Consider a Borel set $E$ in the space $(\mathbb{R}/\mathbb{Z})^3$. Define a translation by $x=(x_1,x_2,x_3)$ with the coordinates of $x$ being $\mathbb{Q}$-linealy independent. If $E$ is invariant under translation, $E$ must have Lebesgue measure 1 or 0.
I just have an idea that this is closely related to Lebesgue density theorem, also, for any point $a$ we pick, after repeatedly doing the translation we will get a dense subset of $\mathbb{T}^3$, but then how? Any help?
If $E$ has positive measure, for some $x > 0$ and point $a$ we have $d_z(a) > \frac{9}{10}$ if $z < x$. If $E$ has measure less than $1$, then for some $y > 0$ and point $b$ we have $d_z(b) < \frac{1}{10}$ if $z < y$. Take $z = \min(x, y) / 2$.
We have $\mu(E \cap B_z(a)) > \frac{9}{10} \cdot V_z$. So also $\mu(E \cap B_{\frac{9z}{10}}(a)) > \frac{9}{10}\cdot V_z - (V_z - V_\frac{9z}{10}) > \frac{1}{10}V_z$. And $\mu(E \cap B_z(b)) < \frac{1}{10} V_z$.
As image of any point under repeating translation is dense, for some $n$ we will have $\rho(T^n a, b) < \frac{z}{10}$. Then we have $T^n\left(B_{\frac{9z}{10}}(a)\right) \subseteq B_z(b)$, but $\mu\left(E \cap T^n\left(B_{\frac{9z}{10}}(a)\right)\right) = \mu\left(E\cap B_{\frac{9z}{10}}(a)\right) > \frac{1}{10}V_z > \mu(E\cap B_z(b))$ - contradiction, as measure of subset can't be greater than measure of set.