This is the last part of my homework assignment for Measure Theory.
Let $\lambda^d$ be the Lebesgue measure on the $\sigma$-algebra $\Lambda^d$ of Lebesgue measurable sets of $\mathbb{R}^d$. We know that it is translation invariant and we know that \begin{eqnarray} \lambda^d(A)&=&\sup\{\lambda^d(K)|K\subset A,K \text{ compact}\}\\ &=&\inf\{\lambda^d(U)|A\subset U,U \text{ open}\} \end{eqnarray} for any $A\in \Lambda^d$. Let let $E\in \Lambda^1$ with $\lambda^1(E)>0$. We showed that for every $0<\alpha <1$ there is an open interval $I$ such that $\lambda^1(E\cap I)>\alpha \lambda^1(I)$. We also showed that $E-E:=\{x-y|x,y\in E\}$ contains an open interval centered at $0$. The next question asks us to prove (under the assumption of the Axiom of Choice) that $E$ contains a subset that is not Lebesgue measurable. I actually have no idea how to approach this. Any help on getting started on this is much appreciated. Please tell me if anything is unclear. The only hint which is given says that we should think of the proof that there cannot be a translation invariant measure $\mu$ on $\mathbb{R}$, defined on all subsets, such that $\mu([0,1])=1$.
Presumably, you already know the Vitali construction - that for any (nonempty open) interval $(a, b)$, we may partition the interval into countably many sets $V_i$ such that no $V_i$ has a Lebesgue-measurable subset of positive measure. (This is a slight strengthening of how it's usually stated, but the proof is basically unchanged.)
OK, so fix some interval $(a, b)$ such that $E\cap (a, b)$ has positive measure, and let $V_i$ be as above. Then for some $i$, $V_i\cap E$ has positive (outer) measure! Why? Well, otherwise, $E\cap (a, b)=E\cap(\bigcup V_i)$ is the union of countably many null sets, and hence null.
So look at that $V_i$, and let $X=V_i\cap E$. What do you know about $X$? (HINT: what do you know about subsets of $V_i$?)