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Let $(X,\mathcal{B},\mu,T)$ be a measure-preserving system. A $T$-invariant sub-sigma algebra of $\mathcal{B}$ is a collection $\mathcal{A}$ of elements of $\mathcal{B}$ that forms a sigma-algebra itself and satisfies $T^{-1}\mathcal{A} = \mathcal{A}$ mod $\mu$, meaning $\forall A \in \mathcal{A} \hspace{1mm} \exists B \in \mathcal{A}$ such that $\mu(T^{-1}A\Delta B) = 0$.

A problem I am working on discusses the smallest T-invariant sub-sigma algebra satisfying a certain property. The existence of such a thing presumably is based on the fact that the arbitrary intersection of $T$-invariant sub-sigma algebras is itself a $T$-invariant sub-sigma algebra.

Question: Is the arbitrary intersection of $T$-invariant sub-sigma algebras a $T$-invariant sub-sigma algebra?

Thoughts: Clearly the arbitrary intersection is a sub-sigma-algebra. The issue is showing $T$-invariance. If "$T$-invariance" meant "$T^{-1}\mathcal{A} = \mathcal{A}$", then $T$-invariance of the arbitrary intersection is obvious. The "mod $\mu$" is the issue.

mathworker21
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  • Would it be possible to modify the situation somewhat and consider equivalence classes of sets? The relation $A \sim B$ defined by $\mu(A \bigtriangleup B) = 0$ is an equivalence relation. If so, and if the equivalence class $[A] \in \bigcap \mathfrak A$ and $\mathcal A \in \mathfrak A$ then $[A] \in \mathcal A$ and $T^{-1}[A] \in \mathcal A.$ – md2perpe Aug 05 '17 at 06:07
  • @md2perpe Do you think the statement is even true? I'm not sure it is true... – mathworker21 Aug 06 '17 at 06:35
  • Is anything known about any kind of continuity of $T$? Does $A \sim B$ imply that $T^{-1}A \sim T^{-1}B$? – md2perpe Aug 06 '17 at 22:11
  • @md2perpe continuity? We are in a measure space. $T$ is measure preserving. – mathworker21 Aug 06 '17 at 23:12
  • The function $d(A, B) = \mu(A \bigtriangleup B)$ is a pseudometric, and this can be used to define continuity. – md2perpe Aug 07 '17 at 05:30
  • @md2perpe see my answer below. I don't know why I previously thought the answer "yes" for countable intersections. – mathworker21 Sep 17 '19 at 15:37

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No. Consider $(X,\mathcal{B},\mu,T) = \left(\{1,2,3,4\},\mathcal{P}(\{1,2,3,4\}),\mu,T\right)$ where $\mu(\{1\}),\mu(\{2\}),\mu(\{4\}) = \frac{1}{3},\mu(\{3\}) = 0$, and $T : 1 \mapsto 2, 2 \mapsto 1, 3\mapsto 3, 4 \mapsto 4$. Let $\mathcal{A}_1 = \{\{1\},\{2\},\{3,4\},\{1,2\},\{2,3,4\},\{1,3,4\},\emptyset,X\}$ and $\mathcal{A}_2 = \{\{1\},\{2,3\},\{4\},\{2,3,4\},\{1,4\},\{1,2,3\},\emptyset,X\}$. These are $T$-invariant sub-sigma algebras. However, their intersection is not $T$-invariant, because $T^{-1}\{1\} = \{2\}$ has no equivalent element (the only element of $\mathcal{A}_1$ it is equivalent to is $\{2\}$, and the only element of $\mathcal{A}_2$ it is equivalent to is $\{2,3\}$).

mathworker21
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