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.
Best Answer
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.