This question arose a few years back when I was an assistant teacher on a course of basic (Lebesgue) measure theory, but I didn't find an answer or anyone able to solve the problem. The setting of the problem is as follows:

We say that $A$ has full outer measure in the unit interval $[0,1]$ if $$m^\ast(A) = m^\ast([0,1]).$$ How many disjoint subsets of the unit interval with full outer measure can there exist? By the addivity property of the Lebesgue measure these sets have to be non-measurable. In this MO post Gerald Edgar shows that by using the Bernstein set you can compose any measurable subset $A$ of $\mathbb{R} ^n$ into two disjoint subsets, both with full outer measure in $A$. This gives a partial answer of "at least two" to my question.

Question: Can the 'Bernstein set -construction' be modified (or is there some other method) to create $n$ disjoint subsets of $[0,1]$ with full outer measure in the unit interval? Can there exist infinitely many disjoint subsets of the unit interval with full outer measure? If so, then can this family be uncountable?

  • $\begingroup$ On the flip side of this, Shelah (Sh:619) has shown that possibly there is some non null subset $X$ of $[0, 1]$, which cannot be partitioned into uncountably many non null sets! $\endgroup$ – Ashutosh Jan 26 '14 at 17:32

In this article, J. Cichon shows that the real line (and hence the unit interval) can be partitioned into continuum many Bernstein subsets. Of course, any Bernstein set has full outer measure.

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    $\begingroup$ Thank you, "it is easy when you know how to do it". I have to say that I find the answer quite astonishing. I would have found the answer "Any finite number can be found." almost natural, especially after the Banach-Tarski paradox, and I even might have accepted the answer "Countably infinite amount of such sets exist." with some peace of mind. But having a continuum of such sets feels just uncanny. I am confused and perplexed, but if I am to believe the words of Eliezer Yudkowsky, acknowledging confusion is an important part in finding understanding. $\endgroup$ – Rami Luisto Jan 23 '13 at 13:56

In 1917 Luzin and Sierpinski proved there exists continuum many pairwise disjoint subsets of the interval $[0,1]$ such that each of these subsets has outer Lebesgue measure $1.$

Nikolai N. Luzin and Waclaw Sierpinski, Sur une décomposition d'un intervalle en une infnité non dénombrable d'ensembles non mesurables [On a decomposition of an interval into nondenumerably many nonmeasurable sets], Comptes Rendus Académie des Sciences (Paris) 165 (1917), 422-424.

See Gallica site for C. R. Paris volumes or Internet Archive copy of volume 165

Although I've mentioned this paper in various internet groups several times over the past 10+ years, I don't think I've ever mentioned why I find it so fascinating. First, that such an amazingly strong result was published so early -- little more than a decade after non-measurable sets were known to exist. Second, that such a result is so rarely mentioned in texts that deal with measure theory, despite that fact that very little background is needed to state the result, which can also easily be done in a one-sentence footnote.

  • $\begingroup$ I have to say that I am also impressed by this paper, among other things for the reasons you mentioned. I find it odd that I've never seen it even mentioned. I have to give a hint to my professors that write their own lecture notes for the courses on measure theory. $\endgroup$ – Rami Luisto Jan 24 '13 at 10:34

I can solve this assuming the continuum hypothesis. (Edit: CH isn't needed, see below.) Lemma: if $A$ is a countable set and $(S_m)$ is a sequence of uncountable sets then we can find a sequence of disjoint countable sets $(T_n)$ such that $A \cap T_n = \emptyset$ for all $n$ and $S_m \cap T_n \neq \emptyset$ for all $m$ and $n$. [Proof: Choose a countable subset $S_1'$ of $S_1 \setminus A$, enumerate it, and put the $n$th element in $T_n$. Then choose a countable subset $S_2'$ of $S_2 \setminus (A \cup S_1')$, enumerate it, and put the $n$th element in $T_n$. Proceed in this way. Each $S_k'$ is countable so the difference $S_{k+1} \setminus (A \cup S_1' \cup \cdots \cup S_k')$ is always uncountable.]

Now there are $2^{\aleph_0}$ open subsets of $[0,1]$ because any open subset is a union of rational intervals and there are only countably many rational intervals. So there are only $2^{\aleph_0}$ closed subsets of $[0,1]$. Assume CH and enumerate the closed subsets of $[0,1]$ of positive measure as $C_\alpha$ for $\alpha < \aleph_1$. Observe that each $C_\alpha$ is uncountable. Now we construct disjoint countable sets $T_{\alpha,\beta}$ for $\beta < \alpha < \aleph_1$ by recursion on $\alpha$ as follows. At step $\alpha$ let $A = \bigcup_{\beta' < \alpha' <\alpha} T_{\alpha',\beta'}$ (all the $T$s constructed so far, a countable union of countable sets) and apply the lemma to this $A$ and the sets $C_\beta$ for $\beta < \alpha$. There are only countably many $\beta < \alpha$ so we can do this, and we can relabel the resulting sets $T_n$ as $T_{\alpha,\beta}$ for $\beta < \alpha$.

After this process is complete, for each $\beta$ let $T_\beta = \bigcup_{\alpha > \beta} T_{\alpha,\beta}$. Then the sets $T_\beta$ are disjoint, there are $\aleph_1$ of them, and each one intersects every closed subset of $[0,1]$ of positive measure, so each of them has full outer measure.

Edit: actually this doesn't need CH. Every closed set is the union of a countable set and a perfect set, so if it has positive measure then it contains $2^{\aleph_0}$ elements. That's enough to keep the induction going for $\alpha < 2^{\aleph_0}$ since each $T_{\alpha,\beta}$ will have cardinality $< 2^{\aleph_0}$.

  • $\begingroup$ Thank you for your answer, I'd like to tag both your and Ramiro's answer as an accepted answer, but I have to choose one. Ramiro's answer was more on the spot (as the question mentioned generalizing Bernstein set constructions) but I do like your answer as it is more constructive. (Even though I have to admit that I have spend some time reading Cichon's article before I can say with certainty whether or not the constructions are fundamentally different.) $\endgroup$ – Rami Luisto Jan 23 '13 at 13:46

One could try to find a Vitali-type set with full outer measure.

  • $\begingroup$ The Vitali construction will give you $\aleph_0$ disjoint sets of full outer measure. But (as we now know) this would not be the best result. $\endgroup$ – Gerald Edgar Jan 23 '13 at 14:46
  • $\begingroup$ Yes I got it. I wrote this before Ramiro posted the reference to the very nice result of Cichon. Anyway, I am now wondering how to construct a Vitali set with full outer measure in R. $\endgroup$ – Carlo Mantegazza Jan 23 '13 at 15:01
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    $\begingroup$ I assume that by "Vitali set" you mean a set that contains exactly one member from each coset of the reals mod the rationals. It's not difficult to obtain such a set with full outer measure. Use the usual transfinite-induction construction of a Bernstein set, but make all choices from different cosets mod the rationals. (At the end, if you haven't got elements from all cosets, throw in additional elements, which can only increase the outer measure.) $\endgroup$ – Andreas Blass Jan 26 '14 at 14:58

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