=== Necessary Condition === Suppose that $f$ is Darboux integrable. We need to prove that the set of all discontinuities of $f$ has measure $0$. Let, for some positive real number $s$: :$A_s = \set {x \in \closedint a b: \map {\omega_f} x > s}$ where :$\map {\omega_f} x = \displaystyle \inf \set {\map {\omega_f} I: I \in N_x}$ where :$\map {\omega_f} I = \displaystyle \sup \set {\size {\map f y - \map f z}: y, z \in I \cap \closedint a b}$ :$N_x$ is the set of open subset neighborhoods of $x$ $\map {\omega_f} x$ and $\map {\omega_f} I$ are called the oscillations of $f$ at a point and on a set respectively. By Real Function is Continuous at Point iff Oscillation is Zero we know that :$A_0 = \set {x \in \closedint a b: \map {\omega_f} x > 0}$ is the set of all discontinuities of $f$. We need to prove that $\map \mu {A_0} = 0$. If $A_0$ is empty, it follows by Measure of Empty Set is Zero that $\map \mu {A_0} = 0$. This proves what we wanted for this case. Now assume that $A_0$ is non-empty. This means that $\closedint a b$ contains a point $x$ with $\map {\omega_f} x > 0$. This means, in turn, that there is a strictly positive real number $s$ for which $A_s$ is non-empty. Choose such a number $s$. Let $\epsilon > 0$ be given. By Condition for Darboux Integrability, there exists a subdivision $P = \set {x_0, x_1, x_2, \ldots, x_n}$ of $\closedint a b$ such that: :$\map U P – \map L p < \epsilon$ where: :$\map U P = \displaystyle \sum_{i \mathop = 1}^n \map {M_i} {x_i − x_{i − 1} }$ :$\map L P = \displaystyle \sum_{i \mathop = 1}^n \map {m_i} {x_i − x_{i − 1} }$ where: :$M_i = \displaystyle \sup \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} }$ :$m_i = \displaystyle \inf \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} }$ Let $J$ be the subset of $\set {1, 2, \ldots, n}$ that satisfies:: :$i \in J$ {{iff}} $\openint {x_{i - 1} } {x_i}$ contains a point in $A_s$. First, we intend to prove that $\displaystyle \sum_{i \mathop \in J} \paren {x_i − x_{i - 1} } < \dfrac \epsilon s$. Let $x^{'}_i$ signify a point in $\openint {x_{i - 1} } {x_i} \cap A_s$ whenever $i \in J$. We have: {{begin-eqn}} {{eqn | l = \epsilon       | o = >       | r = \map U P – \map L P }} {{eqn | r = \sum_{i \mathop = 1}^n M_i \paren {x_i − x_{i − 1} } - \sum_{i \mathop = 1}^n m_i \paren {x_i − x_{i − 1} }       | c = Definitions of $\map U P$ and $\map L P$ }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {M_i - m_i} \paren {x_i − x_{i − 1} } }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {\sup \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} } - \inf \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} } } \paren {x_i − x_{i − 1} }       | c = Definitions of $M_i$ and $m_i$ }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {\sup \set {\size {\map f x - \map f y}: x, y \in \closedint {x_{i - 1} } {x_i} } } \paren {x_i − x_{i − 1} }       | c = Supremum of Absolute Value of Difference equals Difference between Supremum and Infimum }} {{eqn | r = \sum_{i \mathop = 1}^n \map {\omega_f} {\closedint {x_{i - 1} } {x_i} } \paren {x_i − x_{i − 1} }       | c = Definition of $\omega_f$ for Oscillation on Real Subset }} {{eqn | o = \ge       | r = \sum_{i \mathop \in J} \map {\omega_f} {\closedint {x_{i - 1} } {x_i} } \paren {x_i − x_{i − 1} }       | c = as $J$ is a subset of $\set {1, 2, \ldots, n}$ }} {{eqn | o = \ge       | r = \sum_{i \mathop \in J} \map {\omega_f} {\closedint {x_{i - 1} } {x_i} } \paren {x_i − x_{i − 1} }       | c = as $\map {\omega_f} {\openint {x_{i - 1} } {x_i} } \le \map {\omega_f} {\closedint {x_{i - 1} } {x_i} }$ by Supremum of Set of Real Numbers is at least Supremum of Subset }} {{eqn | o = \ge       | r = \sum_{i \mathop \in J} \inf \set {\map {\omega_f} I: I \in N_{x^{'}_i} } \paren {x_i − x_{i − 1} }       | c = as $\set {\map {\omega_f} I: I \in N_{x^{'}_i} }$ contains (as an element) $\map {\omega_f} {\openint {x_{i - 1} } {x_i} }$ }} {{eqn | r = \sum_{i \mathop \in J} \map {\omega_f} {x^{'}_i} \paren {x_i − x_{i − 1} }       | c = Definition of $\omega_f$ for Oscillation at Point }} {{eqn | o = >       | r = \sum_{i \mathop \in J} s \paren {x_i − x_{i − 1} }       | c = as $\map {\omega_f} {x^{'}_i} > s$ since $x^{'}_i \in A_s$ }} {{eqn | r = s \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} }       | c =  }} {{end-eqn}} Therefore: :$\displaystyle \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} } < \dfrac \epsilon s$ Next, we intend to prove that $\map \mu {A_s} = 0$. We have: {{begin-eqn}} {{eqn | l = A_s       | o = \subseteq       | r = \paren {\bigcup_{i \mathop \in J} \closedint {x_{i - 1} } {x_i} } \bigcup \bigcup_{i \mathop = 0}^n \set {x_i} }} {{eqn | o = \subset       | r = \paren {\bigcup_{i \mathop \in J} \closedint {x_{i - 1} } {x_i} } \bigcup \bigcup_{i \mathop = 0}^n \openint {x_i - \frac \epsilon {s \paren {n + 1} } } {x_i + \frac \epsilon {s \paren {n + 1} } }       | c = as $\set {x_i} \subset \openint {x_i - \dfrac \epsilon {s \paren {n + 1} } } {x_i + \dfrac \epsilon {s \paren {n + 1} } }$ }} {{end-eqn}} We can now define an upper bound for $\map \mu {A_s}$: {{begin-eqn}} {{eqn | l = \map \mu {A_s}       | o = \le       | r = \map \mu {\paren {\bigcup_{i \mathop \in J} \openint {x_{i - 1} } {x_i} } \bigcup \bigcup_{i \mathop = 0}^n \openint {x_i - \frac \epsilon {s \paren {n + 1} } } {x_i + \frac \epsilon {s \paren {n + 1} } } }       | c = Measure is Monotone }} {{eqn | o = \le       | r = \map \mu {\bigcup_{i \mathop \in J} \paren {x_{i - 1} } {x_i} } + \map \mu {\bigcup_{i \mathop = 0}^n \openint {x_i - \frac \epsilon {s \paren {n + 1} } } {x_i + \frac \epsilon {s \paren {n + 1} } } }       | c = Measure is Subadditive }} {{eqn | o = \le       | r = \sum_{i \mathop \in J} \map \mu {\openint {x_{i - 1} } {x_i} } + \sum_{i \mathop = 0}^n \map \mu {\openint {x_i - \frac \epsilon {s \paren {n + 1} } } {x_i + \frac \epsilon {s \paren {n + 1} } } }       | c = Measure is Subadditive }} {{eqn | r = \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} } + \sum_{i \mathop = 0}^n \frac {2 \epsilon} {s \paren {n + 1} }       | c = Measure of Interval is Length }} {{eqn | o = <       | r = \frac \epsilon s + \sum_{i \mathop = 0}^n \frac {2 \epsilon} {s \paren {n + 1} }       | c = as $\displaystyle \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} } < \frac \epsilon s$ }} {{eqn | r = \frac \epsilon s + \frac {2 \epsilon} s }} {{eqn | r = \frac {3 \epsilon} s }} {{end-eqn}} Therefore: :$\map \mu {A_s} < \dfrac {3 \epsilon} s$ Since $\epsilon$ is an arbitrary number greater than zero: :$\map \mu {A_s} = 0$ Next, we intend to prove that $\map \mu {A_0} = 0$. Consider the set sequence $\sequence {A_{1 / n} }_{n \mathop \in \N_{>0} }$. This sequence is increasing since $A_{1 / n} \subseteq A_{1 / \paren {n + 1} }$ for every $n \in \N_{>0}$ by the definition of $A_s$. By the definition of limit of increasing sequence of sets, the limit of $\sequence {A_{1 / n} }_{n \mathop \in \N_{>0}}$ equals $\displaystyle \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$. Every set $A_{1 / n}$ contains only points of discontinuity of $f$. Therefore, $\displaystyle \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$ is a subset of $A_0$, the set of all discontinuities of $f$. Also, every point of discontinuity of $f$ belongs to some set $A_{1 / n}$. Therefore, $A_0$ is a subset of $\displaystyle \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$. Accordingly: :$\displaystyle \bigcup_{n \mathop \in \N_{>0} } A_{1 / n} = A_0$ Furthermore, since $\displaystyle \bigcup_{n \mathop \in \N_{>0}} A_{1 / n}$ equals the limit of $\sequence {A_{1 / n} }_{n \mathop \in \N_{>0} }$, $A_0$ equals the limit of $\sequence {A_{1 / n} }_{n \mathop \in \N_{>0} }$. In other words, $A_{1 / n} \uparrow A_0$ as defined in Limit of Increasing Sequence of Sets. We have by property (3) of Characterization of Measures:  {{begin-eqn}} {{eqn | l = \map \mu {A_0}       | r = \lim_{n \mathop \to \infty} \map \mu {A_{1 / n} }       | c = as $A_{1 / n} \uparrow A_0$ }} {{eqn | r = \lim_{n \mathop \to \infty} 0       | c = as $\map \mu {A_{1 / n} } = 0$ for every $n$ in $\N_{>0}$ since $\map \mu {A_s} = 0$ for every $s > 0$ }} {{eqn | r = 0 }} {{end-eqn}} In other words, $A_0$ is a $\mu$-null set. Therefore, the set of all discontinuities of $f$ is a $\mu$-null set as $A_0$ is the set of all discontinuities of $f$. {{qed|lemma}} === Sufficient Condition === {{ProofWanted}} {{qed}} {{Namedfor|Georg Friedrich Bernhard Riemann|name2 = Henri Léon Lebesgue|cat = Riemann|cat2 = Lebesgue}}