Is there a dense subset of [0,1] of measure 1/2 whose complement is also dense?

Question

I want to find a set $A \subset [0,1]$ so that:

1. $A$ is dense in $[0,1]$
2. $A^c$ is dense in $[0,1]$
3. $A$ is Lesbesgue measure $1/2$ (Failing this....I want both sets to be positive measure)

My first thought is to take the set of irrational numbers, and then use some criteria to subdivide them into two sets, so that each set is still dense. The subdivision will be "symmetric" in some way so that each set is still measure 1/2. Alternivly, maybe there is a construction using something from Ergodic theory?

Remark: What I really want is a set $A$ so that:

$$\int_A f(x) d\lambda(x) - \int_{A^c} f(x) d\lambda(x) = 0$$ for every continuous function $f:[0,1]\to\mathbb{R}$. I think the above conditions will be close to doing that.

Answer 1

I can give you an example which is not very symetric. It is classical that there are Cantor sets with arbitrary Lebesgues measure (between 0 and 1). To construct it, consider the sequence $K_n$ of compacts subsets of $[0,1]$ obtained by the following method : let $K_0 = [0,1]$. To obtain $K_{n+1}$, remove from each connected component of $K_n$ an interval centered at the middle of this connected component of length $\alpha_n l_n$, where $0<\alpha_n<1$ and $l_n$ is the length of said connected component. Now let $K = \cap_n K_n$. It is a non-vacuous compact subset of $[0,1]$ and by chosing an appropriate sequence of $\alpha_n$, you can have $m(K)=\frac{1}{2}$ where $m$ is the Lebesgues measure.

Now $K$ the only problem is that $K$ is not dense (its complement is). But that is easy to remedy : just let $\tilde{K} = K \cup (\mathbb{Q} \cap [0,1])$. It does not change the measure of $K$ and its complement since the rationals are countable, but now both $\tilde{K}$ and its complement are dense.

Answer 2

• We can find an example of a set which satisfies 1., 2. and 3., for example $$A:=(\Bbb Q^c\cap ([0,1/4)\cup[3/4,1)))\cup (\Bbb Q\cap [1/4,3/4)).$$

• However, there is no set $A$ which satisfies the latest condition in the OP. Indeed, fix $x\in A$ and $r>0$. We can approximate the characteristic function of $B(x,r)$ pointwise by continuous functions. By dominated convergence theorem, we would have that $2\lambda(A\cap B(x,r))=\lambda(B(x,r))$, and we conclude by a similar argument as here.

Answer 3

A simple example for the first three conditions is to define $S=[0,1] \cap (\mathbb {R \setminus Q})$, then $A=([0,1/2] \cap S) \cup ((1/2,1] \cap \mathbb Q)$. It clearly does not satisfy the integral condition. Maybe it is more anti-symmetric than symmetric.