The original question is:
Let $A$ be Lebesgue measurable in $\mathbb{R}$ with positive measure. Show that it is not true that there must exist a sequence $\{x_n\}^\infty_{n=1}$ such that the complement of $\bigcup^\infty_{n=1}(A+\{x_n\})$ in $\mathbb{R}$ is measure zero.
To disprove the statement, I want to take $A$ to be the fat rational set and argue that for any sequence $\{x_n\}$ in $\mathbb{R}$, the complement of the set stated above must have positive measure.
Let $\{q_n\}$ be an enumeration of $\mathbb{Q}$, fix $\epsilon\in(0,1)$ and let \begin{equation} I_n=(q_n-\frac{\epsilon}{2^{n+1}},q_n+\frac{\epsilon}{2^{n+1}}). \end{equation} Let $G=\cup_nI_n$, and $A=[0,1]\setminus G$, then clearly any rationals cannot be in $A$ and it follows that $A$ is nowhere dense. It is easy to show that $A$ is Lebesgue measurable and has measure \begin{equation} m(A)= 1-\sum^\infty_{n=1}\frac{\epsilon}{2^n}=1-\epsilon\in(0,1), \end{equation} so $A$ satisfies the hypothesis of the question. Since $A$ is nowhere dense so $\bigcup^\infty_{n=1}(A+\{x_n\})$ is nowhere dense also for any sequence $\{x_n\}$ in $\mathbb{R}$.
However, I found it difficult to prove that the complement of $\bigcup^\infty_{n=1}(A+\{x_n\})$ must have positive measure for arbitrary sequence $\{x_n\}$