Let $G$ be a simple linear algebraic group acting on a projective variety $X$ through rational maps. Let $x\_0\in X$ with stabilizer group $H$ and assume that $G/H$ in not compact and carries a $G$-invariant measure $\mu$. Let $z\in X$ be in the boundary of the orbit $G.x\_0$ and let $U\subset X$ be an open neighborhood of $z$. Let $G(U)\subset G/H$ be the set of all $gH\in G/H$ such that $g.x\_0\in U$. Is it true that $\mu(G(U))=\infty$? 



Haar measure of algebraic orbits
Not neccessarily. Take $G=SL(2,\mathbb R)$ acting on $X=\mathbb R^2$. Take $x\_0=(1,0)$ and $z=(0,0)$. The measure is Lebesgue measure. Then clearly $z$ has neighborhoods of finite volume.


