Locally finite vs. Borel measures on $\sigma$-compact Polish spaces Let $E$ be a Polish space, and let $\mu$ be a measure on $E$. Define the following properties:

*

*$E$ is $\sigma$-compact if $E$ is the countable union of compact sets.

*$E$ is locally compact if every $x \in E$ has an open neighborhood $U$ whose closure is compact.

*$\mu$ is locally finite if for every $x \in E$, there is an open set $U \subset E$ containing $x$ with $\mu(U) < \infty$.

*$\mu$ is a Borel measure if for every compact $K \subset E$, we have $\mu(K) < \infty$.

Clearly a locally finite measure is Borel. And if $E$ is locally compact (not even necessarily separable or complete), Borel measures are necessarily locally finite. But are there more general conditions for which Borel measures are locally finite?
Question: If $E$ is a $\sigma$-compact Polish space and $\mu$ is a Borel measure, is $\mu$ locally finite?
I can’t think of a counter example to this, but I’m having trouble proving it. My original strategy was to prove that a $\sigma$-compact Polish space is locally compact. However, as the comments demonstrate, $\sigma$-compact Polish spaces are not necessarily locally compact, so that strategy doesn’t work. A counter example is the subset $X \subset \ell^2$ given by the union of the lines $X_k = \{\lambda e_k : \lambda \in \mathbb R\}$, where $\{e_k\}_{k \geq 1}$ is the standard orthonormal basis of $\ell^2$. Then $X$ is $\sigma$-compact, but not locally compact (at the origin specifically).
But I’m not sure of a measure $\mu$ on this space that is Borel, but not locally finite; the problem is there are compact subsets of $X$ containing $0$ and intersecting infinitely many of the $X_k$. Can anyone think of a counter example?
 A: I think that indeed the idea from the common thread works:
Let $e_n, n \in \Bbb N$ be the standard orthonormal base of Hilbert space $\ell^2$. Let $X = \{\lambda e_n \mid n \in \Bbb N, \lambda \in \Bbb R\}$. Then $X$ is $\sigma$-compact Polish but not locally compact (at $0$).
Let $L_k = \{\lambda e_k: \lambda \neq 0\}$ and $\delta_A$ be the Dirac measure with carrier $A$: $\delta_A(B) = 1$ iff $A \cap B \neq \emptyset$ and $0$ otherwise, we can define $\mu = \displaystyle_{k=1}^\infty \delta_{L_k}$, which is a Borel measure (not $\sigma$-finite, and not locally finite at $0$), but (I think) finite at compacta.
I haven't checked all the nitty gritty details. Others might feel inclined to add to this.
A: Following the answer I wrote in your follow up thread, I would like to try and propose a counter example for this more restrictive question. I think this is in the same spirit as @Henno Brandsma's answer.
Consider the space of nested rectangles in the real plane. This is a subspace of $\mathbb{R}^2$ with the induced topology, consisting of $L_1=\{x=1\}$, $L_2=\{ x=-1 \}$ and a sequence of boundary of rectangles $R_n=\partial B_n$, given by $B_n=\{ -\frac{n}{n+1} \leq x\leq \frac{n}{n+1} \}\cap \{ -n \leq  y\leq n \}$. This is a Polish space which is $\sigma$-compact but not locally compact.
Define the measure $\mu=\sum_{n=1}^\infty \delta_{ R_n}$. Any neighborhood of a point on $L_1$ has to intersect infinitely many $\partial R_n$, and so such a neighborhood would have infinite measure.
I think that any set intersecting $L_1$ or $L_2$ nontrivially would not be compact. So any compact subset in the space would have to intersect finitely many of the $R_n$, because of the topology induced from $\mathbb{R}^2$.
