I'm having some trouble with a certain step for the proof of the following theorem
Suppose $f\in\mathcal{R}(\alpha)$ on $[a,b]$, $m\le f\le M$, $\phi$ is continuous on $[m,M]$, and $h(x)=\phi(f(x))$ on $[a,b]$. Then $h\in\mathcal{R}(\alpha)$
The proof is as follows
Choose $\epsilon>0$. Since $\phi$ is uniformly continuous on $[m,M]$, there exists $\delta>0$ s.t. $\delta<\epsilon$ and $|\phi(s)-\phi(t)|<\epsilon$ if $|s-t|\le\delta$ and $s,t\in[m,M]$.
Since $f\in\mathcal{R}(\alpha)$, there is a partition $P=\{x_0,\dots,x_n\}$ of $[a,b]$ s.t.$$U(P,f,\alpha)-L(P,f,\alpha)<\delta^2$$Let $M_i,m_i$ be the suprenum and infinium of the function on a given interval, and let $M_i^*,m_i^*$ be the analogous numbers for $h$. Divide the numbers $1,\dots,n$ into two cases, $i\in A$ if $M_i-m_i<\delta$, $i\in B$ if $M_i-m_i\ge\delta$.
For $i\in A$, our choice of $\delta$ shows that $M_i^*-m_i^*\le\epsilon$
What I'm not sure about is why the final inequality is less than or equal. My understanding is that the value of $M_i^*-m_i^*$ taken with $|M_i-m_i|<\delta$ would be less than or equal to that taken with $|M_i-m_i|\le\delta$. However the text seems to suggest that it would instead be even greater. This doesn't make sense to me since we are discarding a potential maximum/minimum at the boundary. What have I missed?
