How to show: $ \lambda(E)=0$, $E\subset V$ ,V:Vitali nonmeasurable set

Question

Let $V$ be the Vitali nonmeasureable set, $\lambda$ :the Lebesgue measure on the line.

If E is a Lebesgue measurable subset of V,

How to show: $ \lambda(E)=0$

Answer 1

Notice that for vitaly non-measurable set you can look at it's "move" by a rational number $q$. if you look at the union over all the movements of $V$ by rationals between -1 and 1 you'll get a set that contained in [-1,2], moreover every move by a rational is disjpint to an other move by a rational. Therfore, if is $E\subset V$ then $\bigcup_{q\in[-1,1]}(E+q) \subset \bigcup_{q\in[-1,1]}(V+q)\subset[-1,2]$ so now you can use the fact that since $E\subset V$ then $E+q \ne E+r$ (as disjoint) and the sigma-additivity and move invariant of lesbegue measure.

Answer 2

This seems like a duplicate of

The subset of non-measurable set

Answer 3

First, let's remember that $V$ has the following property:

$(*)$For each $r\in\mathbb{R}$, there exists only one $v\in V$ such that $r-v\in\mathbb{Q}$.

Suppose that $E\subseteq V$ is measurable. To show that $\lambda(E)=0$, let's assume that $1\not\in E$ (because a single point doesn't affect measure). Let $n\in\mathbb{N}$. For $i=0,\ldots,n-1$, let $E_i=E\cap [i/n,(i+1)/n)$, so $E=\sum_{i=0}^{n-1}E_i$, where $\sum$ denotes disjoint union.

By property $(*)$, we have $(E_i-i/n)\cap (E_j-j/n)$ for $i\neq j$, so, by translation invariance of $\lambda$: $$\lambda(E)=\sum \lambda(E_i)=\sum\lambda(E_i-i/n)=\lambda(\sum (E_i-i/n))\subseteq\lambda[0,1/n)=1/n.$$ Letting $n\rightarrow\infty$, we obtain $\lambda(E)=0$.