# Is the inverse of a measurably parametrised family of bijections between standard Borel spaces measurably parametrised?

It is known that a measurable bijection $$f \colon [0,1] \to [0,1]$$ has a measurable inverse. (Here, all measurability is simply with respect to the Borel $$\sigma$$-algebra of $$[0,1]$$.)

Now fix an arbitrary measurable space $$(\Omega,\mathcal{F})$$, and let $$(f_\omega)_{\omega \in \Omega}$$ be a family of bijections $$f_\omega \colon [0,1] \to [0,1]$$ such that the map $$(\omega,x) \mapsto f_\omega(x)$$ is $$(\mathcal{F}\otimes\mathcal{B}([0,1]),\mathcal{B}([0,1]))$$-measurable. Is it necessarily the case that the map $$(\omega,x) \mapsto f_\omega^{-1}(x)$$ is $$(\mathcal{F}\otimes\mathcal{B}([0,1]),\mathcal{B}([0,1]))$$-measurable?

• It's certainly true if $(\Omega, \mathcal{F})$ is standard Borel. Not sure if that's of any use to you. Oct 16, 2019 at 4:05
• Thank you; yes, for that case it should presumably follow from the non-random case by considering the map $(\omega,x) \mapsto (\omega,f_\omega(x))$. But I really want not to assume that $(\Omega,\mathcal{F})$ is standard. Oct 16, 2019 at 11:35
• If this is true, I would try to reconstruct the proof that injective Borel images of Borel sets are Borel (which involves the Lusin Separation Thm), using that the $f_\omega$ are “uniformly measurable” (i.e., the complexity of $f_\omega^{-1}(X)$ with varying $\omega$ is bounded because of the measurability of $(\omega,x) \mapsto f_\omega(x)$). But it's just a hunch. Oct 31, 2019 at 13:47