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Stated in a non-oneliner way:

Let $X\subset\mathbb{R}$ be Lebesgue-measurable with positive measure.

Is it possible to find sets $A, B$ with the following properties?

  1. $A\cap B = \emptyset$
  2. $A\cup B = X$
  3. all measurable subsets of $A$ are null sets
  4. all measurable subsets of $B$ are null sets

These properties of course imply that if such a decomposition exists, $A$ and $B$ must be non-measurable.

Continuing my descent into madness measure theory, my current research topic somehow hinges on the above question. My intuition tells me, and I sincerely hope, that such a decomposition is impossible, but I have learnt to be quite afraid of non-measurable sets.

Betta George
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    This article (users.math.msu.edu) seems to be helpful. Specifically, look at Theorem 4.1. I haven’t read it carefully myself though. But I think it means such a decomposition of $[0,1)$ is possible. – X-Rui Feb 01 '24 at 14:44
  • Thank you @X-Rui, the results in the article do indeed answer the question (in the affirmative). Would you like to post it as an answer? Otherwise, I shall answer my own question for future visitors. – Betta George Feb 01 '24 at 15:04

1 Answers1

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Using Theorem 4.1 from Non-measurable sets (hosted on joelshapiro.org) by Joel H. Shapiro, we can find such a decomposition of any set of positive measure.

In that article, Shapiro gave a non-measurable subset $\mathit{NM}$ of $[0,1)$ such that any measurable subset of it has measure $0$, and any measurable superset ($\subseteq[0,1)$) of it has measure $1$. Its complement in $[0,1)$, $\mathit{NM}' = [0,1) \setminus \mathit{NM},$ has the same properties; specifically, any measurable subset of it has measure $0$. They already constitute a desired decomposition of $[0,1)$. We can use these two sets to decompose any subset of $\mathbb{R}$ of positive measure.

Let $A, B \subseteq \mathbb{R}$ be $$A = \mathit{NM} + \mathbb{Z} = \bigcup_{n \in \mathbb{Z}} (\mathit{NM}+n),\quad\text{and}\quad B = \mathit{NM'} + \mathbb{Z}.$$ We have $A \cup B = \mathbb{R}$ and $A \cap B = \emptyset$. More importantly, they both have inner measure $0$. Let $E$ be a measurable subset of $A$. $$\begin{split} \mu(E)&=\mu\left(\bigcup_{n\in\mathbb{Z}}E\cap[n,n+1)\right)\\ &=\sum_{n\in\mathbb{Z}}\mu(E\cap[n,n+1))\\ &=\sum_{n\in\mathbb{Z}}\mu(\underbrace{(E-n)\cap[0,1)}_{\subseteq\mathit{NM}})\\ &=\sum_{n\in\mathbb{Z}}0=0. \end{split}$$ Similarly, we can show $B$ has inner measure $0$. Now, given any subset $X$ of $\mathbb{R}$ of positive measure, we can simply decompose it into its intersections with $A$ and $B$, i.e. $X \cap A$ and $X \cap B$. $\square$

X-Rui
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