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Just having trouble with this problem. First, it says to prove that if $E$ is Lebesgue Measurable, and $\epsilon>0$ is arbitrary, then there is an open $O$ such that $E \subset O$ and $m(O\setminus E)<\epsilon$. Now for this part, since $E$ is Lebesgue measurable, I was able to take the definition of being Lebesgue outer measurable (the infimum of open coverings of $E$) and easily construct $O$ from that, and it works out.

But now I want to show that there is an $F$ closed with $F \subset E$ and $M(E\setminus F)<\epsilon$. The problem is, I can't figure out how to construct this $F$! The idea seems obvious intuitively, but I feel like I am given no tools for constructing a closed subset of an arbitrary set that satisfies these conditions. Am I missing something obvious here?

Thanks!!

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Kirsten Janowicz
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  • Think about complements. – ncmathsadist Oct 07 '12 at 17:33
  • Someone else I know told me this same advice, but I wasn't sure how to proceed. What do you mean? – Kirsten Janowicz Oct 07 '12 at 17:41
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    @KirstenJanowicz I think you cannot delete since there's an answer. This prevents, for example, someone from getting an answer to his/her homework and then hiding the evidence. – Quinn Culver Oct 07 '12 at 18:52
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    You cannot delete your post because of two reasons: (1) There exists an upvoted answer, attempted vandalism will not remove the upvotes; (2) your user is unregistered, and unregistered users cannot delete their posts. – Asaf Karagila Oct 14 '12 at 20:15
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    @AsafKaragila I wonder why Kirsten wants to delete the question and answer so badly. – Ayman Hourieh Oct 14 '12 at 20:57

1 Answers1

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If $E$ is measurable, then its complement $E^\complement$ is also measurable and we can find an open set $W$ so that $E^\complement \subset W$ and $m(W - E^\complement) < \varepsilon$. Let $F = W^\complement$ be a closed set. Since $E^\complement \subset W$, we have $F \subset E$. Moreover, $W-E^\complement = E-F$. Hence $m(E-F) < \varepsilon$.

Ayman Hourieh
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