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?
Best Answer
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.