Baire Class 1 Functions

For $n \ge 2$ let $f_n(x) = \sum \frac{1}{q}\,\mathrm{tri}(n^3(x - p/q))$ , summed over reduced fractions with $1 \le q \le n$ , $0 \le p \le q$ , $\gcd(p,q) = 1$ , where $\mathrm{tri}(t) = \max(0, 1 - |t|)$ . Prove each $f_n$ is continuous on $[0, 1]$ and converges pointwise to Thomae's function $T$ ( $1/q$ at reduced $p/q$ , $0$ at irrationals, $T(0) = T(1) = 1$ ): $f_n(x) \to T(x)$ for every $x$ .

f_n(x) = \sum_{\substack{1 \le q \le n \\ 0 \le p \le q \\ \gcd(p,q)=1}} \frac{1}{q}\,\mathrm{tri}\!\left(n^3\left(x - \tfrac{p}{q}\right)\right)

\text{tent width } \frac{2}{n^3} < \frac{1}{n(n-1)} \le \text{ spacing of reduced fractions} \;\Longrightarrow\; \text{disjoint tents}

N = 12

: 47 pairwise disjoint tents.

f_{12}(1/2) = 0.5

,

f_{12}(\sqrt{2}/2) = 0

,

f_{12}(1/3) = 1/3

— the probes already match Thomae exactly. The proof below is the machine behind this pointwise convergence.

Show proof

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. $f_n(x) \to 0 = T(x)$ .

Prove Thomae's function $T$ is discontinuous at every rational and continuous at every irrational. That is, the continuity set of $T$ is exactly the irrationals — a dense $G_\delta$ .

T(x) = \begin{cases} 1/q & x = p/q \text{ in lowest terms} \\ 0 & x \text{ irrational} \end{cases}

\{x : T \text{ continuous at } x\} = \mathbb{R} \setminus \mathbb{Q} = \bigcap_{q=1}^{\infty} \bigcup_{p \in \mathbb{Z}} \left(\mathbb{R} \setminus \left\{\tfrac{p}{q}\right\}\right)

N = 5

:

f_5

is continuous, built from 11 tents. The limit

T

spikes (

0.5, 1/3

) at the rationals

1/2, 1/3

but sinks to

0

at irrationals — the proof below is why the split happens.

Show proof

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 is open, and $\mathbb{R} \setminus \mathbb{Q} = \bigcap_{q \ge 1}\bigcup_{p}(\mathbb{R} \setminus \{p/q\})$ is a countable intersection of open sets — a $G_\delta$ . The irrationals are dense, so it is a dense $G_\delta$ .

Let $f: \mathbb{R} \to \mathbb{R}$ be Baire class 1 (a pointwise limit of continuous functions). For the oscillation $\omega_f(x) = \inf\{\mathrm{diam}\,f(U) : U \text{ a neighborhood of } x\}$ , prove each set $\{x : \omega_f(x) < 1/k\}$ is open and dense, and conclude via the Baire category theorem that the continuity set of $f$ contains a dense $G_\delta$ .

\omega_f(x) = \inf_{U \ni x} \mathrm{diam}\,f(U), \qquad \omega_f(x) = 0 \iff f \text{ continuous at } x

\{x : f \text{ continuous at } x\} \supseteq \bigcap_{k=1}^{\infty} \left\{x : \omega_f(x) < \tfrac{1}{k}\right\}, \text{ a dense } G_\delta

N = 5

: Thomae's function is Baire-1 yet discontinuous — but its continuity points (the irrationals) still form a dense

G_\delta

. The proof below shows this holds for every Baire-1 function.

Show proof

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).
$U_0$ , open in the complete space $\mathbb{R}$ , is itself a Baire space (the usual nested-closed-ball proof of Baire's theorem works verbatim inside $U_0$ , using completeness of $\mathbb{R}$ ). Hence some $H_{n_0,j_0}$ has nonempty interior in $U_0$ : there is a nonempty open $W \subseteq U_0$ with $W \subseteq H_{n_0,j_0}$ . 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 $G_\delta$ . ∎

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