Theorem: Riemann-Lebesgue Theorem

Let $f: \closedint a b \to \R$ be a bounded mapping.
Let $\mu$ be a one-dimensional Lebesgue measure.
{{explain|Link to definition of one-dimensional in this specific context}}
Then $f$ is Darboux integrable {{iff}} the set of all discontinuities of $f$ is a $\mu$-null set.

Proof:
=== 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
:$\map {\omega_f} x = \ds \inf \set {\map {\omega_f} {I \cap {\closedint a b}}: I \in N_x}$
where
:$\map {\omega_f} {I \cap {\closedint a b}} = \ds \sup \set {\size {\map f y - \map f z}: y, z \in I \cap \closedint a b}$
:$N_x$ is the set of open subset neighborhoods of $x$
$\map {\omega_f} x$ and $\map {\omega_f} {I \cap {\closedint a b}}$ 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:
:$\map U P = \ds \sum_{i \mathop = 1}^n \map {M_i} {x_i − x_{i − 1} }$
:$\map L P = \ds \sum_{i \mathop = 1}^n \map {m_i} {x_i − x_{i − 1} }$
where:
:$M_i = \ds \sup \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} }$
:$m_i = \ds \inf \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} }$
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 $\ds \sum_{i \mathop \in J} \paren {x_i − x_{i - 1} } < \dfrac \epsilon s$.
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} {\openint {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 \cap {\closedint a b} }: I \in N_{x^{'}_i} } \paren {x_i − x_{i − 1} }
      | c = as $\set {\map {\omega_f} {I \cap {\closedint a b} }: I \in N_{x^{'}_i} }$ 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:
:$\ds \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} } < \dfrac \epsilon s$
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 $\ds \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} } < \frac \epsilon s$
}}
{{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 $\ds \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$.
Every set $A_{1 / n}$ contains only points of discontinuity of $f$.
Therefore, $\ds \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$ 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 $\ds \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$.
Accordingly:
:$\ds \bigcup_{n \mathop \in \N_{>0} } A_{1 / n} = A_0$
Furthermore, since $\ds \bigcup_{n \mathop \in \N_{>0}} A_{1 / n}$ 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}}