Exercise 76Calculus › Single-variable calculus › Integration › Riemann integralL1Added 5 Nov 2026

A monotone function on a fine grid

#monotone functions#Riemann integrability#step approximation

Statement

Let $f:[a,b]\to\mathbb{R}$ be monotone (say nondecreasing). Show that for every $\varepsilon>0$ there exist step functions $\varphi\le f\le\psi$ with $\int(\psi-\varphi)<\varepsilon$ .

Hint

Take a subdivision with small mesh $\eta$ ; define $\varphi$ and $\psi$ constant on each interval with values $f(x_{k-1})$ and $f(x_k)$ .

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Solution

Subdivision. Since $f$ is bounded (being monotone on a compact interval), let $M=f(b)-f(a)$ . Choose a subdivision $\sigma=\{a=x_0<\cdots<x_n=b\}$ with $\max_k\Delta_k<\varepsilon/M$ .

Step functions. Define $\varphi(x)=f(x_{k-1})$ and $\psi(x)=f(x_k)$ on $]x_{k-1},x_k[$ , with arbitrary values at the points $x_k$ . Monotonicity ensures $\varphi\le f\le\psi$ .

Integral difference.
$\int_a^b(\psi-\varphi)=\sum_k(f(x_k)-f(x_{k-1}))\Delta_k<\eta\sum_k(f(x_k)-f(x_{k-1}))=\eta\cdot M<\varepsilon.$

Conclusion. $f$ is Riemann integrable (by the squeeze criterion).

A monotone function on a fine grid

Statement

Solution

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