I am reading the book "Introduction to arithmetic groups" by Dave Witte-Morris, and I am in Chpater 14 on ergodic theory. In particular I am stuck on exercise 14.2#16 which reads;
Assume $\Gamma$ is an irreducible lattice, and let $H$ be a closed, non-comapct subgroup of $G$. Show, for a.e. $x\in G/\Gamma$, that $Hx$ is dense in $G/\Gamma$.
I know from previous work that in this setting the action of $H$ on the quotient $G/\Gamma$ is ergodic, (and in fact the action of $\Gamma$ on $G/H$ is also ergodic), so my thought was to consider the indicator function $\chi$ of $\overline{Hx}$ for some $x\in G/\Gamma$.
Then $\chi:G/\Gamma\to\mathbb{R}$ is an $H$-invariant measurable function, hence is essentially constant by ergodicity. Since $\chi$ only takes two values we have that either $\chi=1$ or $\chi=0$.
However both cases seem like they can happen so I don't know how to conclude that $\overline{Hx}=G/\Gamma$ from here.
Also I am wondering how you prove that this holds for a.e. $x\in G/\Gamma$ rather than all such $x$?
Let $\{U_n\}_n$ be a countable base (of open sets). Then, $Hx$ is dense if and only if for each $n$, $x \in HU_n$. So it suffices to show $\cap_n HU_n$ has measure $1$, and so it suffices to show each $HU_n$ has measure $1$. But $H(HU_n) = HU_n$, so by ergodicity, the measure of $HU_n$ is $0$ or $1$, so it's $1$.