$\mu(E\setminus (E+x))=0$ for all $x\in\mathbb{R}$. Prove that $\mu(E)=0$ or $\mu(\mathbb{R}\setminus E)=0$ I am preparing myself to the mini-exam in measure theory by solving problems from lecturer's notes and I have encountered some difficulties I cannot overcome. I would appreciate if you could solve the following (or if you could give me some huge hints at least):

Let $\mu$ be the Lebesgue measure on $\mathbb{R}$ and let $E$ be a Borel subset of $\mathbb{R}$ such that $$\mu(E\setminus (E+x))=0$$ for any $x\in\mathbb{R}$, where $E+x=\{z+x\mid z\in E\}$. Prove that $\mu(E)=0$ or $\mu(\mathbb{R}\setminus E)=0$.

I know that $\mu(E)=\mu(E+x)$ and I suppose I have to prove the statement by assuming the opposite, but I really do not know how. Thanks in advance!
 A: If $\mu(E) \not=0$ and $\mu(E^c) \not=0$ (here $E^c = \mathbb R \setminus E$), we can choose intervals $I$ and $J$ such that $\mu(I\cap E) \geq (3/4) \mu(I)$ and $\mu(J\cap E^c) \geq (3/4) \mu(J)$. By shrinking either $I$ or $J$ if necessary, we can take $\mu(I) = \mu(J)$, i.e. $I$ and $J$ have the same length. Then we can choose $x$ such that $J+x = I$. But then $I \cap E \cap (E^c + x)$ is a subset of $E\cap (E^c +x) = E\setminus (E+x)$, and we have 
$$\mu(I \cap E \cap (E^c + x)) \geq \mu(I \cap E) + \mu(I \cap (E^c + x)) - \mu(I) ,$$
by elementary set theory and since $I\supset (I \cap E) \cup(I \cap (E^c + x))$. We have $\mu(I\cap (E^c + x)) = \mu(J\cap E^c)$, and so
$$\mu(I \cap E \cap (E^c + x)) \geq (3/4) \mu(I) + (3/4)\mu(I) - \mu(I) = (1/2)\mu(I) > 0 .$$
Thus $E\setminus (E+x)$ has positive measure. 
A: I would like to give another proof I have found recently after being told to make use of Fubini's theorem.
Since $\mu(E\setminus E+x)=0$ for all $x\in\mathbb{R}$, we have
$$\begin{align}
0=\int\limits_{\mathbb{R}}\mu(E\setminus (E+x))\,\text{d}\mu(x)=&\int\limits_{\mathbb{R}}\int\limits_{\mathbb{R}}\chi_{E\setminus (E+x)}(s)\, \text{d}\mu(s)\,\text{d}\mu(x)\\=&\int\limits_{\mathbb{R}}\int\limits_{\mathbb{R}}\chi_E(s)\left(1-\chi_{E+x}(s)\right)\,\text{d}\mu(s)\,\text{d}\mu(x)\\=&\int\limits_{\mathbb{R}}\int\limits_{\mathbb{R}} \chi_E(s)\left(1-\chi_{E+x}(s)\right)\,\text{d}\mu(x)\,\text{d}\mu(s)\\=&\int\limits_{\mathbb{R}}\chi_E(s)\left(\int\limits_{\mathbb{R}}\left(1-\chi_{E-s}(x)\right)\,\text{d}\mu(x)\right)\,\text{d}\mu(s)\\=&\int\limits_{\mathbb{R}}\chi_E(s)\left(\int\limits_{\mathbb{R}}\chi_{(E-s)^c}(x)\,\text{d}\mu(x)\right)\,\text{d}\mu(s)=\int\limits_{\mathbb{R}}\chi_E(s)\mu\left((E-s)^c\right)\,\text{d}\mu(s)\\=&\int\limits_{\mathbb{R}}\chi_E(s)\mu\left(E^c\right)\,\text{d}\mu(s)=\mu\left(E^c\right)\int\limits_{\mathbb{R}}\chi_E(s)\,\text{d}\mu(s)=\mu(E^c)\mu(E).
\end{align}$$
Therefore either $\mu(E)$ or $\mu(\mathbb{R}\setminus E)$ is $0$.
Notice that we used Fubini's theorem only once while changing the order of integration in equality no. 4. The eighth equality follows from the translation invariance of Lebesgue mesure.
