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 $m(A) \geq m(A \cap E) + m(A \setminus E)$.
 A: 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.
A: Let me know if this works.
If $m_*(E)=\infty$, then taking $O=\mathbb{R}$ give the desired result. Suppose then that $m_*(E)<+\infty$ and let $\epsilon>0$ be given. By definition of the exterior measure we are guaranteed the existence of a countable collection of closed cubes $\{I_j\}_j$ with
$$E \subseteq \bigcup_j I_j$$
such that
$$\sum_j \vert I_j \vert \leq \vert E \vert + \frac{\epsilon}{2}.$$
For each $j$, let $I^*_j$ be a closed cube such that
$$I_j \subseteq \mathrm{int}(I^*_j).$$
and
$$\vert I^*_j \vert \leq \vert I_j \vert + \frac{\epsilon}{2^{j+1}}.$$
Then we can take our open set to be
$$O = \bigcup_j \mathrm{int}(I^*_j)$$
which is open as arbitrary unions of open are open and it is clear to see $E \subset O.$ We can thus compute
\begin{eqnarray*}
m_*(O) &=& \inf \sum_j \vert \mathrm{int}(I^*_j)\vert\\
&\le&   \sum_j \vert I^*_j \vert \\ &\le&  \sum_j (\vert I_j \vert + \frac{\epsilon}{2^{j+1}}) \\
&\le& \vert E \vert + \epsilon,
\end{eqnarray*}
as needed.
A: Hossien proved that there is an open set $O$ that contains $E$ such that $m(O\backslash E)<\epsilon$. Let me try to prove that there is a compact set $K$ contained in $E$ such that $m(E\backslash K)<\epsilon$.
Since $E$ is measurable and measurable sets are an algebra, the complement of $E$, $E^c$, is also measurable. Then, because Hossien proved it, there exists an open set $O$ such that $m(O\backslash E^c)<\epsilon$. But the complement of $O$, $O^c$, is a closed set contained in $E$. Therefore it is compact because it is closed and bounded and we are in $\mathbb{R}$. Moreover,
$$ m(E\backslash O^c) = m(E\cap O) = m(O\cap E) = m(O\backslash E^c) < \epsilon $$
And then we proved that $K=O^c$ is that set.
