For each $n\in\Bbb Z$, let $f_n\colon [n,n+1]\to\Bbb R$ be a function such that $f_{n}(n+1)=f_{n+1}(n+1)$ and $f_n$ is $0$ on $\Bbb R\setminus [n,n+1]$ and $f^{-1}_{n}(U)$ is $F_\sigma$ ( a countable intersection of closed sets) for every $n\in\mathbb Z$ and for any open set $U\subset\Bbb R$
Now, define $f$ as follows $$f=\sum_{n\in\Bbb Z} f_n$$ Clearly, $f$ is a function from $\Bbb R$ to $\Bbb R.$ My question
Does $f^{-1}(U)$ is $F_\sigma$ for any open set $U\subset\Bbb R$?
My guess is yes. $$f^{-1}(U)=\{x\in\Bbb R\colon (\sum_{n\in\mathbb Z}f_n)(x)\in U\}$$ Notice that $(\sum_{n\in\mathbb Z} f_n)(x)$ behaves like $f_n$ for some $n\in\mathbb Z$. But I did not know how I can go further further. Any idea will be appreciated greatly
From all those definitions we can see that $f \equiv f_n$ on $(n, n+1)$.
Then $f^{-1}(U) = \bigcup_{n\in\mathbb{Z}} (f_n^{-1}(U)\cap (n, n+1)) \cup \{n\in\mathbb{Z} : f(n)\in U\}$.
Now note that by assumption $f_n^{-1}(U)$ is $F_\sigma$ and $(n, n+1)$ is $F_\sigma$ as an open set, and finite intersection of $F_\sigma$ sets is also $F_\sigma$. So $f_n^{-1}(U)\cap (n, n+1)$ is $F_\sigma$. But a countable union of $F_\sigma$ sets is also $F_\sigma$, so their whole union $\bigcup_{n\in\mathbb{Z}} (f_n^{-1}(U)\cap (n, n+1))$ is $F_\sigma$. But $\{n\in\mathbb{Z} : f(n)\in U\}$ is clearly $F_\sigma$ too, so $f^{-1}(U)$ is $F_\sigma$ as an union of two $F_\sigma$ sets.