The comments in the post

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?

  • 5
    $\begingroup$ 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). $\endgroup$ Dec 14, 2019 at 12:30

1 Answer 1


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 $\omega\in\Omega$.


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