Let $\mu(dx,dy)$ be a Borel measure with compact support on $\left\{(x,y) \mid x>0, y>0 \right\}$ and let $\mu(dx,dy)$ satisfy $$ \mu(dx,dy) = \lambda^{\alpha} \mu \left( d \frac{x}{\lambda}, d\frac{y}{\lambda} \right), \tag{1} $$ for any $\lambda>0$ and for some fixed $\alpha>0$. Is it true that $\mu(dx,dy) \equiv 0$?
If measure is absolutely continuous: $\mu(dx,dy) = a(x,y) dxdy$ then the equation $(1)$ can be rewritten as $$ a(x,y) = \lambda^{\alpha-2} a \left( \frac{x}{\lambda}, \frac{y}{\lambda} \right). $$ Now let $\lambda \to \infty$ if $\alpha \leqslant 2$ or $\lambda \to 0$ if $\alpha \geqslant 2$ to obtain $a(x,y) \equiv 0$. So the case of absolutely continuous measures is clear. Then the question is if it is true also for arbitrary Borel measures with compact support.
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$\begingroup$Fix $S\subset\{(x,y),x>0,y>0\}$ a measurable set and let $K$ be the support of $\mu$. Then for all $\lambda>0$, we have $$\mu(S)=\lambda^\alpha\mu(\alpha^{-1}S)\leqslant \lambda^\alpha\mu(K).$$ Now let $\lambda\to 0$.
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