Products of Blackwell spaces and regular conditional probabilities (Q1068227)

[Cf. the review above for notations and definitions.] - In previous work, the author investigated the Blackwell property for sets \(X\subseteq {\mathbb{R}}\) such that \({\mathbb{R}}\setminus X\) is totally imperfect (contains no Cantor-like sets). Here, the following is proved: Proposition: Let \(X\subseteq {\mathbb{R}}\) have totally imperfect complement. Let A be an analytic set. For each \(n\geq 1\), the following are equivalent: 1) \(X^ n\) has the strong Blackwell property; 2) \(X^ n\) has the Blackwell property; 3) \(X^ n\times {\mathbb{R}}\) has the strong Blackwell property; 4) \(X^ n\times {\mathbb{R}}\) has the Blackwell property; 5) \(X^ n\times A\) has the strong Blackwell property; 6) \(X^ n\times A\) has the Blackwell property. An example is given of a probability space whose Borel structure is strong Blackwell but for which regular conditional probabilities do not always exist. This answers a question of D. Ramachandran.