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  such that $\pi^{-1}(U)$ is isomorphic to $U\times\mathbb{R}^n$?
 A: 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.
