Riemann-Lebesgue Theorem

Theorem
Let $f: \closedint a b \to \R$ be a bounded mapping.

Let $\mu$ be a one-dimensional Lebesgue measure.

Then $f$ is Darboux integrable the set of all discontinuities of $f$ is a $\mu$-null set.

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


 * $\map {\omega_f} x = \ds \inf \set {\map {\omega_f} I: I \in N_x}$

where


 * $\map {\omega_f} I = \ds \sup \set {\size {\map f y - \map f z}: y, z \in I \cap \closedint a b}$


 * $N_x$ is the set of open subset neighborhoods of $x$

$\map {\omega_f} x$ and $\map {\omega_f} I$ 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::
 * $i \in J$ $\openint {x_{i - 1} } {x_i}$ 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:

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:

We can now define an upper bound for $\map \mu {A_s}$:

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:

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 $f$.