Subset of $\mathbb{I}\cap [0,1]$ (irrationals in [0,1]) that is closed in $\mathbb{R}$ and has measure $\epsilon \in (0,1)$ Measure theory guarantees that every Lebesgue finite measurable set $E$ has a closed subset $F$ such that $m(E \backslash F)<\epsilon$  for small $\epsilon$. 
But today I saw in some text that for all small $\epsilon >0$ there exist a subset of $\mathbb{I}\cap [0,1]$ (irrationals in [0,1]) that is closed in $\mathbb{R}$ and has Lebesgue measure less than $\epsilon$? Note the set is required to be closed in $\mathbb{R}$, not in $\mathbb{I}\cap [0,1]$. Can someone give me an example / a proof that such thing does not exist?
 A: In fact, we can choose a set which have exactly measure $\varepsilon$.
For a fixed $\delta>0$, consider the set $S_{\delta}:=\bigcup_{n\in\mathbb N}(q_n-2^{-n}\delta,q_n+\delta 2^{-n})$, where $\{q_n,n\in\Bbb N\}$ is an enumeration of the rationals of $[0,1]$. Then $S_{\delta}$ is open and dense in $[0,1]$, since it contains all the rationals of this interval. The maps $f\colon\delta\mapsto \lambda(S_{\delta}\cap (0,1))$ is Lipschitz-continuous. Indeed, if $\delta_1\leq\delta_2$, we have 
\begin{align*}f(\delta_2)-f(\delta_1)&=\lambda(S_{\delta_2}\setminus S_{\delta_1})\\\
&\leq \lambda\left((0,1)\cap \bigcup_{n=0}^{+\infty}(q_n-2^{-n}\delta_2,q_n+\delta_2 2^{-n})\setminus (q_n-2^{-n}\delta_1,q_n+\delta_1 2^{-n})\right)\\\
&\leq \sum_{n=0}^{+\infty}\lambda((q_n-2^{-n}\delta_2,q_n+\delta_2 2^{-n})\setminus (q_n-2^{-n}\delta_1,q_n+\delta_1 2^{-n}))\\\
&=2(\delta_2-\delta_1)\sum_{n=0}^{+\infty}2^{-n}.
\end{align*}
Now we use the intermediate value theorem to pick $\delta$ such that $\lambda(S_{\delta}\cap (0,1))=1-\varepsilon$, and we consider the complement in $[0,1]$ of $S_{\delta}$. 
A: Let $\epsilon > 0$.  Enumerate the rationals in $[0,1]$ with the sequence $\{q_n\}_{n=1}^\infty$.  Now for each $n$, choose an open interval $I_n\subseteq [0,1]$ containing $q_n$ with length less than $\epsilon/2^n$  If $q_n = 0$ or $q_n = 1$, choose a half-open interval.  Said half-open
intervals are still relatively open in $[0,1]$.
Put $G = \bigcup_n I_n$.  The Lebesgue measure of $G$ is at most $\epsilon$.  Now put $F = [0,1] - G.$
The set $G$ is open.  The set $F$ is closed.  And its measure is at least $1 - \epsilon$.
Since $G$ contains all of the rationals in $[0,1]$, $F$ contains only irrationals.
