3

Is it possible to partition the unit interval $[0,1]$ into two disjoint sets $A$ and $B$ such that:

  1. $A$ and $B$ are totally disconnected
  2. $\lambda(A)>0$ and $\lambda(B)>0$, where $\lambda$ is the Lebesgue measure

Of course you can have a partition like this: $A=\mathbb{Q}\cap[0,1]$ and $B=A^C\cap[0,1]$, but in this case $\lambda(A)=0$.

And from this answer https://math.stackexchange.com/a/2729230, I sense using fat Cantor sets may not be a good idea, since they are closed.

So, I'm not able to come up with such partition nor prove that it does not exist, and that's what I'm wondering: is it possible?

francescop21
  • 1,124

1 Answers1

4

A simple example is like this: $$A=\left(\Bbb Q\cap \left[0,\frac12\right]\right)\cup\left(\left[\frac12,1\right]\setminus\Bbb Q\right)\\B=\left(\Bbb Q\cap \left(\frac12,1\right]\right)\cup\left(\left[0,\frac12\right]\setminus\Bbb Q\right)$$

An example such that $\lambda(A\cap (a,b))>0$ and $\lambda(B\cap (a,b))>0$ for all $a,b$ would be more interesting.

Notice that if you consider any $A$ such that $0<\lambda(A\cap I)<\lambda(I)$ for all intervals $I$ (see "Creating a Lebesgue measurable set with peculiar property." and/or "Construction of a Borel set with positive but not full measure in each interval"), this $A$ is necessarily totally disconnected: if $x,y\in A$ were in the same connected component, then $[x,y]\subseteq A$ and $\lambda(A\cap (x,y))=\lambda(x,y)$. Moreover, the complement $B=[0,1]\setminus A$ of any such set satisfies the same properties.