# The most explicit way of partitioning the reals into two dense subsets with positive measure

In @UmbertoP's response to the question, "Partition of real numbers into dense subsets of positive measure," the answer is understandable to a advanced undergraduate; however, I have inadequate knowledge of measure and set theory.

I tried to create a more explicit example here but it's too complicated.

Question: Is there a more explicit version of Umberto P's answer that's understandable to an average undergraduate?

Maybe you'll like this one a bit better. For simplicity, I'll do my construction for $$I = [0,1)$$ instead of $$\mathbb R$$: to get a partition of $$\mathbb R$$, just repeat my $$A$$ and $$B$$ periodically.
At stage $$n$$ (for positive integers $$n$$) I'll partition $$I$$ into two sets $$A_n$$ and $$B_n$$, each a union of finitely many half-intervals. Start with $$A_1 = [0,1/2)$$ and $$B_1 = [1/2, 1)$$.
Given $$A_n$$ and $$B_n$$: for each interval $$[a,b)$$ in one of these sets of length $$s = b-a$$, remove an interval of length $$2^{-n} s$$ from the centre of the interval and give it to the other set. Thus from $$[0, 1/2)$$ in $$A_1$$, we remove $$[1/8, 3/8)$$ and put it in $$B_2$$, while $$A_2$$ keeps $$[0,1/8)$$ and $$[3/8, 1/2)$$, and from $$[1/2, 1)$$ in $$B_1$$, we remove $$[5/8, 7/8)$$ and put it in $$A_2$$, resulting in $$A_2 = [0,1/8) \cup [3/8,1/2) \cup [5/8, 7/8)$$ and $$B_2 = [1/8, 3/8) \cup [1/2, 5/8) \cup [7/8, 1)$$.
Note that in going from stage $$n$$ to stage $$n+1$$, the measure of the points transferred is $$2^{-n}$$. Since $$\sum_n 2^{-n}$$ is finite, almost every point is transferred only finitely many times. Since sets of measure $$0$$ are negligible, I'll define $$A$$ to consist of the points that are eventually in $$A_n$$, and $$B$$ as its complement the points that are in $$B_n$$ for infinitely many $$n$$, i.e. $$A = \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k, \ B = [0,1) \backslash A = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty B_k$$
• @RobertIsrael Never mind, $A$ and $B$ don't have equal measure, but what are the measure of the sets? Aug 3 at 21:09
• @RobertIsrael I’m not sure how to repeat your $A$ and periodically to get what I want. (I placed a +100 bounty on the post.) Aug 7 at 4:56
• What I mean is $x \in A$ iff $x - \lfloor x \rfloor \in A$. Aug 7 at 14:14