The comments in the post 


[Almost sure identity](https://mathoverflow.net/questions/348332/almost-sure-identity?noredirect=1#comment872443_348332)


has led me to the following question:


Suppose that $(\Omega,\mathcal{F},P)$ is a probability space. 
Denote by $\mathcal{B}(\mathbb R)$ the Borel sigma-algebra on $\mathbb R$.


Let $A\in\mathcal{F}\otimes\mathcal{B}(\mathbb{R})$ be such that $P\otimes\lambda(A)>0$. 


Is it true that there exists a $\mathcal{F}$-measurable random variable $X\colon \Omega\to\mathbb R$ such that
$$
P(\{\omega\colon(\omega,X(\omega))\in A\})>0?
$$
 If it does not true, under which conditions on $(\Omega,\mathcal F,P)$ it holds?



Measurable selection
Yes! This actually follows from standard measure theory.


Let $\mathcal{R}$ denote the collection of rectangles $U\times V$ where $U\in\mathcal{F}$ and $V\in\mathcal{B}(\mathbb{R})$. The finite disjoint unions of elements of $\mathcal{R}$ form an algebra $\mathcal{A}$ on $\Omega\times\mathbb{R}$ which in turn generates the product $\sigma$-algebra $\mathcal{F}\otimes\mathcal{B}(\mathbb{R})$.


By the approximation lemma, for every $\varepsilon>0$, we can find $A'\in\mathcal{A}$ such that $(P\times\lambda)(A\operatorname{\triangle} A')<\varepsilon$. Choosing $\varepsilon>0$ sufficiently small, we find that $(P\times\lambda)(A')>0$ for some $A'\in\mathcal{A}$. This in turn implies that $(P\times\lambda)(U\times V)>0$ for a rectangle $U\times V\in\mathcal{R}$. In particular, $P(U)>0$.


Now pick an arbitrary $x\_0\in V$ and define $X(\omega):=x\_0$ for every $\omega\in\Omega$.


