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Exercise 75Calculus › Single-variable calculus › Integration › Riemann integralL1Added 5 Nov 2026

Composition with a continuous function

#composition#continuous#integrability preservation

Statement

Let f:[a,b]→Rf:[a,b]\to\mathbb{R} be Riemann integrable and Φ:R→R\Phi:\mathbb{R}\to\mathbb{R} continuous. Show that Φ∘f\Phi\circ f is Riemann integrable.

Hint

Approximate ff by step functions and use uniform continuity of Φ\Phi on a bounded interval.

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Solution

Setup. Since ff is bounded, f([a,b])⊆[−M,M]f([a,b])\subseteq[-M,M] for some M>0M>0; Φ\Phi is uniformly continuous on [−M,M][-M,M].

Approximation. For ε>0\varepsilon>0, choose step functions φ≤f≤ψ\varphi\le f\le\psi with ∫(ψ−φ)<δ\int(\psi-\varphi)<\delta, where δ\delta is chosen so that ∣Φ(s)−Φ(t)∣<ε′|\Phi(s)-\Phi(t)|<\varepsilon' whenever ∣s−t∣<δ|s-t|<\delta on [−M,M][-M,M].

Step functions. Then Φ∘φ≤Φ∘f≤Φ∘ψ\Phi\circ\varphi\le\Phi\circ f\le\Phi\circ\psi (where Φ\Phi is increasing — WLOG, working with Φ\Phi is more subtle; but for uniformly continuous Φ\Phi, one uses Φ∘φ\Phi\circ\varphi and Φ∘ψ\Phi\circ\psi plus bounds). The integral difference is controlled by ε′\varepsilon' and the length of [a,b][a,b].

Conclusion. Φ∘f\Phi\circ f is Riemann integrable.

Remark. The result is standard: composition of Riemann integrable with continuous preserves integrability.