# Approximating measures by open sets and compact sets.

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).
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