When inverse of measurable function also measurable?

Question

Let $(X, \Sigma)$ be a measurable space and $f:X\to X$ be measurable bijection. When can we say that $f^{-1}$ is measurable? I'm aware that full answer may be hard to produce, so anything more than I posted will do.

I was assigned task in which a measurable and invertible function appeared, but it didn't specify if it was invertible in the category of sets, or measurable spaces. It got me wandering, when those are the same.

I know that if $\Sigma$ was family of Borel sets of topology, in which $f$ was homeorphism, then it would be true. But this is quite strong condition, as there are $\sigma$-algebras which are not generated by any topology. Also even when we assume, that $\Sigma$ was generated by some topology, finding one in which $f$ was homeomorphism seems to be non-trivial task (and perhaps not always doable as well).

Broader approach (from which first one emerged) may be to say, that $f^{-1}$ is measurable with respect to $\sigma$-algebra generated by $f$: $$ A\in \Sigma \implies f(B) \in \Sigma $$ Where $B = f^{-1}(A)$ is general form of set in $\sigma(f)$. Now we have, that if $\sigma(f) = \Sigma$, then $f^{-1}$ is measurable. But I don't know, how to easily check if a function generates whole $\Sigma$.

If $B=f^{-1}(A)$ then the impliation is trivial. Do you may be mean $A\in\Sigma\implies f(A)\in\Sigma$ as a criterion for the measurability of $f^{-1}$? — Peter Melech, Oct 17 '22 at 09:59
Non-trivial topic. See Corollary 4.5.5 , p. 154 of A Course on Borel sets by Srivastava. — geetha290krm, Oct 17 '22 at 10:02
Does this answer your question? $f :\mathbb R \to \mathbb R$ be a bijective Lebesgue measurable function , then is $f^{-1}:\mathbb R \to \mathbb R$ Lebesgue measurable? See also Measurability of the inverse of a measurable function. — Anne Bauval, Oct 17 '22 at 10:03
@PeterMelech later I write about $\sigma(f) = \Sigma$, where difficulty hides — Esgeriath, Oct 17 '22 at 10:15
@AnneBauval that seems helpful, but not quite what I was looking for. — Esgeriath, Oct 17 '22 at 10:24
@Esgeriath Ok, but with $B=f^{-1}(A)$ the implication you have written there is just $A\in\Sigma\implies A\in\Sigma$, on the other hand $A\in\Sigma\implies f(A)\in\Sigma$ is just the definition of the measurability of $f^{-1}$ together with $(f^{-1})^{-1}=f$. — Peter Melech, Oct 17 '22 at 10:28
Esgeriath, what you are precisely looking for is unclear to me. — Anne Bauval, Oct 17 '22 at 10:29
@PeterMelech this trivial implication shows that $f^{-1}: (X, \Sigma) \to (X, \sigma(f))$ is measurable. So equality $\sigma(f) = \Sigma$ does the job. — Esgeriath, Oct 17 '22 at 10:33
@AnneBauval I'm looking for characterization of measurable bijections, whose inverses are measurable as well. Various examples are welcome, but are not full answers. — Esgeriath, Oct 17 '22 at 10:37
@Esgeriath Fair enough, I see, but as you already said it will not be easy to check if $\sigma(f)=X$, so you have just changed one difficulty for a simliar difficulty. The question remains open.. — Peter Melech, Oct 17 '22 at 11:14

score 1 · Accepted Answer · answered Dec 28 '23 at 08:39

1

If $(X, \mathcal X)$ and $(Y, \mathcal Y)$ are standard Borel spaces, and if $f : X \to Y$ is bijective and Borel measurable, then $f^{-1}$ is also Borel measurable.

This is an immediate consequence of corollary 15.2 on p.89 of "Classical Descriptive Set Theory" by Alexander Kechris (Springer, 1995).

answered Dec 28 '23 at 08:39

Alex M.

35,207

That's a nice result! Back when I asked this question I was dealing with standard probability space though, which more often than not aren't standard Borel spaces. But this definitely answers the question as stated. – Esgeriath Jan 11 '24 at 16:01

When inverse of measurable function also measurable?

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