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I bumped into a definition of measurable bundle from page 101 of an introduction to smooth ergodic theory by Barreira and Pesin.

Let $E$, $X$ be measurable spaces and $\pi:E\to X$ be a measurable transformation. Say $E$ is a measurable bundle over $X$ if there exists a countable collection of measurable sets $Y_i\subset X$ which covers $X$, and measurable maps $\pi_i:X\times\mathbb{R}^n\to X$ with $\pi_i^{-1} (Y_i) = Y_i\times \mathbb{R}^n$.

Is the condition above implying a local isomorphism from the section of a neighborhood of $x$ in E to the bundle $X\times \mathbb{R}^n$? Would this definition imply $E$ is a fiber bundle over $X$?

(Edit: ) will there be an open set 𝑈 such that $\pi^{-1}(U)$ is isomorphic to $U\times\mathbb{R}^n$?

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  • $\begingroup$ @MarianoSuárez-Álvarez Definition by fiber bundles? I can't find any formal definition for the term "measurable bundle" on google, so I'd be curious to know. $\endgroup$
    – duang
    Dec 27, 2022 at 21:25
  • $\begingroup$ @MarianoSuárez-Álvarez Thanks for pointing out. I'll edit. $\endgroup$
    – duang
    Dec 27, 2022 at 21:29
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    $\begingroup$ I don't understand the definition. How are the $\pi_i$ related to $\pi$? $\endgroup$ Dec 27, 2022 at 21:53
  • $\begingroup$ @JasonDeVito That's what the book has. I thought it meant an isomorphism. $\endgroup$
    – duang
    Dec 27, 2022 at 21:55
  • $\begingroup$ Something must be wrong, for otherwise, every $E$ is a bundle over every $X$. Take $\pi$ to be a constant, select $Y_i = X$, and defined $\pi_i$ to be the obvious projection $X\times \mathbb{R}\rightarrow X$. $\endgroup$ Dec 29, 2022 at 15:26

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Yes, the condition above implies a local isomorphism from the section of a neighborhood of x in E to the bundle X×Rn. This is because the measurable maps πi are defined to be isomorphisms between the sections of E and the bundle X×Rn in the neighborhoods Yi.

However, this definition alone does not imply that E is a fiber bundle over X. A fiber bundle is a special type of bundle that has additional structure, such as a continuous surjective map from the bundle to the base space, a continuous right inverse map, and a smooth structure on the fibers. In order to show that E is a fiber bundle, it would be necessary to show that these additional conditions are satisfied.


The definition you provided for a measurable bundle does not guarantee the local product structure. Specifically, the condition you provided only states that there exists a countable cover of X by measurable sets $Y_i$ and measurable maps $\pi_i: X \times \mathbb{R}^n \to X$ such that the inverse image of $Y_i$ under $\pi_i$ is equal to $Y_i \times \mathbb{R}^n$. This does not necessarily imply that there exists an open set $U$ in X such that the inverse image of $U$ under $\pi$ is isomorphic to $U \times \mathbb{R}^n$.

However, if you further assume that the measurable bundle is a fiber bundle, then it will have the local product structure. A fiber bundle is a type of bundle that has a locally trivializable projection map and a fiber, which is a topological space that is homeomorphic to the inverse image of a point under the projection map. In the case of a measurable fiber bundle, the projection map and the fiber are required to be measurable as well. The local trivializability condition guarantees that there exists an open set $U$ in X such that the inverse image of $U$ under the projection map is isomorphic to $U \times F$, where $F$ is the fiber. This implies that the local product structure holds for measurable fiber bundles.

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  • $\begingroup$ I'm most interested in knowing whether the bundle will have the local product structure. That is, will there be an open set $U$ such that $\pi^{-1}(U)$ is isomorphic to $U\times\mathbb{R}^n$? $\endgroup$
    – duang
    Dec 28, 2022 at 2:39
  • $\begingroup$ Ive edited my original response with an answer to your comment $\endgroup$ Dec 28, 2022 at 3:36

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