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?
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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 \(f(x)=\sum_{n=0}^{\infty} a^{n} \cos \left(b^{n} \pi x\right)\) | ? | \(\emptyset\) | \({\mathbf{R}}\) | ? | ? |
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 \(\underline{\int_0^1} f \lneq 0 \lneq \overline{\int_0^1} f\).