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).
[Math] Does a positively measurable set contains a closed set of a positive measure
measure-theory
Best Answer
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.