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In the proof of Theorem 6.11, $\varphi$ is uniformly continuous and hence for arbitrary $\epsilon > 0$ we can pick $\delta > 0$ s.t. $\left|s-t\right| \leq \delta$ implies $\left|\varphi\left(s\right)-\varphi\left(t\right)\right|<\epsilon$. However, I do not understand why he claims that $\delta < \epsilon$. Help?

Statement of the theorem: Suppose $f\colon\left[a,b\right]\rightarrow \mathbb{R}$ is Riemann integrable on $\left[a,b\right]$, $m\leq f \leq M$ (for $m,M\in\mathbb{R}$), $\varphi\colon \left[m,M\right]\rightarrow \mathbb{R}$ is continuous on $\left[m,M\right]$. Let $h\equiv\varphi\circ f$. Then $h$ is Riemann integrable on $\left[a,b\right]$.

tediso
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  • Added the statement of the theorem. Since $\varphi$ is continuous on a compact set, $\varphi$ is uniformly continuous. – tediso Mar 23 '14 at 17:49
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    I don't have the text to hand, but if some $d\gt \epsilon$ will do the trick, then $\delta \lt \epsilon$ will do it too because then $\delta \lt d$ – Mark Bennet Mar 23 '14 at 17:50

2 Answers2

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If the assertion holds for some $\delta$ it also holds for $\delta'=\min\{\delta,\varepsilon\}$. Since $\delta'\leqslant\varepsilon$, this entails that one can choose $\delta\leqslant\varepsilon$ without loss of generality.

If the inequality is $\delta\lt\varepsilon$, consider $\delta''=\min\{\delta,\frac12\varepsilon\}$.

Did
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If we can pick $\delta>0$ s.t. $\left|s-t\right| \leq \delta$ implies the inequality $$\left|\varphi\left(s\right)-\varphi\left(t\right)\right|<\epsilon$$ then with $\delta'=\min(\delta,\frac\epsilon2)<\epsilon$ : $\left|s-t\right| \leq \delta'$ implies also this inequality since $$(\delta'\le\delta)\Rightarrow \left(\left\{(s,t)\quad|\quad \left|s-t\right| \leq \delta'\right\}\subset \left\{(s,t)\quad|\quad \left|s-t\right| \leq \delta\right\}\right)$$