Let $f$ be a real-valued bounded function. Consider $D_f$ is the set of discontinuity points of f. I want to prove that the $D_f$ is a Borel set.
I tried the following, but I don't know it if is correct: $D_f$ must be countable, and I know that each of these is a borel set, but I don't know why I can say $D_f$ must be countable.
The function $f$ is contiuous at $a$ iff $$\forall \epsilon>0\;\exists\delta>0\;\forall x\;|x-a|<\delta\to|f(x)-f(a)|<\epsilon.$$ We can let countability shine through this condition if we translate it to $$\forall n\in \Bbb N\;\exists m\in\Bbb N\;\forall x\; |x-a|<\frac1m\to|f(x)-f(a)|<\frac1n.$$ So if we define $$ A_{n,m}=\{\,a\in\Bbb R:\forall x\;|x-a|<\frac1m\to |f(x)-f(a)|<\frac1n\,\}$$ Then $D_f$ is the complement of $$ \bigcap_n\bigcup_m A_{n,m}.$$ If $A_{n.m}$ were open (or closed) we'd be done, but sometimes life is not that easy. However, if $a\in A_{n.m}$ then $b\in A_{n,m+1}$ for all $b$ with $|b-a|<\frac1m-\frac1{m+1}=\frac1{m^2+m}$.