# Borel selections of partitions of the space of probability measures

The environment:

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).

Let $p_1,\dots,p_N\in\Delta(X)$.

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.

Let us endow $T=\Delta(X) \setminus S$ with the subspace topology. It is also Polish.

Now define an equivalence relation (which maybe also be viewed as a partition) $\Pi$ on $T$.

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 $q = (\sum_{k=1}^{N} \alpha_k p_k)+\alpha_{N+1}r$.

Main Question: Does $\Pi$ admit a Borel selector?

Related Questions that may help resolve the main question:

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 $U^*=\bigcup_{u\in U} \lbrace t\in T : u \Pi t \rbrace$.

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?

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 $q\Pi r$?

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I cannot understand the reason to exclude $S$ from $\Delta(X)$. Why don't you take simply $T=\Delta(X)$. –  zhoraster Feb 9 '11 at 15:59
The reason for excluding $S$ from $\Delta(X)$ is to make Related Question 2 easier to deal with. If $\forall n(q_n,r\in T)$ and $q\in S$ and $q_n\to q$ and $\forall n(q_n\Pi r)$ and $\forall n(r_n=r)$ then $\neg(q\Pi r)$. Perhaps a better way to think about it is to define $\Pi$ on $\Delta(X)$ and consider whether $\Pi \cap T\times T$ is closed. –  BSL Feb 10 '11 at 0:14
How does one prove that $\Pi$ is an equivalence relation? A first consequence of $q\Pi r$ as it is written is $r = (\sum_{k=1}^{N} \beta_k p_k)+\alpha_{N+1}^{-1}q$ with $\beta _ k :=-\alpha_k/\alpha_{N+1}$, but I can't see why should the new linear combination of the $p_k$ belong to $\Delta(X)$ as required. Is there a more subtle reason ? –  Pietro Majer Oct 6 '11 at 14:42
The equivalence relation is defined over $\Delta(X)$. To be more precise, the definition should begin with the phrase "Let $q,r\in\Delta(X)$. We will say that $q\Pi r$ if and only if..." –  BSL Oct 6 '11 at 20:39
Er, sorry. I meant to write "Let $q,r\in\Delta(X)\setminus S$..." –  BSL Oct 6 '11 at 20:41