Eigenvalues are measurable Consider an $n\times n$-symmetric matrix A. It is well known that A has n real eigenvalues $\lambda _i$, which we order in an increasing way, i.e. $\lambda_1\leq\lambda_2\leq...\leq\lambda_n$. Now the claim is that the mapping $ A\mapsto \lambda=(\lambda_1,...,\lambda_n)$ is measurable, but I don't see a way to prove it.  
 A: There are several methods. Here is one that is simple to describe. 
Without loss of generality, all eigenvalues are strictly positive. You can guarantee this by adding $(1+\max_{j} \sum_{i} |a_{ij}| ) I_{n} $ to $A$, a continuous function of the entries.
This is because if $\lambda$ is an eigenvalue and $v$ is a corresponding eigenvector, then $\lambda v_i = \sum_j a_{ij} v_j$. Therefore $|\lambda| \sum_{i} |v_i| \le \sum_{i,j} |a_{i,j}||v_j| \le (\max_{j} \sum_{i}|a_{i,j}|)\sum_{j} |v_j|$, and $\sum|v_j|>0$. 
Next, recall that for any $n$ positive  numbers $r_1,r_2,\dots, r_n$, 
$$\max_{i} r_i = \lim_{k\to\infty} (\sum_{i\le n} r_i^k)^{1/k}.$$ 
Therefore you immediately obtain $\lambda_n$ by taking the limit 
$$ \lim_{k\to\infty} (\mbox{Tr} (A^k))^{1/k}.$$ 
This is clearly a measurable function of the entries: it's a point-wise  limit of continuous functions of the entries. 
Continue inductively, and define for $1\le j\le n-1$: 
$$\lambda_{n-j-1} = \lim_{k\to\infty} (\mbox{Tr} (A^k)-\lambda_{n-j}^k-\lambda_{n-j+1}^k-\dots -\lambda_n^k)^{1/k}.$$
We're done. I hope. 
