Proof
(related to Proposition: Monotonic Real Functions on Closed Intervals are Riemann-Integrable)
- Let \([a,b]\) be a closed real interval.
- Let \(f:[a,b]\mapsto\mathbb R\) be a monotonic real function.
- We will show by construction that $f$ is Riemann-integrable.
- By setting $$t_k:=a+k\frac{b-a}n,\quad k=0,1,\ldots, n,$$ we get an equidistant partition of the closed interval $[a,b]$ with $a=t_0 < t_1 < \ldots < t_n=b.$
- By setting $$\begin{array}{rcl}\phi(x)&:=&f(t_{k-1}),\quad t_{k-1}\le x < t_k\\ \psi(x)&:=&f(t_{k}),\quad t_{k-1}\le x < t_k\\\phi(b)&:=&f(b)\\\psi(b)&:=&f(b)\end{array}$$ for \(k=1,2,\ldots n\) we achieve that $\phi(x)\le f(x)\le \psi(x)$ for all $x\in[a,b]$.
- Furthermore, we have by construction
$$\begin{array}{rcl}\int_a^b\psi(x)dx-\int_a^b\phi(x)dx&=&\sum_{k=1}^n f(t_k)(t_k-t_{k-1})- \sum_{k=1}^n f(t_{k-1})(t_k-t_{k-1})\\&=&\frac{b-a}n\left(\sum_{k=1}^n f(t_{k})-\sum_{k=1}^n f(t_{k-1})\right)\\&=&\frac{b-a}n(f(t_n))-f(t_0)).\quad\quad ( * )\end{array}$$
- Because of the existence of arbitrarily small positive rational numbers, $( * )\le\epsilon,$ for sufficiently big natural number $n$.
- This means that $f$ is Riemann-integrable.
∎
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
