Exercise 7.4.4

I claim that there exists some non-empty interval $I=[c,d]\subseteq [a,b]$ and some $𝜖>0$ such that for all non-empty subintervals $J\subseteq I$, there exists $x\in J$ with $f(x)\ge 𝜖$. Note that once we prove this claim, we can readily show ${\int }_a^{b}f\ge {\int \; <!--\; nolimits\; -->}_a^{b}g=𝜖(d-c)$, where

We’ll prove this claim by contradiction. Suppose that for all non-empty intervals $I\subseteq [a,b]$ and for all $𝜖>0$, there exists a non-empty subinterval of $J\subseteq I$ where $f(x)<𝜖\forall <!--\; FUNCTION\; APPLICATION\; -->x\in J$.

Now let $I_0=[a,b]$, and for $n>0$, let ${𝜖}_n=1∕n$ and $I_n\subseteq I_{n-1}$ satisfying $f(x)<{𝜖}_n\forall <!--\; FUNCTION\; APPLICATION\; -->x\in I_n$. Now by the Nested Interval Property, $\exists <!--\; FUNCTION\; APPLICATION\; -->s\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{I}_n$. Now, $s$ satisfies $f(s)<1∕n$ for all $n$, implying $f(s)\le 0$ — a contradiction since $f(x)>0$ for all $x\in [a,b]$.