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.

Another approach. By replacing $E$ by $\bigcup_{r\in\Bbb{Q}} (r+E)$, we may assume WLOG that $E$ is invariant under rational translation. Then $t+E$ contains a rational number if and only if $t+E$ contains all rational numbers.
Now can you show that $t+E$ does not contain $0$ for some $t$?
