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Is there a set $A \subset [0,1]$ such that for every Borel set $B \subset [0,1]$ of positive Lebesgue measure, both $B \cap A$ and $B \setminus A$ are non-empty?

This is, in a sense, the "measure-theoretic analogue" of the more obvious topological question: is there a set $A$ such that for every non-empty open $U \subset [0,1]$, both $U \cap A$ and $U \setminus A$ are non-empty? (For which the answer is obviously $A:=\mathbb{Q}$.)

Now it is clear that $A$ cannot itself be a Borel set of non-trivial Lebesgue measure (just take $B=A$). My intuition is that $A$ cannot be any Lebesgue-measurable set.

I thought about taking $A$ to be a set consisting of one point from every orbit of $x \mapsto x+\sqrt{2} \; \mathrm{mod} \; 1$, or at least a union of such sets. But I'm not sure if this gets anywhere.

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Julian Newman
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As shown in this question there are $\mathfrak c$ Borel sets, which can be extended to the fact that there are $\mathfrak c$ Borel sets of positive measure. Each one with positive measure must have continuum many points. Put them in a well order in type $\mathfrak c$, so each one has $\lt \mathfrak c$ predecessors. Take each set in turn and find two points that have not been accounted for yet. Put one in $A$ and one in $\overline A$. As by the time you get to each set you have accounted for less than $\mathfrak c$ points, you can always find two new points to work with. If you have any points not accounted for when you are done with the positive Borel sets, put them wherever you want.

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Ross Millikan
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  • This in fact proves that $A$ can be taken such that $A$ and $[0,1] \setminus A$ intersect every Borel set of size $\mathfrak{c}$. – Julian Newman Mar 11 '17 at 18:42
  • If you only run this construction for perfect sets (rather than Borel sets of positive measure) you get what is known as a Bernstein set. It's a very similar beast, and it is enough to answer the question. In fact, a Bernstein set meets every Borel set of size $\frak c$, not just those of positive measure. This is because every Borel set is countable, or has a perfect subset. – Asaf Karagila Mar 13 '17 at 07:53
  • @AsafKaragila: I didn't use the fact that the sets were Borel at all. I presume this needs AC to get the well order of the Borel sets. I wondered if there was a way to use the Borel nature to avoid AC, but nobody has shown one. – Ross Millikan Mar 13 '17 at 15:04
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    Ross, of course not. As noted, such a set cannot be measurable, and it is consistent that all sets are measurable. In fact, it is consistent that all sets are Borel (but then measure theory goes down the drain anyway, so let's not think about that scenario). – Asaf Karagila Mar 13 '17 at 15:05
  • @AsafKaragila: Why does measure theory go down the drain, may I ask? If we just take ZF+DC, we still get most of measure theory. Does ZF+DC prove that there is a non-Borel set? – user21820 Mar 25 '17 at 13:43
  • @user21820: If the real numbers are a countable union of countable sets, then the Borel measure is already not a $\sigma$-additive measure. You can correct for this by requiring that only Borel sets with codes are measured, and this does let you salvage a significant part of measure theory. But it is no longer as "nice" to work with, since Borel codes mean we do things effectively, which has a far more constructive flavor to it. With even just countable choice, though, you can prove every Borel set has a code, so certainly there will be non-Borel sets. So, yes, ZF+DC will do. – Asaf Karagila Mar 25 '17 at 13:45
  • @AsafKaragila: Thanks for that. It's clearly my lack of topology knowledge showing here; I didn't know about Borel codes, which I found described here, together with some extra information about handling measure theory with them. =) – user21820 Mar 25 '17 at 16:03