# measurable function induces a measurable bundle

Greetings I am preparing a work on bundles and I found this statement

Let $V$ a topological vector space with $\dim V=n$ and $(E,\pi,M)$ a vector bundle continuous over $M$ (compact space).

If $G_k(V)=\{ W: W$ is subspace of $V, \dim W=k \}$ and consider in $G_k(V)$ one topology $\tau$ where $U\in \tau$ is open iff the set $\widehat{U}=\lbrace v: v\in W\backslash \lbrace 0\rbrace, \mbox{for some} \ W\in U \rbrace$ is open in $V$.

Assume that the application $x \mapsto F_x$ is borel-measurable, where $x\in M$ and $F_x \in G_k(V)$. If further $F_x \subset E_x$ for all $x\in M$ then

For all $x\in M$ there set measurable $U_x\subset M$ where $x\in U_x$ and $v_i: U_x \rightarrow V$ measurable for all $1\leq i\leq k$ and such that $\lbrace v_1(z),\ldots,v_{k}(z)\rbrace$ is an orthonormal basis of $F_z$ for all $z\in U_x$.

I would appreciate if you could give me some guidelines to prove the statement

Note: This implies that $F=\{ F_x\}$ induce a measurable sub bundle of $E$ over $M$.

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I just edit the question increasing a condition on $F$ – helmonio Apr 27 '14 at 2:56