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 $\nu(K \in \mathcal{N}(X):p \in K)=0$.

nowhere densesubsets of $X$. $\endgroup$