# Does a set of positive outer measure contain a *measurable* set of positive measure?

Given a complete measure space $$(X, \mathcal{X}, \mu)$$, and a subset $$A \subseteq X$$ of positive outer measure, does there necessarily exist a subset $$E$$ of a $$A$$ which is measurable for which $$\mu(E) > 0$$?

This feels intuitively like it should be true, but I don't see how to show it, since it involves inner approximations and outer measure is generally defined by reference to outer approximations. One thought I had was to exploit the idea of inner regularity for Borel measures, but I think that definition is typically only stated as the ability to approximate measurable sets from within. However, I want to find a positive-measure and measurable subset of a potentially non-measurable set.

EDIT: Since somebody asked, by the outer measure I mean $$\mu^*(A) : = \inf \left\{ \sum_{n = 1}^\infty \mu(E_n) : E_n \in \mathcal{X}, A \subseteq \bigcup_{n = 1}^\infty E_n \right\} .$$

• What outer measure are you using? $\mu^*(E) = \inf\{\sum_{n=1}^\infty \mu(A_n) : (A_n) \subseteq \mathcal{X} \text{ and } E \subseteq \bigcup_{n=1}^\infty A_n\}$? Commented Apr 30, 2023 at 17:03
• @NickF Yes, I thought that was the standard way of defining an outer measure from a measure.
– AJY
Commented Apr 30, 2023 at 17:15
• In general, no. Let $X= [0,1]$ with Lebesgue measure. A Bernstein set $A\subseteq [0,1]$ has Lebesgue outer measure $1$ and inner measure $0$. So any subset $E \subseteq A$ has inner measure $0$. In particular, $E$ is not measurable with positive measure. Commented Apr 30, 2023 at 17:23

Not necessarily. In fact, we can prove that for all non-measurable $$A$$ such that $$\mu^*(A)<\infty$$ there is some non-measurable $$B\subseteq A$$ such that all measurable subsets of $$B$$ are null sets and (necessarily) $$\mu^*(B)>0$$.

Let $$A$$ be as in the hypothesis and let $$s=\sup_{C\subseteq A\\ C\in \mathcal X}\mu^*(C)$$, consider $$s_n\nearrow s$$ and $$C_n\subseteq A$$ measurable such that $$\mu^*(C_n)=s_n$$. Call $$C=\bigcup_{n\in\Bbb N} C_n$$. $$C\in \mathcal X$$ and it's clear by monotonicity and by the definition of $$s$$ that $$\mu^*(C)=\mu(C)=s$$.

Now, let's call $$B=A\setminus C$$. $$B\notin \mathcal X$$, otherwise $$A=C\cup B$$ would be too. In particular, $$\mu^*(B)>0$$. By Carathéodory's condition, $$\mu^*(B)=\mu^*(A)-\mu^*(C)=\mu(A)-s$$

If $$B$$ had a measurable subset $$H$$ of positive measure, then $$C\cup H$$ would be a measurable subset of $$A$$ of measure $$\mu(C\cup H)=\mu(C)+\mu(H)-\mu(C\cap H)=\mu(C)+\mu(H)-\mu(\emptyset)=\mu(C)+\mu(H)>s$$ Which is against the definition of $$s$$.

The classical Vitali set in the Lebesgue measure space $$(\mathbb{R}, \mathcal{M},m)$$ serves as a counterexample for what you describe. The details are laid in DiBenedetto's Real Analysis, 2nd ed, exercises from section 13c, if you want to see for yourself. I can provide a brief summary here.

If $$A \subseteq [0,1]$$ is a Vitali set, then every Lebesgue measurable $$E \subseteq A$$ must have $$m(E) = 0$$. For if we enumerate $$\mathbb{Q} \cap [0,1) = \{r_n : n\in\mathbb{Z}^+\}$$, and we take the modulo-$$1$$ translates of $$E$$ by the enumeration, i.e., $$E_n = \{x+r_n : x \in E \cap [0,1-r_n)\} \cup \{x+r_n-1 : x \in E \cap [1-r_n,1)\},$$ then each $$E_n$$ is measurable with $$m(E_n) = m(E)$$. However, $$\bigcup_{n=1}^\infty E_n \subseteq [0,1]$$, hence $$\sum_{n=1}^\infty m(E) = \sum_{n=1}^\infty m(E_n) \leq m\left(\bigcup_{n=1}^\infty E_n\right) \leq m([0,1]) = 1,$$ so $$m(E)$$ cannot be positive lest the far LHS be infinite. Thus, $$m(E) = 0$$.

However, $$m^*(A) > 0$$. For if $$m^*(A) = 0$$, then $$A$$ is a Lebesgue null-set and hence measureable, a contradiction. Therefore, $$A$$ is a set with positive outer measure which does not contain any measurable subset with positive measure.

• Good point. I hadn't considered that subsets of a Vitali set could be reasoned about in the same way we reason about the Vitali set.
– AJY
Commented May 2, 2023 at 17:25