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

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 $t$?


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?


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?

Tags:

Rational Numbers

Real Analysis

Measure Theory

Lebesgue Measure

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