Talking about an example of an open dense subset of $[0,1]$ with measure $1/2$ I heard talking about 'fat rationals' but I don't find anything on the internt.. Does someone know about this?
Dense subset of $[0,1]$ and fat rationals
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3'fat rationals' probably means a set of the form $A = \bigcup_{n\in \mathbb N} B_{\epsilon_n}(q_n)$ where $\{q_n\}_{n\in \mathbb N}$ is an enumeration of the rationals and the \epsilon_n>0 are chosen in such a way that the measure of $A$ is $1/2$. – 2012-04-17
2 Answers
For a fixed $\varepsilon$, consider the set $S_{\varepsilon}:=\bigcup_{n\in\mathbb N}(q_n-2^{-n}\varepsilon,q_n+\varepsilon 2^{-n})$, where $\{q_n\}$ is an enumeration of the rationals. Then $S_{\varepsilon}$ is open and dense in $[0,1]$, since it contains all the rationals of this interval. The maps $f\colon\varepsilon\mapsto \lambda(S_{\varepsilon})$ is Lipschitz-continuous. Indeed, if $\varepsilon_1\leq\varepsilon_2$, we have \begin{align*}f(\varepsilon_2)-f(\varepsilon_1)&=\lambda(S_{\varepsilon_2}\setminus S_{\varepsilon_1})\\\ &\leq \lambda\left(\bigcup_{n=0}^{+\infty}(q_n-2^{-n}\varepsilon_2,q_n+\varepsilon_2 2^{-n})\setminus (q_n-2^{-n}\varepsilon_1,q_n+\varepsilon_1 2^{-n})\right)\\\ &\leq \sum_{n=0}^{+\infty}\lambda((q_n-2^{-n}\varepsilon_2,q_n+\varepsilon_2 2^{-n})\setminus (q_n-2^{-n}\varepsilon_1,q_n+\varepsilon_1 2^{-n}))\\\ &=2(\varepsilon_2-\varepsilon_1)\sum_{n=0}^{+\infty}2^{-n} \end{align*}
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1The continuity consideration of $\varepsilon \mapsto \lambda(S_\varepsilon)$ implies of course that by suitably choosing $\varepsilon$ you can obtain any desired real value $\gt 0$ as measure of $\lambda(S_\varepsilon)$. – 2012-04-17
As an alternative to Davide Giraudo's proposal which avoids having to order the rationals, try
$S_k = [0,1] \cap \bigcup_{a\in\mathbb Z, b\in \mathbb Z^+}\left(\frac{a}{b}-\frac{k}{b 2^{b}},\frac{a}{b}+\frac{k}{b 2^{b}}\right).$
For $k\approx 0.32184058\ldots$ this will have measure $\frac12$.
This has the same continuity considerations as Davide's and I suspect that as a function of $k$ the measure of $S_k$ has a zero derivative for all rational $k$, though I have not checked.
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0That page has now moved to http://www.se16.info/hgb/nowhere.htm – 2016-03-02