Inherence of full measure sets from a Lie group to its homogeneous space. Let $G$ be a Lie group with Haar measure $\mu$ and $\Gamma$ be a lattice of $G$. Namely $\Gamma$ is a closed discrete subgroup of $G$ with $G/\Gamma$ admitting a left $G$-invariant probability measure $\nu$. Let $U$ be any subset of $G$ with full Haar measure, namely $\mu(G-U)=0$. Let $x\in G/\Gamma$, a point in the homogeneous space. I wonder if it is necessarily true that $Ux$ also has full $\nu$ measure.
$$\nu(G/\Gamma - Ux)=0.$$
Or at least, $\nu (G/\Gamma - Ux)=0$ for $\nu$-almost every $x\in G/\Gamma$?
 A: I have a proof when $\Gamma$ is a unimodular lattice in the Lie group $G$:
Since $G$ is a Lie group, thus second countable, as a discrete subgroup $\Gamma$ must be countable.
Let $U$ be any subset of $G$ with full Haar measure. Now by the Theorem 1.5.3 (Quotient Integral Formula, $f$ there can be any $L^1$ function on $G$. Take $f=1_{(G-U)g_0}$, noticing that $\mu((G-U)g_0)=0$ because of the unimodularity (left and right invariance) of $\mu$) in the Deitmar and Echterhoff's book: Principles of Harmonic Analysis, and Fubini's theorem:
$$0=\int_G f d\mu = \int_{G/\Gamma} \int_{\Gamma} f(g\gamma) d\gamma ~d\nu(g\Gamma) =  \int_{\Gamma}\int_{G/\Gamma} f(g\gamma)d\nu(g\Gamma)~ d\gamma.$$
Since $\Gamma$ is countable, the Haar measure on it must be a scalar multiple of counting measure. So we must have for every $\gamma \in \Gamma$, in particular for $\gamma = e$,
$$0=\int_{G/\Gamma} f(g)d\nu(g\Gamma)= \nu((G-U)g_0 \Gamma).$$
Since $g_0\in G$ is arbitrary, and by the transitivity of the action, $G/\Gamma - Ux$ is contained in $(G-U).x$, we are done.
Please let me know if this is correct.
