Let $(E, \tau)$ be a topological space. We say a measure $\mu$ on the Borel $\sigma$-algebra $\mathcal B$ is:

  • locally finite if for all $x \in E$, there is an open $U \ni x$ for which $\mu(U) < \infty$;
  • a Borel measure if $\mu(K) < \infty$ for every compact $K \subset E$.

Clearly locally finite measures are Borel. And if $E$ is also locally compact (i.e., every $x \in E$ has a neighborhood $U \ni x$ for which $\overline U$ is compact), then clearly all Borel measures are locally finite.

Question 1: Is there a condition more general than local compactness under which Borel measures are not locally finite?

I'm particularly interested in $\sigma$-compact spaces, i.e., topological spaces that are countable unions of compact sets.

Question 2: Is a Borel measure on a $\sigma$-compact space locally finite?

I've asked this question before here, and a counterexample to Question 2 is proposed, but the counterexample actually doesn't work (see comments in that answer). Is anyone aware of a proof/counterexample to Question 2, or an answer to Question 1?

EDIT: I'll also accept a reference to an answer to either of these questions.


1 Answer 1


I'll try a second time at a counter example.

Consider the Smirnov deleted sequence topology. It is $\mathbb{R}$ with open sets of the form $U\setminus B$, where $U$ is open in the standard topology and $B\subseteq \{ \frac{1}{n}: n\in \mathbb{N} \}$.

This space is $\sigma$-compact but not locally compact with $0$ being a point with no compact neighborhood. Consider the measure $\mu= \sum_{n=1}^\infty \delta_{ A_n }$, where the summands are Dirac measures on $A_n:= \big( \frac{1}{n+1},\frac{1}{n} \big)$.

For any set $B\subseteq \mathbb{R}$, satisfying $B\cap (0,\epsilon)=\emptyset$ for some $\epsilon>0$, you have that $\mu(B)<\infty$.

Since any neighborhood of $0$ should intersect infinitely many $A_n$, we get that $\mu(U)=\infty$ if $0\in U$ and $U$ is open. Any compact set containing must intersect the $A_n$ finitely many times, and so has to have finite measure.

  • $\begingroup$ By "Dirac measures on $A_n$," do you mean that for each $A_n$, you choose a point $p_n\in A_n$, and that $\delta_{A_n} = \delta_{p_n}$? $\endgroup$
    – Alex Ortiz
    Commented Feb 8, 2023 at 15:17
  • $\begingroup$ I mean more of an indicator function, or Dirac measure in the sense of en.wikipedia.org/wiki/Dirac_measure. $\endgroup$ Commented Feb 8, 2023 at 15:23
  • $\begingroup$ Ah… So for example, $[-1,1]$ is not compact because $U_n := (-2,2)\setminus\{k^{-1} : k > n\}$ is an open cover of $[-1,1]$ without a finite subcover. In fact it seems every compact space containing $0$ must have empty interior near $0$, in a certain sense… more precisely, $0$ must be an isolated point of any compact set. Am I on target? $\endgroup$
    – D Ford
    Commented Feb 8, 2023 at 16:42
  • 1
    $\begingroup$ @DFord Well, I think $[-1,0]$ is compact, but $0$ is not isolated in it. It definitely can't have $0$ in the interior. Though I could be wrong. $\endgroup$ Commented Feb 8, 2023 at 16:57
  • 1
    $\begingroup$ @DFord It of course does not answer your original question in the other thread, since this is not a Polish space. Though maybe searching for a Polish not locally compact but $\sigma$-compact in topology.pi-base.org, can help you with that counter-example. $\endgroup$ Commented Feb 9, 2023 at 9:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .