Measure, absolutely continuous on boundary

Question

Let $\mu$ be a finite nonnegative Borel measure on $\mathbb R^2_+=[0,+\infty) \times [0,+\infty)$ such that $\mu( \partial \mathbb R^2_+)=0$, i.e. $\mu$ is absolutely continuous on boundary. Is it true that there exist such $\alpha > 0$, $\beta>0$ that $$ \int\limits_{\Large\mathbb R^2_+} \frac{1}{x^\alpha y^\beta} d\mu < \infty $$ or we can find a counterexample?

Answer 1

No, not necessarily. For a counterexample, let $\mu$ place mass $2^{-n}$ at the point $(2^{-2^n},1)$ for $n=1$, $2$, $3$, $\dots$ and no mass elsewhere. Then $\mu$ satisfies the given conditions but for any positive $\alpha$ and $\beta$, $$ \int\limits_{\Large\mathbb R^2_+} \frac{1}{x^\alpha y^\beta} d\mu = \sum_n 2^{-n} 2^{2^n \alpha}=\sum_n 2^{-n+2^n\alpha}=\infty. $$