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I'm having trouble starting two similar proofs:

Let $\epsilon > 0$. And let $E$ be a measurable set of finite measure.

Prove that there is an open set $U$ containing $E$ such that $m(U \setminus E) < \epsilon$.

Similarly, prove there is a compact set $K$ contained in $E$ such that $m(E \setminus K) < \epsilon$.

Any hints are much appreciated.

NOTE: $m$ is the Lebesgue outer measure. And $E \subseteq \mathbb{R}$ is measurable if $m(A) \geq m(A \cap E) + m(A \setminus E)$.

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I presume $E$ is a subset of $\mathbb{R}$ or $\mathbb{R}^n$ and the measure is Lebesgue measure? – t.b. May 2 '12 at 22:26
and that $m$ denotes Lebesgue measure? – user12014 May 2 '12 at 22:27
The first assertive is kind of a definition the second one is the definition on the complementar of E. – checkmath May 2 '12 at 22:27
And your definition of measurable set is? – Michael Greinecker May 2 '12 at 22:28
Yes, $m$ is the Lebesgue measure. And $E \subseteq \mathbb{R}$ is measurable if $m(A) \geq m(A \cap E) + m(A \setminus E)$. – Mike C. May 2 '12 at 22:34

Assuming you're talking about Lebesgue measure on $\mathbb{R}$ or $\mathbb{R^n}$, then here's one approach.

Without loss of generality, we may assume that $E$ is contained in a bounded subset of $\mathbb{R^n}$ (by splitting it up into countably parts if need be).

Recall the method of defining the Lebesgue measure of a set by considering outer measure, where we try to find a minimal covering by your (open) generating sets. I think you can see this by looking at a proof of Caratheodory's Extension theorem. From this it follows that you can approximate from outside by open sets. In a bounded set, we may take complements without worry about infinite measure cropping up, so we can use the same logic to show that we can approximate from inside by closed sets. Recalling that closed, bounded sets are compact, this shows the result for bounded E, which we can then extend by countable additivity.

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how can one say that assuming $E $ is bounded does not sacrifice generality? – Guldam Sep 12 '15 at 11:13
It seems the author of the above reply assumed $\sigma$-finiteness of the space (which of course is true for Euclidean space). – ZMI Mar 11 at 14:10

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