Any $F_\sigma$ set in $\mathbb R$ is a set of discontinuity of some function $f$.

Question

Consider the following lemma:

Lemma

Let $A\subset \mathbb R$ be an $F_\sigma$ set,then $\exists f:\mathbb R\to \mathbb R$ such that $D(f)=A$.

Proof: $A=\bigcup\limits_{n\in \mathbb N} A_n$ where $A_n$ are all closed.

Define $f_n:=\large\chi_{A_n}$,characteristic function $A_n$.Define $f(x)=\sum\limits_{n=1}^{\infty} 2^{-n}f_n(x)$ and this function will satisfy the requirement.

This proof was given in stackexchange.But I cannot figure out why $x\in A$ should imply $x\in D(f)$.Can someone help me out of this?The same stack exchange article says that $\exists y$ arbitrarily close to $x\in A$ such that $f_n(x)\neq f_n(y)$ but I do not get it.What if $(x-\epsilon,x+\epsilon)\subset A_n$ for some $n\in \mathbb N$ and for some $\epsilon>0$.

I don't think this works : every closed set is an $F_{\sigma}$ set, so let us take, say $A = [0,1]$ , with all the $A_n$ then equalling $[0,1]$. Then, $f(x) = 0$ for $x \notin A$ and $f(x)= \sum_{n=1}^\infty 2^{-n}=1$ for $x \in A$. This function is basically $1$ on $[0,1]$ and $0$ elsewhere, so its set of discontinuity is ${0,1}$, not $A$. Hence the idea and answer above is wrong ; we have to be more creative. — Sarvesh Ravichandran Iyer, Apr 24 '20 at 04:36
For a proof of the result in question, see theorem 7.2 in Oxtoby's Measure and Category. — Reveillark, Apr 24 '20 at 05:03
@KaviRamaMurthy https://math.stackexchange.com/questions/629926/function-whose-discontinuity-points-are-a-prefixed-f-sigma-set-in-mathbbr — Kishalay Sarkar, Apr 24 '20 at 05:50
Be careful when you quote other's proofs. You are obviously misquoting the proof. — Kavi Rama Murthy, Apr 24 '20 at 05:52
What would make this work much easier is if you could express $A$ as a countable union of pairwise disjoint closed sets, but in general this is not possible (the "pairwise disjoint" part). — Dave L. Renfro, Apr 24 '20 at 09:21

Any $F_\sigma$ set in $\mathbb R$ is a set of discontinuity of some function $f$.

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