Let $\mathcal{K}=\{-K, -K+1, \cdots ,-1, 0, 1, \cdots, K-1, K\}$, where $K$ is a positive integer.
Let $\phi(x)=\displaystyle\sum_{k \in \mathcal{K}}c_k \, e^{\mathrm{i}k x}$, where $\mathrm{i}=\sqrt{-1}$ and $c_k \in \mathbb{C},\, \forall k \in \mathcal{K}$.
I would like to know if the set defined below has zero Lebesgue measure in $\mathbb{R}^N$, where $N > 2K+1$:
$A = \{(x_1, x_2, \cdots, x_N): \phi(x_n) = x_n , \text{ for } n \in \{1, 2, 3, \cdots, N\}, \text{ every } x_n \in (0, 2\pi), \text{ and } x_1 < x_2, \cdots < x_N\}.$
$A$ is the set for all $\phi$ of degree at most $K$.
EDIT: Are the following two approaches correct?
Approach 1: Can we consider the problem as counting the number of zero-crossings of the double derivative of $\phi$ with respect to $x$? Since the maximum frequency content of $\phi$ is $K$ rad/sec, $\phi$ can have at most $2K$ zero crossings. Since $N>2K+1$ in our problem, the set $A$ turns out to be an empty set.
Approach 2: Let's look at the problem as the number of ordinates at which $\phi(x)$ intersects the linear function $x$.
If $K=0$, $\phi(x)=c_0$. The set $A$ contains $N$-tuples (in this case $N>1$) such that $\phi(x_n)=x_n$. Hence, $c_0 \in \mathbb{R}$. For any c_0, since it can not intersect the linear function $x$ at not more than 1 ordinate, $A$ is an empty set.
Similarly, for $K=1$ and assuming $c_{-1}=c_1$, we get $\phi(x)=c_0+2c_1 \cos(x)$. Now, $\phi(x)$ can intersect the funciton $x$ at not more than 2 ordinates in the interval $(0, 2\pi)$. Again $A$ (it contains N-tuples, where $N >3$) turns out to be an empty set.
Can we make a similar argument for any $K$?
I am not sure how to formalise these approaches.