If $E_P(X) \in $ span($E_{P_i}(X)$), $1 \leq i \leq n$, for all integrable $X$, then is $P$ a convex combination of the $P_i$? Let $(\Omega, \mathfrak{F}, P)$ be a probability space and let $R = \{P_1,...,P_n \}$ be a finite set of probability measures on $(\Omega, \mathfrak{F})$, each of which is absolutely continuous with respect to $P$. I'm wondering if the following result holds.

Suppose that for all random variables $X$ which are integrable with respect to $P$ and $P' \in R$ we have that $E_P(X)$ lies in the interval spanned by $E_{P_i}(X)$, $1 \leq i \leq n$. Then $P$ is a convex combination of the members of $R$.

In this paper the result is shown for the case where $\Omega$ is finite. The proof relies on identifying random variables and probability measures with members of $\mathbb{R}^n$ and proceeds by appealing to some basic facts about convex sets. How might one approach this problem for general $(\Omega, \mathfrak{F})$?

Addendum. Here is what I've come up with. I would appreciate feedback.
We begin with a general measurable space $(\Omega, \mathfrak{F})$ and the space $V$ of all bounded random variables on $(\Omega, \mathfrak{F})$. The space $V$ is a (real) vector space with respect to pointwise addition and scalar multiplication, and if we equip $V$ with the $\sup$-norm, it becomes a Banach space (it is complete). We now identify probability measures $P$ on $(\Omega, \mathfrak{F})$ with continuous linear functionals in the dual space $V'$ via the embedding $P \mapsto \int (\cdot) dP$. Hence, for $X \in V$, we have $P(X) = E_P(X)$, where $E_P$ is expected value with respect to $P$. In order to consider sets of probability measures with certain topological properties, we endow $V'$ with its weak*-topology, which is the topology generated by the evaluation functionals $\lambda_X$ in the double dual $(V')'$, defined by $\lambda_X(\phi) = \phi(X)$, where $X \in V$ and $\phi \in V'$. That is, the weak*-topology is the weakest topology that makes the evaluation functionals continuous, and, moreover, a linear functional $\lambda$ on $V'$ is continuous if and only if it is an evaluation functional. Now, $V'$ is a locally convex topological vector space in the weak*-topology, and the set of probability measures on $(\Omega, \mathfrak{F})$ is weak*-compact in $V'$. It follows that any weak*-closed set of probability measures is weak*-compact. This much is just my attempt to summarize some facts that I learned by studying chapter 14 of Royden's Real Analysis, together with the thought that probability measures can be identified with continuous linear functionals in $V'$ (I also consulted Chapter 3 and Appendices D and E of Walley's Statistical Reasoning with Imprecise Probabilities). If what I have said so far is acceptable, then I believe that I've answered my question here.
Theorem
Let $(\Omega, \mathfrak{F}, P)$ be an arbitrary probability space and $R$ an arbitrary set of probability measures on $(\Omega, \mathfrak{F})$ with $|R| \geq 2$. Then, the following two assertions are equivalent. (a) $E_P(X)$ lies strictly inside the interval spanned by $\{E_{P'}(X) \}_{P' \in R}$, i.e.,
$$\inf_{P' \in R} \{E_{P'}(X) \} < E_P(X) < \sup_{P' \in R} \{E_{P'}(X) \}.$$
(b) $P$ is in the interior of the convex weak*-closure of $R$, i.e. $P \in \text{int}(\text{co}R).$
Proof. We note that the convex weak*-closure $\text{co}R$ of $R$ is a convex, weak*-compact subset of $V'$ with non-empty interior, because $|R| \geq 2$. We begin by supposing that $P \notin \text{int}(\text{co}R)$. By a standard separating hyperplane result, there exists a continuous linear functional $\lambda$ on $V'$ and $\alpha \in \mathbb{R}$ such that $\lambda(P) \geq \alpha$ and $\lambda(P') \leq \alpha$ for all $P' \in \text{co}R$. But the continuous linear functionals on $V'$ consist of all and only the evaluation functionals. So for some $X \in V$, we have $\lambda = \lambda_X$, and hence $\lambda_X(P) \geq \alpha \geq \lambda_X(P')$ for all $P' \in \text{co}R$. But by the definition of the evaluation functional, this implies 
\begin{equation}\label{eqn: spanning violation}
E_P(X)=P(X) \geq \alpha \geq P'(X) = E_{P'}(X)
\end{equation}
for all $P' \in \text{co}R$. Hence, $E_P(X) \geq \sup_{P' \in \text{co}R}\{E_{P'}(X) \} \geq \sup_{P' \in R}\{E_{P'}(X) \}$, and (a) is violated.
Conversely, suppose that $P \in \text{int}(\text{co}R)$. Then there exists a collection $\{P_i \}_{i=1}^n$, $n \geq 2$, of probability measures in $R$ such that $P = \sum_{i=1}^n \beta_i P_i$, with $\beta_i > 0$ and $\sum_{i=1}^n \beta_i = 1$ (I'm not sure I can justify this step; suggestions here are would be appreciated). For an arbitrary random variable $X \in V$, we have
    \begin{align*}
    E_P(X) &= \int X dP = \int X d(\sum_{i=1}^n \beta_i P_i) \\
    &= \sum_{i=1}^n \beta_i \int X dP_i = \sum_{i=1}^n \beta_i E_{P_i}(X).
    \end{align*}
We note that this calculation relies on the fact that the $P_i$ are finite measures. We now see that $E_P(X)$ is a strict convex combination of the $E_{P_i}(X)$, so (a) is satisfied.
 A: I believe that you may reduce the first problem to the finite dimensional case in the following way: 
Let $\Delta_n =\{\lambda\in {\Bbb R}^n : \sum_{i=1}^n \lambda_i=1, \lambda_1,...,\lambda_n\geq 0\}$
For $A\in \mathfrak{F}$ define $Z(A)=\{ \lambda \in \Delta_n^n: P(A)=\sum_{i=1}^n \lambda_i P_i(A)\}$
Then by hypothesis each $Z(A)$ is a non-empty, compact subset of $\Delta_n$. If $Z(\mathfrak{F})=\bigcap_{A\in \mathfrak{F}} Z(A)$ is non-empty then any $\lambda\in Z(\mathfrak{F})$ allows to write $P(A)=\sum_i \lambda_i P_i(A)$ on measurable sets which suffices.
So suppose instead that $Z(\mathfrak{F})=\emptyset$ then by the finite intersection principle for compact sets there must be a finite number of sets $A_1,...,A_n$ for which $Z(A_1)\cap ...\cap Z(A_n)=\emptyset$. Let ${\cal B}=\sigma(A_1,...,A_n)$ be the (finite) $\sigma$ algebra generated by this finite collection and consider the restriction of $P$ and the $P_i$'s to ${\cal B}$.
We are then dealing with probability measures on a finite space which will satisfy the same hypotheses but  $Z({\cal B})=\emptyset$ which apparently is in contradiction with the paper you cite.
Regarding the more general problem, let $E$ denote the vectorspace of
signed measures on $(\Omega, \mathfrak{F})$ with the variation norm
and $\Lambda=\overline{B}_1(L^\infty(\Omega, \mathfrak{F}))$, i.e.
the set of measurable maps: $X:\Omega \rightarrow [-1,1]$.
The variation norm of $\mu\in E$  is then 
$\|\mu\|= \sup_{\Lambda\in \Lambda} |\mu(X)|$ showing that $\Lambda$
separates points in $E$. The set $\Lambda$ gives rise to
a weak topology on $E$.
Given $\mu\in E$, $X_1,...,X_m\in \Lambda$, 
$\epsilon>0$ we set 
$$ \{ \nu \in E: |\nu(X_i)-\mu(X_i)| < \epsilon, i=1,...,m\}$$
Then the sets of this type forms a
base (neighborhood of $\mu$) for the $\Lambda$-topology on $E$.
In general it is weaker than the weak topology generated by the dual $E'$
since $\Lambda$ may be viewed as a subset of $E'$.
Now let $M\subset E$ be the subset of probability measures and
consider a subset $R\subset M$ and $\mu\in M$. We set $S=co(R)$, the
closed convex hull of $R$, closed for the $\Lambda$ topology.
If $\mu\notin S$ then by definition there is an open neighborhood $B$ of
$\mu$ that does not intersect $S$.
Thus, $|\nu(X_i)-\mu(X_i)|<\epsilon$, $i=1,...,m$. But by a finite
dimension argument 
(see e.g. S Lang: Real and Functional Analysis, Thm 1.2, p. 85) there
are real constants $c_1,...,c_m$ such that 
$X=c_1X_1+ \ldots c_m X_m$ separates $\mu$ from $S$, in other words, there
is $a\in {\Bbb R}$ s.t. 
$$ \mu(X)  > a \geq \nu(X)$$
 for every $\nu\in S$.  In your case, the condition you impose on $P$
tells us that the above can not happen, so 
$P\in co(R)$. (This follows pretty much the same line of arguments you give).
Now, the implications of this are not so obvious. In general you
can not hope for $P$ to be a finite linear combination of elements in $R$.
If e.g. you take $P_1,P_2,...$ to be a sequence in $M$ with disjoint
support, then $P = \sum_k 2^{-k} P_k$ is in the closed convex hull but is not
finitely generated.
Worse, if $R$ is uncountable,
you can not in general hope to write $P$
as a countable convex combination of elements in $R$.
Choquet's Theorem states that if $S$ is a metrizable compact convex
subset of a locally compact Hausdorff topol vector space and
$P\in S$ then there is a measure supported on the extremal
points of $S$ (which here should be a subset of $R$ it seems) which
represents $P$ (so you may write $P$ as an 'integral' over $R$).
Now, what you need to metrize is the $\Lambda$ topology and
this is not possible in our general setup.
(It would be if $\Omega$ was compact Hausdorff 
and we looked at Borel measures.)  I think that in the general setup
you can not obtain further information on how to decompose $P$
with respect to $R$.
