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](https://www.encyclopediaofmath.org/index.php/Measure_algebra_(measure_theory)) but I don't know much about them. 



Measure support decomposition that "tends to infinity"
I think the following is a counterexample.


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.


