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

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    $\begingroup$ Won't a random Cantor set of a.s. positive measure have this property? $\endgroup$ Dec 11, 2015 at 16:27
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    $\begingroup$ In particular, a random translate of a given positive measure Cantor set on the circle. $\endgroup$ Dec 11, 2015 at 16:38
  • $\begingroup$ Choosing $\nu$ to be the Dirac measure at $\{X\}$ gives a trivial "yes". What are the implicit assumptions on $\nu$? $\endgroup$
    – YCor
    Dec 28, 2015 at 13:27
  • $\begingroup$ @YCor: It is stated in the question that $\mathcal{N}(X)$ is the set of non-empty closed nowhere dense subsets of $X$. $\endgroup$ Dec 28, 2015 at 15:45
  • $\begingroup$ Thanks, I missed this one (it's just "empty interior" here). $\endgroup$
    – YCor
    Dec 28, 2015 at 16:11

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Yes. As in the comments: take $X=\mathbb{S}^1$; and let $\nu$ be the law of the random set constructed by taking a positive-Lebesgue-measure Cantor set $K \subset \mathbb{S}^1$ and rotating $K$ through a random angle selected according to the uniform distribution.

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