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In part of the proof of a problem I am trying to solve I need the following fact (assume that $\mu$ is the Lebesgue measure):

If $E\subseteq \mathbb{R}$ is measurable and $\delta>0$ then there exists open set $U\subseteq \mathbb{R}$, such that $E\subseteq U$ and $\,$ $\delta \mu(U)<\mu(E)$.

I know and have proven the following fact:

Suppose $E \subseteq \mathbb{R}$. Then for each $\epsilon>0$ there exists an open set $U\subseteq \mathbb{R}$ such that $E\subseteq U$ and $\mu(U)< \mu(E)+\epsilon$.

I am pretty sure I can use the second fact to prove the first fact, but I keep getting a value of $\epsilon$ that is in terms of $\mu(U)$, which isn't good because $U$ should depend on $\epsilon$, not the other way around. Some help?

Ana
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    As written the problem is trivial: you can take $U=\emptyset$ independently of $E$. Presumably you want some constraint on $U$ in terms of $E$. – Ian Mar 19 '17 at 20:33
  • @Ian You are absolutely right. I forgot to mention that $E\subseteq U$. I edited my post. Thank you. – Ana Mar 19 '17 at 20:39
  • @Ian http://math.stackexchange.com/questions/103306/for-a-set-of-positive-measure-there-is-an-interval-in-which-its-density-is-high I was reading the first solution in this post and the comment that followed. – Ana Mar 19 '17 at 20:48
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    Sorry, I got it backwards: your inequality can only hold for $\delta<1$. Then you work backwards: you want $\delta \mu(U)< \mu(E)$, so $\mu(U) < \delta^{-1} \mu(E) = \mu(E) + (\delta^{-1}-1)\mu(E)$. Now take $(\delta^{-1}-1) \mu(E)$ to be your $\epsilon$ – Ian Mar 19 '17 at 20:56

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Let me expand Ian's comment.

First, the inequality can only hold for $\delta<1$. For example, if $E=[0,1]$ and $\delta\geq1$, there never is such a $U$. But the claim is indeed true for all $\delta\in (0,1)$.

Take any such $\delta$ and work backwards: You want $\delta \mu(U)< \mu(E)$. That is, you want $$\mu(U) < \delta^{-1} \mu(E) = \mu(E) + (\delta^{-1}-1)\mu(E). $$ Now take $\epsilon=(\delta^{-1}-1) \mu(E)$ in your lemma. Then $\mu (U)<\mu (E)+\epsilon $ for some open $U \supset E$. Since this is the estimate you want, $U $ is the set you want.

Joonas Ilmavirta
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