# 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. – Gae. S. Feb 22 at 7:56
• By Baire a $\sigma$-compact Polish $X$ is somewhere locally compact. So your idea holds for homogeneous spaces. – Henno Brandsma Feb 22 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. – Henno Brandsma Feb 22 at 10:21
• @HennoBrandsma isn't $X$ simply $\ell^2$ in your example? – D Ford Feb 22 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. – Henno Brandsma Feb 22 at 15:37

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.
• 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 $\mu(K) = \infty$. – D Ford Feb 23 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 $L_k$.) – D Ford Feb 23 at 0:20