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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?

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What is $\mathbb{I}$? –  Chris Eagle Jun 19 '12 at 10:28
    
@ChrisEagle Irrationals. I have fixed it. –  Polymorpher Jun 19 '12 at 10:29
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Such a thing does indeed exist. See this answer, for example. –  Dejan Govc Jun 19 '12 at 10:44
    
@DejanGovc Thanks! Can you make it an answer so that people can see this question is answered? –  Polymorpher Jun 19 '12 at 11:09
    
In what text did you see this and what page? –  Thomas E. Jun 19 '12 at 11:55

2 Answers 2

up vote 4 down vote accepted

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 $S_{\delta}$.

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And with a bit more work we can find a more "constructive" construction of such a set (at least, one that doesn't rely on the intermediate value theorem). Let $U_0 = \emptyset$, and $U_n = U_{n-1} \cup (q_n - r, q_n + r)$ where $r$ is chosen so that $\lambda(U_n) = (1-2^{-n})(1-\epsilon)$. Since $U_{n-1}$ is a finite union of intervals, $\lambda(U_n)$ is a piecewise-linear function, so this can be done explicitly. –  Robert Israel Jun 24 '12 at 6:58

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.

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Sorry I messed up a little bit in my question. What I really meant was for all small $\epsilon$, the subset can be made to have measure less than $\epsilon$. The measure of F being constructed is bounded below by $\epsilon$, not above. –  Polymorpher Jun 19 '12 at 11:45

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