# Are Radon measures on Polish spaces $\sigma$-finite?

If $$\Omega$$ is a Polish space and $$\mu$$ is a Radon measure on $$\Omega$$ (i.e. an inner-regular Borel measure), is $$\mu$$ $$\sigma$$-finite?

I know that Radon measures in general need not be $$\sigma$$-finite, and $$\sigma$$-finite measures need not be Radon. The standard counterexample to the former is an uncountable set with the discrete topology and counting measure (which is Radon, but not $$\sigma$$-finite), and one counterexample to the latter is $$\Omega = \mathbb R$$ with $$\mu$$ the counting measure on $$\mathbb Q$$ (which is $$\sigma$$-finite, but not Radon, or even Borel).

But what if $$\Omega$$ is Polish? If $$Q \subset \Omega$$ is countable and dense, since $$\mu$$ is Borel, every $$q \in Q$$ has an open neighborhood $$U_q \ni q$$ for which $$\mu(U_q) < \infty$$. But a priori, there's no guarantee that $$W:=\bigcup_{q \in Q} U_q = \Omega$$. We know that by density of $$Q$$ in $$\Omega$$, we have that $$W^c$$ has empty interior, but $$W^c$$ could still have infinite measure in principle (e.g. let $$\mu = \lambda^2 + \lambda^1$$ on $$\mathbb R^2$$, where $$\lambda^2$$ is the $$2$$-dimensional Lebesgue measure on $$\mathbb R^2$$ and $$\lambda^1$$ is the $$1$$-dimensional Lebesgue measure on the $$x$$-axis, and take $$Q = \left\{(p,q) \in \mathbb Q^2 : p \neq 0\right\}$$). And taking the closures $$\overline U_q$$ might not work because if $$\Omega$$ is an infinite-dimensional Banach space, for example, $$\overline U_q$$ need not be compact, so $$\mu\left(\overline U_q\right) = \infty$$ is possible.

I'm really not sure one way or the other about the answer to this question. I can't think of a counterexample, but I can't think of a proof, either. Anything I'm not thinking of?

For every point $$x$$ of the space $$X$$ pick a neighbourhood $$U_x$$ with $$\mu(U_x)<\infty$$. Since Polish spaces are separable they are also Lindelöf, hence you can find a countable $$I\subset X$$ such that $$\{U_y\mid y\in I\}$$ still covers $$X$$, and this cover witnesses that $$\mu$$ is $$\sigma$$-finite.
Also note that in hereditarily Lindelöf spaces (so separable metric spaces for example) every Radon measure is not only $$\sigma$$-finite, but also moderated, which is a slightly stronger property.