Suppose that $(X,d)$ is a complete metric space and $\mu$ is a $\sigma$-finite regular Radon measure on $X$ which does not assign measure 0 to non-empty open subsets of $X$.
If $Y\subseteq X$ is a dense $G_\delta$ set the must it have positive $\mu$-measure?
Not even for Lebesgue measure on $\mathbb{R}$. Enumerate the rationals as $q_n$ and for each $\epsilon > 0$ let $U_\epsilon = \bigcup_{n=1}^\infty (q_n - \frac{\epsilon}{2^n}, q_n + \frac{\epsilon}{2^n})$. Clearly $U_\epsilon$ is open and dense, and its Lebesgue measure is at most $2\epsilon$. Now consider $G = \bigcap_{k=1}^\infty U_{1/k}$. Then $G$ is $G^\delta$ and it still contains $\mathbb{Q}$, so it's dense, but its measure is less than that of any $U_{1/k}$, so it has measure zero.
A similar construction would work for a non-atomic Radon measure on any separable metric space.