Let $l>0$ and $\mathbb{Q}^c$ denotes set of irrationals. We define $$ E=\cap_{x\in(-l,l)}( \mathbb{Q}^c+x), $$ where $\mathbb{Q}^c+x=\{q+x:q\in\mathbb{Q}^c\}.$
Is it easy to show that $E$ cannot contains any rationals. Also $E$ cannot contains any real numbers from $(-l,l).$ So definitely $E\subseteq \mathbb{Q}^c\cap (-l,l)^c$.
My question:
- Is $E$ non-empty?
- Is $E$ has zero Lebesgue measure?
$E$ is empty. Let $\alpha$ be irrational. Since the rationals are dense, there exists $x$, $-\ell<x<\ell$, such that $\alpha-x$ is rational. Then $\alpha$ is not in ${\bf Q}^c+x$, so it's not in $E$.
A bit more detail: choose $n$ so $10^{-n}<\ell$, take $x<10^{-n}$ to have decimal expansion agreeing with that of $\alpha$ from the $n$th decimal place on. Then $\alpha-x$ is a terminating decimal, hence, rational.