Borel selections of partitions of the space of probability measures - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:03:14Z http://mathoverflow.net/feeds/question/54812 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54812/borel-selections-of-partitions-of-the-space-of-probability-measures Borel selections of partitions of the space of probability measures BSL 2011-02-08T20:34:06Z 2011-02-08T20:34:06Z <p><strong>The environment:</strong></p> <p>Suppose that $X$ is a Polish space and $\Delta(X)$ the set of Borel probability measures over $X$ (given the topology of weak convergence, with the Prohorov metric).</p> <p>Let $p_1,\dots,p_N\in\Delta(X)$.</p> <p>Let $S=\lbrace \sum_{k=1}^N \alpha_k p_k \in\Delta(X):\forall k(\alpha_k \in\mathbb{R})\rbrace$. That is, $S$ is the set of probability measures that are linear combinations of $p_1,\dots,p_N$. In some sense, $S$ is like the span of $p_1,\dots,p_N$. $S$ should be a closed set.</p> <p>Let us endow $T=\Delta(X) \setminus S$ with the subspace topology. It is also Polish.</p> <p>Now define an equivalence relation (which maybe also be viewed as a partition) $\Pi$ on $T$.</p> <p>Def: Let $q \Pi r$ if and only if $\exists \alpha_1,\dots,\alpha_{N+1}\in\mathbb{R}$ such that $\alpha_{N+1} > 0$ and $q = (\sum_{k=1}^{N} \alpha_k p_k)+\alpha_{N+1}r$.</p> <p><strong>Main Question:</strong> Does $\Pi$ admit a Borel selector?</p> <p><strong>Related Questions</strong> that may help resolve the main question:</p> <p>1) Is the saturation of every open set in $T$ with respect to $\Pi$ a Borel set? A saturation of an open set $U \subseteq T$ is $U^*=\bigcup_{u\in U} \lbrace t\in T : u \Pi t \rbrace$.</p> <p>1') Equivalently, we might think of all open balls (using the Prohorov distance) $U\subseteq \Delta(X)$ such that $U\cap S=\emptyset$. Is $U^*$ a Borel set?</p> <p>2) Is $\Pi$ a closed relation? That is, is it a closed set in $T \times T$? If $\forall n(q_n \Pi r_n)$ and $q_n\to q$ and $r_n\to r$, then is it the case that $q\Pi r$?</p>