Proof that the set of discontinuous points of a Riemann integrable function has measure zero This comes from proving Lebesgue's Theorem for Riemann integrability as specified in Abbott's Understanding Analysis, Theorem 7.6.5. The theorem is as follows:

Let $f$ be a bounded function defined on the interval $[a, b]$. Then, $f$ is Riemann-integrable if and only if the set of points where $f$ is not continuous has measure zero.

I've followed along with the exercises and have been able to prove the <= direction. I've been stuck on one core point in the proof of the => direction.
The specific area I am struggling with is as follows. The proof starts:
Let $\epsilon > 0$ be arbitrary, and fix $\alpha > 0$. Because $f$ is Riemann-integrable, there exists a partition $P_\epsilon$ of $[a, b]$ such that $U(f, P_\epsilon) - L(f, P_\epsilon) < \alpha \epsilon$.
Prove that $D^\alpha$ has measure zero. Where $D^\alpha = \{ x \in [a, b] : f \textrm{ is not }\alpha\textrm{-continuous at }x\}$.
Edit: Including definition of $\alpha$-continuous:

Let $f$ be defined on $[a, b]$, and let $\alpha > 0$. The function $f$ is $\alpha$-continuous at $x\in[a, b]$ if there exists $\delta > 0$ such that for all $y, z \in (x - \delta, x + \delta)$ it follows that $| f(y) - f(z) | < \alpha$.

If $f$ is uniformly $\alpha$-continuous on $[a, b]$, there is a single $\delta > 0$ such that $| x- y | < \delta \implies |f(x) - f(y)| < \alpha$. It follows that if $f$ is $\alpha$ continuous at every point on a compact set $K$, then $f$ is uniformly $\alpha$ continuous on $K$.

My attempt so far has taken me down the path of writing:
$$U(f, P_\epsilon) - L(f, P_\epsilon) < \alpha \epsilon$$
$$\sum_k (M_k - m_k)\Delta x_k < \alpha \epsilon$$
Where $M_k$ and $m_k$ are the supremum and infimum, respectively, of $f$ on the interval $[x_{k-1}, x_k]$. Then I attempt to split the intervals into those that contain points in $D^\alpha$ and those that do not. Since the intervals are compact, those intervals containing no points in $D^\alpha$ are uniformly $\alpha$ continuous, which bounds $M_k - m_k < \alpha$ on those intervals (this only holds within some $\delta$, but we can refine the intervals to have smaller length than $\delta$ and I believe this point works out fine).
From here I'm at a bit of a loss, I can't figure out how to take this argument further. I feel that I'm likely missing something relative obvious, so I'd appreciate any nudging in the right direction.
 A: Commentary.
This needs a bit more.  The only "answer" at the moment needs some bit of editing. More importantly there is more to learn here other than the details of a simple proof.

*

*For any function $f:[a,b]\to\mathbb R$ and any $\alpha>0$ we can (as described here) characterize points of continuity and discontinuity this way:
$C^\alpha$ is the set of points $x\in [a,b]$ with the property that
there exists $δ>0$ such that for all $y,z∈(x−δ,x+δ)\cap [a,b]$ it follows that $|f(y)−f(z)|<α$.

Observe:  Each set $C^\alpha$ is open relative to $[a,b]$.  If $C$ is the set of points at which $f$ is continuous then
$$C = \bigcap_{\alpha>0} C^\alpha = \bigcap_{n=1}^\infty C^{\frac1n} \tag{1}.$$
This shows that the set of continuity points of an arbitrary function is the intersection of a sequence of open sets.  We call such sets $\cal G_\delta$ sets.
Each set $D^\alpha=[a,b]\setminus C^\alpha$ is closed relative to $[a,b]$.  If $D$ is the set of points at which $f$ is not continuous then
$$D= \bigcup_{\alpha>0} D^\alpha = \bigcup_{n=1}^\infty D^{\frac1n} \tag{2}.$$
This shows that the set of discontinuity points of an arbitrary function is the union of a sequence of closed sets.  We call such sets $\cal F_\sigma$ sets.
The open sets are called $\cal G$ sets and the closed sets called $\cal F$ (from German and French words respectively).  So these classes of continuity/discontinuity points are pretty simple.  They stand at the very beginning of a huge hierarchy of sets called "Borel sets."


*The theorem alluded to in the problem is this:


Theorem 1. [19th century version]  A bounded function
$f:[a,b]\to\mathbb R$ is Riemann integrable if and only if the set of
points $D$ at which $f$ is discontinuous is the union of a sequence of
closed sets, each of content [measure] zero.
Theorem 2. [20th century version]  A bounded function
$f:[a,b]\to\mathbb R$ is Riemann integrable if and only if the set of
points $D$ at which $f$ is discontinuous is   measure zero.

It is most curious that many students know only Theorem 2 and give credit to Lebesgue for it.  But Theorem 1 was well-known long before Lebesgue and his teacher Borel reworked measure theory, replacing the older theory of content.   Theorem 2 follows immediately and without more than ten seconds of thought from Theorem 1.  Besides Lebesgue was not the only one to point out this completely trivial fact; W.H. Young and Vitali (among others?) mentioned it.  One needs only to remember  that, for bounded closed sets, content and measure are exactly the same.


*Proof of the theorems:  Following the same reasoning and notation as in the comments and the other answer apply Riemann's criterion for a Riemann integrable function.  If $f$ is Riemann integrable and $\epsilon>0$ and $\alpha>0$,  there is a partition $a=x_0<x_1<\dots<x_n=b$ so that  $$\sum_k (M_k-m_k)(x_k-x_{k-1}) < \alpha \epsilon . \tag{3}$$
Let $A$ be the collection of indexes $k$ for which $(x_{k-1},x_k)$ contains a point of $D^\alpha$.  Clearly $(M_k-m_k) \geq \alpha$ if $k\in A$.  The union
$$\bigcup_{k\in A} (x_{k-1},x_k)$$
is an open set with finitely many components that contains all of $D^\alpha$ excepting at most a finite number of points.  The content of that open set is less than $\epsilon$ because
$$ \alpha \sum_{k\in A} (x_k-x_{k-1}) \leq  \sum_{k\in A} (M_k-m_k)(x_k-x_{k-1}) < \alpha \epsilon.$$
It follows that each set $D^\alpha$ has content [measure] less than $\epsilon$ and so must have content [measure] equal to zero.  The set of discontinuity points of $f$ is then the union of a sequence $\{D^{\frac1n}\}$ of closed sets of content zero.
The other direction is slightly more challenging.  But the OP wanted this one and hence this discussion.  The Riemann criterion goes both ways and is the basis for the proof in both of the directions.
A: Answer found thanks to Stephen Donovan in the comments.
Let $\epsilon > 0$ be arbitrary and fix $\alpha > 0$. As $f$ is Riemann-integrable, there exists a partition $P_\epsilon$ such that:
$$U(f, P_\epsilon) - L(f, P_\epsilon) < \alpha \epsilon$$
This can be written as:
$$\sum_{k}  (M_k - m_k) \Delta x_k < \alpha \epsilon$$
Now consider $A = \{ k : (x_{k - 1}, x_k ) \cap D^\alpha \not = \emptyset\}$, the indexes of intervals whose interiors contain points in $D^\alpha$. If we consider summing only over those intervals, we see:
$$\sum_{k\in A} (M_k - m_k)\Delta x_k \leq \sum_{k}  (M_k - m_k) \Delta x_k < \alpha \epsilon$$
As $k \in A$ it follows that $M_k - m_k \geq \alpha$ and therefore:
$$\alpha \sum_{k\in A} \Delta x_k \leq \sum_{k\in A} (M_k - m_k)\Delta x_k  < \alpha \epsilon \implies \sum_{k \in A} \Delta x_k < \epsilon$$
This shows the set $\cup_{k\in A}(x_{k-1},x_k)$ has measure zero.
Recalling that we left off the interval endpoints in our construction of $A$, we consider $B = D^\alpha \cap \{x_0, x_1, ..., x_n\}$, the set of endpoints in $D^\alpha$. This set is finite and therefore has measure zero. The union of measure zero sets is also measure zero and we note:
$$D^\alpha \subseteq  \left(\cup_{k\in A}(x_{k-1},x_k)\right) \cup B$$
Therefore $D^\alpha$ has measure zero.
