# Is $|x|^{-\alpha}$ integrable for polynomially bounded measures on $\mathbb{R}^n$

We know that $$|x|^{-\alpha}$$ is in $$L^1 (x\in \mathbb{R}^n:|x| \ge 1)$$ 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 $$L^1 (\{x\in \mathbb{R}^n:|x| \ge 1\},\mu)$$ 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^{-n\alpha} \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^{-n\alpha} \mu\{|x| \le 2^{n+1}\} \le C \sum_{n=0}^\infty 2^{-n\alpha} (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 $$N >0$$.