It seems to be a well known fact that a subset $D\subset\mathbb{R}$ that contains exactly one element in $\mathbb{Q}+x$ for any $x\in\mathbb{R}$ is not measurable. But trying to proof it I stumbled across the following argument $$1=\lambda(\left[ 0,1\right]\cap\mathbb{R} )=\lambda((\underset{q\in\mathbb{Q}}{\bigcup}D+q)\cap[0,1])\overset{?}{=}\underset{q\in\mathbb{Q}}{\sum}\lambda(D).$$ I don't quite understand the last equation. He argues that every element in $D$ can get pushed injectively onto the $[0,1]$ intervall and I see that, but I don't see why the resulting set (everything from $D$ projected onto $[0,1]$ calculating modulo 1) has the same masure as $D$. For sure two sets do not have te same measure becouse they are bijective only?
Thanks in advance! I am blank on measure theory, so it may be obvious, but I still don't see it.
Lebesgue measure $\lambda$ is invariant by translation so $\lambda(D+q) = \lambda(D)$.
Lebesgue measure is also countably additive. Hence
$$\lambda((\underset{q\in\mathbb{Q}}{\bigcup}D+q)\cap[0,1])= \underset{q\in\mathbb{Q}}{\sum}\lambda(D + q) = \underset{q\in\mathbb{Q}}{\sum}\lambda(D)$$
For more details see here.