Let $\Omega$ be the set of $(x,y) \in \mathbb R^2$ such that $0<y(1+x^8)<1$. For which $p \in [1,+\infty)$ the function $f(x,y)=x^2+y^2$ is in $L^p(\Omega)$?
The solution I was provided states that clearly $0< y < \dfrac{1}{1+x^8}$ so $x^2<f(x,y)<x^2+1$ and says that this implies that $f \in L^p(\Omega)$ $\iff$ $x^2 \in L^p(\Omega)$ $\iff$ $p<\dfrac{7}{2}$. I can't understand why, shouldn't I evaluate $\int_{\Omega}|x^2+1|^p dx$ instead of $\int_{\Omega}|x^2|^p dx$?
If $f,g\in L^p(\Omega)$ then also $f+g\in L^p(\Omega)$. Now the function $g=1\in L^p(\Omega)$ if $\Omega$ is bounded. You can see that this is the case by considering the Cauchy distribution and noting that $x^2>x^8$ for $|x|<1$