# 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 $t$?

• If $0\in t+E, \forall t$, then $-t\in E, \forall t$. That means $\mathbb{R}\subseteq E$, contradiction. Aug 25, 2016 at 2:54
• Ooh, nice, I like this one! Note the duality between "a countable intersection of measure-1 sets has measure 1" (used in my answer) and "a countable union of measure-0 sets has measure 0" (used in this answer). +1 Aug 25, 2016 at 2:56
• Nice proof. I was puzzled by this Q. Aug 25, 2016 at 5:25

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?

• What if the statement of the question is generalized to the following: "Suppose $Q$ and $E$ are two sets (of reals) both with measure zero. Show that there exists a $t\in\mathbb{R}$ such that $E+t$ contains no point of $Q$." So in this generalization, $Q$ may be uncountable. Is this more general statement true or false? Aug 25, 2016 at 12:55
• Without the restriction that one set be countable, the claim is false. Let $K$ be the usual Cantor set. Take $Q$ and $E$ to be $\bigcup\limits_{i\in\mathbb{Z}}\,(K+i)$. Then, both $Q$ and $E$ are of measure $0$ (noting that $K-K=[-1,1]$). However, $Q-E=\mathbb{R}$, whence for any $t\in\mathbb{R}$, $t+E$ intersects $Q$. Aug 25, 2016 at 13:02

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?

• I devised a proof but failed to complete it,;will you please help me complete it???Proof: Since $m(E)=0\implies m(E+t)=0\forall t\in \Bbb R$. Assume that $E+t$ contains a rational number for each $t\in \Bbb R$ .It is obvious that $E+t_i$ and $E+t_j$ contains a common rational since rational numbers are countable. Aug 25, 2016 at 3:00
• But I am unable to derive a contradiction from here?Do you mind helping me out Aug 25, 2016 at 3:00
• @BobWilson That route doesn't really work - in particular, all you can derive is that there is some rational $q$ and uncountably many reals $t_i$ ($i\in I$) such that $q\in E+t_i$. But that doesn't in itself lead to a contradiction. You need to use some properties of measure. Look at my answer: what can you conclude about the measure of the set $S_r$? (Note that $S_r$ is not a translate of $E$ - read the definition carefully!) Aug 25, 2016 at 3:14
• @NoahSchweber, I'm not quite sure how to go about this, how does one interpret the sets $S_r$? Oct 19, 2023 at 21:28
• @nolemonnomelon Given a real number $r$, some translates of $E$ "avoid" $r$ (= don't contain $r$ as an element) and other translates don't. $S_r$ is the set of ways to translate $E$ so that you don't wind up containing $r$. For instance, if $E$ is the Cantor set then ${1\over 2}\in S_1$ since if I shift $E$ over by $1\over 2$ the "middle gap" includes $1$. Oct 20, 2023 at 0:10