There exists a subset of $\mathbb{R}$ which is not Lebesgue measurable

Question

From Mathematical Analysis by Andrew Browder;

In the last paragraph they said: for any $x \in [0,1)$,there exists $a \in A$ such that $x \sim a$ i.e. $x-a $ is rational ;then $x=a \bigoplus q$,where $q=x-a$or $q=x-a+1$ is a rational in $[0,1)$.I am unable to show this.

I know: for any $x \in [0,1)$,there exists $a \in A$ such that $x \sim a$ i.e. $x-a $ is rational. But how to show $x=a \bigoplus q$?

Thanks in advance!

Answer 1

Hint: $x,y \in [0,1) $ implies that $x-a \in [-1,1)$. So $x-a \in [0,1)$ or $x-a \in [-1,0)$ and in the second case $x-a+1 \in [0,1)$.