Exercise 6.6

Following the hint, recall that $P=\cup E_n$ where $E_n$ is a set of $2^n$ disjoint close intervals of length $3^{-n}$ obtained by removing the middle thirds of the intervals in $E_{n-1}$. The total length of the intervals of $E_n$ is ${(2∕3)}^n$ and so $\to 0$ as $n\to \infty$. We can replace the closed intervals of $E_n$, $[a_{i,n},b_{i,n}]$ with slightly larger open intervals $(a_{i,n}-\delta ∕2^n,b_{i,n}+\delta ∕2^n)$, with total length ${(2∕3)}^n+\delta$, so that we can cover $P$ with a set of disjoint open intervals with total length as small as possible.

Now proceeding as in the proof of Theorem 6.10, let $M=\sup <!--\; FUNCTION\; APPLICATION\; -->|f(x)|$, and cover $P$ with a collection of open intervals $(u_j,v_j)$ with total length less than $𝜀$. The complement $K$ of the union of the open intervals in $[a,b]$ is compact. Then $f$ is uniformly continuous on $K$ and there exists $\delta >0$ such that $|f(s)-f(t)|<𝜀$ if $s,t\in K$, $|s-t|<\delta$.

Form a partition $Q=\{x_0,\dots ,x_n\}$ of $[a,b]$ such that each $u_j$ and $v_j$ occurs in $Q$, no point of any segment $(u_j,v_j)$ occurs in $Q$, and if $x_{i-1}$ is not one of the $u_j$, then $\mathrm{\Delta }{x}_i<\delta$.

Note that $M_i-m_i\le 2M$ for every $i$, and that $M_i-m_i\le 𝜀$ if $x_{i-1}$ is not one of the $u_j$. Hence

Hence $f\in R$ by Theorem 6.6.