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I have this exercise:

$M$ is the lebesgue measurable sets on $\mathbb{R}$. Suppose that $E \in M$. Show that for each $\epsilon > 0$, there is a closed set $F$, with $F \subset E$ and $\lambda(E \backslash F) <\epsilon$.

Here $\lambda$ is the Lebesgue measure.

I am stuck on this exercise. Using the definition of the Lebesgue measure, I know that there is a sequence of open intervals $\{I_n\}$, $E \subset \bigcup I_n$, so that for instance $\lambda(E) > \sum \lambda(I_n)-\epsilon/2$. I thought that maybe I could shrink the intervals to get closed intervals, however that didn't work because I don't know how much they have to be shrinked in order to get containment in E, and also there is a problem that an infinite union of closed sets might not be closed itself.

Another option is to use the Caratheodory criterion somehow, but I don't see how that helps?

Bach
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user119615
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  • To go from open to closed, we usually take complements. So, I would try to apply the result about open intervals to $E^c$, and see what happens. By the way, closed sets can be more complicated than unions of closed intervals; you should not be thinking in terms of intervals when dealing with them. –  Aug 28 '14 at 04:19

1 Answers1

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It seems you have an idea of how to prove outer regularity (the analogous statement for open sets), so let us assume that for any $\epsilon > 0$ there is an open $U$ containing $E \in M$ such that $\mu(U\setminus E) < \epsilon$ ($\mu$ is Lebesgue measure).

Step 1: Suppose $E$ is not closed (otherwise the statement is trivial), is bounded (will generalise later) and let $\epsilon > 0$. Choose $U$ open containing $\bar{E}\setminus E$ such that $\mu(U\setminus (\bar{E}\setminus E)) < \epsilon$. Set $K = \bar{E} \setminus U$. Note that $K \subset E$, $K$ is compact (hence closed), and show that $\mu(K) > \mu(E) -\epsilon$, i.e. $\epsilon > \mu(E \setminus K)$.

Step 2: Extend to the case where $E$ is neither bounded nor closed by performing Step 1 on $E_n := E \cap (n,n+1]$ with accuracy $\epsilon 2^{-n}$ for any given $\epsilon$, and note that $\{E_n\}$ is a disjoint family.

Source: Folland's real analysis, Thm. 1.18, p. 36.

snar
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  • Thanks. But for the analogous statement for open sets. I am able to prove that if the measure of E is finite, but not infinite. Can you help me with that special case? I guess you do not need this case for your proof?, but it would be interesting if you knew it? – user119615 Aug 28 '14 at 05:03
  • You're trying to show $\mu(E) = \inf{\mu(U) : E \subset U, U\text{ open} }$, with equality in the case of $\mu(E) = \infty$ meaning that both sides are infinite. In this case $\geq$ holds trivially. But for any $E \subset U$, $\mu(E) \leq \mu(U)$. Therefore, the infimum is at least... – snar Aug 28 '14 at 15:07