# Measure support decomposition that “tends to infinity”

I would like to know the answers to the following two questions.

Let $$S$$ be a locally compact Hausdorff space, $$\mu$$ be a regular Borel measure with non-compact support $$M$$. Denote $$\mathscr{H}=\{\mathcal{H}\subset 2^M: \mathcal{H}\mbox{ is a disjoint family of Borel sets of positive measure}\}$$ Note that for measures with the continuous part or infinitely many atoms there is an infinite family $$\mathcal{H}\in\mathscr{H}$$.

Question #1. Does there exist an infinite family $$\mathcal{H}\in\mathscr{H}$$ such that for any compact $$K\subset M$$ only finitely many elements of $$\mathcal{H}$$ have positive measure intersection with $$K$$?

A few side notes:

• I know, how to prove this in the case where $$S$$ is $$\sigma$$-compact;
• There is an obvious example without $$\sigma$$-compactness - counting measure on an uncountable set;
• From [342B, Measure theory. Vol 3. Measure algebras. D. H. Fremlin] I know how to construct at most countable disjoint family $$\mathcal{H}$$ consisting of compacts of positive measure.

Maybe it would be easier to characterize measures with the opposite property.

Question #2. What can we say about $$S$$ or $$\mu$$ if for any countable $$\mathcal{H}\in \mathscr{H}$$ there exists a compact $$K$$ such that infinitely many elements of $$\mathcal{H}$$ have positive measure intersection with $$K$$.

It looks like these questions fit well into the realm of measure algebras but I don't know much about them.

• Perhaps you could give your definition of "regular measure", since this term is not always consistently defined in the literature. – Nate Eldredge Feb 26 at 23:21
• @NateEldredge, measures that come from continuous linear functionals on $C_0(S)$ – Norbert Feb 26 at 23:27
• Does "$C_0$" mean compactly supported, or vanishing at infinity? I guess the former, otherwise you only get finite measures. – Nate Eldredge Feb 26 at 23:57
• $C_0$ means continuous vanishing functions – Norbert Feb 27 at 8:23

Consider the Stone–Čech compactification $$\beta \mathbb{N}$$. Fix some $$x \in \beta \mathbb{N} \setminus \mathbb{N}$$ and set $$S = \beta \mathbb{N} \setminus \{x\}$$, which is locally compact Hausdorff and not compact. Let $$\mu$$ be the Borel measure on $$S$$ which puts mass $$2^{-n}$$ on each point $$n \in \mathbb{N} \subset S$$. This measure is regular, and its support is all of $$S$$ (which is non-compact), since $$\mathbb{N}$$ is dense in $$S$$.
Suppose $$\mathcal{H}$$ is an infinite disjoint family of Borel sets having positive measure. Then each $$H \in \mathcal{H}$$ must contain at least one point of $$\mathbb{N}$$, so write $$\mathcal{H} = \{H_1, H_2, \dots\}$$, and for each $$k$$ choose some $$n_k \in H_k \cap \mathbb{N}$$. Set $$A = \{n_1, n_3, n_5, \dots\}$$ and $$B = \{n_2, n_4, n_6, \dots\}$$. I claim that either $$A$$ or $$B$$ is contained in a compact set $$K$$.
To see this, take some function $$f : \mathbb{N} \to \{0,1\}$$ which is $$0$$ on $$A$$ and $$1$$ on $$B$$, and extend it to a continuous $$\hat{f} : \beta \mathbb{N} \to \{0,1\}$$. Suppose that $$\hat{f}(x) = 1$$. Then the set $$K = \hat{f}^{-1}(\{0\})$$ is closed in $$\beta \mathbb{N}$$, hence compact, and contains $$A$$ but not $$x$$. So $$K$$ is also compact in $$S$$, and since it contains $$A$$, it intersects $$H_1, H_3, H_5, \dots$$ with positive measure. If instead we had $$\hat{f}(x) = 0$$, then just interchange $$A$$ and $$B$$, and get a compact set $$K$$ intersecting $$H_2, H_4, H_6,\dots$$.
It is true if $$S$$ is metrizable. Fix a compatible metric $$d$$. Since $$M$$ is not compact, there is a sequence $$x_n \in M$$ with no convergent subsequence; since $$M$$ is closed, no subsequence converges in $$S$$ either. Let $$B_n$$ be disjoint open balls centered at $$x_n$$, having radius at most $$1/n$$. Since each $$x_n$$ is in the support, each $$B_n$$ has positive measure, so take $$\mathcal{H} = \{B_n\}$$. Now if $$K$$ meets infinitely many $$B_n$$ (at all), then there is a sequence $$y_{n_k} \in K \cap B_{n_k}$$. If this has a convergent subsequence $$y_{n_{k_j}} \to y$$, then $$x_{n_{k_j}} \to y$$ as well, a contradiction. So $$K$$ is not compact.