Baire Class 1 Functions

First, why the sum is over reduced fractions only.

Summing over all pairs $(p, q)$ counts each rational many times ($1/2 = 2/4 = 3/6 = \cdots$): at $x = 1/2$ the stacked heights $1/2 + 1/4 + 1/6 + \cdots$ would diverge, so the sum could never converge to Thomae.

With reduced fractions each rational appears exactly once.

Each tent $x \mapsto \mathrm{tri}(n^3(x - p/q))$ is continuous (a composition of continuous maps), and the sum is finite (at most $\sum_{q \le n}(q+1)$ terms), so $f_n$ is continuous. Also $f_n \ge 0$.

The tents are pairwise disjoint.

For distinct reduced fractions $p/q \ne r/s$ with denominators $\le n$, $\left|\frac{p}{q} - \frac{r}{s}\right| = \frac{|ps - qr|}{qs} \ge \frac{1}{qs}$.

If $qs \le n(n-1)$ this is $\ge 1/(n(n-1))$; if $qs = n^2$ then $q = s = n$, so $|ps - qr| = n|p - r| \ge n$ and the quotient is $\ge 1/n \ge 1/(n(n-1))$.

Hence the spacing is always $\ge 1/(n(n-1))$.

Each tent's support has width $2/n^3$, and $2/n^3 < 1/(n(n-1))$ (since $2(n-1) < n^2$, i.e. $(n-1)^2 + 1 > 0$). So no two tent supports overlap: at each $x$ at most one tent is nonzero.

For rational $x = a/b$ in lowest terms ($0 \le a \le b$): if $n \ge b$, the tent at $(a, b)$ contributes $\frac{1}{b}\,\mathrm{tri}(0) = 1/b$.

Every other reduced $p/q$ ($q \le n$) is at distance $\ge 1/(n(n-1)) > 1/n^3$ from $x$, so its tent vanishes at $x$.

Hence $f_n(a/b) = 1/b = T(a/b)$ — exact from $n \ge b$ onward.

For irrational $x$: given $\varepsilon > 0$, pick an integer $Q_0 \ge 1$ with $1/Q_0 < \varepsilon$.

The reduced rationals in $[0, 1]$ with denominator $\le Q_0$ are finitely many, all different from $x$, so $c := \min|x - p/q| > 0$.

For $n$ with $n^3 c > 1$, the tents with denominator $\le Q_0$ miss $x$; every other tent has height $\le 1/(Q_0+1) < \varepsilon$; and at most one tent is nonzero at $x$.

Hence $0 \le f_n(x) < \varepsilon$ — i.e. $f_n(x) \to 0 = T(x)$.

At a rational $r = a/b$ in lowest terms, $T(r) = 1/b > 0$.

The irrationals are dense, so there is a sequence of irrationals $x_k \to r$ with $T(x_k) = 0 \not\to 1/b = T(r)$.

Hence $T$ is discontinuous at $r$.

At an irrational $x_0$, $T(x_0) = 0$.

Given $\varepsilon > 0$, pick an integer $M > 1/\varepsilon$.

The rationals with $1/b \ge \varepsilon$, i.e. denominator $b \le M$, are finitely many in $[x_0 - 1, x_0 + 1]$.

Let $\delta$ be the minimum distance from $x_0$ to this finite set ($\delta > 0$ since $x_0$ is irrational).

Then for $|x - x_0| < \delta$: if $x$ is rational its denominator exceeds $M$, so $T(x) < \varepsilon$; if irrational, $T(x) = 0$.

Thus $|T(x) - T(x_0)| < \varepsilon$ — $T$ is continuous at $x_0$.

Hence the continuity set is exactly the irrationals $\mathbb{R} \setminus \mathbb{Q}$.

Each singleton $\{p/q\}$ is closed, so its complement $\mathbb{R} \setminus \{p/q\}$ is open and dense.

For fixed $q$ every $p$ must be removed, so the inner operation is an intersection: $\mathbb{R} \setminus \mathbb{Q} = \bigcap_{q \ge 1}\bigcap_{p \in \mathbb{Z}}(\mathbb{R} \setminus \{p/q\})$ — the pairs $(p, q)$ are countable, so this is a countable intersection of open sets, a $G_\delta$.

The irrationals are dense, so it is a dense $G_\delta$.

First, $\omega_f(x) = 0$ iff $f$ is continuous at $x$.

($\Rightarrow$) If $\omega_f(x) = 0$, every $\varepsilon > 0$ admits a neighborhood $U$ with $\mathrm{diam}\,f(U) < \varepsilon$, so $|f(y) - f(x)| \le \mathrm{diam}\,f(U) < \varepsilon$ for $y \in U$.

($\Leftarrow$) If $f$ is continuous at $x$, every $\varepsilon > 0$ admits a neighborhood $U$ with $f(U) \subseteq (f(x) - \varepsilon, f(x) + \varepsilon)$, so $\mathrm{diam}\,f(U) \le 2\varepsilon$ and $\omega_f(x) \le 2\varepsilon$; $\varepsilon$ arbitrary gives $\omega_f(x) = 0$.

$\{x : \omega_f(x) < 1/k\}$ is open: if $\omega_f(x_0) < c < 1/k$, some neighborhood $U$ has $\mathrm{diam}\,f(U) < c$; for every $x \in U$, $U$ is itself a neighborhood of $x$, so $\omega_f(x) \le \mathrm{diam}\,f(U) < c < 1/k$. Thus $U \subseteq \{\omega_f < 1/k\}$.

Now density.

Let $f_n \to f$ pointwise with $f_n$ continuous.

Cover $\mathbb{R}$ by closed intervals $V_j$ ($j \in \mathbb{Z}$) of length $< 1/(2k)$ whose interiors also cover $\mathbb{R}$ (e.g. $V_j = [\frac{j}{8k} - \frac{1}{8k}, \frac{j}{8k} + \frac{1}{8k}]$.

Fix a nonempty open $U_0$.

For each $x \in U_0$, $f(x)$ lies in some $\mathrm{int}(V_j)$, and since $f_m(x) \to f(x)$ there is $n$ with $f_m(x) \in V_j$ for all $m \ge n$.

Hence $U_0 = \bigcup_{n,j} H_{n,j}$, where $H_{n,j} = \{x \in U_0 : f_m(x) \in V_j \ \forall m \ge n\}$ is relatively closed in $U_0$ ($f_m$ continuous, $V_j$ closed).

$U_0$ itself is not complete, so Baire's theorem cannot be applied to $U_0$ directly — restrict to a closed ball instead.

Since $U_0$ is open and nonempty, some closed ball $\overline{B}(x_0, r) \subseteq U_0$ exists, and $\overline{B}(x_0, r)$ is complete.

Now $\overline{B}(x_0, r) = \bigcup_{n,j} (H_{n,j} \cap \overline{B}(x_0, r))$ with each $H_{n,j} \cap \overline{B}(x_0, r)$ closed in this complete space, so by Baire's theorem some $H_{n_0,j_0} \cap \overline{B}(x_0, r)$ has nonempty relative interior in $\overline{B}(x_0, r)$.

Intersecting with the open ball $B(x_0, r)$ gives a nonempty open $W \subseteq H_{n_0,j_0}$ with $W \subseteq U_0$ (an open set meeting $\overline{B}(x_0, r)$ in a relative-interior piece also meets its interior).

For $y \in W$, $f_m(y) \in V_{j_0}$ for all $m \ge n_0$, and $V_{j_0}$ closed gives $f(y) \in V_{j_0}$.

So $f(W) \subseteq V_{j_0}$ and $\mathrm{diam}\,f(W) < 1/(2k)$.

Since $W$ is a neighborhood of each of its points $x$, $\omega_f(x) \le \mathrm{diam}\,f(W) < 1/k$.

Thus $W \subseteq U_0 \cap \{\omega_f < 1/k\} \ne \varnothing$ — $\{\omega_f < 1/k\}$ is dense.

Each $\{\omega_f < 1/k\}$ is open and dense, so by the Baire category theorem $G := \bigcap_{k=1}^{\infty}\{\omega_f < 1/k\}$ is a dense $G_\delta$. By the first step, $G \subseteq \{\omega_f = 0\} =$ the continuity set of $f$, so the continuity set of $f$ contains a dense $G_\delta$. ∎