Given a compact metrisable topological space $X$, we write $\mathcal{N}(X)$ for the set of non-empty closed nowhere dense subsets of $X$, which is a Polish space under the topology induced by the Hausdorff distance.
Does there exist a compact metrisable topological space $X$ and a Borel probability measure $\nu$ on $\mathcal{N}(X)$ such that for all $p \in X$, $\,\nu(K : p \in K)>0$?
(If anyone has a reference for this, that would be particularly useful.)
Remark: Intuitively, I expect that the Baire category theorem will somehow imply that the answer is no. To prove that the answer is no, it would be sufficient to prove the following assertion:
Conjecture. Let $X$ be a compact metrisable topological space, let $Y$ be a Polish space, and let $G$ be a closed subset of $X \times Y$. Suppose that for every countable set $S \subset Y$ there is a dense set $D \subset X$ such that for every $p \in D$ and $y \in S$, $(p,y) \not\in G$. Then for every Borel probability measure $\nu$ on $Y$ there exists $p \in X$ such that $\,\nu(y \in Y : (p,y) \in G)=0$.
To see this: Suppose the conjecture is true. Let $X$ be a compact metrisable space, take $Y:=\mathcal{N}(X)$, and take $G:=\{(p,K): p \in K\}$. For every countable $S \subset \mathcal{N}(X)$, the Baire category theorem gives that $\,\bigcup S\,$ has empty interior; so set $D:=X \setminus \bigcup S$. Then $D$ is dense and for any $p \in D$ and $K \in S$, $p \not\in K$ and so $(p,K) \not\in G$. Hence for every probability measure $\nu$ on $\mathcal{N}(X)$ there exists $p \in X$ such that $\nu(K \in \mathcal{N}(X):p \in K)=0$.
