Measurable vector bundles trivial I hope you can help:
If $E$ is a measurable vector bundle over a compact metric space $(X,\mu)$ then there is a subset $Y\subset X$ such that $\mu(Y)=1$ and $\pi ^{-1}(Y)$ is isomorphic to a trivial vector bundle.
There is a suggestion that I do not quite understand:
Since $\bigcup_{r>0}\partial B(x,r)$ is an uncountable disjoint union, there $r_x>0$ such that:
1- $\mu(\partial B(x,r_x))=0$.
2- $\pi| B(x,r_x)$ is measurably isomorphic to $B(x,r_x)\times \mathbb{R}^{n}$.
Thanks for your attention.
 A: Let $x \in X$ be arbitrary. There exsits an open neighborhood $U$ of $x$ such that $E$ restricted to $U$ is trivial. Let $r > 0$ such that $B_r(x) \subset U$. 
Claim: There exists $0 < s \leq r$ such that $\mu (\partial B_s(x)) = 0$.
proof: For $0 < s \neq \tilde s \leq r$ we have $\partial B_s(x) \cap \partial B_{\tilde s}(x) = \emptyset$. Thus for each $n \geq 1$ the set $I_n := \{0 < s \leq r | \mu(\partial B_s(x)) \geq 1/n\}$ is finite since 
$$\infty > \mu(X) \geq \mu (B_r(x)) \geq \Sigma_{s \in I_n} \mu(\partial B_s(x)) \geq 1/n |I_n|.$$
It follows that $I := \{0 < s \leq r | \mu(\partial B_s(x)) > 0\} = \bigcup_{n \geq 1}I_n$ is countable. In particular $I^c = \{0 < s \leq r | \mu(\partial B_s(x)) = 0\}$ is nonempty $\square$
It follows that for all $x \in X$ there exists an $s(x) > 0$ such that $\mu(\partial B_s(x)) = 0$ and $E$ restricted to $B_{s(x)}(x)$ is trivial. Clearly $X = \bigcup_{x \in X}B_{s(x)}(x)$. Since $X$ is compact there exists a finite number of points $x_1, \dots , x_k$ such that $X = \bigcup_{1 \leq i \leq k} B_{s(x_i)}(x_i)$. Let $B_i = B_{s(x_i)}(x_i)$ and define inductively $U_1 = B_1$ and $U_k = B_k \setminus (\overline U_1 \cup \dots \cup \overline U_{k - 1})$. Then the $U_i$ are disjoint, open and $E$ restricted to each $U_i$ is trivial. Hence $E$ restricted to $\bigcup_i U_i$ is trivial. Further by construction $X \setminus \bigcup_iU_i \subseteq \bigcup_i \partial B_i$. Thus 
$$\mu(X \setminus \bigcup U_i) \leq \mu(\bigcup \partial B_i) = 0.$$
