(related to Proposition: Continuous Real Functions on Closed Intervals are Riemann-Integrable)

- Let \([a,b]\) be a closed real interval.
- Let \(f:[a,b]\mapsto\mathbb R\) be continuous.
- We want to show that $f$ is Riemann-integrable.
- According to the approximability of continuous real functions on closed intervals by step functions, there exist for a fixed \(\epsilon > 0\) step functions $\phi,\psi$ with $$\phi(x)\le f(x)\le \psi(x),\quad\text{ and }\quad |\psi(x)-\phi(x)|\le \frac\epsilon{b-a}$$ for all $x\in[a,b].$
- From the linearity and monotony of the Riemann integral for step functions, it follows that $$\int_a^b\psi(x)dx-\int_a^b\phi(x)dx=\int_a^b(\psi(x)-\phi(x))dx\le \int_a^b\frac\epsilon{b-a}dx=\epsilon.$$
- Thus a sufficient condition for Riemann-integrable functions is fulfilled.
- Thus, $f$ is Riemann-integrable.∎

**Forster Otto**: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983