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.

HINT: For a fixed real number $r$, let $S_r$ be the set of shifts which avoid $r$: $$S_r=\{t: r\not\in E+t\}.$$


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*If $E$ has measure zero, what can you say about $S_r$?

*The rationals are countable, and $$\{t: E+t\cap\mathbb{Q}=\emptyset\}=\bigcap_{q\in\mathbb{Q}}S_q.$$ Do you see why the intersection of countably many sets with the property above is nonempty?
