Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $E_i$ be a $\mathbb R$-Banach space and $\mathcal E_i:=\mathcal B(E_i)$
- $\mu_i$ be a measure on $(E_i,\mathcal E_i)$
- $X_i:\Omega\to E_i$ be $(\mathcal A,\mathcal E_i)$-measurable with $$(X_1,X_2)_\ast\operatorname P=f(\mu_1\otimes\mu_2),$$ i.e. the pushforward measure of $(X_1,X_2)$ with respect to $\operatorname P$ equals the measure with density $f$ with respect to $\mu_1\otimes\mu_2$, for some $\mathcal E_1\otimes\mathcal E_2$-measurable $f:E_1\times E_2\to[0,\infty)$
Now, let $$f_1(x):=\int f(x,\;\cdot\;)\:{\rm d}\mu_2\;\;\;\text{for }x\in E_1.$$
How can we show that $$f_1>0\;\;\;(X_1)_\ast\operatorname P\text{-almost everywhere}\tag1$$ and that $f_1^{-1}$ is the density of the absolutely continuous part of $\mu_2$ with respect to $(X_1)_\ast\operatorname P$?
Especially in the first part, I don't see why $(1)$ should hold, for example, in the case $X\equiv0$.