2

Let $\mu$ be a Borel probability measure on $\mathbb{R}$ and let $S\subseteq\mathbb{R}$ be a Borel set such that $\mu\left(S\right)>0$. Is it true that for any $A_1,A_2\subset\mathbb{R}$ that are disjoint and whose union is $S$, there exist $\widetilde{A_1}\subseteq A_1$ and $\widetilde{A_2}\subseteq A_2$ which are measurable, and either $\mu\left(\widetilde{A_1}\right)>0$ or $\mu\left(\widetilde{A_2}\right)>0$?

user754663
  • 81
  • How about taking an unmeasurable set $U \subset [1,2]$ and $V= U^c \cap [1,2]$. Then have $A_1=[0,1) \cup U$ and $A_2 = V \cup (2,3]$ so $A_1 \cup A_2=[0,3]$? – Henry Aug 07 '23 at 15:53
  • In that case, both $A_1$ and $A_2$ have measurable subsets of positive measure. I meant to ask if this is true generally. I'll edit so that it is more clear. Thanks! – user754663 Aug 07 '23 at 15:57
  • OK - I read your original "is it possible to find" as "are there any examples" while you seem to have intended "is it always the case" – Henry Aug 07 '23 at 16:00

1 Answers1

2

This is false. Take $\mu$ to be the Lebesgue measure on $\mathbb R$ restricted to the Borel $\sigma$-algebra and $S = [0, 1]$. Let $A_1 = V$ be a Vitali set in $[0, 1]$ of outer measure $1$ (constructed here) and $A_2 = [0, 1] \setminus V$. Then $A_1$ and $A_2$ partition $S$.

Consider any measurable sets $\tilde A_1 \subseteq A_1$ and $\tilde A_2 \subseteq A_2$. Since any measurable subset of a Vitali set has measure zero, we must have $\mu(\tilde A_1) = 0$. Then note that $E := [0, 1] \setminus \tilde A_2 \supseteq [0, 1] \setminus A_2 = V$ is a measurable set containing $V$.

By the definition of Lebesgue measurability, for any $\varepsilon > 0$, there is some open set $O_\varepsilon \supseteq E$ with $\mu(O_\varepsilon \setminus E) < \varepsilon$. Thus, letting $\mu_*$ denote the outer measure, we have $$\mu_*(V) = \inf_{\text{open } O \supseteq V} \mu(O) \leq \mu(O_\varepsilon) = \mu(E) + \mu(O_\varepsilon \setminus E) < \mu(E) + \varepsilon.$$ As this holds for all $\varepsilon > 0$, we must have $\mu_*(V) \leq \mu(E) \leq \mu([0, 1]) = 1$. But since $\mu_*(V) = 1$, this implies that $\mu(E) = 1$ and hence $\mu(\tilde A_2) = \mu([0, 1]) - \mu(E) = 1 - 1 = 0$.

Joey
  • 113