If $f(\cdot, \cdot)$ is measurable in each variable and $g(\cdot)$ is measurable, then $x \mapsto f(x,g(x))$ is measurable?

Question

Let $ f:\mathbb R^2 \to \mathbb R , g: \mathbb R \to \mathbb R $ be such that $f$ is Borel measurable in each variable and $g$ is Borel measurable.

Is it true that $$x \mapsto f(x,g(x))$$ is measurable?

I suppose it's not, but cannot find a counterexample.

Any help is appreciated.

Adapt this example (set $g(x)=x$ instead). – Gabriel Romon Sep 05 '19 at 09:22 — Gabriel Romon, Sep 05 '19 at 09:22

Jonathan Hole · Accepted Answer · 2019-10-28T17:00:56.317

3

Begin with an unmeasurable subset of $\mathbb{R}$ say $A$. Then $A\times \{0\}$ is also unmeasurable. Now rotate $A\times \{0\}$ by $45$ degrees counter-clockwise about the origin, then scale it by a factor of $\sqrt{2}$ and call the set you get for $B$. Then $B$ lies on the diagonal in $\mathbb{R}^2$ and is also unmeasurable since rotation is a measurable transformation. Let now $g: \mathbb{R} \rightarrow \mathbb{R}$ be the identity function and $f$ the characteristic function of $B$. Each section of $f$ is either identically $0$ or the characteristic function of a singleton so measurable, and moreover $x \mapsto f(x, g(x))=f(x,x)$ is just the characteristic function of $A$, hence unmeasurable.

edited Oct 28 '19 at 17:00

answered Sep 05 '19 at 09:22

Jonathan Hole

3,564

1

+1. $f(x,x)$ is the characteristic function of $A$, (considered as a subset if $\mathbb R$) right? – Kavi Rama Murthy Sep 05 '19 at 09:30
Yes, thank you! – Jonathan Hole Sep 05 '19 at 09:34

If $f(\cdot, \cdot)$ is measurable in each variable and $g(\cdot)$ is measurable, then $x \mapsto f(x,g(x))$ is measurable?

1 Answers1