roots of polynomial with measurable coefficients If I have polynomial:
$$a_n(\omega) x^n+\cdots+a_0(\omega)$$
$a_i:\Omega\to\mathbb R$ are measurable functions 
Assume that for every $\omega$ there are $n$ real roots to the polynomial (including multiplicity)
given in order $\xi_1(\omega)\leq\xi_2(\omega)\leq\cdots\leq\xi_n(\omega)$
Are these roots measurable functions?
my attempt:
$f:\Omega \times \mathbb R\to \mathbb R$ is continuous in the second slot (fixed $\omega$) and measurable in the first slot so it is measurable hence $f^{-1}(0)$ is measurable set $A$.
But now I want to project this $A$ to $\Omega $ and I don't know whether it is still measurable.
thank you
 A: Let $p(\omega,x)$ be a random polynomial, that is, 
for each fixed $x\in\mathbb{R}$ the map $\omega\mapsto p(\omega,x)$
is measurable, and for fixed $\omega\in\Omega$ the map $x\mapsto p(\omega,x)$
is a polynomial.  
Define $\xi(\omega)=\inf(x: p(\omega,x)=0)$ so that 
$\xi:\Omega\to [-\infty,\infty]$. That $\xi$ is 
measurable follows from 
$$\{\omega :\xi(\omega)\leq x\}=\left\{\omega: \bigcap_{k=1}^\infty 
\bigcup_{s\leq x, s\in\mathbb{Q}} |p(\omega,s)|<1/k\right\}.$$
Therefore $\xi(\omega)$ is the smallest root of 
$x\mapsto p(\omega,x)$ if a root exists, $\xi(\omega)=\infty$
 if there is no root, and $\xi(\omega)=-\infty$ if the function 
is the zero polynomial.    
Now define a new random polynomial on $\Omega\times \mathbb{R}$
by 
$$q(\omega,x)
=\lim_{k\to \infty} k\, p(\omega,x+1/k)\,{\bf 1}_{[\xi(\omega)=x]}
+{p(\omega,x)\over x-\xi(\omega)}\,{\bf 1}_{[\xi(\omega)\in\mathbb{R}\setminus\{x\}]}.$$ 
This gets rid of the root $\xi(\omega)$, and you can now continue the process.
If the original random polynomial $p$ has $n$ roots for each $\omega$, then 
continuing the process $n$ times gives  random variables 
 $\omega\mapsto (\xi_1(\omega),\xi_2(\omega),\dots,\xi_n(\omega))$
which is  the ordered list of roots, with multiplicities. 
