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Consider a measure space $(X,\mathcal{A},\mu)$ and a measure-preserving transformation $\phi \colon X\rightarrow X$, that is, $\phi$ is measurable and $\phi_*\mu = \mu$. My intuition tells me that we should not loose any dynamical information by looking at the restriction $(X,\mathcal{A}',\mu') := (X,\phi^{-1}(\mathcal{A}),\mu|_{\phi^{-1}(\mathcal{A})})$. To make this intuition more rigorous, I am trying to show the following claim: for any $A \in \mathcal{A}$ there exists some $A' = \phi^{-1}(B) \in \mathcal{A}'$ ($B \in \mathcal{A})$ such that their symmetric difference $A \triangle A'$ has $\mu$-measure zero.

In my case, $\mathcal{A}$ is the Borel sigma-algebra of the Polish space $X$, $\mu$ is a Radon measure, and $\phi$ is continuous. Thus, there exists a nested sequence of compact subsets $K_n$ of $A$ with $\mu(A \cap K_n^c) \rightarrow 0$. In general, it does not hold that $\mu(A \cap \phi^{-1}(\phi(K_n))) \rightarrow 0$, but I am hoping that this does hold with a smart choice for $K_n$ that uses the $\phi$-invariance of $\mu$. Then, $B = \bigcup_n \phi(K_n)$ would do the job.

Any idea for a way to show this or a counterexample? Alternatively, is there another way to formalise the idea that $(X,\mathcal{A}',\mu')$ captures all the relevant measure-theoretic dynamics?

I am not assuming that $\phi$ is ergodic, but I added the tag as ideas from ergodic theory may help.

Cheers!

Consider a measure space $(X,\mathcal{A},\mu)$ and a measure-preserving transformation $\phi \colon X\rightarrow X$, that is, $\phi$ is measurable and $\phi_*\mu = \mu$. My intuition tells me that we should not lose any dynamical information by looking at the restriction $(X,\mathcal{A}',\mu') := (X,\phi^{-1}(\mathcal{A}),\mu|_{\phi^{-1}(\mathcal{A})})$. To make this intuition more rigorous, I am trying to show the following claim: for any $A \in \mathcal{A}$ there exists some $A' = \phi^{-1}(B) \in \mathcal{A}'$ ($B \in \mathcal{A}$) such that their symmetric difference $A \mathbin\triangle A'$ has $\mu$-measure zero.

In my case, $\mathcal{A}$ is the Borel sigma-algebra of the Polish space $X$, $\mu$ is a Radon measure, and $\phi$ is continuous. Thus, there exists a nested sequence of compact subsets $K_n$ of $A$ with $\mu(A \cap K_n^c) \rightarrow 0$. In general, it does not hold that $\mu(A \cap \phi^{-1}(\phi(K_n))) \rightarrow 0$, but I am hoping that this does hold with a smart choice for $K_n$ that uses the $\phi$-invariance of $\mu$. Then, $B = \bigcup_n \phi(K_n)$ would do the job.

Any idea for a way to show this or a counterexample? Alternatively, is there another way to formalise the idea that $(X,\mathcal{A}',\mu')$ captures all the relevant measure-theoretic dynamics?

I am not assuming that $\phi$ is ergodic, but I added the tag as ideas from ergodic theory may help.

Consider a measure space $(X,\mathcal{A},\mu)$ and a measure-preserving transformation $\phi \colon X\rightarrow X$, that is, $\phi$ is measurable and $\phi_*\mu = \mu$. My intuition tells me that we should not loose any dynamical information by looking at the restriction $(X,\mathcal{A}',\mu') := (X,\phi^{-1}(\mathcal{A}),\mu|_{\phi^{-1}(\mathcal{A})})$. To make this intuition more rigorous, I am trying to show the following claim: for any $A \in \mathcal{A}$ there exists some $B \in \mathcal{A}'$ such that their symmetric difference $A \triangle B$ has $\mu$-measure zero.

In my case, $\mathcal{A}$ is the Borel sigma-algebra of the Polish space $X$, $\mu$ is a Radon measure, and $\phi$ is continuous. Thus, there exists a nested sequence of compact subsets $K_n$ of $A$ with $\mu(A \cap K_n^c) \rightarrow 0$. In general, it does not hold that $\mu(A \cap \phi^{-1}(\phi(K_n))) \rightarrow 0$, but I am hoping that this does hold with a smart choice for $K_n$ that uses the $\phi$-invariance of $\mu$. Then, $B = \bigcup_n \phi(K_n)$ would do the job.

Any idea for a way to show this or a counterexample? Alternatively, is there another way to formalise the idea that $(X,\mathcal{A}',\mu')$ captures all the relevant measure-theoretic dynamics?

I am not assuming that $\phi$ is ergodic, but I added the tag as ideas from ergodic theory may help.

Cheers!

Consider a measure space $(X,\mathcal{A},\mu)$ and a measure-preserving transformation $\phi \colon X\rightarrow X$, that is, $\phi$ is measurable and $\phi_*\mu = \mu$. My intuition tells me that we should not loose any dynamical information by looking at the restriction $(X,\mathcal{A}',\mu') := (X,\phi^{-1}(\mathcal{A}),\mu|_{\phi^{-1}(\mathcal{A})})$. To make this intuition more rigorous, I am trying to show the following claim: for any $A \in \mathcal{A}$ there exists some $A' = \phi^{-1}(B) \in \mathcal{A}'$ ($B \in \mathcal{A})$ such that their symmetric difference $A \triangle A'$ has $\mu$-measure zero.

In my case, $\mathcal{A}$ is the Borel sigma-algebra of the Polish space $X$, $\mu$ is a Radon measure, and $\phi$ is continuous. Thus, there exists a nested sequence of compact subsets $K_n$ of $A$ with $\mu(A \cap K_n^c) \rightarrow 0$. In general, it does not hold that $\mu(A \cap \phi^{-1}(\phi(K_n))) \rightarrow 0$, but I am hoping that this does hold with a smart choice for $K_n$ that uses the $\phi$-invariance of $\mu$. Then, $B = \bigcup_n \phi(K_n)$ would do the job.

Any idea for a way to show this or a counterexample? Alternatively, is there another way to formalise the idea that $(X,\mathcal{A}',\mu')$ captures all the relevant measure-theoretic dynamics?

I am not assuming that $\phi$ is ergodic, but I added the tag as ideas from ergodic theory may help.

Cheers!

Consider a measure space $(X,\mathcal{A},\mu)$ and a measure-preserving transformation $\phi \colon X\rightarrow X$, that is, $\phi$ is measurable and $\phi_*\mu = \mu$. My intuition tells me that we should not loose any dynamical information by looking at the restriction $(X,\mathcal{A}',\mu') := (X,\phi^{-1}(\mathcal{A}),\mu|_{\phi^{-1}(\mathcal{A})})$. To make this intuition more rigorous, I am trying to show the following claim: for any $A \in \mathcal{A}$ there exists some $B \in \mathcal{A}'$ such that their symmetric difference $A \triangle B$ has $\mu$-measure zero.

In my case, $\mathcal{A}$ is the Borel sigma-algebra of the Polish space $X$, $\mu$ is a Radon measure, and $\phi$ is continuous. Thus, there exists a nested sequence of compact subsets $K_n$ of $A$ with $\mu(A \cap K_n^c) \rightarrow 0$. In general, it does not hold that $\mu(A \cap \phi^{-1}(\phi(K_n))) \rightarrow 0$, but I am hoping that this does hold with a smart choice for $K_n$ that uses the $\phi$-invariance of $\mu$. Then, $B = \bigcup_n \phi(K_n)$ would do the job.

Any idea for a way to show this or a counterexample?

I am not assuming that $\phi$ is ergodic, but I added the tag as ideas from ergodic theory may help.

Cheers!

Consider a measure space $(X,\mathcal{A},\mu)$ and a measure-preserving transformation $\phi \colon X\rightarrow X$, that is, $\phi$ is measurable and $\phi_*\mu = \mu$. My intuition tells me that we should not loose any dynamical information by looking at the restriction $(X,\mathcal{A}',\mu') := (X,\phi^{-1}(\mathcal{A}),\mu|_{\phi^{-1}(\mathcal{A})})$. To make this intuition more rigorous, I am trying to show the following claim: for any $A \in \mathcal{A}$ there exists some $B \in \mathcal{A}'$ such that their symmetric difference $A \triangle B$ has $\mu$-measure zero.

In my case, $\mathcal{A}$ is the Borel sigma-algebra of the Polish space $X$, $\mu$ is a Radon measure, and $\phi$ is continuous. Thus, there exists a nested sequence of compact subsets $K_n$ of $A$ with $\mu(A \cap K_n^c) \rightarrow 0$. In general, it does not hold that $\mu(A \cap \phi^{-1}(\phi(K_n))) \rightarrow 0$, but I am hoping that this does hold with a smart choice for $K_n$ that uses the $\phi$-invariance of $\mu$. Then, $B = \bigcup_n \phi(K_n)$ would do the job.

Any idea for a way to show this or a counterexample? Alternatively, is there another way to formalise the idea that $(X,\mathcal{A}',\mu')$ captures all the relevant measure-theoretic dynamics?

I am not assuming that $\phi$ is ergodic, but I added the tag as ideas from ergodic theory may help.

Cheers!