I was studying the "Fundamental lemma of calculus of variations" and I have fully understood the proof to it except for one little step. The problem is:
$f$ is a contineous function in $\Omega \subset \mathbb{R}^n$ and we assume that for $x_0 \in \Omega$ $f$ holds $f(x_0)\neq0$. Since $f$ is contineous for a small $\delta>0$ there exists a neighborhood $V_\delta$ with $f(x)>0$ for all $x\in V_\delta(x_0)$. Than we define the function $\psi(x)$
$\psi(x) := \begin{cases} (\| x-x_0\|-\delta)^2 \cdot (\| x-x_0\|-\delta)^2 & \text{for } x\in V_\delta(x_0),\\ 0 & \, \text{otherwise}. \end{cases}$
The problem now is, that in the book I am learning from the author states that $\psi\in L^1(\Omega)$ without any proof or calculation. Why is it so that $\psi\in L^1(\Omega)$? Do I need to show $\int_\Omega\psi(x)\text{ d}x<\infty$ or is there any easier way to show that.
I would appreciate any further help :)