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 $N >0$.

  • $\begingroup$ Notice you can bypass the dyadic partitioning by setting up the first sum over dyadic ranges. $\endgroup$ – Guacho Perez Nov 30 '18 at 15:47
  • $\begingroup$ I have simplified the proof with regard to your comment. Now the proof is simpler. $\endgroup$ – p4sch Nov 30 '18 at 22:09
  • $\begingroup$ nice proof, (+1). $\endgroup$ – Guacho Perez Nov 30 '18 at 22:11

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