The observed density is the latent marginal under Tonelli

Fix $\theta$ and let ${\mathcal{D}}_{\theta }$ be the collection of subsets $E\subseteq \mathrm{\Theta }\times \mathsf{X}\times \mathsf{Z}$ whose section $E_{\theta }:=\{(x,z):(\theta ,x,z)\in E\}$ belongs to $\mathcal{A}\otimes \mathcal{C}$. Sections commute with complements and countable unions, so ${\mathcal{D}}_{\theta }$ is a sigma-algebra. It contains each measurable rectangle $T_0\times A_0\times C_0$, whose section is $A_0\times C_0$ if $\theta \in T_0$ and empty otherwise. By [F12], every product-measurable set has a measurable section. Applying this to preimages of Borel sets shows that $p_{\theta }(x,z):=p(\theta ,x,z)$ is product-measurable. For complex $h$, first compose with the continuous Borel coordinate maps $\mathrm{Re},\mathrm{Im}:\mathbb{C}\to \mathbb{R}$; the same section argument makes both $\mathrm{Re}{h}_{\theta }$ and $\mathrm{Im}{h}_{\theta }$ measurable, hence $h_{\theta }$ is complex-measurable.

By [F13], choose a measurable exhaustion $Z_n\uparrow \mathsf{Z}$ with $\mu (Z_n)<\mathrm{\infty }$. On $S:=\mathrm{\Theta }\times \mathsf{X}$, the constant assignment $K(s,B):=\mu (B)$ is a measure kernel: each section is $\mu$ and each evaluation map is constant. The same exhaustion makes it uniformly sigma-finite by [F2].

For fixed $\theta$, step 1.1 makes $p_{\theta }$ product-measurable, so Tonelli [F4] gives ${\int }_{\mathsf{X}}m(\theta ,x)\nu (dx)={\int }_{\mathsf{X}\times \mathsf{Z}}{p}_{\theta }(x,z)d(\nu \otimes \mu )=1$ by [F1]. Thus $m(\theta ,\cdot )$ is a probability density. Also $E_{\theta }:=\{(x,z):p_{\theta }(x,z)=\mathrm{\infty }\}$ is measurable. For every positive integer $n$, $n{\mathbf{1}}_{E_{\theta }}\le p_{\theta }$, so [F6] gives $n(\nu \otimes \mu )(E_{\theta })\le 1$. Since this holds for every $n$, [F7] implies $(\nu \otimes \mu )(E_{\theta })=0$.

The function $p$ is nonnegative and product-measurable by [F1]. Under the finite-product identification $S\times \mathsf{Z}=(\mathrm{\Theta }\times \mathsf{X})\times \mathsf{Z}$ from [F12], apply [F3] to the constant kernel $K$ from step 1.2. It follows that $s\mapsto \int p(s,z)\mu (dz)=m(s)$ is measurable on $S$.

For fixed $\theta$, write $u_{\theta }=\mathrm{Re}{h}_{\theta }$ and $v_{\theta }=\mathrm{Im}{h}_{\theta }$. Define the finite-valued measurable function ${\tilde{p}}_{\theta }:=p_{\theta }{\mathbf{1}}_{\{p_{\theta }<\mathrm{\infty }\}}$ and then $H_{\theta }:=u_{\theta }{\tilde{p}}_{\theta }+i{v}_{\theta }{\tilde{p}}_{\theta }\mathrm{.}$ By [F11] this is complex-valued product-measurable; it is defined to be zero on the null set where $p_{\theta }=\mathrm{\infty }$. It agrees with $h_{\theta }{p}_{\theta }$ off that null set, and the assumed absolute integrability gives $\int \mathrm{\mid }{H}_{\theta }\mathrm{\mid }d(\nu \otimes \mu )<\mathrm{\infty }$. Fubini [F5] applies to this $L^1$ representative and gives ${\int }_{\mathsf{X}\times \mathsf{Z}}{H}_{\theta }d(\nu \otimes \mu )={\int }_{\mathsf{X}}({\int }_{\mathsf{Z}}{H}_{\theta }(x,z)\mu (dz))\nu (dx)={\int }_{\mathsf{Z}}({\int }_{\mathsf{X}}{H}_{\theta }(x,z)\nu (dx))\mu (dz),$ with its zero assignment on exceptional null sections. Since $H_{\theta }=hp$ almost everywhere on the fixed-$\theta$ product space, these are the asserted exchanges for $hp$.

Let $E:=\{x:m(\theta ,x)=\mathrm{\infty }\}$, measurable by step 2.2. For each positive integer $n$, $n{\mathbf{1}}_E\le m(\theta ,\cdot )$, so [F6] and step 2.1 give $n\nu (E)\le 1$. Since this holds for every $n$, [F7] forces $\nu (E)=0$. If $0<m(\theta ,x)<\mathrm{\infty }$, division by that scalar and the definition of $m$ give $\int q(z\mid \theta ,x)\mu (dz)=m(\theta ,x)\mathrm{/}m(\theta ,x)=1$. The valid-fibre set $V=m^{-1}((0,\mathrm{\infty }))$ is measurable by step 2.2 and [F8]. For measurability, define $\rho :[0,\mathrm{\infty }]\to [0,\mathrm{\infty }]$ by $\rho (t)=1\mathrm{/}t$ for finite $t>0$ and $\rho (0)=\rho (+\mathrm{\infty })=0$. For $a<0$, $\{\rho >a\}=[0,\mathrm{\infty }]$; for $a=0$ it is $(0,\mathrm{\infty })$; and for $a>0$ it is $(0,1\mathrm{/}a)$. These sets are Borel by [F8], so [F9] makes $\rho$ Borel. By [F10], $\rho \circ m$ is measurable on $\mathrm{\Theta }\times \mathsf{X}$. Pulling it back along the measurable coordinate projection from the finite product space, which is measurable by [F12], gives a product-measurable function; multiplying by $p$ preserves measurability by [F11]. This product restricts to $q=p\mathrm{/}m$ on $V\times \mathsf{Z}$, so $q$ is measurable on the valid-fibre set.

Steps 2.2, 2.1, 3.2, and 3.1 prove the marginal measurability, normalization, valid-fibre density, and integrable-exchange clauses.