A zero-evidence fiber has no normalized fiberwise update

The half-half set function ${\mu }_0$ is a probability: in any disjoint Borel sequence each of $0$ and $1$ occurs in at most one member, so the mass of the union equals the sum of the masses, and ${\mu }_0(\mathbb{R})=1$. Since ${\mu }_0(D)=1$, it is a prior on $D$ by [F1]. The constant map $F(u)=0$ is a forward map on the nonempty truth set $U=D$ by [F12], and its restriction is measurable because every preimage is $D$ or $\mathrm{\varnothing }$. For each $u\in D$, $Q(u,\cdot )={\delta }_0$ is a probability measure by [F4], and every evaluation $u\mapsto Q(u,B)$ is measurable on finite $D$; hence [F7] makes $Q$ a probability kernel. Thus [F3] gives a proper Bayesian model.

With counting measure on $\mathsf{Y}$, the density is $L(u,y)={\mathbf{1}}_{\{y=0\}}$ for both parameter values; its sum over $y\in \{0,1\}$ is one, so it represents the stated kernel. The potential and density are jointly measurable: each singleton rectangle $\{u\}\times \{y\}$ is measurable, so every subset of finite $D\times \mathsf{Y}$ is a finite union of measurable rectangles by [F13]. Integrating the kernel against the half-half prior gives data law concentrated at $0$, and the same density gives $Z(0)=\frac{1}{2}+\frac{1}{2}=1$ and $Z(1)=0$. At $y=1$, [F9] rules out any finite positive scaling of the zero likelihood measure as a probability. The potential equals $+\mathrm{\infty }$ on $D$ at this datum, so the finite-section hypothesis in [F11] fails.

The explicit dominated kernel from [F8] uses the normalized update at $y=0$ and prior filler ${\mu }_0$ at $y=1$. The data law gives $\{1\}$ probability zero by step 2.1, so [F10] allows replacing that filler by ${\delta }_1$ without changing the conditional identities. Both fillers are probability measures on $D$ by [F4], but neither is obtained by normalizing the zero-evidence fiber.