequence $(x_i)$ in $E_j$ tending to $\pm \infty$, a contradiction. Hence, $u$ is integrable.

Answer 2

More generally, let $u$ be a measurable function such that $$\lim_{\pm\infty}u=0\quad\text{and}\quad m{\bf1}_E\le|u|\le M{\bf1}_E$$ for some measurable set $E$ and positive constants $m,M.$ Since $\lim_{\pm\infty}|u|=0<m,$ there exists a constant $A$ such that for all $x\in\Bbb R,$ $$|x|>A\implies|u(x)|<m.$$ For such an $A$, we have $E\subseteq[-A,A]$ and $$\int|u|\,d\lambda\le M\,\lambda(E)\le2AM<\infty$$ hence $u$ is integrable.