Integrability of power of mod function

Question

Let $f:\Omega\to\mathbb{R}$ be defined by $f(x)=|x|^\alpha$ for some $\alpha\in\mathbb{R}$, where $\Omega$ is a bounded open subset of $\mathbb{R}^N$. Then if $\alpha<N$, one has $|x|^\alpha\in L^1(\Omega)$.

Just to be alert, $\Omega$ may or may not contain origin. First thing I want to ask whether the above results is correct or not? If so, how to prove this result.

Your help is very much appreciated.

Thanks.

Answer 1

Abridged solution.

If $\Omega$ is bounded and $\inf\limits_{x \in \Omega} \|x\| > 0$ then $f$ will be integrable whatever $\alpha$ may be. The same is true if $\alpha > 0$ as the function will be bounded.

On the other hand, if $\inf\limits_{x \in \Omega} \|x\| = 0,$ and $\alpha < 0$ then you can use "polar coordinates" to get $\int\limits_{B(0; r)} \|x\|^\alpha dx = \int_0^r r^\alpha r^N \mathrm{V}_N dr,$ where $\mathrm{V}_N$ is the volume of the $N$-dimension unit ball (a finite constant). This says that when $\alpha + N > -1$ the function is integrable on $\Omega.$ Q.E.D.