# Measurable selection

Almost sure identity

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?

• This is not a selection theorem! On the contrary, it follows directly from Fubini's theorem: write $P \otimes \lambda(A)$ as the $\lambda$-integral of $f(x) = P(\{\omega : (\omega, x) \in A\})$, choose $x$ such that $f(x) > 0$, and set $X(\omega) = x$ (as in Algernon's answer). Dec 14, 2019 at 12:30

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 $$\omega\in\Omega$$.