Does a positively measurable set contains a closed set of a positive measure? Everywhere I looked it was referred as "a well known fact", but how can I show that every measurable set of a positive measure in a polish space contains a closed subset of a positive measure? (measure is non-atomic).
 A: The trick is to simultaneously consider approximation from the inside by closed sets and from the outside by open sets. If $S$ is a metric space, and $\mu$ a finite measure on $(S,{\cal B}(S))$, define
$$ {\cal D}=\left\{B\in{\cal B}(S):\mu B=\sup_{F\subset B}\mu F=\inf_{G\supset B}\mu G \right\},$$ with $F$ and $G$ restricted to the classes of closed and open subsets of $S$, respectively. 
Since every open set $G$ is $F_\sigma$, all open sets belong to $\cal D$. Now use Dynkin's $\pi-\lambda$ theorem to conclude that all Borel sets belong to $\cal D$. 
The space $S$ doesn't have to be Polish, only metric. It doesn't matter whether $\mu$ is atomic or not. 
Reference: Lemma 1.34 (page 18) Foundations of Modern Probability (2nd edition) by Olav Kallenberg. 
A: I am not familiar with general measure theory, but the process in the Lebesgue measure is the following.
Theorem: Given a measurable set $S$ and $\varepsilon > 0$ there is open $O$ with $S \subset O$ and $m(O) < m(S) + \varepsilon$.
Proof: By definition of the Lebesgue measure there are (countably many) covering open intervals of total length within $\varepsilon$ of $m(S)$. Their union is open and has measure between the measure of $S$ and the total length of the covers. The result follows.
Corollary: There is closed $C$ with $C \subset S$ and $m(C) > m(S) - \varepsilon$.
Proof: Use the above theorem on $S^c$ and take complements.
