Let G be a group of algebraic transformations of
**R**^{n}, i.e., the group of functions generated by
bijections
of **R**^{n} of the form (f_{1},...,f _{n})
where each f_{i} 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 **R**^{n}
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**(X_{1},...,X_{n}))^{n} such that
\mu({x:h_{1}(x)=h_{2}(x)})=0 for all
different h_{1},h_{2}\in H.
In particular if G is any uncountable subgroup of affine
transformations of **R**^{n},
\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
**R**^{n}.
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.**