Baire's theorem: a Baire class one function on a closed bounded interval $[a,b]$ is continuous at the points of a dense subset of $[a,b]$ that is the trace of a $G_{\delta }$ set, so its set of discontinuities is meager

Refinement claim. Let $[c,d]\subseteq [a,b]$ with $c<d$ and let $\epsilon >0$ be real. For $N\in \mathbb{N}$ put $E_N:=\{x\in [c,d]:\mathrm{\mid }{f}_n(x)-f_m(x)\mathrm{\mid }\le \epsilon \mathrm{/}4\text{ for all }n,m\ge N\}$.

Claim 1. Fix a real $\epsilon >0$ and let $D_{\epsilon }:=\{x\in [a,b]:{\omega }_f(x)\ge \epsilon \}$. It is closed in $\mathbb{R}$, being $[a,b]\cap G$ with $G$ closed and $[a,b]$ closed.

Claim 4. The set of continuity points of $f$ on the domain $[a,b]$ is $[a,b]\cap V$ for a $G_{\delta }$ subset $V\subseteq \mathbb{R}$.

Claim 3. Let $x\in [a,b]$ and let $\rho >0$ be real. The set $[a,b]\cap N_{\rho }(x)$ contains a nondegenerate closed interval $[c,d]$ with $c<d$, because $a<b$: taking $c:=\max \{a,x-\rho \mathrm{/}2\}$ and $d:=\min \{b,x+\rho \mathrm{/}2\}$ gives $[c,d]\subseteq [a,b]\cap N_{\rho }(x)$ and $c<d$: if $x=a$ then $c=a$ and $d=\min \{b,a+\rho \mathrm{/}2\}>a$; if $x=b$ then $d=b$ and $c=\max \{a,b-\rho \mathrm{/}2\}<b$; and if $a<x<b$ then $c<x<d$.

Each $E_N$ is closed: for fixed $n,m$ the set $\{x\in [c,d]:\mathrm{\mid }{f}_n(x)-f_m(x)\mathrm{\mid }\le \epsilon \mathrm{/}4\}$ is the preimage under the continuous function $\mathrm{\mid }{f}_n-f_m\mathrm{\mid }$ of the closed set $\{y:y\le \epsilon \mathrm{/}4\}$, hence of the form $G\cap [c,d]$ with $G$ closed, hence closed since $[c,d]$ is; and $E_N$ is the intersection of that nonempty family of closed sets over the pairs $n,m\ge N$.

$[c,d]={\bigcup }_{N\in \mathbb{N}}{E}_N$: for $x\in [c,d]$ the sequence $(f_k(x))$ converges to $f(x)$, so there is $N$ with $\mathrm{\mid }{f}_k(x)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}8$ for all $k\ge N$, and then $\mathrm{\mid }{f}_n(x)-f_m(x)\mathrm{\mid }\le \mathrm{\mid }{f}_n(x)-f(x)\mathrm{\mid }+\mathrm{\mid }f(x)-f_m(x)\mathrm{\mid }<\epsilon \mathrm{/}4$ for all $n,m\ge N$.

By the interval form of Baire category applied to $[c,d]$ and the sequence $(E_N)$, there are $N\in \mathbb{N}$ and reals $u^{\mathrm{\prime }}<v^{\mathrm{\prime }}$ with $[u^{\mathrm{\prime }},v^{\mathrm{\prime }}]\subseteq E_N\cap [c,d]=E_N$.

For every $x\in [u^{\mathrm{\prime }},v^{\mathrm{\prime }}]$ one has $\mathrm{\mid }{f}_N(x)-f(x)\mathrm{\mid }\le \epsilon \mathrm{/}4$. Indeed, let $\eta >0$ be real; since $f_m(x)\to f(x)$ there is $m\ge N$ with $\mathrm{\mid }{f}_m(x)-f(x)\mathrm{\mid }<\eta$, and then $\mathrm{\mid }{f}_N(x)-f(x)\mathrm{\mid }\le \mathrm{\mid }{f}_N(x)-f_m(x)\mathrm{\mid }+\mathrm{\mid }{f}_m(x)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}4+\eta$; as $\eta >0$ was arbitrary this gives $\mathrm{\mid }{f}_N(x)-f(x)\mathrm{\mid }\le \epsilon \mathrm{/}4$.

Put $x_0:=(u^{\mathrm{\prime }}+v^{\mathrm{\prime }})\mathrm{/}2$, so $u^{\mathrm{\prime }}<x_0<v^{\mathrm{\prime }}$. Since $f_N$ is continuous at $x_0$ there is a real $\delta >0$ with $\mathrm{\mid }{f}_N(x)-f_N(x_0)\mathrm{\mid }<\epsilon \mathrm{/}4$ for every $x\in [a,b]$ with $\mathrm{\mid }x-x_0\mathrm{\mid }<\delta$. Put $u:=\max \{u^{\mathrm{\prime }},x_0-\delta \mathrm{/}2\}$ and $v:=\min \{v^{\mathrm{\prime }},x_0+\delta \mathrm{/}2\}$, so that $u<x_0<v$ and $[u,v]\subseteq [u^{\mathrm{\prime }},v^{\mathrm{\prime }}]$ with $\mathrm{\mid }x-x_0\mathrm{\mid }<\delta$ for every $x\in [u,v]$.

For $x,y\in [u,v]$: $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le \mathrm{\mid }f(x)-f_N(x)\mathrm{\mid }+\mathrm{\mid }{f}_N(x)-f_N(x_0)\mathrm{\mid }+\mathrm{\mid }{f}_N(x_0)-f_N(y)\mathrm{\mid }+\mathrm{\mid }{f}_N(y)-f(y)\mathrm{\mid }\le \epsilon \mathrm{/}4+\epsilon \mathrm{/}4+\epsilon \mathrm{/}4+\epsilon \mathrm{/}4=\epsilon$. Hence ${\omega }_f([u,v])\le \epsilon$, $\epsilon$ being an upper bound of the set whose supremum that is.

The refinement claim is proved: for every $[c,d]\subseteq [a,b]$ with $c<d$ and every real $\epsilon >0$ there are $u<v$ with $[u,v]\subseteq [c,d]$ and ${\omega }_f([u,v])\le \epsilon$. Moreover every $x$ with $u<x<v$ satisfies ${\omega }_f(x)\le \epsilon$, since $[a,b]\cap N_{\rho }(x)\subseteq [u,v]$ for $\rho :=\min \{x-u,v-x\}>0$ and ${\omega }_f$ is monotone under inclusion.

$D_{\epsilon }$ contains no nondegenerate closed interval. Were $[c,d]\subseteq D_{\epsilon }$ with $c<d$, the refinement claim applied to $[c,d]$ and to the positive real $\epsilon \mathrm{/}2$ would give $u<v$ with $[u,v]\subseteq [c,d]$ and ${\omega }_f(x)\le \epsilon \mathrm{/}2<\epsilon$ for every $x$ with $u<x<v$; such an $x$ lies in $[c,d]\subseteq D_{\epsilon }$ and so satisfies ${\omega }_f(x)\ge \epsilon$, which is impossible.

Hence $D_{\epsilon }$ is nowhere dense: it is closed, so it equals its own closure, and its interior is empty, since an interior point would have a neighbourhood $N_{\rho }(x)\subseteq D_{\epsilon }$ and then $[x-\rho \mathrm{/}2,x+\rho \mathrm{/}2]$ would be a nondegenerate closed interval inside $D_{\epsilon }$.

Claim 2. $D={\bigcup }_{n\in \mathbb{N}}{D}_{1\mathrm{/}\iota (n+1)}$, and each $D_{1\mathrm{/}\iota (n+1)}$ is nowhere dense by step 8.1, so $D$ is a union of a sequence of nowhere dense sets, that is, meager.

Suppose $[c,d]\subseteq D$ with $c<d$ as in step 1.4. Then $[c,d]$ is covered by the sequence $(D_{1\mathrm{/}\iota (n+1)})$ of closed sets, so by the interval form of Baire category some $D_{1\mathrm{/}\iota (n+1)}\cap [c,d]$ contains a nondegenerate closed interval, contradicting step 7.1. So $[c,d]\subseteq ̸D$, and any point of $[c,d]\setminus D$ is a point of $C$ inside $[a,b]\cap N_{\rho }(x)$.

Claims 1, 2, 3 and 4 are therefore proved: claim 1 by steps 1.2, 7.1 and 8.1, claim 2 by step 9.1, claim 3 by steps 1.4 and 10.1, and claim 4 by step 1.3.