Say that a measurable subset $A$ of $[0,1]$ is ${\mathbb Q}$-stable if $a+q\in A$ whenever $a\in A,q\in{\mathbb Q}$ and $a+q\in [0,1]$. Obviously, $\emptyset$ is ${\mathbb Q}$-stable and has measure zero, and $[0,1]$ is ${\mathbb Q}$-stable and has measure one. Are there ${\mathbb Q}$-stable subsets with measure other than $0$ or $1$ ?
EDIT : Let $A$ be ${\mathbb Q}$-stable. It is easy to see that if $[a,b[$ and $[c,d[$ are two subintervals of $[0,1]$ with rational endpoints and with the same length ($d-c=b-a$), then $A\cap [c,d[=(c-a)+A\cap [a,b[$ whence $\mu(A\cap [c,d[)= \mu(A\cap [a,b[)$. It follows that there is a function $\alpha : [0,1] \to {\mathbb R}_{+}$ such that $\mu(A\cap [a,b[)=\alpha(b-a)$ for any rational $a<b$ in $[0,1]$. The map $\alpha$ is obviously nondecreasing. Also, if $x,y \in {\mathbb Q}$ are nonnnegative with $x+y\leq 1$ then $(A\cap [0,x+y[)$ is the disjoint union of $(A\cap [0,x[)$ and $(A\cap [x,x+y[)$, and hence $\alpha(x+y)=\alpha(x)+\alpha(y)$.
It follows that there is a constant $c$ such that $\alpha(t)=ct$ for any $t\in{\mathbb Q}\cap [0,1[$. By monotonicity the inequality stays true if we take $t\in [0,1[$. Not sure about how to finish from here.
Finally I found a solution to my own question. By what is shown in the edit, the density of $A$ is equal to $c$ (or does not exist) at every point. But according to the Lebesgue density theorem, if $\mu(A)>0$ then this density exists and is equal to $1$ a.e. This forces $c=1$ and $\mu(A)=1$.