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Proof

(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.
    ∎

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983