This is the theorem:
Let $m$ denote the Lebesgue measure. Let $A \subset \mathbb{R}:m(A)=0$. Then, there exists $r \in \mathbb{R} : A \cap (r + \mathbb{Q})=\emptyset$
$x + A$ is the translation of set $A$ by $x$. Namely $x+A=\{x+a:a\in A\}$
I was able to prove this, but now I was asked to generalize it somehow. I don't know how I can further generalize this statement. I assumed I could use this fact (proven in the proof of the theorem) that $\cup_{q \in \mathbb{Q}}q+A$ has either all rationals (if there is at least one rational number in $A$) or no rationals.
We can repalce $\mathbb Q$ by any countable set. Let $C$ be at most countbale. Then $m(A-c)=0$ for any $c \in C$ so $m(\cup_{c \in C} (A-c)=0$. It follows that the real line is not contained in $\cup_{c \in C} (A-c)$. Let $r$ be a real number which is not in this union. Then $A \cap (r+C)$ is empty.