# 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?

• The space in the link is a closed subset of $\Bbb R^2$, therefore it should be locally compact. However, I'm not sure right now what a completely metrizable separable $\sigma$-compact and not locally cmpact space should be. I had an idea, but I don't seem to be able to make it work right now.
– user239203
Feb 22, 2021 at 7:56
• By Baire a $\sigma$-compact Polish $X$ is somewhere locally compact. So your idea holds for homogeneous spaces. Feb 22, 2021 at 9:19
• In $\ell^2$ let $e_k, k \in \Bbb N$ be the standard orthonormal base. Then $X=\{\lambda e_k\mid k \in \Bbb N, \lambda \in \Bbb R\}$ in the subspace topology is $\sigma$-compact, Polish but not locally compact. Feb 22, 2021 at 10:21
• @HennoBrandsma isn't $X$ simply $\ell^2$ in your example? Feb 22, 2021 at 15:35
• No that would include linear combinations (sums) of the $e_k$ as well. This is just a collection of lines through the origin. Feb 22, 2021 at 15:37

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.

• Actually I’m not sure that this is finite for compact $K \subset X$. Let $K_k = \left\{\lambda e_k : 0 \leq \lambda \leq k^{-1}\right\} \subset \overline L_n$. Clearly each $K_k \subset X$ is compact. Let $K = \bigcup_{k=1}^\infty K_k$. Then if $(x_n) \subset K$ is any sequence, it clearly has a finite subsequence if the sequence lies in finitely many $L_k$; and if the sequence lies in infinitely many $L_k$, then $x_n$ converges to $0$. So $K$ is sequentially compact, and hence compact, even though $\mu(K) = \infty$. Feb 23, 2021 at 0:13
• (Or at least, in the above example, $x_n$ has a subsequence that converges to $0$ if $x_n$ lies in infinitely many $L_k$.) Feb 23, 2021 at 0:20
• @DFord Then maybe another sum of Dirac measures might work? Being finite on compacta is necessary for a Borel measure in your definition? Feb 23, 2021 at 10:35
• I've been playing with a couple sums of Dirac measures, and I usually run into similar issues. I'll keep thinking about it. (Or maybe the answer to the original question is yes after all). But yes, by the definition I'm relying on, Borel measures must be finite on compact sets (at least that's the property I care about); I want to know if that's equivalent to local finiteness. Feb 23, 2021 at 16:31

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 $$\mathbb{R}^2$$.

• Yes, after you suggested looking at $\pi$-base with the other problem, I found this example and thought of something similar. Thanks very much! Feb 9, 2023 at 19:58