I've seen the (centered) Hardy-Littlewood maximal function defined as follows: Let $X$ be a metric space, and $\mu$ a Borel measure on $X$. For $f \in L^1(\mu)$, define $$M^* f(x) : = \sup_{r > 0} \mu(B(x, r))^{-1} \int_{B(x, r)} f \mathrm{d} \mu .$$ Now my understanding is that the prototype for this is Euclidean space with the Lebesgue measure, which is of course fully supported (i.e. every nonempty open ball around any point will have positive measure). So I can see that this function is measurable as a function from $\operatorname{supp}(\mu)$ to $\mathbb{C}$, and even continuous under some mild additional assumptions. But my question is how the maximal function is defined for $x \not \in \operatorname{supp}(\mu)$.
I assume the answer is supposed to be that $\mu(X \setminus \operatorname{supp}(\mu)) = 0$, so we just follow the custom of throwing up our hands and shrugging off a null set because this is measure theory and we ignore null sets in measure theory, but it's not clear to me what conditions are needed to ensure that $\mu(X \setminus \operatorname{supp}(\mu)) = 0$. The Wikipedia page on support says there are examples of nonzero Borel measures on compact Hausdorff spaces for which the support has measure $0$.
So how is the Hardy-Littlewood maximal function defined for measures without full support?
Thanks!