Exercise 7.6.12

Now, the sums of the lengths of elements of $S$ cannot be greater than $𝜖$ (otherwise $U(f,P_{𝜖})-L(f,P_{𝜖})\ge \mathit{\alpha 𝜖}$). So we can easily create a collection of open intervals $O_1$, whose length is no more than $2𝜖$, whose union is a superset of the union of all elements of $S$. A value greater than $𝜖$ is needed because subintervals are closed intervals, so we need to go a little beyond the endpoint in each case.

Now, all elements of $D^{\alpha }$ appear in some set in $O_1$, except possibly for elements in $P_{𝜖}$. However, there are only finitely many elements of $P_{𝜖}$, so we can just have a second collection of open intervals $O_2$ with total length $𝜖$ covering all elements of $P_{𝜖}$. The union of $O_1$ and $O_2$ give us a collection of open intervals, whose length is no more than $3𝜖$, and whose union contains all elements of $D^{\alpha }$. Since $𝜖$ was arbitrary, this argument can readily be extended to show $D^{\alpha }$ has measure zero.