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Context

I would like to find sufficient conditions such that $\upsilon^\psi\ll\upsilon$ where $\upsilon$ is a sigma-finite measure and $\psi$ is an invertible, measurable function.

Idea/Attempt

Let $(\mathsf{Z}, \mathcal{Z})$ be a measurable space, and $\upsilon$ be a sigma-finite measure on it. Let $\psi:\mathsf{Z}\to\mathsf{Z}$ be $\mathcal{Z}$-measurable and invertible, with inverse $\psi^{-1}:\mathsf{Z}\to\mathsf{Z}$. We wish to show that for any $\mathsf{A}\in\mathcal{Z}$ we have $$ \upsilon(\mathsf{A}) = 0 \implies \upsilon^\psi(\mathsf{A}). $$ This reduces to showing that the preimages of $\upsilon$-null sets are $\upsilon$-null sets. I have browsed a few related questions such as this and this but they are much less general than my question. Here I am not dealing with $\upsilon$ being a Lebesgue measure and I am not assuming any Jacobian (although, that is a case that I am very much interested in).

I feel that geometric measure theory could come in handy, but after browsing for a bit, I could not find anything at this level of generality.

Euler_Salter
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    It seems to me that your question is equivalent to asking when the inverse $\psi^{-1}:Z\to Z$ has the Lusin $\mathsf{N}$-property with respect to $\mu$, that is, when $\mu(A)=0\implies \mu(\psi^{-1}(A))=0$. This is discussed in Bogachev. – Snoop Feb 06 '24 at 17:44
  • @Snoop Thanks that's very helpful! However, in Bogachev $\mu$ has to be a Lebesgue, whereas in my case it's more general than that. – Euler_Salter Feb 07 '24 at 21:07

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