Standard Borel probability spaces (a Polish space with a measure on it's Borel sigma algebra) satisfy the following:
(*) For any Borel set $E$ of positive measure and any $\epsilon>0$, there is $F\subseteq E\subseteq U$ with $F$ closed and $U$ open such that $m(E\setminus F)<\epsilon$ and $m(U\setminus E)<\epsilon$ (where $m$ is the measure).
Can we have the same property as above except with $F$ open and $U$ closed (call this new property (~))?
As far as I can see, we can prove (~) because it is true for open balls and is closed under countable unions and complementation.
Further question: Are any two standard Borel spaces $(X,\mathcal{B},m)$, $(Y,\mathcal{B}',m')$ such that there are null $M, N$ in their respective spaces, with $X\setminus M$ homeomorphic to $Y\setminus N$? This question comes up from the proof of the Borel Isomorphism Theorem (the statement of the theorem is as above except $X\setminus M$ is isomorphic as a measure space to $Y\setminus N$ ) if you try to prove it using (~) instead of the usual way with (*).
Any set of positive measure with empty interior is a counterexample. ("Fat Cantor set", for example.)