Exercise 11.21 in Real Analysis for... by Richard Bass says: Suppose $f: \mathbb{R}^n : \rightarrow \mathbb{R}$ is measurable, $c>0$, and $p<n$? If $|f(x)| \leq c|x|^{-p} \chi_{B(0,1)}(x)$ a.e, prove that $f$ is integrable.
First of all, am I right to assume that $|x|$ means $||x||$?
In what follows, I will choose $||\cdot|| = || \cdot ||_1$
To get some intuition, I tried the case where $n=2$ and $p=1$.
We have that:
$\int_{\mathbb{R^2}} |f(x)| \, dm(x) \leq c\int_{B(0,1)} ||x||_1 dm(x) = c\int_{-1}^{1} [ \int_{-1}^{1} \frac{1}{|x_1| + |x_2|}dm(x_1)]\, dm(x_2)$
and
$\int_{-1}^{1} \frac{1}{|x_1| + |x_2|}dm(x_1) \leq \int_{-1}^{1} \frac{1}{|x_1|}dm(x_1)=2 \int_{0}^{1} \frac{1}{x_1} dm(x_1)$
But $\int_{0}^{1} \frac{1}{x_1} dm(x_1)$ is not finite.
So, even in the easiest case I am kind of stuck.
Can someone help me out ?
Anwer to your previous question :
Yes. $|\cdot|$ is used sometimes to avoid the heaviness of $||\cdot ||$.
Plus, remember that all norms are equivalent on $\mathbb R^n$, so (in this case) you do not really care which one it is !
(Partial) answer to the new one :
Your estimation is far too braod : $\frac{1}{|x|+|y|}$ is integrable for $x$ between $-1$ and $+1$, except when $y = 0$. So you should not estimate this by using the case $y=0$, and on the contrary integrate this away from the line $y = 0$ (which is negligible). Then you find an integrable function of $y$.