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We know that $|x|^{-alpha}$ is in $L^1 (xin mathbb{R}^n:|x| ge 1)$ with the normal Lebesgue measure for $alpha > n$. But what if we had a measure $mu$ on $mathbb{R}^n$ which is polynomially bounded, i.e., $mu(|x|le A) le C(1+A^N)$ where $C,N$ are fixed constants, then would we have something like $|x|^{-alpha}$ is in $L^1 ({xin mathbb{R}^n:|x| ge 1},mu)$ for $alpha >N$?

We split the integral into dyadic pieces in the following way:
begin{align}
int_{{|x| ge 1}} |x|^{-alpha} , mathrm{d}mu(x) &= sum_{n=0}^infty int_{{2^{n+1} > |x| ge 2^n}} |x|^{-alpha} , mathrm{d}mu(x) \
&le sum_{n=0}^infty 2^{-nalpha} mu{2^n le |x| < 2^{n+1}}
end{align}
Now we can use the polynomially growth bound $mu(|x| le 2^{n+1}) le C(1+2^{(n+1)N})$ to get that the last term is bounded by
begin{align}
sum_{n=0}^infty 2^{-nalpha} mu{|x| le 2^{n+1}} le C sum_{n=0}^infty 2^{-nalpha} (1+ 2^{(n+1)N}).
end{align}
Here the last sum is convergent if and only if $alpha >N$ and $alpha >0$. Thus, in this more general case, we have also integrability provided that $alpha > N,$ where I supposed that $N >0$.