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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?

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    $\begingroup$ It's certainly true if $(\Omega, \mathcal{F})$ is standard Borel. Not sure if that's of any use to you. $\endgroup$
    – Nate Eldredge
    Commented Oct 16, 2019 at 4:05
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    $\begingroup$ 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. $\endgroup$
    – Julian Newman
    Commented Oct 16, 2019 at 11:35
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    $\begingroup$ 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. $\endgroup$
    – Pedro Sánchez Terraf
    Commented Oct 31, 2019 at 13:47

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