Let $A\subseteq\mathbb{R}$ Lebesgue measurable such that $\mu(A\backslash{}A+x)=0$ for every $x\in\mathbb{R}$. Show that $\mu(A)=0$ or $\mu(\mathbb{R}\backslash{}A)=0$.
Since $A+x$ is measurable I can conclude that $\mu(A)=\mu(A\cap{}A+x)+\mu(A\backslash{}A+x)=\mu(A\cap{}A+x)=\mu(A+x)$.
In fact for every measurable $I$ we have that $0=\mu(A\backslash{}A+x)=\mu((A\backslash{}A+x)\cap{}I)+\mu((A\backslash{}A+x)\backslash{}I)=\mu((A\cap{}I)\backslash{}((A+x)\cap{}I))$ which for $\mu(I)<\infty$ gives $\mu((A+x)\cap{}I)=\mu(A\cap{}I)$ for every $x\in\mathbb{R}$ or equivalently $\mu(A\cap{}(I+x))=\mu(A\cap{}I)$.
Taking $I=(0,1)$ we have that $\mu(A)=\mu(\cup_{n=-\infty}^{\infty}A\cap{}(I+n))=\sum_{n=-\infty}^{\infty}\mu(A\cap{}I)$ and thus if $\mu(A)<\infty \implies \mu(A)=\mu(A\cap{}I)=0$.
I cannot handle the case $\mu(A)=\infty$. Is my proof above correct? How can I proceed?
This requires a more careful analysis when $\mu (A)=\infty$.
Let $\mu (B) <\infty$. We have $\int_{B \cap A} 1 dy= \int_{B \cap A} I_{A+x} (y) dy$ for every $x$ since $B\cap A \setminus B \cap A \cap (A+x) \subset A \setminus (A+x)$. Since $I_{A+x} (y) \leq 1$ this implies $I_{A+x} (y)=1$ for almost every $y \in B \cap A$ and this is true for every $x$. By a standard Fubini argument we can now conclude that for almost every $y \in B\cap A$ we have $I_{A+x} (y)=1$ for almost every $x$. If $\mu (B \cap A) >0$ the there is at least one $y$ for which $I_{A+x} (y)=1$ for almost every $x$. But this implies that $\mu (\mathbb R \setminus A)=0$. Now apply this to $B=(-n,n)$ and vary $n$ to finish the proof.