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Is there a set $S\subset[0,1]$ such that:

  • $\forall s_1,s_2\in S:\exists x\not\in S, s_1<x<s_2$

  • $\forall x_1,x_2\not \in S:\exists s\in S, x_1<s<x_2$

And that if $Y\sim U(0,1)$ is a uniformly distributed random variable, $$0<Pr[Y\in S]<1$$

R B
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    By "randomly distributed" you mean what? Uniformly distributed? How about Letting $S$ consist of the rational numbers in $(0,1/2)$ and the irrational numbers in $(1/2,1)$? – bof Dec 08 '14 at 08:05
  • @bof The $U$ in $U(0,1)$ means uniform. And your proposal doesn't work I am pretty sure, because for any rational in the lower half there are infinite irrationals in the upper half. So the PDF would not be uniform but rather a Heaviside step at $1/2$. – GDumphart Dec 08 '14 at 08:06
  • Does "if $Y\sim U((0,1)$ is a uniformly distributed random variable, $0\lt Pr[Y\in S]\lt1$" mean something different from "$0\lt m(S)\lt1$ where $m$ is the Lebesgue measure"? – bof Dec 08 '14 at 08:18
  • @bof indeed, it is the same. – R B Dec 08 '14 at 08:21
  • @bof - I think your solution is valid. – R B Dec 08 '14 at 08:49

2 Answers2

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Take $S$ to be $(\Bbb Q\cap(0,\frac14))\cup((\frac14,\frac34)\setminus\Bbb Q) \cup(\Bbb Q\cap(\frac34,1))$.

Really the idea is to partition the interval into three smaller intervals, take dense co-dense sets in each, and ensure not to get a full probability. The rationals and irrationals give us the dense sets, just break out the probability.

The set I gave can be simplified using these ideas, and you should come up with simpler sets like that.

Asaf Karagila
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If you want the relative measure of $S$ in every interval to be $\frac12$, that sort of behavior is pretty strongly ruled out by the Lebesgue Density Theorem, which states that if $S$ is measurable, its density at almost every point in $[0,1]$ is either $0$ or $1$. See further discussion in questions 20170 and 89424.

Community
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Chris Culter
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    Right, though I don't think the OP asked anything like that. On the other hand if the OP just wants a measurable set $S$ with $0\lt m(S\cap I)\lt m(I)$ for every interval $I$, that can be done, and I'm pretty sure it's been asked and answered here at least once. – bof Dec 08 '14 at 09:30
  • @bof Good point. I guess my answer is more aimed at RB's comment on Asaf's answer, with the $\sim$, than the original question. – Chris Culter Dec 08 '14 at 20:31