Algebraically invariant extensions of $\sigma$-finite measures on Euclidean space

Abstract:

Let $G$ be a group of algebraic transformations of ${{\mathbf {R}}^n}$, i,e., the group of functions generated by bijections of ${{\mathbf {R}}^n}$ of the form $({f_1}, \ldots ,{f_n})$ where each ${f_i}$ is a rational function with coefficients in ${\mathbf {R}}$ in $n$-variables. For a function $\gamma :G \to (0,\infty )$ we say that a measure $\mu$ on ${{\mathbf {R}}^n}$ is $\gamma$-invariant when $\mu (g[A]) = \gamma (g)\cdot \mu (A)$ for every $g \in G$ and every $\mu$-measurable set $A$. We will examine the question: "Does there exist a proper $\gamma$-invariant extension of $\mu ?$ We prove that if $\mu$ is $\sigma$-finite then such an extension exists whenever $G$ contains an uncountable subset of rational functions $H \subset {({\mathbf {R}}({X_1}, \ldots ,{X_n}))^n}$ such that $\mu (\{ x:{h_1}(x) = {h_2}(x)\} ) = 0$ for all ${h_1},{h_2} \in H,{h_1} \ne {h_2}$. In particular if $G$ is any uncountable subgroup of affine transformations of ${{\bf {R}}^n},\gamma (g{\text {)}}$ is the absolute value of the Jacobian of $g \in G$ and $\mu$ is a $\gamma$-invariant extension of the $n$-dimensional Lebesgue measure then $\mu$ has a proper $\gamma$-invariant extension. The conclusion remains true for any $\sigma$-finite measure if $G$ is a transitive group of isometries of ${{\mathbf {R}}^n}$. An easy strengthening of this last corollary gives also an answer to a problem of Harazisvili.