My question:
If $A \subset \mathbb{R}$ is a measurable subset and $t_n \rightarrow t$ is a convergent sequence such that $A +t_n = A$ mod $\mu$, does $A+t=A$ mod $\mu$ hold?
Here $A = B$ mod $\mu$ means that $\mu(A \Delta B)=0$ where $A \Delta B$ is the symmetric difference between $A$ and $B$ and $\mu$ is the Lebesgue measure on $\mathbb{R}$.
Let $A'$ be an intersection of $A+kt_n$ for all $k\in\mathbb Z,n\in\mathbb N$. Since all those sets are equal up to sets of measure zero, $A=A'\mod\mu$ and moreover $A'=A'+t_n$ for each $n$. Unless $t_n$ is eventually constant (in which case the statement is trivial), this implies that $A'$ is invariant under translations by a subgroup of $\mathbb R$ which is dense. Standard ergodicity results imply that a set $A'$ invariant under such set of translations either has measure zero or its complement has measure zero. It follows that $A'=A'+t\mod\mu$ for any $t\in\mathbb R$, in particular for $t=\lim_{n\to\infty}t_n$.