A Can the Unit Interval be Measurably Split into Two Sets of Equal Length?

[mathman]

TL;DR Summary

Splitting Unit interval into two measurable sets.

Can the unit interval be split into two measurable sets (A and B each measure 1/2), so that for any sub-interval [c,d] the intersections $A\cap [c,d]$ and $B\cap [c,d]$ each measure half the length of [c,d]? I doubt it, but I would like to see a proof, one way or the other.

 

[Infrared]

The Lebesgue density theorem shows this to be impossible (https://en.wikipedia.org/wiki/Lebesgue's_density_theorem)

 

[andrewkirk]

I think it could be done for the rationals, but perhaps not for the reals. A partition of the rational unit interval $J\triangleq [0,1]\cap\mathbb Q$ that I think might work is as follows:

Let $f:\mathbb N\to\mathbb Q$ be the enumeration of the rationals that considers them in maximally reduced form (no common factors of numerator and denominator) and orders them first by denominator, then numerator, viz: 0/1, 1/1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5,...
Consider the partition $P:\{1,2\}\to 2^J$ such that $q\in J$ is in $P(1)$ if $f^{-1}(q)$ is odd, otherwise it is in $P(2)$. So our two parts are, showing just the numerators in columns two and three:

Denominator	Numerators in P(1)	Numerators in P(2)
1	0	1
2	1
3	2	1
4	3	1
5	2, 4	1, 3
6	3	1,5
7	1,3,5	2,4,6
...

Neither part of this partition contains any intervals and I cannot see any biases in it that would cause accumulation points of other material variations in density for either P(1) or P(2). There would be work in proving it satisfied the sought criterion, but I guess that it could be done.

For the real case, I would imagine trying a non-constructive proof by contradiction. Let $\mathscr P$ be the set of all two-way partitions of $I$, $\mathscr K$ be the set of all sub-intervals of [0,1] and define $h:\mathscr P\to [0,1]$ by
$$h(P) =\sup_{\alpha\in\mathscr K} \frac{\max(|\alpha\cap P(1)|, |\alpha\cap P(2)|)}{|\alpha|}$$
where $|S|$ indicates the measure of set S, and
$$r =\inf_{P\in\mathscr P} h(P)$$

If $r>0.5$ we try to construct a partition P such that $0.5<h(P)<r$, thereby giving a contradiction and proving that $r=0.5$.

If we can prove that $r=0.5$ then we try to prove that the inf is a min by constructing the min, and that min will be a partition that satisfies the criterion. I think that would be the hardest bit, and the most likely place where the attempt fails.

I'm moderately confident we could make $\mathscr K$ the set of intervals with rational endpoints without invalidating the proof. That could make it easier because then $\mathscr K$ will be enumerable and we might use the enumeration function for things like constructing maxima or minima or generating contradictions.

EDIT: I didn't see the post about the Lebesgue Density Theorem until I had finished posting this. It sounds like we will have $r=1$.

 

[Infrared]

@andrewkirk What do you mean by $|\alpha\cap P(1)|$? These sets will be infinite, so it doesn't make sense to just count elements. But there's also not a good measure to put on $[0,1]\cap\mathbb{Q}$ since it's countable and infinite.

 

[andrewkirk]

@andrewkirk What do you mean by $|\alpha\cap P(1)|$? These sets will be infinite, so it doesn't make sense to just count elements. But there's also not a good measure to put on $[0,1]\cap\mathbb{Q}$ since it's countable and infinite.

I am using the mod signs to indicate the measure. I meant to write that in but forgot. Will put it in now.

 

[Infrared]

What measure are you using? You can't use the usual measure if you're only working with rationals (or else everything will have measure zero)

 

[zinq]

This is something I thought about a long time ago. It's very interesting that the half-open interval [0,1) is not the disjoint union of two measurable sets that intersect each subinterval [a,b) in half its measure, (b-a)/2.

I believe that by using the Axiom of Choice it is possible to partition [0,1) into two disjoint isometric dense sets A and B , that are in fact metrically homogeneous (in the sense that for any L > 0 they each intersect each interval of length L in isometric sets A ∩ L and B ∩ L). But this cannot be done measurably.