Endow $\{0,1\}$ with the fair coin measure $\mu(\{0\})=1/2=\mu(\{1\})$, and consider the infinite product measure $m$ on $\mathcal{C}=\{0,1\}^\mathbb{N}$. The $\mathcal{C}$ can be seen as an abelian group under coordinatewise addition mod 2. Then, the set $Q$ of sequences with at most finitely many ones is a dense subgroup $\mathcal{C}$. I'd like to show that the action of $Q$ on $\mathcal{C}$ given by $x\mapsto x+q$, $q\in Q$ is ergodic with respect to the measure $m$, i.e. given a measurable set $E\subset \mathcal{C}$,
$$ \forall q\in Q(q+E=E)\implies m(E)=0 \vee m(E)=1.$$
This is easy to see for cylinder sets, so I tried to show that the collection of sets with this property, say $\mathcal{B}$, forms a $\sigma$-algebra. Its fairly straightforward to see that $\mathcal{B}$ is closed under complementation, but I do not see how to show that it is closed under countable unions.
If we take $B_n\in\mathcal{B}$ for $n\in\mathbb{N}$, and let $B=\bigcup_n B_n$, then the fact that $q+B=B$ for all $q\in Q$ does not necessarily mean that $q+B_n=B_n$ for every $n$ and every $q$, or at least the implication is not immediate.
This is a general argument. Let $T$ be given by $T(\omega) = q+\omega$, and take $A$ such that $TA = A$. Take $\epsilon > 0$ and $I$, a finite union of disjoint cylinders so that $\mu(A\Delta I) < \epsilon$. Then, take $k$ large so that $\mu(T^k I \cap I) = \mu(I)^2$ (which is possible). Then, $\mu(A) = \mu(T^k A \cap A) \approx_\epsilon \mu(T^k I \cap I) = \mu(I)^2 \approx_\epsilon \mu(A)^2$. This implies (once you fill in the details), that $\mu(A) \in \{0,1\}$.