Consider the equivalence relation in $[0,1)$ given by$$ x\sim y \quad \Leftrightarrow\quad x-y\in\mathbb Q$$ and let $\mathcal F$ be the family of all equivalence classes of $\sim$ . Let $V$ be a section of $\mathcal F$ i.e. put in $V$ an element for each equivalence class of $\sim$ (notice that we are using the axiom of choice).

Given $q\in\mathbb Q\cap [0,1)$ define$$ V_q=((V+q)\cap [0,1))\cup((V+q-1)\cap[0,1))$$ that is $V_q$ is obtained translating $V$ by a quantity $q$ to the right and then cutting the piece which goes beyond the point $1$ and putting it on the left, starting from $0$ .

Now notice that given $x\in[0,1)$ there exists $y\in V$ such that $x\sim y$ (because $V$ is a section of $\sim$ ) and hence there exists $q\in \mathbb Q\cap[0,1)$ such that $x\in V_q$ . So$$ \bigcup_{q\in\mathbb Q\cap[0,1)} V_q = [0,1).$$

Moreover all the $V_q$ are disjoint. In fact if $x\in V_q\cap V_p$ then $x-q$ (modulus $[0,1)$ ) and $x-p$ are both in $V$ which is not possible since they differ by a rational quantity $q-p$ (or $q-p+1$ ).

Now if $V$ is Lebesgue measurable, clearly also $V_q$ are measurable and $\mu(V_q)=\mu(V)$ . Moreover by the countable additivity of $\mu$ we have$$ \mu([0,1)) = \sum_{q\in \mathbb Q\cap[0,1)} \mu(V_q) = \sum_q \mu(V).$$ So if $\mu(V)=0$ we had $\mu([0,1))=0$ and if $\mu(V)>0$ we had $\mu([0,1))=+\infty$ .

So the only possibility is that $V$ is not Lebesgue measurable.