Reductions of Bayesian experiments: regular conditional experiment I'm reading Florens et al.'s 'Elements of Bayesian Statistics', currently working through chapter 1, 'Reduction of Bayesian Experiments'. I find most of it clear, except the definition of regular conditional experiments (p. 51) and everything that relies on this definition. I'll appreciate help in deciphering it. I assume those who would respond to my post are familiar with this book and its terms and notations, but i'll be happy to provide more details per request.

The conditional experiment $\mathcal{E}^\mathcal{T}$ is said to be regular if there exists a regular version of $\Pi^\mathcal{T}$ such that there exist regular versions of $\mu^\mathcal{S}$ and $P^{\mathcal{A}\vee\mathcal{T}}$, in this case we then have:
  $$\left(1.4.9\right)\hspace{10mm}\Pi^\mathcal{T}=\mu^\mathcal{T}\otimes P^{\mathcal{A}\vee\mathcal{T}}=P^\mathcal{T}\otimes\mu^\mathcal{S}$$

What i don't understand is the expression $\mu^\mathcal{T}\otimes P^{\mathcal{A}\vee\mathcal{T}}$. The way the $\otimes$ operator is defined (Theorem 0.3.10, p. 18), in the expression $Q\otimes W$ $Q$ is a probability on $\left(M,\mathcal{M}\right)$ and $W$ is a transition from $\left(M,\mathcal{M}\right)$ to $\left(N,\mathcal{N}\right)$. However, in the present case, for every $\left(a,s\right)\in A\times S$, $\mu^\mathcal{T}\left(a,s\right)$ is a probability on the $\sigma$-algebra $\mathcal{A}\times S$, whereas $P^{\mathcal{A}\vee\mathcal{T}}$ is a transition from $\mathcal{A}\vee\mathcal{T}$ to $A\times\mathcal{S}$. So the expression $\mu^\mathcal{T}\otimes P^{\mathcal{A}\vee\mathcal{T}}$ is not well defined, hence meaningless.
 A: Let us consider a more general case with arbitrary $\sigma$-fields $\mathcal{B}\subseteq\mathcal{A}$, $\mathcal{T}\subseteq\mathcal{S}$ and $\mathcal{M}\subseteq\mathcal{A}\otimes\mathcal{S}$.
The equations
$$
\left(1.4.16\right)\hspace{10 mm}\Pi^\mathcal{M}_{\mathcal{B}\vee\mathcal{T}}=
\mu^\mathcal{M}_\mathcal{B}\otimes P^{\mathcal{B}\vee\mathcal{M}}_\mathcal{T}=
P^\mathcal{M}_\mathcal{T}\otimes \mu^{\mathcal{M}\vee\mathcal{T}}_\mathcal{B}
$$
from the book (which are the generalized counterpart of equations $\left(1.4.9\right)$ mentioned in the question) are, technically speaking, incorrect (not well-structured), since, as per the product measure theorem (Theorem $0.3.10$, p. $18$), in the expression $\mu^\mathcal{M}_\mathcal{B}\otimes P^{\mathcal{B}\vee\mathcal{M}}_\mathcal{T}$, the subscript of the left term should be identical to the superscript of the right term, and the subscript of the result should be the product of the subscripts of the two terms. The expression $P^\mathcal{M}_\mathcal{T}\otimes \mu^{\mathcal{M}\vee\mathcal{T}}_\mathcal{B}$ is likewise incorrect.
Using the notation introduced here, these equations should be interpreted as though they were written thus
$$
\left(1.4.16'\right)\hspace{10 mm}
\begin{array}{lcl}
\Pi^\mathcal{M}_{\left(\mathcal{B}\vee\mathcal{M}\right),\left(\mathcal{T}\vee\mathcal{M}\right)} & = & \mu^\mathcal{M}_{\mathcal{B}\vee\mathcal{M}}\otimes P^{\mathcal{B}\vee\mathcal{M}}_{\mathcal{T}\vee\mathcal{M}}\hspace{10mm}\left[\Xi,\Pi_\mathcal{M}\right]-\mathrm{a.s.} \\
\Pi^\mathcal{M}_{\left(\mathcal{T}\vee\mathcal{M}\right),\left(\mathcal{B}\vee\mathcal{M}\right)} & = & P^\mathcal{M}_{\mathcal{T}\vee\mathcal{M}}\otimes \mu^{\mathcal{T}\vee\mathcal{M}}_{\mathcal{B}\vee\mathcal{M}}\hspace{10mm}\left[\Xi,\Pi_\mathcal{M}\right]-\mathrm{a.s.}
\end{array}
$$
The following guidelines from p. $58$ should be used to simplify the result:

One may identify the conditional experiments $\mathcal{E}^\mathcal{M}_{\mathcal{B}\vee\mathcal{T}}$ and $\mathcal{E}^\mathcal{M}_{\left(\mathcal{B}\vee\mathcal{M}\right)\vee\left(\mathcal{T}\vee\mathcal{M}\right)}$. In other words, in conditional experiments, the product structure $\mathcal{B}\vee\mathcal{T}$ is considered unaffected if the parameters involve observable variables included in the conditioning $\sigma$-field, or if the statistics involve parameters included in the conditioning $\sigma$-field.

More accurately, the experiment we identify $\mathcal{E}^\mathcal{M}_{\mathcal{B}\vee\mathcal{T}}$ with, via embedding, is
$$
\mathcal{E}^{\mathcal{M}\otimes\mathcal{M}}:=\left(\Delta'_1,\Pi^{\mathcal{M}\otimes\mathcal{M}}\right)
$$
The following trivial observations can also be used to simplify the expressions appearing in $\left(1.4.16'\right)$.
$$
\begin{array}{lcl}
\Pi_{\mathcal{A}\otimes\mathcal{S}} & = & \Pi \\
\mu_\mathcal{A} & = & \mu \\
P_\mathcal{S} & = & P \\
\mathcal{T}\subseteq\mathcal{S} & \implies & \mathcal{T}\vee\mathcal{S}=\mathcal{S} \\
\mathcal{B}\subseteq\mathcal{A} & \implies & \mathcal{B}\vee\mathcal{A}=\mathcal{A}
\end{array}
$$
as well as the conventions that $\mathcal{A}$ may stand for $\left\{D\times S:\mid D\in\mathcal{A}\right\}$ and that $\mathcal{S}$ may stand for $\left\{A\times E:\mid E\in\mathcal{S}\right\}$ (see Remark on p. $27$).
Thus, with $\mathcal{B}:=\mathcal{A}$, $\mathcal{T}:=\mathcal{S}$, $\mathcal{M}:=\mathcal{T}'$ (for some sub-$\sigma$-algebra $\mathcal{T}'\subseteq\mathcal{S}$), equations $\left(1.4.16'\right)$ simplify to
$$
\Pi^\mathcal{T'}=\mu^\mathcal{T'}\otimes P^{\mathcal{A}\vee\mathcal{T'}}=P^\mathcal{T'}\otimes \mu^\mathcal{S}\tag{*}
$$
with the implicit understandings that the equalities hold $\left[\Xi,\Pi_\mathcal{M}\right]$-almost surely and that we identify the spaces $\Delta'_1$ and $\Delta'_2$ in the natural way.
Equations $(*)$ are precisely equations $\left(1.4.9\right)$ appearing in the question.
