# When is a Borel measure locally finite?

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.

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.

• 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 $\delta_{A_n} = \delta_{p_n}$? Commented Feb 8, 2023 at 15:17
• I mean more of an indicator function, or Dirac measure in the sense of en.wikipedia.org/wiki/Dirac_measure. Commented Feb 8, 2023 at 15:23
• 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? Commented Feb 8, 2023 at 16:42
• @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. Commented Feb 8, 2023 at 16:57
• @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. Commented Feb 9, 2023 at 9:52