The set of discontinuous points is countable union of closed sets

Solution 1:

1. Yes: if $\omega_f(a)\ge\frac1n$, then certainly $\omega_f(a)\ge\frac1{n+1}$, so $F_n\subseteq F_{n+1}$ for each $n$.

2. The statement that $D(f)=\bigcup_{n\ge 1}F_n$ really does require proof, though the proof is very easy. First off, it’s clear that if $\omega_f(a)\ge\frac1n$ then $\omega_f(a)>0$, so $F_n\subseteq D(f)$ for each $n\ge 1$. Thus, $D(f)\supseteq\bigcup_{n\ge 1}F_n$. Now suppose that $a\in D(f)$. Then $\omega_f(a)>0$, so there is some positive integer $k$ such that $\frac1k<\omega_f(a)$. But then $a\in F_k\subseteq\bigcup_{n\ge 1}F_n$, and since $a$ was any element of $D(f)$, this implies that $D(f)\subseteq\bigcup_{n\ge 1}F_n$. Combining the inclusions, we see that $D(f)=\bigcup_{n\ge 1}F_n$, as claimed.

3. Do you mean in the equality $\{a:\omega_f(a)>0\}=\{a:\omega_f(a)\ge\epsilon\text{ for some }\epsilon>0\}$? That step could just as well be omitted: in (2) above I showed directly that $D(f)=\bigcup_{n\ge 1}F_n$ without using $\{a:\omega_f(a)\ge\epsilon\text{ for some }\epsilon>0\}$ at all. I don’t think that it contributes anything to one’s understanding of the argument.

4. $\{a:\omega_f(a)<r\}$ is a set of real numbers, a subset of $\Bbb R$; $\{I:\omega(f;I)<r\}$ is a set of bounded intervals in $\Bbb R$. These two sets don’t even have the same kind of elements: the elements of the first set are individual real numbers, while the elements of the second set are bounded intervals of real numbers. Thus, the two sets are very different: not only is it impossible for them to be equal to each other, it’s impossible for either of them to be a subset of the other.