Let $f$ be a bounded function on a compact interval $J$, and let $I(c,r)$ denote the open interval centered at $c$ of radius $r>0$. Let $osc(f,c,r)=\sup|f(x)-f(y)|$, where the supremum is taken over all $x,y\in J\cap I(c,r)$, and define the oscillation of $f$ at $c$ by $osc(f,c)=\underset{r\rightarrow 0}{\lim}osc(f,c,r)$. Clearly, $f$ is continuous at $c\in J$ if and only if $osc(f,c)=0$.

(a) For every $\epsilon>0$, the set of points $c$ in J such that $osc(f,c)\geq \epsilon$ is compact.

(b) If the set of discontinuities of $f$ has measure $0$, then $f$ is Riemann integrable.

I've already proved part (a) so this is my attempt at part (b)

My proof attempt:

Proof. Fix $\epsilon>0$. Define \begin{equation*} D_\epsilon=\{c\in J: osc(f,c)>\epsilon\} \end{equation*} Then $D_\epsilon$ is the set of discontinuities of $f$. By hypothesis, $D_\epsilon$ is measure zero. We will prove that $f$ is Riemann integrable by showing that there exists a partition $P$ such that $|U(P,f)-L(P,f)|<\epsilon$, where $U(P,f)$ and $L(P,f)$ are the upper and lower sums of $f$ over partition $P$, respectively.

By Observation 3 of the Exterior Measure in Stein, there exists an open set $\mathcal{O}$ such that $D_\epsilon\subset \mathcal{O}$ and $m(\mathcal{O})\leq \epsilon.$ By theorem 1.3, $\mathcal{O}$ can be written as a disjoint countable union of open intervals $(I_j)_j$. By Part (a), $D_\epsilon$ is compact. Since $(I_j)_j$ covers $D_\epsilon$, we need only a finite sub-cover to cover $D_\epsilon$, say $(I_j)^{k}_{1},$ and denote the collection as $\mathcal{F}$. Then \begin{equation*} m(D_\epsilon)\leq m(\underset{I\in \mathcal{F}}{\bigcup}I)\leq m(\mathcal{O})\leq \epsilon \end{equation*} Hence, we now have a finite cover of $D_\epsilon$ whose "total length" is $\leq \epsilon$.

Since $J$ is closed and $I\in \mathcal{F}$ is open, it follows that $J\setminus \underset{I\in \mathcal{F}}{\bigcup}I$ is closed. Since $J\setminus \underset{I\in \mathcal{F}}{\bigcup}I\subset J$, it is also compact. Any continuous $f$ over a compact interval is uniformly continuous. Therefore, there exists $\delta>0$ such that for all $x,y\in J\setminus \underset{I\in \mathcal{F}}{\bigcup}I$ satisfying \begin{equation*} |x-y|<\delta \implies |f(x)-f(y)|<\epsilon. \end{equation*} Let $P_1$ be a partition of $J$ such that every pair of consecutive points is within $\delta$ of each other, e.g. one way of constructing such a set is to partition $J$ into $\frac{|J|}{2^n}$ length intervals for large enough $n\in \mathbb{N}$. Now define $\hat{P}_1$ to be a partition of $J$ such that \begin{equation*} \hat{P}_1=P_1\setminus \{x\in P_1: x\in \underset{I\in \mathcal{F}}{\bigcup}I\}. \end{equation*} Next, define $P_2$ to be a partition of $J$ such that $P_2$ is the set of the end points of $I_1, I_2,...,I_k$, and $J$. Finally, let $P$ be the common refinement of $P_1$ and $P_2$, i.e. $P=P_1\cup P_2$. We will now show that $P$ is the desired partition of $J$.

Enumerate the points of $P$ as $x_1,x_2,...,x_N$. Denote $\Delta x_i=|x_{i+1}-x_i|$ for $i=1,...,N-1$. Denote as well the interval $[x_{i+1},x_i]$ as $\nabla x_i.$ If two consecutive points in $P$ are also in $P_2$, denote the length instead by $\Delta y_i$. Moreover, for each $\Delta x_i$, let $m_i$ and $M_i$ be the infimum and supremum of $f$ over $\nabla x_i$, respectively. Define \begin{equation} C:=2\cdot \sup\{|f(x)|:x\in J\}<\infty \quad \text{ Since $f$ is bounded.} \end{equation} Finally, observe \begin{align*} |U(P,f)-L(P,f)|&=|\underset{x_i\in P}{\sum} M_i\Delta x_i- \underset{x_i\in P}{\sum} m_i\Delta x_i| \\ &=|\underset{x_i\in P}{\sum} (M_i-m_i)\Delta x_i| \\ &=|\underset{x_i\in P_1}{\sum} (M_i-m_i)\Delta x_i + \underset{x_i\in P_2}{\sum} (M_i-m_i)\Delta y_i| \\ &\leq \underset{x_i\in P_1}{\sum} |(M_i-m_i)|\Delta x_i + \underset{x_i\in P_2}{\sum} |(M_i-m_i)|\Delta y_i \\ &\leq \epsilon|J| + C\epsilon \\ &\text{ By Uniform Continuity of $f$ over $J\setminus \underset{I\in \mathcal{F}}{\bigcup}I$} \\ &=\epsilon(|J|+C) \\ \end{align*} As niether $|J|$ nor $C$ depend on $\epsilon$, this completes our proof.

The definition of the Riemann Integrable I use is from Rudin's PMA. Any corrections of the proof or comments on style are highly appreciated.

I recognize the proof is really long and I appreciate the time of anybody who parses through it.



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