TITLE: Irrational Cantor set?
QUESTION [14 upvotes]: Can someone help me?
How can I prove that exists a number $k \in \mathbb R$ that $$A = \{x + k;\ x \in \text{Cantor set} \} \subset\text{ Irrationals}\;?$$

REPLY [5 votes]: We had a similar problem in one of our Real Analysis assignments. It was a generalization of the given problem.
Let $E \subset \mathbb{R}$ be a set with Lebesgue outer measure $0$. Then there exists $h \in \mathbb{R}$ such that $E+h \subseteq \mathbb{Q}^c$. 
We can prove this using the identity Brian derived. 
If $\Bbb Q\cap(E+h)\ne\varnothing$ for every $h\in\Bbb R$, then
$$\Bbb R=\bigcup_{h\in\Bbb Q}(h-E)\;.$$
$$m(\Bbb R) \leq \sum_{h \in \Bbb Q}{} m^*(h-E)=0$$
which gives us a contradiction. So, $\Bbb Q\cap(E+h)=\varnothing$ for some $h \in \Bbb R$.