Let $\pi: E \to M$ be an orientable vector bundle. Suppose that $\{U, f_{\alpha}\}$ consists of oriented trivializations(meaning the determinant $f_{\alpha}f_{\beta}^{-1}$ is positive on fibers. Is Euler class dependent on the choice of such trivializations?
Suppose that $\{U, g_{\alpha}\}$ consists of another set of oriented trivializations. But the determinant of $g_{\alpha} f_{\beta}^{-1}$ is negative. Then in my opinion integration along each fiber will give me opposite answers with respect those two choices simply by change of variables theorems(thus leading to opposite Thom class and Euler class). But this does not seem to be correct. What is wrong here?
PS: everything is over $\mathbb R$
Both Thom classes and Euler classes depend on orientation. You could see this more intuitively by using the properties of the Thom class and Euler class with respect to integration. For instance, the Thom class $\Phi_E\in H_{cv}(E)$ satisfies that integration along every fiber is $1$. However, if fibers are oriented with the opposite orientation, integration along the fiber would yield $-1$ in every fiber. Therefore, $\Phi_{-E} = -\Phi_E$.