Let $A \subset \Bbb R$ be an $F_\sigma$ set. Construct a function $f$ such that it precisely has $A$ as it's set of discontinuity. Let $A \subset \Bbb R$ and $A$ is an $F_\sigma$ set. An $F_\sigma$ set is a countable union of close sets. My task is to construct a function $f : \Bbb R \to \Bbb R$ such that it precisely has $A$ as it's set of discontinuity.
My attempt :
I have tried for some relatively simple special cases.
Like if A has no limit points (say something like $\Bbb Z$), then we could simply define, $$f(x):= x , x \in \Bbb R \text{ \ } A \\ :=x+2 ,x\in A $$
If $A$ was of the form , $A = \cup_{n\in \Bbb N} {[a_n,b_n]}$, where $b_n > a_n, \forall n \in \Bbb N$, consider $A_1 = A \cap \Bbb Q$ and $A_2 = A \text{ \ } A_1$ and define,$$f(x):= \chi_{A_1} - \chi_{A_2}$$ (where $\chi_X$ is the indicator function on X) .
But I am unable to construct $f$ when $A$ is in full generality.
Thanks in advance for help.
 A: I will prove the equivalent statement that if $A$ is a $G_\delta $ set then there is a function which is continuous precisely at points of $A$. Notation: $\omega (f,x)= \inf_{\delta >0} \sup \{|f(x)-f(y)|:|x-y| \leq \delta\}$. $f$ is continuous at $x$ iff $\omega (f,x)=0$. Let $A=\cap_{n=1}^{\infty }G_{n}$ with $G_{n}$ open and $%
G_{n+1}\subset G_{n}$ for all $n$. Let $f_{n}=I_{C_{n}\backslash E_{n}\text{ 
}}$where $C_{n}=G_{n}^{c}$ and $E_{n}=%
\cap C_{n}^{0}$. Let $f=\sum_{n=1}^{\infty }\frac{1}{n!}f_{n}$. We
claim that $f$ has the desired properties. First let $x\in A$. Then $%
f_{n}(x)=0$ for all $n$. In fact, for each $n$, $f_{n}$ vanishes in a
neighbourhood of $x$. Hence each $f_{n}$ is continuous at $x$. By uniform
convergence of the series defining $f$ we see that $f$ is also continuous at 
$x$. Now let $x\in A^{c}.$ Let $k$ be the least positive integer such that $%
x\in C_{k}$. If $x\in C_{k}^{0}$ then, in sufficiently small neighbourhoods
of $x,$ $f_{k}$ take both the values $0$ and $1$ and so its oscillation at $%
x $ is $1$. We claim that the oscillation of $f_{j}$ at $x$ is $0$ for each $%
j<k:$ since $x\notin C_{j}$ it follows that points close to $x$ are all in $%
C_{j}^{c}$ and hence $f_{j}$ vanishes at those points. Now $\omega (f,x)\geq 
\frac{1}{k!}\omega (f_{k},x)-\sum_{j=k+1}^{\infty }\frac{1}{j!}$
since $\omega (f_{j},.)\leq 1$ everywhere. Thus $\omega (f,x)\geq \frac{1}{k!%
}-\sum_{j=k+1}^{\infty }\frac{1}{j!}\geq \frac{1}{k!}%
[1-\sum_{j=k+1}^{\infty }\frac{1}{(k+1)(k+2)...(j)}]>\frac{1}{k!}%
[1-\sum_{j=k+1}^{\infty }\frac{1}{2^{j-k}}]=0.$
