8
$\begingroup$

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?

$\endgroup$
13
$\begingroup$

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 $\mathbb{R}^n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.