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?


1 Answer 1


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.

  • $\begingroup$ I was wondering for so long how I could get a countable subcover, then I forgot about the Lindelöf property! You did not use inner regularity anywhere, would I be right to say that? $\endgroup$ Feb 18, 2021 at 6:57
  • $\begingroup$ @TeresaLisbon Yes, indeed, but I suspect that a locally finite Borel measure on a Polish space might be Radon hence inner regular for free anyway. This is surely true for finite Borel measures on Polish spaces, but I'm not sure whether it holds for locally finite measures right now $\endgroup$ Feb 18, 2021 at 7:06
  • $\begingroup$ Indeed, I think it holds. A strongly Radon space is a space where every locally finite Borel measure is a Radon measure i.e. an inner regular, outer regular and locally finite measure. It is known that every Polish space is Suslin, and every Suslin space is strongly Radon. I saw this on the Wikipedia article titled "Radon measure". $\endgroup$ Feb 18, 2021 at 7:11
  • $\begingroup$ You can try this related problem here when you are free. Some changes were made, but in the comments to that question an answer/clarification to EDIT 1 was requested. $\endgroup$ Feb 19, 2021 at 4:40

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