Let G be a group of algebraic transformations of Rn, i.e., the group of functions generated by bijections of Rn of the form (f1,...,f n) where each fi is a rational function with coefficients in R in n-variables. For a function \gamma:G-->(0,\infty) we say that a measure \mu on Rn is \gamma-invariant when \mu(g[A])=\gamma(g)\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 (R(X1,...,Xn))n such that \mu({x:h1(x)=h2(x)})=0 for all different h1,h2\in H. In particular if G is any uncountable subgroup of affine transformations of Rn, \gamma(g) 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 Rn. An easy strengthening of this last corollary gives also an answer to a problem of Harazisvili.
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Last modified January 7,2002.