Let $X$ be some (Lebesgue-)measurable subset of $\mathbf{R}$ such that, for any rational $q \neq 0$, we have $qX=X$. Assume that the Lebesgue measure $\mu(X)$ of $X$ is $>0$. Does it hold then that the Lebesgue measure of the complement of $X$, $\mu(\complement X)$, is $0$?
In other word, is Lebesgue measure on the real line $\mathbf{Q}^*$-ergodic? (Lebesgue measure is quasi-invariant under $\mathbf{Q}^*$ so ergodicity has a clear meaning here)
This is related to Countable family of pairwise disjoint sets of full outer measure
I shall try to answer my own question.
Lemma. Let $H$ be some locally compact group and $\Gamma$ be a dense subgroup. Then the usual operation of $\Gamma$ on $H$ (endowed with Haar measure) is ergodic.
Proof. Let $A$ be some $\Gamma$-invariant measurable set. The idea of the proof is that the characteristic function $1_A$ is $\Gamma$-invariant as well. Since $H$ acts continuously on $L^\infty(H)$ (for the weak-*-topology), and since $\Gamma$ is dense, we see that $1_A$ (more precisely, element of $L^\infty(H)$ corresponding to $1_A$) is $H$-invariant. Then a standard lemma shows that there is some measurable $A'$, that is equal mod. 0 to $A$, and is $H$-invariant.
This lemma is basically taken from the book of Zimmer. Now take $H=\mathbf{R}^*$ and $\Gamma=\mathbf{Q}^*$. Since the Haar measure of $H$ is equivalent to Lebesgue measure we see that the answer to my question is: yes.