9
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ I am not sure if I understand your question. Given any subset of a non-trivial measure space, there will be a set of strictly positive measure containing it (namely the entire space). $\endgroup$ Jan 22 at 11:00
  • $\begingroup$ @ThusleGadelankz I believe the OP is asking about the existence of a measurable subset (of positive measure) of a non-measurable set within a measure space, instead of containing the said non-measurable set. $\endgroup$
    – David Gao
    Jan 22 at 11:07
  • $\begingroup$ Why do you say "with the property that there exists no zero measure set which contains all of $A$? That seems redundant, since you have already said that $A$ is nonmeasurable. A subset of a null set is a null set, right? $\endgroup$
    – user14111
    Jan 22 at 12:22
  • 2
    $\begingroup$ @user14111 Technically, a subset of a null set could be non-measurable if the measure space isn’t complete. $\endgroup$
    – David Gao
    Jan 22 at 12:27
  • 1
    $\begingroup$ @user14111 I was trying to accomodate for non-Lebesgue measures, which may not be complete. $\endgroup$ Jan 22 at 18:39

3 Answers 3

12
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ Fair enough! I am accepting David Gao's answer because it illustrates nicely what exactly "goes wrong", but I like the simplicity of your argument. $\endgroup$ Jan 22 at 18:45
  • $\begingroup$ Why is $\mathcal B$ countable? I'd think of the usual "choose a rational in every set" argument to show the countability of a disjoint set of intervals with non-empty interior, but non-null, measurable sets can have empty interior. $\endgroup$
    – LSpice
    Jan 22 at 20:40
  • 2
    $\begingroup$ @LSpice $\mathcal{B}$ contains pairwise disjoint non-null measurable sets. We fix a sequence of finite measure sets $A_i$ whose union is $\mathbb{R}$. No $A_i$ can have non-null intersections with uncountably many elements of $\mathcal{B}$, as $A_i$ has finite measure and the sum of uncountably many positive numbers is infinite. So there are only countably many elements of $\mathcal{B}$ in total that can intersect with any of the $A_i$ in a non-null way… $\endgroup$
    – David Gao
    Jan 22 at 23:10
  • 2
    $\begingroup$ @LSpice $\cup_i A_i = \mathbb{R}$, so everything else in $\mathcal{B}$ must be null, and since we are assuming $\mathcal{B}$ contains only non-null sets, there’s actually nothing else, i.e., $\mathcal{B}$ must be countable. $\endgroup$
    – David Gao
    Jan 22 at 23:12
  • 3
    $\begingroup$ @LSpice: More generally, in any finite or $\sigma$-finite measure space, any collection of disjoint measurable sets with nonzero measure is necessarily at most countable. The proof given here generalizes easily. $\endgroup$ Jan 22 at 23:26
6
$\begingroup$

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 $A$.


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.

$\endgroup$
2
  • $\begingroup$ Nice! Can you explain as if I were stupid why we need CH for that? Why does any well-ordering provided by AC not suffice? $\endgroup$ Jan 22 at 11:55
  • 2
    $\begingroup$ @BettaGeorge Thinking about this more, you’re right. The argument can work without CH. The argument is just more complicated in that case because we could have neither iterated integrals exists (and that, even if it does exist, it’s no longer straightforward to argue it’s 0 and 1), so it becomes harder to argue all measurable subsets of $A$ are null. In the presence of CH, the horizontal sections are all countable and the vertical sections are all co-countable, so the iterated integrals both exist and are straightforward to compute. $\endgroup$
    – David Gao
    Jan 22 at 12:04
5
$\begingroup$

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 $\beta = 0$.

$\endgroup$
1
  • 1
    $\begingroup$ And the same goes for a Bernstein set, which is another commonly discussed type of nonmeasurable set. A Bernstein set by definition contains no uncountable closed (or Borel) set, and therefore contains no set of positive Lebesgue measure. $\endgroup$
    – user14111
    Jan 23 at 1:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .