Question:
Let $F:\mathbb{R^2\to R}$ be a continuous function.
Define $p(x)$ as the number of $y$ such that $F(x,y)=0$,i.e. $p(x)=\#\{y\in\mathbb{R}|F(x,y)=0\}$.
Prove that $p(x)$ is (extended) Borel measurable(as $p(x)$ valued in $[0,+\infty]$).
Attempt:
It's not hard to see that $\{(x,y)|F(x,y)=0\}$ and $\{y|F(x_0,y)=0\}$ are both closed set.But I don't know how to deal with the 'counting function' $p(x).$Thanks in advance!
Define $I_Q$ to be all the closed intervals in $\mathbb{R}$ with rational endpoints. Fix a positive integer $n \in \mathbb{Z}^+$, for any $I_1, I_2,...,I_n$ in $I_Q$, pairwise disjoint, we can define $$E_{I_1,I_2...,I_n} = \{x\in \mathbb{R}|\exists y_1\in I_1,...y_n \in I_n, st. F(x,y_1) = F(x,y_2) =... =F(x,y_n) = 0\}$$Hence, $E_{I_1,I_2...,I_n} \subset \{p(x)\ge n\}$. Since for n arbitrary real numbers, they can always be covered in n pairwise disjoint intervals respectively in $I_Q$, it is easy to prove that $\bigcup_{I_1,...,I_n\in I_Q, pairwise\ dijoint}E_{I_1,...I_n} = \{p(x)\ge n\}$. Since this is a countable union, we only need to prove that $E_{I_1,I_2...,I_n}$ is an open or a closed set.
In fact, $E_{I_1,I_2...,I_n}$ is a closed set. Foy any limit point of $E_{I_1,I_2...,I_n}$, denoted by $x_0$, there is a sequence $\{x_m\}$ in $E_{I_1,I_2...,I_n}$ which converges to $x_0$. Meanwhile, by definition, there are sequences $\{y_m^{(1)}\},...,\{y_m^{(n)}\}$ such that $F(x_m,y_m^{(1)}) =...= F(x_m, y_m^{(n)}) = 0$. Since these are all bounded sequences, we can easily take n convergent subsequences one after another, so just assume that they are all convergent and denote their limits by $y_1,...y_n$. Since $F$ is continuous, we must have $F(x_0,y_1) = ... =F(x_0, y_n) = 0$. Hence, $x_0 \in E_{I_1,I_2...,I_n}$. Finishing the proof.
Another slightly different proof:
1.$\{p(x)\geqslant 1\}=\cup_{n=1}^{+\infty}\{x|\exists y:|y|\leqslant n,F(x,y)=0\}:=\cup_{n=1}^{+\infty}A_n^{(1)}$,and $A^{(1)}_n$ is closed since any sequence has a limit point in a compact set;
2.For general $n\in\mathbb{N^*}$,$\{p(x)\geqslant n\}=\cup_{N=1}^{+\infty}\cup_{k=1}^{+\infty}\{x|\exists y_1,...,y_n:|y_i|\leqslant n,|y_i-y_j|\geqslant {1\over k}(i\neq j),F(x,y_i)=0\}:=\cup_{N=1}^{+\infty}\cup_{k=1}^{+\infty}A_{N,k}^{(n)}$
and $A_{N,k}^{(n)}$ is closed for the same reason(note that the gap $1\over k$ guarantee that $\{y_n^k\}_{n\in\mathbb{N^*}}$ will have $n$ distinct limit points).