Inverse of disintegration theorem on euclidean spaces i was asking myself if some sort of "inverse" of the disintegration theorem holds (or eventually when).
Let's say I have a family of Borel measures $(\mu_t)_{t \in [0,1]}$ over $R^d$ such that the map $t \rightarrow \mu_t(B)$ is borel measurable for each Borel set $B$, and let's say I want to associate to this family a finite Borel measure $\mu$ over $R^d \times [0,1]$ clearly such that if $f \in C_0^\infty(R^d \times [0,1])$ the following holds:
$\int_{R^d \times [0,1]} f(x,t)d\mu= \int_0^1 \int_{R^d}f(x,t)d\mu_tdt.$
Is this always possible?
edit: the family $\mu_t$ is a family of finite borel measures
 A: Yes, this is always possible under the conditions which you assumed, namely that $t \mapsto \mu_t(B)$ is measurable for all measurable sets $B$ and that the $\mu_t$ are finite.
The notion that I think you're specifically looking for is the product of transition kernels. Note that every measure is trivially a transition kernel which is constant in its first coordinate.
A: As the other states what you are looking for are products or compositions (different authors swap these around confusingly) of transition kernels. However there are different types of products and compositions in the literature, and the one you are looking for does not look exactly like the one provided in the Wiki pages, so will provide some more detail.
The collection $(\mu_t)$ defines a transition kernel $\kappa$ with source $([0,1],\mathcal B([0,1]))$ and target $(\mathbb R^d,\mathcal B(\mathbb R^d))$. In other words $\kappa:[0,1]\times \mathcal B(\mathbb R^d)\to[0,\infty)$ is defined by $\kappa(t,B)=\mu_t(B)$. Furthermore the Lebesgue measure $\lambda$ on $[0,1]$ can be thought of as a transition kernel from a singleton set $S:=\{s\}$ to $([0,1],\mathcal B([0,1]))$. This is all standard so far and contained in the Wiki article. Now what you are looking for is precisely a product(composition) kernel $\lambda\otimes \kappa$ from $S$ to the product space $[0,1]\times \mathbb R^d$ equipped with the product $\sigma$-algebra. This is defined for any $s\in S$ and $B\in \mathcal B([0,1])\otimes\mathcal B(\mathbb R^d)$ by
$$\lambda\otimes \kappa(s,B)=\int_{[0,1]}\int_{\mathbb R^d}\mathbb 1_B(t,x)\kappa(t,dx)\lambda(dt).$$
In our case $S$ is a singleton, so $\lambda\otimes \kappa$ is just a measure, and precisely the $\mu$ you want.
To understand it better we can examine its behaviour on product sets:
$$\mu(A\times B)=\int_{[0,1]}\mathbb 1_A(t)\int_{\mathbb R^d}\mathbb 1_B(x)\kappa(t,dx)\lambda(dt)=\int_{[0,1]}\mathbb 1_A(t)\mu_t(B)\lambda(dt).$$
From this characterisation you can easily prove (via Caratheodory extension theorem) that $\mu$ is indeed a Borel measure on $([0,1],\mathcal B([0,1]))$. Finally you can prove that your function characteristic holds for $\mu$ by taking the limit of step functions.
