1
$\begingroup$

This is a follow up to this question, where I wish to partition the reals into two sets $A$ and $B$ that are dense (with positive measure) in every non-empty sub-interval $(a,b)$ of $\mathbb{R}$.

(In this case, I want to find $\lim_{t\to\infty} \lambda(A\cap [-t,t])/(2t)$ and $\lim_{t\to\infty} \lambda(B\cap [-t,t])/(2t)$, where $\lambda$ is the Lebesgue measure restricting the Lebesgue outer measure $\lambda^{*}$ on sets measurable in the Caratheodory sense.)

I have a rough understanding of the answer; however, I'm unsure of the measures of $A$ and $B$.

Question: What is the measure of $A$ and $B$? If both measures are $1/2$, how do we change the answer so the measures are positive but unequal?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ How could both measures be $1/2$? Doesn't $\mathbb R$ have measure $\infty$? What measure are you using? $\endgroup$ Commented Aug 4, 2023 at 0:43
  • 1
    $\begingroup$ Given a number $\varepsilon\gt0$ we can constract an $F_\sigma$ set $A\subset\mathbb R$ so that (i) for every nonempty open set $U\subseteq\mathbb R$, both $U\cap A$ and $U\setminus A$ have positive Lebesgue measure;, and (ii) the set $A$ has Lebesgue measure less than $\varepsilon$. $\endgroup$ Commented Aug 4, 2023 at 0:48
  • $\begingroup$ @bof I made edits. $\endgroup$ Commented Aug 4, 2023 at 0:53

1 Answer 1

Reset to default
3
$\begingroup$

We can make those limits be any two positive numbers that add to $1$, or we can make them nonconvergent. The reason is that in any interval, we can construct $A$ and $B$ on that interval by using the iterated fat-Cantor set construction (as described here), and we can make $A$ have arbitrarily small measure on that one interval, which is what bof refers to in his comment.

By repeating those small-measure sets on successive intervals, or by using large-measure sets as desired on successive intervals, we can control the limit value of your asyptotic measure so as to realize any desired value or a nonconvergent value.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.