Does there exist a continuous measure'' on a metric space?

Let $X$ be a separable complete metrizable space. Does there exist a complete metric $d$ and a Borel measure $\mu$ such that

(a) $\mu(B_r(x))<\infty$ for every open ball $B_r(x)$ of radius $r>0$ around $x\in X$,

(b) for each $r>0$ the map $x\mapsto \mu(B_r(x))$ is continuous on $X$,

(c) $\mathrm{supp}(\mu)=X$?

• You mean a metric compatible to the topology on $X$? It looks unlikely for, say, $X=\mathbb{Q}$. – Fedor Petrov Jun 22 '16 at 7:00
• Can this question be stated without referring to the metric? For example, using the topology on the family of closed sets? – erz Jun 22 '16 at 7:01
• @Fedor Petrov: you are right. I shoyuld ask for the metric to be complete. I added this. – user1688 Jun 22 '16 at 7:06
• @erz: I don't know, but I doubt it. – user1688 Jun 22 '16 at 7:07
• I guess this is a bit subtle. For instance if you take $X = [0,1] \cup \{2\}$ with its usual metric, then I think no measure will work; you have to charge $\{2\}$ to get full support, and that would violate (b) because you can shift the center of some ball slightly to include or exclude 2. (And I'm pretty sure you can't charge some other points to compensate.) But if you change the metric so that it looks like a half-circle together with its center, then Lebesgue measure on the half-circle and a point mass at the center would seem to work. – Nate Eldredge Jun 22 '16 at 13:46

Take $$X = [-1,0] \cup \{\frac1n\,:\,n \in \mathbb N\}$$ with the induced topology from $\mathbb R$. Each isolated point $\frac1n$ has to have positive measure. Since they accumulate to $0$ you cannot put a distance on $X$ separating them from the interval $[-1,0]$. This should force a discontinuity for a suitable $r>0$ when $x$ travels along $[-1,0]$.