Can two disjoint sets have equal measure on any interval in a given interval?

[dimitri151]

I saw this problem on this site a while back and started to think about it. I can't find the post so I'll start it anew. The problem is: can you have two disjoint sets dense on an interval so that the measure of each set on any interval of that interval is equal? That is, say you have A, B in [0,1] A(intersection),B=emptyset, are there such sets A and B so that if you take x<y, x,y in [0,1], m(A(intersection)[x,y])=m(B(intersection)[x,y])?

If you let A_n={x in [0,1] : x's binary expansion has a zero in the nth place }, then you can show that for any epsilon there is a pos integer N such that m(A_n(intersection)[x,y])-m(~A_n(intersection)[x,y])<epsilon when n>=N.
However this isn't equality. Furthermore what I'd really like to say is the required set A is lim_n->infinityA_n. Only, I'm not sure what {x in [0,1] : x's binary expansion has a zero in the infinitieth place} means. That doesn't mean anything. So how does one proceed?

 

[disregardthat]

Let A = the set of rationals on the form k/2^n in (0,1) where k, n are integers, and B the set of k/3^n in (0,1). These sets are disjoint and both are dense in [0,1]. Being countable, each have measure 0 on any interval.

 

[dimitri151]

Oops, I forgot to add that the sets had to be of positive measure.

 

[disregardthat]

Ok, what about this:

A_n = [0,\frac{1}{2^n}) \cup [\frac{2}{2^n},\frac{3}{2^n}) \cup ... \cup [\frac{2^n-2}{2^n},\frac{2^n-1}{2^n})

B_n = [0,1)-A_n

Let A = \liminf_n A_n = \cup_I \cap_{n \in I} A_n over cofinite subsets I in \mathbb{Z}^+, and B = \liminf_n B_n

If x is in A \cap B, x is in cofinitely many A_n's and cofinitely many B_n's, hence in some A_n \cap B_n which is impossible, so they are disjoint.

It is clear that any dyadic number 0 \leq \frac{k}{2^n} &lt;1 is in A, since \frac{2 k 2^{N-1-n}}{2^N} \leq (=) \frac{k}{2^n} &lt; \frac{2 k 2^{N-1-n}+1}{2^N} for all N \geq n+1. Hence A is dense.

Consider \overline{B} = \overline{\cup_I \cap_{n \in I} B_n} \supseteq \cup_I \overline{\cap_{n \in I} B_n} = \cup_I \cap_{n \in I} \overline{B_n}.

\overline{B_n} = [\frac{1}{2^n},\frac{2}{2^n}] \cup [\frac{3}{2^n},\frac{4}{2^n}] \cup ... \cup [\frac{2^n-1}{2^n},1]

A dyadic number 0&lt;\frac{k}{2^n} \leq 1 is in \overline{B}, since \frac{2 k 2^{N-1-n}-1}{2^N} \leq \frac{k}{2^n} \leq (=) \frac{2 k 2^{N-1-n}}{2^N} for all N \geq n+1, so B is also dense (since \overline{B} is dense).

I think these sets will have positive and equal measure, I will come back to that later. Hope this helps anyway.

EDIT: I calculated the measure of A, it is 0, oh well..

 

[Werg22]

Pick two distinct points p and q in [0,1] and let A = {p} and B = {q}. Cut the interval in half, to get [0,1/2] and [1/2,1] and choose a pair of distinct points from each ({a,b} and {c,d} say). Throw into A one point from each interval (say a and c), and do the same for B (b and d). Repeat this process ad infinitum: at stage n you are selecting 2^n points from 2^n-1 intervals and adjoining a selected point from each interval to A, and throwing the rest in B. Always choose distinct points; this is possible because at each stage there are only a finite number of points to avoid, whereas each interval you are selecting from has an infinite number of points. When 'done', the sets A and B so obtained should be what you are looking for.

 

[dimitri151]

I think those sets are null sets.

 

[Werg22]

I think those sets are null sets.

Of course, since they are countable. But you didn't say they had to have positive measure...

 

[dimitri151]

I added the condition in a subsequent post. Solution of the conditions with countable sets is almost trivial.