# properties of a special set in the proof that there is no Lebesgue measure for all subsets of $\mathbb{R}$

i am working through a proof that there could be no measure on $\mathbb{R}$ such that

• $\lambda([a,b]) = b - a$
• $\lambda(A) = \lambda(A + \{c\})$

First a set $A \subset [0,1]$ is constructed with $$\forall x \in [0,1] ~ \exists ! y \in A ~:~ x - y \in \mathbb{Q}$$ (the notation $\exists !$ means there exists exactly one such element) Then the following set is considered $$B := \bigcup_{r \in [-1:1] \cap \mathbb{Q}} (A + \{r\})$$ and it is claimed that it has the property $$[0,1] \subset B \subset [-1:2]$$ but i don't see why this inclusion-relations hold?

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yes, i changed it! – Stefan Jun 16 '12 at 23:28

Let $x\in[0,1]$. There exists $y\in A\subseteq[0,1]$ such that $x-y\in\mathbb{Q}$. Moreover, since $x,y\in[0,1]$, $|x-y|\leq 1$, which shows that $[0,1]\subseteq B$.
Now take any $x\in A$. It follows that $x\in[0,1]$. Let $r$ be a rational number between $-1$ and $1$. So $x+r\geq x+(-1)=x-1\geq 0-1=-1$. Also $x+r\leq x+1\leq 1+1=2$. So $B\subseteq[-1,2]$.