# Ergodic action of dense subgroup

Let $$G$$ be a group acting ercodically on a probability measure space $$(X, \mu)$$. Let $$\Gamma$$ be a countable dense subgroup of $$G$$. Is the action of $$\Gamma$$ also ergodic?

The case I am interested in is actually when $$G$$ is a Lie group acting smoothly on a compact submanifold of $$\mathbb{R}^n$$ with the normalized Lebesgue measure.

Here by ergodic action I mean that if $$\mu(gA \Delta A) = 0$$ for all $$g \in G$$, then $$\mu(A) = 0$$ or 1.

For the "no" part: In the general case, this is lacking a assumption giving a "link" between the topology of $$G$$ and the action; here is an example of such a discontinuous action:
Let $$G=\mathbb{Z}\oplus\mathbb{Z}\sqrt{2}\oplus \mathbb{Z}\sqrt{3}$$, $$\Gamma=\mathbb{Z}\oplus \mathbb{Z}\sqrt{2}$$. We endow $$G$$ with the induced topology from $$\mathbb{R}$$. In this case, $$\Gamma$$ is a dense subgroup. Now consider an ergodic action of $$G/\Gamma\simeq \mathbb{Z}\sqrt{3}$$ on a measure space (for example, an irrational rotation on the circle). This gives an action of $$G$$ itself, which is ergodic, but the action of $$\Gamma$$ is trivial.
For the "yes" part: the action of $$G$$ on $$X$$ induces a unitary representation on $$L^2(X,\mu)$$ (the Koopman representation) $$\pi:G\to \mathcal{U}(L^2(X)),$$ $$\pi(f)(x)=f(g^{-1}x).$$ A useful property that may (or may not) have this representation is to be strongly continuous, i.e. $$g_n\to g\implies \forall f \in L^2(X), \;\| \pi(g_n)f-\pi(g)f\|\to 0.$$ The relevance of this property to ergodicity of dense subgroup is that in this case, the stabilizer of a characteristic function $$f=1_A$$ $$Stab(f)=\{ g \in G \, : \, \pi(g)f=f \},$$ is then a closed subgroup; so if a set $$A$$ is $$\Gamma$$-invariant, $$f=1_A$$ is $$\Gamma$$-invariant, so is also $$\bar{\Gamma}=G$$-invariant, so $$\mu(A)\in \{0,1\}$$.