Measure with `somewhere dense' support

Question

Let $X$ be a compact Hausdorff (but not necessarily metrizable) space. Is it always true that there exists a probability Borel measure $\mu$ and an open set $U$ such that any nonempty open set $V\subset U$ has positive measure? Which additional hypothesis would help?

Answer 1

No. An almost $P$-space is a topological space where every nonempty $G_{\delta}$-set has a nonempty interior. By the answers to this recent question, there exists compact almost $P$-spaces without any isolated points. However, I claim that if $X$ is a compact $P$-space and $\mu$ is a Borel measure on $X$, then for each nonempty open set $U$ there is a nonempty open subset $V\subseteq U$ with $\mu(V)=0$. Suppose that $U$ is a nonempty open subset of $X$. Then by induction, we can construct a sequence $(U_{n})_{n\in\omega}$ of open sets with $U_{0}\subseteq U,\overline{U_{n+1}}\subseteq U_{n}$ and where $\mu(U_{n+1})<\frac{1}{2}(\mu(U_{n}))$. Let $G=\bigcap_{n}U_{n}$. Then $G$ is a non-empty $G_{\delta}$-set with $\mu(G)=0$. However, since $X$ is an almost $P$-space, there is a nonempty open set $V$ with $V\subseteq G$. Clearly $\mu(V)=0$ but $\emptyset\neq V\subseteq U$.

Answer 2

By turning this result into a Boolean algebra problem, a result by Thomas Jech completely characterizes the compact zero-dimensional spaces which have measures $\mu$ so that $\mu(U)\neq\emptyset$ for each non-empty open set $U$.

$\mathbf{Theorem}$ Let $B$ be a Boolean algebra. Then the following are equivalent.

1. There is a finitely additive probability measure $\mu$ on $B$ so that $\mu(b)>0$ for each $b\in B^{+}$.

2. $B\setminus\{0\}$ is the countable union of some sets $(C_{n})_{n}$ where

   i. For all $n$, there is some natural number $K(n)$ so that if $R\subseteq C_{n}$ and $r\wedge s=0$ for $r,s\in R,r\neq s$, then $|R|\leq K(n)$.

ii. If $a\vee b\in C_{n}$, then $a\in C_{n+1}$ or $b\in C_{n+1}$.

The above theorem is proven in the paper 1.

By transferring the above theorem via Stone duality to compact spaces, we obtain the following theorem.

$\mathbf{Theorem}$

Let $X$ be a compact zero-dimensional space and let $\mathcal{O}(X)$ be the collection of all open subsets of $X$. Then the following are equivalent.

1. There exists a regular Borel probability measure $\mu$ on $X$ so that $\mu(U)>0$ for each non-empty open set $U$.

2. There are subsets $C_{n}\subseteq\mathcal{O}(X)$ so that $\bigcup_{n}C_{n}=\mathcal{O}(X)\setminus\{\emptyset\}$ and where

i. There is some natural number $K(n)$ so that if $R\subseteq C_{n}$ and $U\cap V=\emptyset$ whenever $U,V\in R,U\neq V$, then $|R|\leq K(n)$

ii. if $U,V$ are open sets and $U\cup V\in C_{n}$, then $U\in C_{n+1}$ or $V\in C_{n+1}$.

Although the above theorem is stated in terms of compact zero-dimensional spaces it seems like the above result should hold for all compact Hausdorff spaces, but I am unaware of anyone who has worked out the details.

1. Thomas Jech. Algebraic characterizations of measure algebras Proc. Amer. Math. Soc. 136 (2008), 1285-1294

Answer 3

As it is mentioned in the comments, a condition that would prevent a space from having such a measure is that every open subspace has uncountable cellularity. One can prove that a compact almost $P$-space without isolated points (see Joseph Van Name´s answer) has this property, thus obtaining a different proof that such a space does not admit a "somewhere-dense-supported" measure.

Let me give now a first-countable example (so it cannot be an almost $P$-space):

Let $X=Z^\omega$ where $Z$ is the Alexandroff duplicate of the Cantor space. Since $Z$ is first-countable, zero-dimensional and has a dense set of isolated points, $X$ is homogeneous by a result of D.B. Motorov announced in Zero-dimensional and linearly ordered bicompacta: properties of homogeneity type, Russ. Math. Surv. 44:6 (1989). Also, $X$ is compact and has uncountable cellularity (because $Z$ is compact and has uncountably many isolated points), which implies by homogeneity that any open subspace of $X$ has uncountable cellularity.