Suppose we have a set of states $\Theta$ and a set of messages $M$. The prior measure is $\mu_0 \in \Delta \Theta$.
A signal $\sigma: \Theta \times M \to [0,1]$ is a regular conditional probability (RCP) such that
- $\forall \theta \in \Theta$, $\sigma(\cdot | \theta) \in \Delta M$
- $\forall m \in M$, $\sigma(m | \cdot)$ is measurable
A posterior $q: M \times \Theta \to [0,1]$ is a RCP such that
- $\forall m \in M$, $q(\cdot | m) \in \Delta \Theta$
- $\forall \theta \in \Theta$, $q(m | \cdot)$ is measurable
We also have some integrable function $\delta: \Theta \to \mathbb{R}$
Next, we have two sets $W \subseteq \Theta$ and $\mathcal{M} \subseteq M$ such that
- $\sigma(\mathcal{M} | \theta) = 1$ for every $\theta \in W$
- for all $m \in \mathcal{M}$
- $\int_{\Theta} \delta(\theta) dq(\theta | m) \geq 0$
- $supp \; q(\cdot | m) \subseteq W$
The idea here is that every state in $W$ (and only those states) sends only messages from $\mathcal{M}$. Every message from $\mathcal{M}$ induces a posterior that averages $\delta(\theta)$ to be non-negative.
Claim: $\int_W \delta(\theta) d\mu_0(\theta) \geq 0$.
Proof sketch: now, if it were true that $\int_\Theta \sigma(m | \theta)d\mu_0(\theta) > 0$ for all $m \in \mathcal{M}$, we could apply Bayes rule $q(\cdot | m) \propto \sigma(m | \cdot) \mu_0(\cdot)$ to say the following:
$$\forall m \in \mathcal{M}, \int_\Theta \delta(\theta) dq(\theta | m) = \int_W \delta(\theta) dq(\theta | m) \geq 0 \Longrightarrow \int_W \delta(\theta) \sigma(m | \theta) d\mu_0(\theta) \geq 0$$ (Bayes rule w/ positive denominator). Then, integrate the latter over all $m \in \mathcal{M}$ to get $\int_\Theta \delta(\theta) \sigma(\mathcal{M} | \theta) d\mu_0(\theta) \geq 0$, which is equivalent to $\int_\Theta \delta(\theta) d\mu_0(\theta) \geq 0$ since $\sigma(\mathcal{M} | \theta) = 1$ for every $\theta \in W$.
Now, the issue is that in my model, there definitely exist messages $m \in \mathcal{M}$ for which the Bayes rule above does not apply because $\int_\Theta \sigma(m | \theta)d\mu_0(\theta) = 0$. From what I understand, the Bayes rule I should be using is: $$\int_{\hat{\Theta}}\sigma(\hat{M} | \theta)d\theta = \int_\Theta\int_{\hat{M}}q(\hat{M} | m)d\sigma(m | \theta) d\mu_0(\theta), \text{ for every Borel } \hat{\Theta}\subseteq \Theta \text{ and } \hat{M} \subseteq M$$
Question is, how do I apply the Bayes rule to prove the above claim?