A set with measure $0$ has a translate containing no rational number. 
Suppose $E$ is a set with measure $0$. Show there exists $t\in \mathbb{R}$ such that $E+t$ contains no rational number.

My idea is to find an interval in $E$, then we can get a contradiction. I try to begin with a point in $E$ and then consider if there is an interval containing this point in $E$. But I don't know how to start.
Maybe, we can go by contradiction.  If $E+t$ contains a rational number $q_t$ for every $t\in \Bbb R$, then we have a function $f:\mathbb{R}\to\mathbb{Q}$, $t\mapsto q_t$.  But this idea leads nowhere.

You can also show that $Z:=\bigcup\limits_{q\in\mathbb{Q}}\,(q-E)$ has measure $0$. For a real number $t\notin Z$, can $t+E$ intersect the rationals?
