# Are measurable bundles fiber bundles?

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 $$U\times\mathbb{R}^n$$?

• @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. Dec 27, 2022 at 21:25
• @MarianoSuárez-Álvarez Thanks for pointing out. I'll edit. Dec 27, 2022 at 21:29
• I don't understand the definition. How are the $\pi_i$ related to $\pi$? Dec 27, 2022 at 21:53
• @JasonDeVito That's what the book has. I thought it meant an isomorphism. Dec 27, 2022 at 21:55
• 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 $X\times \mathbb{R}\rightarrow X$. Dec 29, 2022 at 15:26

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.
• 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 $U\times\mathbb{R}^n$? Dec 28, 2022 at 2:39