Dense subset of $[0,1]$ and fat rationals 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?
 A: 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*}
A: 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.
