Equivalence of Relative Entropy for Product Measures In a paper for Transport Inequalities by Nathael Gozlan, the following assertion is made:
Let the relative entropy with respect to $\mu \in P(\mathcal X)$ be defined by
$$
H(\nu \mid \mu) =
\left\{
  \begin{array}{@{}ll@{}}
    \int_\mathcal X \log(\frac{d\nu}{d\mu})d\nu, & \text{if}\ \nu \ll\mu \\
    +\infty, & \text{otherwise}
  \end{array}\right. , \ \nu \in P(\mathcal X)
$$
Now, make $\mu_1$ and $\mu_2$ defined on $\mathcal X_1$ and $\mathcal X_2$, respectively. For a measure $\nu$ on
$\mathcal X_1 \times \mathcal X_2$, write the disintegration of $\nu$ (conditional expectation) with respect to the first coordinate as:
$$
d\nu(x_1,x_2) = d\nu_1(x_1) d\nu^{x_1}(x_2)
$$
Note that the disintegration is pretty much just a formal way of writing the conditional probability formula $P(X=x_1,Y=x_2) = P(X=x_1 \mid Y =x_2)P(Y=x_2)$.
Finally, the author asserts that for the product measure $\mu_1 \otimes\mu_2$ (this is equivalent to $\mu_1 \times \mu_2$, which is a diffrent notation, but with the same meaning), one can prove the following equality:
$$
H(\nu \mid \mu_1 \otimes \mu_2) = H(\nu_1 \mid \mu_1) + \int_{\mathcal X_1}
H(\nu_2^{x_1}\mid \mu_2)d\nu_1(x_1)
$$
My question is how to prove this equality above.
Since the definition of a disintegration is not very common, I will give it here to save people the trouble of hunting it down:
Given two polish (complete and separable) measurable spaces $(\Omega, \mathcal F)$ and $(E, \mathcal A)$. If $P$ is a probability measure in $(\Omega \times E, \mathcal F \otimes \mathcal A)$, and $P_1$ the marginal distribution of the first coordinate. Then, there exists an unique probability kernel $K: \Omega \times \mathcal A \rightarrow [0,1]$, satisfying:
$$
P(A\times B) = \int_A K(\omega, B) P_1(d\omega), \ \forall A \in \mathcal F, \ B \in \mathcal A
$$
In this case, we can define
$$
P[X_2 \in B \mid X_1 = w] := K(w,B)
$$
Where $X_1$ and $X_2$ represent the first and second coordinates respectively.
 A: In order to answer this question, I answered this question on absolute continuity, and now we are ready to discuss the general chain rule of the relative entropy. In a nutshell: The $answer$ to this question is my question.
As discussed in the answer to your question on absolute continuity, the conditional Radon-Nikodym derivative is well-defined for Borel spaces, and thus we get the following. Since the notation $H(\nu|\mu)$ may be easily confused with the conditional entropy, I'll write $D(\nu\|\mu)$ for the relative entropy instead, and $D(\nu\|\mu|\nu_1)$ for the conditional relative entropy (in lack of a better alternative). Notice that we have
\begin{align*}
D(\nu\|\mu)
&=\int\log\left(\frac{\mathrm d\nu}{\mathrm d\mu}(x,y)\right)\nu(\mathrm dx,\mathrm dy)
=\int\int\log\left(\frac{\mathrm d\nu_1}{\mathrm d\mu_1}(x)\frac{\mathrm d\nu_x}{\mathrm d\mu_x}(y)\right)\nu_x(\mathrm dy)\nu_1(\mathrm dx)\\
&=\int\log\left(\frac{\mathrm d\nu_1}{\mathrm d\mu_1}(x)\right)\nu_1(\mathrm dx)
+\int\int\log\left(\frac{\mathrm d\nu_x}{\mathrm d\mu_x}(y)\right)\nu_x(\mathrm dy)\nu_1(\mathrm dx)\\
&=D(\nu_1\|\mu_1)+D(\nu\|\mu|\nu_1).
\end{align*}
The first step, where we replace the derivative by the product of derivatives, was thoroughly discussed in the other answer.
