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

FunctionBounded?\(D_f\)\(D'_f\)\(\mathcal{R}\)?\(\mathcal{L}\)?
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 \(f(x)=\sum_{n=0}^{\infty} a^{n} \cos \left(b^{n} \pi x\right)\)?\(\emptyset\)\({\mathbf{R}}\)??

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.

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*}

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
1.
Riemann integrable by the Lebesgue criterion.