Suppose that $X$ is a Polish metric space (with a distance $d$). We know that for each $\epsilon>0$, $X$ can be partitioned into a countable number of non-empty Borel subsets, $\{B_n\}$, whose diameters do not exceed $\epsilon$. Given a strictly positive measure $\mu$ on $(X,\mathcal{B}(X))$ (i.e., $\operatorname{supp}(\mu)=X$), can we assume that $\mu(B_n)>0$ for each $n$?
I think this is true because if $\mu(B_n)=0$ for some $n$ and $x\in B_n$, there exists another set $B_m$ in the partition such that $x\notin B_m$, $d(x,B_m)=0$, and $\mu(B_m)>0$. (Since $\mu$ is strictly positive, $\mu(B_{\tau}(x))>0$ for any $\tau>0$.)
This paper (an English version was published here) shows that there exists an "almost open partition" into Borel sets such that each $B_n$ has a non-empty interior, which solves the problem.