Here is Theorem 7.48 (Lebesgue's Criterion for Riemann Integrability) in the book Mathematical Analysis - A Modern Approach to Advanced Calculus by Tom M. Apostol, 2nd edition:
Let $f$ be defined and bounded on $[a, b]$ and let $D$ denote the set of discontinuities of $f$ in $[a, b]$. Then $f \in R$ on $[a, b]$ if, and only if, $D$ has measure zero.
By $f \in R$ on $[a, b]$, we mean the Riemann integral $\int_a^b f(x) dx$ exists.
And, here is Apostol's proof:
(Necessity) First we assume that $D$ does not have measure zero and show that $f$ is not integrable. We can write $D$ as the countable union of sets$$D = \bigcup_{r=1}^\infty D_r,$$where$$D_r = \left\{ x : \omega_f(x) \geq \frac1r \right\}.$$If $x \in D$, then $\omega_f(x) > 0$ so $D$ is the union of the sets $D_r$ for $r = 1, 2, 3, \ldots$.
Here$$\omega_f(x) = \lim_{h \to 0+} \Omega_f \big( (x-h, x+h) \cap [a, b] \big) = \inf_{h > 0} \Omega_f \big( (x-h, x+h) \cap [a, b] \big),$$and$$\Omega_f (S) = \sup \{ f(x) - f(y) : x \in S, y \in S \}$$for any subset $S$ of $[a, b]$.
Now if $D$ does not have measure zero, then some set $D_r$ does not (by Theorem 7.46). Therefore, there is some $\epsilon > 0$ such that every countable collection of open intervals covering $D_r$ has a sum of lengths $\geq \epsilon$. For any partition $P$ of $[a, b]$ we have$$U( P, f) - L(P, f) = \sum_{k=1}^n \left[ M_k(f) - m_k(f) \right] \Delta x_k = S_1 + S_2 \geq S_1, $$where $S_1$ contains those terms coming from subintervals containing points of $D$ in their interior, and $S_2$ contains the remaining terms. The open intervals from $S_1$ cover $D_r$ except possibly for a finite subset of $D_r$, which has measure $0$, so the sum of their lengths is at least $\epsilon$. [This is the first sentence which I'm unable to figure out the facts mentioned wherein.] Moreover, in these intervals we have$$M_k(f) - m_k(f) \geq \frac1r \mbox{ and hence } S_1 \geq \frac{\epsilon}{r}.$$This means that$$U(P, f) - L(P, f) \geq \frac{\epsilon}{r},$$for every partition $P$, so Riemann's condition cannot be satisfied. Therefore $f$ is not integrable. In other words, if $f \in R$, then $D$ has measure zero.
Here is the Definition of sets of measure zero:
A set $S$ of real numbers is said to have measure zero if, for every $\epsilon > 0$, there is a countable covering of $S$ by open intervals, the sume of whose lengths is less than $\epsilon$. If the intervals are denoted by $\left( a_k, b_k \right)$, the definition requires that$$S \subseteq \bigcup_k \left( a_k, b_k \right) \qquad \mbox{ and } \qquad \sum_k \left( b_k - a_k \right) < \epsilon. $$If the collection of intervals is finite, the index $k$ in (3) runs over a finite set. If the collection is countably infinite, then $k$ goes from $1$ to $\infty$, and the sum of the lengths is the sum of an infinite series given by$$\sum_{k=1}^\infty \left( b_k - a_k \right) = \lim_{N \to \infty} \sum_{k=1}^N \left( b_k - a_k \right).$$
And, here is Theorem 7.44:
Let $\mathbf{F}$ be a countable collection of sets in $\mathbb{R}$, say$$\mathbf{F} = \left\{ F_1, F_2, F_3, \ldots \right\},$$each of which has measure zero. Then their union$$S = \bigcup_{k=1}^\infty F_k$$also has a measure zero.
PS:
My question is as follows:
Let $P$ be any partition of $[a, b]$. If $D_r$ is a subset of $[a, b]$ such that $D_r$ is not of measure zero and if$$U(P, f) - L(P, f) = \sum_{k=1}^n \left[ M_k(f) - m_k(f) \right] = S_1 + S_2,$$where $S_1$ consists of terms coming from those subintervals containing points of $D$ in their interior and $S_2$ consists of the remaining terms, then how to show that the open intervals from $S_1$ cover $D_r$, except possibly for a finite subset of $D_r$? Here of course$$D_r = \left\{ x : \omega_f(x) \geq \frac1r \right\}$$and $D_r$ is a subset of $D$ such that $D_r$ is not of measure zero.
This question I even posed between square brackets right inside the proof.