I am self-studying measure theory and I have seen this theorem:
If $A$ is a set of positive measure, then there exists a subset $D$ of $A$ that is non measurable.
I am not sure how to prove it. I have read the article about Vitali set in here. If $V(0,1)$ represents the Vitali set constructed in the interval $[0,1]$. Then isn't that $V=\bigcup_{n\in \mathbb Z} V(n,n+1)$ be a non measurable set? If so, whether $D=V\cap A$ is the desirable subset of $A$ which is non measurable?
Best Answer
This presentation is essentially repeated from "Measure and Integral" by Wheeden and Zygmund. The function $\lvert \cdot \rvert_e$ denotes outer measure. It can be proven that if $E$ is a measurable subset of $\mathbb{R}$ with positive measure, then the set of differences $\{ x-y \ \lvert \ x,y \in E \}$ contains an interval at the origin. The proof is in this answer as well as the book referenced: https://math.stackexchange.com/a/104126/35667.
Let $A \subset \mathbb{R}$ be a set with positive outer measure. Let $E_r$ be the Vitali set on the real line translated by $r$, i.e. a set of representatives of equivalence classes of $\mathbb{R} / \mathbb{Q}$. For $r,q$ rational, $r \ne q$, we have $E_r \cap E_q$ is empty, and $\cup_{r \in \mathbb{Q}} E_r = \mathbb{R}$. So $A = \cup_r (A \cap E_r)$, and $\lvert A \rvert_e \leq \sum_r \lvert A \cap E_r \rvert_e$. Now if $A \cap E_r$ is measurable, then it must have measure $0$ by the preceding paragraph since its set of differences contains no interval at the origin (any two elements of this set differ by an irrational number). But since $\lvert A \rvert_e > 0$, we must have $A \cap E_r$ with positive outer measure for some $r$. Then this $A \cap E_r$ is the desired nonmeasurable set.
This can be extended to $\mathbb{R}^n$.