Union of translations of a measurable set with the same measure as the original set Let $\mu$ be the Lebesgue measure on $[0,1]$ and let $A+\Bbb Q=\{a+q\mid a\in A\text{ and } q\in \Bbb Q\text{ and }a+q\in[0,1]\}$ for any $A\subset[0,1]$
I know that $\mu(A)=0$ or $\mu(A)=1$ implies that $\mu(A+\Bbb Q)=\mu(A)$. Is this implication also true in reverse? Does there exist an $A$ such that $0<\mu(A)=\mu(A+\Bbb Q)<1$? If not, does there exist an $A$ such that $0<\mu(A)<\mu(A+\Bbb Q)<1$?
 A: There is no such $A$. I'm going to work with everything on all of $\mathbb{R}$ for a bit, because restricting the Minkowski sum might mean I have to be more careful with some of the integrals I'm about to write.
Claim: Let $E \subseteq \mathbb{R}$ be such that $E + r = E$ for every $r \in \mathbb{Q}$. Then either $E$ or $E^{c}$ has measure $0$.
Before I prove that claim, let's take a look at why it helps. For any set $A$, copper.hat noted in the comments to the question that $A + \mathbb{Q} + r = A + \mathbb{Q}$ for every $r \in \mathbb{Q}$, where these Minkowski sums are not restricted to the unit interval. Thus, either $A + \mathbb{Q}$ or $(A + \mathbb{Q})^{c}$ has measure $0$ in $\mathbb{R}$. Now, we can restrict to the unit interval and see that $(A + \mathbb{Q}) \cap [0, 1]$ either has measure $0$ or $1$. Therefore, if $\mu(A) = \mu((A + \mathbb{Q}) \cap [0, 1])$, then $A$ must have measure $0$ or $1$.
Proof of claim:
Define the auxiliary function
$$f(x) = \int_{\mathbb{R}} 1_{E}(y) 1_{E^{c}}(y + x) \mathrm{d} y.$$
By the continuity of shifting operators with respect to $L^{1}$, $f$ is continuous. Moreover, for any $r \in \mathbb{Q}$,
$$f(r) = \int_{\mathbb{R}} 1_{E}(y) 1_{E^{c}}(y + r) \mathrm{d} y = \int_{E} 1_{E^{c}}(y + r) \mathrm{d} y = 0,$$
since $y + r \in E$ for any $y \in E$. Since $f$ is continuous and $0$ at all rational points, $f \equiv 0$. Thus,
$$0 = \int_{\mathbb{R}} f(x) \mathrm{d} x = \int_{\mathbb{R}} \int_{\mathbb{R}} 1_{E}(y) 1_{E^{c}}(y + x) \mathrm{d} y \mathrm{d} x = \mu(E) \mu(E^{c}),$$
using the Tonelli theorem and the translation invariance of the Lebesgue measure to get to that last equality. Hence, either $E$ has measure $0$ or $E^{c}$ has measure $0$.
