$E$ and $E^c$ dense in $\mathbb{R}$ with measure $>0$

Question

Is it a way to construct such a space $E$ (the measure is intended as the Lebesgue measure), with this additional property :

$$ \forall u,t\in \mathbb{R}, \ (u\neq v), \qquad \lambda(E\cap [u,v] )=\lambda(E^c\cap [u,v])>0 $$

For exemple, is it a simple way to part irrationals in two "equal" and "equiprobable" sets (and both still dense in $\mathbb{R}$)?

I asked myself this question for I wanted to show a function everywhere discontinuous and not equal a.e. to a function piecewise continuous (for exemple $1_{\mathbb{Q}}=1$, a.e).

Edit Thank you for this firsts answer, they have pretty good ideas in it but they made me realize that I should have been more precise (the title is not enough!) : I will then change a bit the initial text.

Answer 1

Why not do something like:

$$f(x)=\begin{cases}0 & x\in\mathbb Q,&x\in [2n, 2n+1)\\1 & x\in\mathbb R\setminus\mathbb Q,&x\in [2n, 2n+1)\\1 & x\in\mathbb Q,&x\in [2n+1, 2n+2)\\0 & x\in\mathbb R\setminus\mathbb Q,&x\in [2n+1, 2n+2)\end{cases}, n\in\mathbb Z$$

and set $E=f^{-1}(0), E^c=f^{-1}(1)$?

It looks to me that it satisfies your conditions, but does not quite "mix" well the elements from $E$ and $E^c$.

Answer 2

Yes, it's possible. Let $$ E = \big(\Bbb Q\cap(-\infty, 0]\big)\cup\big(\Bbb Q^c\cap(0, \infty)\big) $$ In other words, $E$ is the set of all the rational negative numbers and all the irrational positive numbners, giving both $E$ and $E^c$ positive measure.

As for an everywhere discontinuous function which is not a.e. equal to a continuous function, that's an entirely different question. The function $1_E$ will not be equal a.e. to a continuous funciton, because of the jump that such a continuous function would have to have at $0$, but it's close (by which I mean, for any $\epsilon>0$, there is a continuous function which disagrees with this one on a set of measure smaller than $\epsilon$).

I you think this feels like "cheating", then I can understand that feeling. Maybe what you had in mind was a set $E$ such that for any interval $(a, b)\subseteq \Bbb R$, we have $0<\mu(E\cap (a, b))<b-a$ (maybe even $\mu(E\cap (a, b)) = \frac{b-a}2$). I don't know whether such a thing could exist, but if it does, then I suspect we are moving into axiom of choice territory. If anyone can make a concrete construction to prove me wrong, though, that would be cool.

Answer 3

An example of such an $E$ is the non-negative rationals unified with negative irrationals i.e. $$E = (\mathbb{Q}\cap[0,\infty))\cup((\mathbb{R}-\mathbb{Q})\cap(-\infty, 0))$$

If you only want to split the irrationals, then you can do the same construction with $A=\{x+\pi|x\in\mathbb{Q}\}$ and $B = \mathbb{R}-\mathbb{Q}-A$, $$E = (A\cap[0,\infty))\cup(B\cap(-\infty, 0))$$