Prove that if $|g(x)|\leq a\big( 1+x \big)^{-n-b}$ then $g\in L^1(\mathbb{R^n})$

Question

I am asked to prove that if $g:\mathbb{R}^n \to [-\infty, \infty]$ is a measurable function that satisfies $|g(x)| \le a(1+|x|)^{-n-b}$ for some constants $a, b >0$, then $g \in L^1(\mathbb{R}^n)$.

I am thinking this is not true because for example if $n=1$, $a=1$ and $b=3$, then if we pick $g(x)=a(1+|x|)^{-n-b}$ the integral $ \displaystyle \int_{\mathbb{R}}(1+x)^{-4}$ is divergent so $g \notin L^1(\mathbb{R})$. But I am probably missing something so help would be appreciated.

Answer 1

$\int_{\mathbb R} \frac 1 {(1+|x|)^{4}}dx =\int_{-1}^{+1} \frac 1 {(1+|x|)^{4}}dx+\int_{|x| >1} \frac 1 {(1+|x|)^{4}}dx \leq 2+\int_{|x| >1} \frac 1 {|x|^{4}}dx=2+\frac 2 3 <\infty $.

For the integrability of $g$ in the general case you can use polar coordinates in $\mathbb R^{n}$

Answer 2

Hint: consider set $B(0,1) \cup (\cup_{k=1}^\infty B(0,2^k)\setminus B(0,2^{k-1}))$