Question about convex set of probabilities Let $(\Omega, \mathcal{F})$ be a measurable space. Let $\mathbb{P}$ be a set of finitely additive probability measures on $(\Omega, \mathcal{F})$ that is convex and closed in the product topology. 
(To clarify, we view $\mathbb{P}$ as a subset of the function space $[0,1]^{\mathcal{F}}$ equipped with the topology of pointwise convergence.)

Let $A,B \in \mathcal{F}$. Suppose that $P(B)>0$ for all $P \in \mathbb{P}$ and $\inf_{P \in \mathbb{P}}P(A) \leq \inf_{P \in \mathbb{P}} P(A \mid B) =: p$. Is it true that for all $\epsilon>0$ there exists $P' \in \mathbb{P}$ such that $P'(A \mid B) - p < \epsilon$ and $P'(A \mid B) \geq P'(A)$?

I checked a few fairly simple examples and the result seems to hold, but I'm not sure how to prove it in general. Any hints are appreciated.
 A: No. Take $\Omega=\{a,b,c,d\}$ with $A=\{a,b\}$ and $B=\{a,c\}$.  Let us think of probability measures as elements of $[0,1]^4$, since they are determined by the measure of each singleton (the first coordinate is the measure of $\{a\}$, and so on).  Let $P_0=(1/8,1/2,3/8,0)$ and $P_1=(1/4,0,0,3/4)$, and let $\mathbb{P}$ be the convex hull of $P_0$ and $P_1$.  Explicitly, $\mathbb{P}$ consists of the measures $$P_t=\left(\frac{1+t}{8},\frac{1-t}{2},\frac{3-3t}{8},\frac{3t}{4}\right)$$ for each $t\in[0,1]$.
We then have $$P_t(A)=\frac{5-3t}{8}$$ and $$P_t(A\mid B)=\frac{1+t}{4-2t}.$$ In particular, $P_t(A)$ is strictly decreasing and $P_t(A\mid B)$ is strictly increasing, with $P_t(A\mid B)<P_t(A)$ for $t$ close to $0$ and $P_t(A)<P_t(A\mid B)$ for $t$ close to $1$.  Also, they have the same infimum, $P_1(A)=P_0(A\mid B)=1/4$.  In order for $P_t(A\mid B)$ to be within $\epsilon$ of this infimum, $t$ must be close to $0$.  If $\epsilon$ is sufficiently small, this forces $P_t(A\mid B)<P_t(A)$.
