Does a non-measurable set that is not contained in a null set always contain a set of positive measure?

Question

My specific problem is with the Lebesgue measure, but out of curiosity I would also appreciate insights on general measure spaces.

Suppose I am given a non-measurable set $A$ with the property that there exists no zero measure set which contains all of $A$.

Is it then true that there exists a measurable subset of $A$ with non-null measure?

For my specific use case, the underlying space can even be finite-dimensional, in case that makes a difference. I am a graph theorist and woefully uneducated in measure theory.

Answer 1

I assume the answer is no just in $\text{ZFC}$, but the easiest example I could think of assumes the continuum hypothesis. (Martin’s axiom would also work.)

Since we are assuming $\text{CH}$, we may let $\prec$ be a well-ordering on $[0, 1]$ with order type $\omega_1$. Then the set $A = \{(s, t) \in [0, 1] \times [0, 1]: s \prec t\}$ is not Lebesgue-measurable because its indicator function does not satisfy Fubini-Tonelli. It is not contained in any null set since the Lebesgue measure is complete, so any subset of a null set would be automatically measurable and in fact, null. However, all its measurable subset must be null, since one of the iterated integrals of $1_A$ is zero, which by Fubini-Tonelli must be larger than or equal to the measure of any measurable subset of $A$.

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The following is an altered version of the argument that does not require $\text{CH}$:

Again, let $\prec$ be a well-ordering on $[0, 1]$, this time with order type $2^{\aleph_0}$. We again define $A = \{(s, t) \in [0, 1] \times [0, 1]: s \prec t\}$. We claim that $A$ is not measurable. Indeed, were $A$ measurable, by Fubini-Tonelli both iterated integrals of $1_A$ exist and are equal. Recall by Steinhaus theorem that any measurable subset of $[0, 1]$ with positive measure must be of cardinality $2^{\aleph_0}$. Horizontal sections of $A$ are of cardinality smaller than $2^{\aleph_0}$. Since $A$ is assumed measurable, almost all of these sections are measurable, whence they must be null. Thus, the iterated integral of $1_A$ obtained by first integrating w.r.t. the horizontal coordinate must be $0$.

On the other hand, all vertical sections of $A$ have complements of cardinality smaller than $2^{\aleph_0}$. Again, since $A$ is assumed measurable, almost all of these sections are measurable and so are the complements of these sections. Thus, almost all complements of vertical sections are null, so almost all vertical sections are of measure $1$. Thus, the iterated integral of $1_A$ obtained by first integrating w.r.t. the vertical coordinate is $1$. The two iterated integrals do not match, a contradiction. Hence, $A$ cannot be measurable.

Now, if $B \subseteq A$ is measurable. Then horizontal sections of $B$ are contained in horizontal sections of $A$, so they are of cardinality smaller than $2^{\aleph_0}$. Thus, by the same argument as before, the iterated integral of $1_B$ obtained by first integrating w.r.t. the horizontal coordinate is $0$. By Fubini-Tonelli, $B$ is then null, i.e., all measurable subsets of $A$ are null.

Answer 2

In the case of Lebesgue measure there is a very easy counterexample. Let $N$ be any nonmeasurable subset of $\mathbb R$ and let $\mathcal B$ be a maximal collection of disjoint nonnull measurable subsets of $N$. Then $\mathcal B$ is countable, and the set $A=N\setminus\bigcup\mathcal B$ is nonmeasurable and contains no measurable subset of positive measure.

Edit. A commenter asked why $\mathcal B$ is countable. For $m,n\in\mathbb N$ let $\mathcal B_{m,n}=\{B\in\mathcal B:\mu(B\cap[-m,m])\ge\frac1n\}$. Since $\mathcal B=\bigcup_{m,n\in\mathbb N}\mathcal B_{m,n}$, it will suffice to show that $\mathcal B_{m,n}$ is countable. Assume for a contradiction that $\mathcal B_{m,n}$ is uncountable. Choose distinct sets $B_0,B_1,\dots,B_{2mn}\in\mathcal B_{m,n}$. Then $$\mu([-m,m])\ge\mu\left(\bigcup_{k=0}^{2mn}(B_k\cap[-m,m]\right)=\sum_{k=0}^{2mn}\mu(B_k\cap[-m,m])\ge\frac{2mn+1}n\gt2m$$ which is absurd.

Answer 3

The Vitali set, which is the non-measurable set most commonly discussed in measure theory books, is already a counterexample.

Since there are different variants of the construction, I'll sketch the details. Consider the equivalence relation $\sim$ on $[0,1]$ where $x \sim y$ if $x - y \in \mathbb{Q}$. Let $A$ be a set containing exactly one element from each equivalence class (which exists by the axiom of choice).

Let $q_1, q_2, \dots$ is an enumeration of all the rationals in $[-1,1]$. Then the sets $A_n = A + q_n \subset [-1,2]$ are pairwise disjoint, and their union contains all of $[0,1]$. If $A$ were contained in some measurable null set $N$, then letting $N_n = N + q_n$, we would have $m(N_n) = 0$ and $[0,1] \subset \bigcup_{n=1}^\infty A_n \subset \bigcup_{n=1}^\infty N_n$. The monotonicity and countable sub-additivity of Lebesgue measure would then imply $$1 = m([0,1]) \le \sum_{n=1}^\infty m(N_n) = \sum_{n=1}^\infty 0 = 0$$ which is absurd.

Now suppose $B$ is any measurable subset of $A$. I claim $B$ must have measure zero. Indeed, let $\beta = m(B)$. If we let $B_n = B + q_n \subset A_n \subset [-1,2]$, then the sets $B_n$ are pairwise disjoint, and all are measurable with $m(B_n) = \beta$ by the translation invariance of Lebesgue measure. Let $C = \bigcup_{n=1}^\infty B_n$. By countable additivity, $m(C) = \beta + \beta + \dots$. On the other hand, $C \subset [-1,2]$ so $m(C) \le 3$. So we must have $\beta = 0$.