$f_n\rightarrow f$ pointwise, $O$ open subset of $\mathbb{R}$ $\Rightarrow$ $f^{-1}(O)$ is $F_{\sigma}$ Let $E$ be a Banach space and $(f_n)$ a sequence of continuous functions from $E$ to $\mathbb{R}$ that converges pointwise to $f$. $O$ is an open subset of $\mathbb{R}$. I need to prove that $f^{-1}(O)$ is an $F_{\sigma}$ subset (countable union of closed subsets). Clearly $f^{-1}(O)=\bigcup_{N\in \mathbb{N}}\bigcap_{n\ge N}f_n^{-1}(O)$ (Edit : this is not true in general as was mentioned by Danny-Pak Keung in his comment) but I don't know what to do next. Thank you for your help!
 A: There is a sequence $K_m$ of closed sets such that $K_m \subset \text{Int}(K_{m+1})$ and $O = \bigcup_{m=1}^\infty K_m$.  Show that
 $$f^{-1}(O) = \bigcup_N \bigcap_{n \ge N} f_n^{-1}(K_n)$$ 
where $\bigcap_{n \ge N} f_n^{-1}(K_n)$ is closed.
A: It is enough to assume that the domain of $f_{n}$ and $f$ is a topological
space $X$. The linear structure, the norm etc of the Banach space
play no role.
Firstly, consider the case that $O$ is an open interval $O=(a,b)$.
Choose $K\in\mathbb{N}$ be sufficiently large such that $a+\frac{1}{K}<b-\frac{1}{K}$.
For $k\geq K$, define $a_{k}=a+\frac{1}{k}$, $b_{k}=b-\frac{1}{k}$.
Let $I_{k}=(a_{k},b_{k})$. Note that $I_{k}\subseteq\bar{I}_{k}\subseteq I_{k+1}\subseteq\bar{I}_{k+1}\subseteq\ldots\subseteq O$
and $O=\cup_{k}I_{k}=\cup_{k}\bar{I}_{k}$.
For each $k\geq K$,
\begin{eqnarray*}
f^{-1}(I_{k}) & \subseteq & \cup_{N=1}^{\infty}\cap_{n=N}^{\infty}f_{n}^{-1}(I_{k})\\
 & \subseteq & \cup_{N=1}^{\infty}\cap_{n=N}^{\infty}f_{n}^{-1}(\bar{I}_{k})\\
 & \subseteq & f^{-1}(\bar{I}_{k}).
\end{eqnarray*}
Denote $F_{k}=\cup_{N=1}^{\infty}\cap_{n=N}^{\infty}f_{n}^{-1}(\bar{I}_{k})$,
which is a $F_{\sigma}$-set. In short, we have $f^{-1}(I_{k})\subseteq F_{k}\subseteq f^{-1}(\bar{I}_{k})$.
It follows that 
\begin{eqnarray*}
f^{-1}(O) & = & \cup_{k}f^{-1}(I_{k})\\
 & \subseteq & \cup_{k}F_{k}\\
 & \subseteq & \cup_{k}f^{-1}(\bar{I}_{k})\\
 & = & f^{-1}(O).
\end{eqnarray*}
Hence $f^{-1}(O)=\cup_{k}F_{k}$, which is a $F_{\sigma}$-set.
Finally, if $O\subseteq\mathbb{R}$ is an open set, there exists a
sequence of open intervals $(I_{n})$ such that $O=\cup_{n}I_{n}$.
Then $f^{-1}(O)=\cup_{n}f^{-1}(I_{n})$, which is also a $F_{\sigma}$-set.
