Bayes formula for dominated kernels

The measure $\pi$ is finite and therefore sigma-finite; $\nu$ is sigma-finite by hypothesis. For each $\theta$, [F3] and the specified density identity give ${\int }_T\mathrm{\ell }(\theta ,y)\nu (dy)=Q(\theta ,T)=1$. By [F4] the formula $\lambda (C)={\int }_C\mathrm{\ell }d(\pi \times \nu )$ defines a measure. Tonelli [F2] gives $\lambda (E\times T)={\int }_E({\int }_T\mathrm{\ell }(\theta ,y)\nu (dy))\pi (d\theta )={\int }_{E}1d\pi =1.$ Thus it is a joint probability. For a rectangle $A\times B$, the same theorem gives $\lambda (A\times B)={\int }_{A}Q(\theta ,B)\pi (d\theta )$, so its first marginal is exactly the prior.

Apply [F1] with $\mu =\pi$, density $p=\mathrm{\ell }$, and supplied filler $\rho =\pi$. All hypotheses were checked in step 1.1: the product is sigma-finite, the density is jointly measurable and nonnegative, and its total mass is one. The resulting normalizer is exactly m and the resulting kernel is the displayed K. The theorem gives its measurable evaluations, pointwise probability sections, marginal $\beta =md\nu$, and $\beta (\{m=0\}\cup \{m=\mathrm{\infty }\})=0$. It also gives $\lambda (A\times B)={\int }_{B}K(y,A)\beta (dy)$ and hence the conditional law on the coordinate probability space. In particular $K(y,\mathrm{\varnothing })=0$ and $K(y,E)=1$ both on good fibres and on filled fibres. No quotient at either excluded endpoint is used.