TITLE: Theorem 6.11 of Rudin's Principles of Mathematical Analysis
QUESTION [5 upvotes]: 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 $\left[a,b\right]$.

REPLY [3 votes]: 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 $\delta''=\min\{\delta,\frac12\varepsilon\}$.

REPLY [3 votes]: 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
$$(\delta'\le\delta)\Rightarrow \left(\left\{(s,t)\quad|\quad \left|s-t\right| \leq \delta'\right\}\subset \left\{(s,t)\quad|\quad \left|s-t\right| \leq \delta\right\}\right)$$