Let $(A,\mathscr{A},\alpha)$ be a probability measure space and $(B, \mathscr{B})$ be a measurable space. Let also $$ s: A\times \mathscr{B} \to \mathbf{R} $$ be a function such that $s(a,\cdot): \mathscr{B} \to \mathbf{R}$ is a probability measure for all $a \in A$ and $s(\cdot, B): A\to \mathbf{R}$ is measurable for all $B \in \mathscr{B}$. Finally, let $\mu: \sigma(\mathscr{A}\times \mathscr{B}) \to \mathbf{R}$ be the unique probability measure on $A\times B$ which satisfies $$ \forall X \in \mathscr{A}, \forall Y \in \mathscr{B},\quad \int_X s(a,Y) \alpha(\mathrm{d}a)=\mu(X\times Y). $$
Question. Is it true that $$ \int_A\int_B f(a,b)\,s(a,\mathrm{d}y)\,\alpha(\mathrm{d}a)=\int_{A\times B}f(a,b) \,\mu(\mathrm{d}a,\mathrm{d}b)\,\,\, $$ for all bounded measurable functions $f: A\times B\to \mathbf{R}$?
This is true by Disintegration theorem, see e.g. Proposition 452F and the following remark in Fremlin's book here.