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.
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.