$A(t)$ measurable. Does there exists measurable $x(t)\in\ker A(t)$? I have a matrix function $t\mapsto A(t)$, say on $[0,1]$, which is measurable with $A(t)\in\mathbb C^{n\times m}$, $m < n$. If $A(t)$ is not injective a.e., does there exist a non-vanishing measurable function $t\mapsto x(t)$ such that $x(t)\in\ker A(t)$ for a.e. $t\in [0,1]$? Or better: $x\in L^\infty(0,1)$.
One may assume that $m=n$ and that $A(t)$ is selfadjoint since we can replace $A(t)$ be $A(t)^*A(t)$.
If $E$ is the exceptional set where $A$ is not injective and if there exists a measurable set $E'\subset E$ with positive measure such that $\bigcup_{t\in E'}\operatorname{Im}A(t)\neq\mathbb C^m$, then I am done because I can pick $x$ outside this union and set $x(t) := P_{\ker A(t)}x$, where $P_{\ker A(t)}$ is the orthogonal projection onto $\ker A(t)$.
 A: I think I could answer my question myself (without too much algebra ;-)). I assume that $A(t)\in\mathbb C^{m\times m}$ is selfadjoint for all $t\in [0,1]$. Also, I set $N(t) := \ker A(t)$ and $R(t) := \operatorname{Im}A(t)$. In addition $P_{N(t)}$ denotes the orthogonal projection onto $N(t)$.
For $x\in\mathbb C^m$ define $N_x := \{t : P_{N(t)}x\neq 0\}$. This is a measurable set (since $P_{N(t)} = -\frac 1 {2\pi i}\int_{\mathbb T}(A(t)-\lambda I)^{-1}\,d\lambda$, where $\mathbb T$ is the unit circle). Let us assume that $|N_x| = 0$ for all $x\in(\mathbb Q + i\mathbb Q)^m$. Then also $|N|=0$, where $N := \bigcup_{x\in(\mathbb Q + i\mathbb Q)^m}N_x$.
Let $t\notin N$. Then for all $x\in(\mathbb Q + i\mathbb Q)^m$ we have $t\notin N_x$ and hence $P_{N(t)}x = 0$, which means $x\in R(t)$. This proves $(\mathbb Q + i\mathbb Q)^m\subset R(t)$. But as $R(t)$ is closed, it follows that $R(t) = \mathbb C^m$ and thus $N(t) = R(t)^\perp = \{0\}$. Therefore, $A(t)$ is invertible for all $t\notin N$ (i.e., for a.e. $t$). As we excluded this case, there exists $x\in\mathbb C^m$ such that $|N_x|>0$. Then, setting $x(t) := P_{N(t)}x$ is a function as desired. As $\|x(t)\|\le\|x\|$ also $x\in L^\infty$ holds.
EDIT: This reasoning can actually be raised to any separable Hilbert space.
A: This should be true, and it follows from the way the system $A \cdot x = 0$ is solved. Consider all $t$ for which $A(t)$ has the rank $r$ equal to the rank of a certain $r\times r$ minor. This is a measurable set. ( the minor is $\ne 0$ and all the larger minors are $0$). On this set, we know how to solve the system, and we can make the solution depend rationally on the entries of $A$ ( and some free parameters, which can be chosen constant). So we can make a measurable choice for a solution of the system. 
