An impossible observation has zero evidence and only a filler filter value

With counting measure on $F_1$, set $g_1(x,y)=G_1(x,\{y\})$; it is jointly measurable on the finite discrete spaces, sums to one over $y$, and gives $G_1(x,C)={\sum }_{y\in C}{g}_1(x,y)$. The predictive state law is ${\delta }_0$, so the evidence is $z_1(a)=1$ and $z_1(b)=0$; the update at $a$ is ${\delta }_0$ and [F2] and [F4] make the update at $b$ the predictive filler ${\delta }_0$. The factorized law [F3] gives $P(Y_1=b)=0$.

Let $\rho$ be any probability measure on $E_1$ and define $q(a,\cdot )={\delta }_0$ and $q(b,\cdot )=\rho$. This is a measurable probability kernel because the observation space is finite discrete. For every $A\subseteq \{0,1\}$ and $C\subseteq \{a,b\}$, the path law gives $P(X_1\in A,Y_1\in C)={\mathbf{1}}_A(0){\mathbf{1}}_C(a)$, while $E[{\mathbf{1}}_{\{Y_1\in C\}}q(Y_1,A)]={\mathbf{1}}_C(a){\delta }_0(A)+0\cdot {\mathbf{1}}_C(b)\rho (A)={\mathbf{1}}_A(0){\mathbf{1}}_C(a)$. These are all conditioning events in the finite observation space, so every probability $\rho$ at $b$ gives the same conditional identities.