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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.

Netchaiev
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    https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval –  Feb 05 '18 at 14:20
  • @Leonhard: thank you! Looking to the link you gave! – Netchaiev Feb 05 '18 at 14:26

3 Answers3

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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 $E^c$.

  • Yes, it is a right answer to my initial question, but of course I wanted something who will "mix" the hell of both sets! But you gave something really interesting : maybe with $$f_m(x)=\begin{cases}0 & x\in\mathbb Q,&x\in [2n/m, (2n+1)/m)\1 & x\in\mathbb R\setminus\mathbb Q,&x\in [2n/m, (2n+1)/m)\1 & x\in\mathbb Q,&x\in [(2n+1)/m, (2n+2)/m)\0 & x\in\mathbb R\setminus\mathbb Q,&x\in [(2n+1)/m, (2n+2)/m)\end{cases}, n\in\mathbb Z$$ and with $f=\lim_m f_m$ ? – Netchaiev Feb 05 '18 at 14:14
  • @Netchaiev What I suppose you really want is a set $E$ such that, for every interval $[a,b]$, you have $\mu(E\cap [a,b])=\frac{1}{2}(b-a)$. However, this is not possible: see Lebesgue density theorem ( https://en.m.wikipedia.org/wiki/Lebesgue%27s_density_theorem ). –  Feb 05 '18 at 15:37
  • @Netchaiev ... and this has been discussed on Math.SE before: https://math.stackexchange.com/q/596676/491874 –  Feb 05 '18 at 15:41
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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.

Arthur
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  • Yes : you made me change my question :) I don't want either my function equal a.e to a piecewise continuous function! – Netchaiev Feb 05 '18 at 14:08
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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 $B = \mathbb{R}-\mathbb{Q}-A$, $$E = (A\cap[0,\infty))\cup(B\cap(-\infty, 0))$$

  • The problem is that one of them have a null measure... – Netchaiev Feb 05 '18 at 14:07
  • No, both $E$ and $\mathbb{R}-E$ have infinite measure. –  Feb 05 '18 at 14:09
  • Sorry, you are right, my mistake. It is correct for my first question, but you guys made me modify it, since I had something more precise in mind and you made me realize it... – Netchaiev Feb 05 '18 at 14:16