The set of Riemann integrable functions

First suppose that $f$ is continuous at $x \in D\text{.}$ Given $\epsilon > 0\text{,}$ there exists a $\delta > 0$ such that for $y \in B_D(x,\delta)\text{,}$ we have $\babs{f(x)-f(y)} < \epsilon\text{.}$ Therefore, if $y_1,y_2 \in B_D(x,\delta)\text{,}$ then

Take the supremum over $y_1$ and $y_2$ to find

As $o(f,x) \leq o(f,x,\delta) \leq 2\epsilon\text{,}$ and $\epsilon > 0$ was arbitrary, $o(f,x) = 0\text{.}$

On the other hand, suppose $o(f,x) = 0\text{.}$ Given $\epsilon > 0\text{,}$ find a $\delta > 0$ such that $o(f,x,\delta) < \epsilon\text{.}$ If $y \in B_D(x,\delta)\text{,}$ then

Equivalently, we want to show that $G \coloneqq \bigl\{ x \in D : o(f,x) < \epsilon \bigr\}$ is open in the subspace topology. Consider $x \in G\text{.}$ As $\inf_{\delta > 0} o(f,x,\delta) < \epsilon\text{,}$ find a $\delta > 0$ such that

Take any $\xi \in B_D(x,\nicefrac{\delta}{2})\text{.}$ Notice that $B_D(\xi,\nicefrac{\delta}{2}) \subset B_D(x,\delta)\text{.}$ Therefore,

So $o(f,\xi) < \epsilon$ as well. As this is true for all $\xi \in B_D(x,\nicefrac{\delta}{2})\text{,}$ we get that $G$ is open in the subspace topology, and $D \setminus G$ is closed as claimed.

Let $S \subset R$ be the set of discontinuities of $f\text{,}$ that is, $S = \bigl\{ x \in R : o(f,x) > 0 \bigr\}\text{.}$ Suppose $S$ is a measure zero set: $m^*(S) = 0\text{.}$ The trick to proving that $f$ is integrable is to isolate the bad set into a small set of subrectangles of a partition. A partition has finitely many subrectangles, so we need compactness. If $S$ were closed, then it would be compact and we could cover it by finitely many small rectangles. Unfortunately, $S$ itself is not closed in general, but the following set is. Given $\epsilon > 0\text{,}$ define

By Proposition 10.4.2, $S_\epsilon$ is closed, and as it is also a subset of the bounded $R\text{,}$ $S_\epsilon$ is compact. Moreover, $S_\epsilon \subset S$ and $S$ is of measure zero, so $S_\epsilon$ is of measure zero. Via Proposition 10.3.7, there exist finitely many open rectangles $O_1,O_2,\ldots,O_k$ that cover $S_\epsilon$ and $\sum_{j=1}^\infty V(O_j) < \epsilon\text{.}$

The set $T \coloneqq R \setminus ( O_1 \cup \cdots \cup O_k )$ is closed, bounded, and thus compact. As $o(f,x) < \epsilon$ for all $x \in T\text{,}$ for each $x \in T\text{,}$ there is a $\delta > 0$ such that $o(f,x,\delta) < \epsilon\text{,}$ so there exists a small closed rectangle $T_x \subset B(x,\delta)$ with $x$ in the interior of $T_x\text{,}$ such that

The interiors of the rectangles $T_x$ cover $T\text{.}$ As $T$ is compact, finitely many such rectangles $T_1, T_2, \ldots, T_m$ cover $T\text{.}$ Construct a partition $P$ out of the endpoints of the rectangles $T_1,T_2,\ldots,T_m$ and $O_1,O_2,\ldots,O_k$ (ignoring those that are outside the endpoints of $R$). The subrectangles $R_1,R_2,\ldots,R_p$ of $P$ are such that every $R_j$ is contained in $T_\ell$ for some $\ell$ or the closure of $O_\ell$ for some $\ell\text{.}$ Order the rectangles so that $R_1,R_2,\ldots,R_q$ are those that are contained in some $T_\ell\text{,}$ and $R_{q+1},R_{q+2},\ldots,R_{p}$ are the rest. See Figure 10.12. So

The second estimate holds because the $R_j$ that are subsets of $\widebar{O}_\ell$ give a partition of $\widebar{O}_\ell$ and hence their volumes sum to $V(O_\ell)\text{.}$ Let $m_j$ and $M_j$ be the inf and sup of $f$ over $R_j$ as usual. If $R_j \subset T_\ell$ for some $\ell\text{,}$ then $M_j-m_j < \epsilon\text{.}$ Let $B \in \R$ be such that $\babs{f(x)} \leq B$ for all $x \in R\text{,}$ so $M_j-m_j \leq 2B$ over all rectangles. Then

We can make the right-hand side as small as we want, and hence $f$ is integrable.

For the other direction, suppose $f$ is Riemann integrable on $R\text{.}$ Let $S$ be the set of discontinuities of $f$ again. Consider the sequence of sets

Fix a $k \in \N\text{.}$ Given an $\epsilon > 0\text{,}$ find a partition $P$ with subrectangles $R_1,R_2,\ldots,R_p$ such that

Suppose $R_1,R_2,\ldots,R_p$ are ordered so that the interiors of $R_1,R_2,\ldots,R_{q}$ intersect $S_{1/k}\text{,}$ while the interiors of $R_{q+1},R_{q+2},\ldots,R_p$ are disjoint from $S_{1/k}\text{.}$ Let $R_j^\circ$ denote the interior of $R_j\text{.}$ Suppose $j \leq q$ and consider $x \in R_j^\circ \cap S_{1/k}\text{.}$ Let $\delta > 0$ be small enough so that $B(x,\delta) \subset R_j\text{.}$ As $x \in S_{1/k}\text{,}$ we get $o(f,x,\delta) \geq o(f,x) \geq \nicefrac{1}{k}\text{,}$ which, along with $B(x,\delta) \subset R_j\text{,}$ implies $M_j-m_j \geq \nicefrac{1}{k}\text{.}$ Then

In other words, $\sum_{j=1}^q V(R_j) < k \epsilon\text{.}$ Let $G$ be the set of all boundaries of all the subrectangles of $P\text{.}$ The set $G$ is of measure zero (it can be covered by finitely many sets from Example 10.3.5). We find

As $G$ can also be covered by open rectangles of arbitrarily small volume, $S_{1/k}$ must be of measure zero. As

and a countable union of measure zero sets is of measure zero, $S$ is of measure zero.