We will use
(1.3.1) for the only if part.
Suppose that
\(f\) is Riemann integrable on
\(I\text{.}\) For each
\(k\in \bbN\text{,}\) we will prove that the closed set
\(D_{k}:=\{y: \osc(f)(y)\ge \frac 1k\}\) is a set of content
\(0\text{.}\)
Given any \(\epsilon >0\text{.}\) There exists a partition \(\cP=\{I_{\alpha}\}\) of \(I\) such that
\begin{equation*}
\sum_{\alpha} \osc(f, I_{\alpha})\vert I_{\alpha} \vert < \frac{\epsilon}{2k}.
\end{equation*}
The intervals in \(\cP\) are divided into two subgroups: the subgroup \(\cL_{k}\) consisting those \(I_{\alpha}\) such that \(\osc(f, I_{\alpha})\ge \frac{1}{2k}\text{,}\) and the subgroup \(\cS_{k}\) consisting those \(I_{\alpha}\) such that \(\osc(f, I_{\alpha})< \frac{1}{2k}\text{.}\) Then it follows from
\begin{equation*}
\frac{1}{2k} \sum_{I_{\alpha}\in \cL_{k} } \vert I_{\alpha} \vert \le
\sum_{I_{\alpha}\in \cL_{k} } \osc(f, I_{\alpha})\vert I_{\alpha} \vert < \frac{\epsilon}{2k}
\end{equation*}
that
\begin{equation*}
\sum_{I_{\alpha}\in \cL_{k} } \vert I_{\alpha} \vert < \epsilon.
\end{equation*}
We now claim that
\begin{equation*}
D_{k}\subset \cup_{I_{\alpha}\in \cL_{k} } I_{\alpha}.
\end{equation*}
This will show that \(D_{k}\) is a set of content \(0\text{.}\)
If the claim were not true, there would exist some
\(x \in D_{k}
\setminus \cup_{I_{\alpha}\in \cL_{k} } I_{\alpha}\text{.}\) Thus
\(x \in \cup_{I_{\alpha}\in \cS_{k} } I_{\alpha}\text{.}\) Since the complement of
\(\cup_{I_{\alpha}\in \cL_{k} } I_{\alpha}\) is open, there exists some interval
\(I(x, r)
\subset \cup_{I_{\alpha}\in \cS_{k} } I_{\alpha}\text{.}\) If
\(x\in \text{ interior}(I_{\alpha})\) for some
\(I_{\alpha} \in \cS_{k}\text{,}\) it would force
\(\frac 1k \le \osc(f)(x)\le \osc(f, I_{\alpha}) < \frac{1}{2k}\text{,}\) which would be a contradiction. So
\(x\) can only be on the boundary of one or more
\(I_{\alpha} \in \cS_{k}\text{.}\) We can choose
\(r>0\) small enough such that any
\(y\in I(x, r)\) and
\(x\) will be in one such common interval. Therefore,
\(\vert f(y)-f(x)\vert < \frac{1}{2k}\text{.}\) This would lead to
\(\osc(f)(x)\le \osc(f, I(x, r)) < \frac 1k\text{,}\) contradicting
\(x \in D_{k}\text{.}\)
For the if part, take any \(\epsilon > 0\text{,}\) then choose \(k\) such that \(k^{-1} < \epsilon\text{.}\) The closed set \(D_{k}=\{y: \osc(f)(y)\ge \frac 1k\}\) is a subset of the negligible set \(D\text{,}\) so can be covered by a union \(G\) of a finite number of intervals such that \(|G| < \epsilon\text{.}\) Every point \(x\in [a, b]\setminus G\) has an open interval \(I(x, r)\) such that \(\osc(f, I(x, r)) < k^{-1}\text{.}\) The closed set \([a, b]\setminus G\) can be covered by the union of a finite number of such intervals and one can find a fine enough partition \(\cP=\{I_{i}\}\) of \([a, b]\) such that if any \(I_{i}\) contains a point of \(D_{k}^{c}\text{,}\) then \(\osc(f, I_{i})< k^{-1}\text{;}\) and the union of those \(I_{i}\subset D_{k} \) has their sum of lengths \(< \epsilon\text{.}\) This would give
\begin{equation*}
\sum_{i} \osc(f, I_{i})|I_{i}| \le k^{-1}(b-a) + 2\epsilon \sup_{[a, b]}|f| ,
\end{equation*}
which proves that \(f\) satisfies the Riemann integrability criterion.
