A set with measure $0$ has a translate containing no rational number.

Question

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.

Answer 1

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 $t$?

Answer 2

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

• 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?

Answer 3

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?