Let $S,T\subseteq \mathbb R^n$ be disjoint such that for every compact $K\subseteq \mathbb R^n$ we have that $K\cap S$ is a null set if and only if $K\cap T$ is a null set. Can both have positive Lebesgue measure?
Intuitively, if one of them had positive measure which is „located“ somewhere, there would not be any „place“ for the other set to co-exist with its positive measure.
It is easy to see that $S$ and $T$ must have empty interior: If one of them (wlog $T$) contained a nonempty open set $U\subseteq T$, there is a compact set $K$ with positive measure, hence $\lambda(K\cap T) = \lambda(K)>0$, but $\lambda(K\cap S) = \lambda(\emptyset)=0$.
This isn't really helpful however, because lots of things have positive measure but empty interior (think $\mathbb R\setminus \mathbb Q$).
I tried to prove various things which turned out to be wrong (e.g. every adhesion point of $S$ is a limit point of $T$ and vice versa) or I didn't manage to prove them (e.g. that every rectangular cover of $S$ cover $T$ modulo null set).
Is this even true?
This follows easily from regularity of Lebesgue measure $m$. If $m(S) >0$ then there is a compact subset $K$ of $S$ with $m(K)>0$. But $K \cap T$ is empty so (by hypothesis) $K\cap S$ must have measure $0$, a contradictiom. This proves that both $S$ and $T$ have measure $0$.