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.

Thanks in advance for your help!

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

  • $\begingroup$ What outer measure are you using? $\mu^*(E) = \inf\{\sum_{n=1}^\infty \mu(A_n) : (A_n) \subseteq \mathcal{X} \text{ and } E \subseteq \bigcup_{n=1}^\infty A_n\}$? $\endgroup$
    – Nick F
    Commented Apr 30, 2023 at 17:03
  • $\begingroup$ @NickF Yes, I thought that was the standard way of defining an outer measure from a measure. $\endgroup$
    – AJY
    Commented Apr 30, 2023 at 17:15
  • 1
    $\begingroup$ 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. $\endgroup$
    – GEdgar
    Commented Apr 30, 2023 at 17:23

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – AJY
    Commented May 2, 2023 at 17:25

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