Counterexamples

Let all of the following integrals to be over a compact interval $[a, b]$ with $0 \leq a < b$ .

Questions to ask:

Is $f$ bounded?
What is the discontinuity set $D_f$ ?
What is the non-differentiability locus $D'_f$ ?
Is $f \in \mathcal{R}$ , i.e. Riemann integrable?
If $f \in \mathcal{L}$ , i.e. Lebesgue integrable?
- If so, what is $\int_{\mathbf{R}}f$ ?

Note that $D_f \in F_\sigma$ !

Weierstrass Function

Function	Bounded?	$D_f$	$D'_f$	$\mathcal{R}$ ?	$\mathcal{L}$ ?
Dirichlet $\chi_{\mathbf{Q}}(x)$	✅, ${\left\lvert {f} \right\rvert} \leq 1$	${\mathbf{R}}$	${\mathbf{R}}$	❌	✅, $\int f=0$
Dirichlet 2 $x\chi_{\mathbf{Q}}(x)$	❌	${\mathbf{R}}\setminus\left\{{0}\right\}$	${\mathbf{R}}$	❌, $U(f) > 1/4 > 0 = L(f)$	?
Dirichlet 3 $x^2\chi_{\mathbf{Q}}(x)$	❌	${\mathbf{R}}\setminus\left\{{0}\right\}$	${\mathbf{R}}\setminus\left\{{0}\right\}$	❌	?
Dirichlet 4 $f(x) = x\qty{ \chi_{\mathbf{Q}}(x) - \chi_{{\mathbf{Q}}^c}(x)}$	❌	${\mathbf{R}}\setminus\left\{{0}\right\}$	${\mathbf{R}}$	❌	?
Thomae $(x={p\over q} \mapsto {1\over q})\chi_{{\mathbf{Q}}}(x)$	✅	${\mathbf{Q}}$	${\mathbf{R}}$	✅, $\int f = 0$ ¹	✅
Weierstrass $f(x)=\sum_{n=0}^{\infty} a^{n} \cos \left(b^{n} \pi x\right)$	?	$\emptyset$	${\mathbf{R}}$	?	?

remark:

Full definition of the Weierstrass function:

$\begin{align*} f(x)=\sum_{n=0}^{\infty} a^{n} \cos \left(b^{n} \pi x\right) && a \in (0, 1), b \in {\mathbf{Z}}_{\geq 0}, ab > 1 + {3\pi \over 2} .\end{align*}$

Note that this series converges uniformly since it’s bounded above by $\sum {\left\lvert {a^n} \right\rvert}$ , which is geometric.

remark:

Full definition of the Thomae function:

$\begin{align*} f ( x ) = \begin{cases} \frac 1 q, & x = \frac p q \in {\mathbf{Q}},~(p,q) = 1 \\ 0, & \text{else} \end{cases} \end{align*}$

proof (of non-integrability of Dirichlet 4):

Restrict attention to ${\left[ {\frac 1 2, 1} \right]}$ $\begin{align*} \overline{\int_0^1} f &= \inf \left\{{ \sum \sup f(x) (x_i - x_{i-1}) }\right\} \\ \sup f(x) = x_i \implies \sum \sup f(x) (x_i - x_{i-1}) &= \sum x_i (x_i - x_{i-1}) \\ &> \sum \frac 1 2 (x_i - x_{i-1}) \\ &= \frac 1 2 \left(\frac 1 2\right) = \frac 1 4 \\ \implies \overline{\int_0^1} f &\geq \frac 1 4 \end{align*}$ and $\begin{align*} \underline{\int_0^1} f &= \sup \left\{{ \sum \inf f(x) (x_i - x_{i-1})}\right\} \\ \inf f(x)= -x_i \implies \sum \inf f(x) (x_i - x_{i-1}) &= \sum -x_i (x_i - x_{i-1}) \\ &< -\sum \frac 1 2 (x_i - x_{i-1}) \\ &= -\frac 1 2 \left( \frac 1 2 \right) = -\frac 1 4 \\ \implies \underline{\int_0^1} f &\leq -\frac 1 4 \end{align*}$ So we have $\underline{\int_0^1} f \lneq 0 \lneq \overline{\int_0^1} f$ .

Footnotes

Riemann integrable by the Lebesgue criterion.