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. 
 A: 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.
