Riemann-Lebesgue Theorem

Theorem
Let $f: \left[{a \,.\,.\, b}\right] \to \R$ be a bounded mapping.

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

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

Necessary Condition
Suppose that $f$ is Riemann integrable.

We need to prove that the set of all discontinuities of $f$ has measure $0$.

Let, for some positive real number $s$:


 * $A_s = \left\{{x \in \left[{a \,.\,.\, b}\right]: \omega_f \left({x}\right) > s}\right\}$

where


 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

where


 * $\omega_f \left({I}\right) = \displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap \left[{a \,.\,.\, b}\right]}\right\}$


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

$\omega_f \left({x}\right)$ and $\omega_f \left({I}\right)$ 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


 * $A_0 = \left\{{x \in \left[{a \,.\,.\, b}\right]: \omega_f \left({x}\right) > 0}\right\}$

is the set of all discontinuities of $f$.

We need to prove that $\mu \left({A_0}\right) = 0$.

If $A_0$ is empty, it follows by Measure of Empty Set is Zero that $\mu \left({A_0}\right) = 0$.

This proves what we wanted for this case.

Now assume that $A_0$ is non-empty.

This means that $\left[{a \,.\,.\, b}\right]$ contains a point $x$ with $\omega_f \left({x}\right) > 0$.

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 Riemann Integrability, there exists a subdivision $P = \left\{{x_0, x_1, x_2, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$ such that


 * $U \left({P}\right) – L \left({P}\right) < \epsilon$

where


 * $U \left({P}\right) = \displaystyle \sum_{i = 1}^n M_i \left({x_i − x_{i − 1}}\right)$


 * $L \left({P}\right) = \displaystyle \sum_{i = 1}^n m_i \left({x_i − x_{i − 1}}\right)$

where


 * $M_i = \displaystyle \sup \left\{{f \left({x}\right): x \in \left[{x_{i - 1} \,.\,.\, x_i}\right]}\right\}$


 * $m_i = \displaystyle \inf \left\{{f \left({x}\right): x \in \left[{x_{i - 1} \,.\,.\, x_i}\right]}\right\}$

Let $J$ be the subset of $\left\{{1, 2, \ldots, n}\right\}$ that satisfies: $i \in J$ $\left({x_{i - 1} \,.\,.\, x_i}\right)$ contains a point in $A_s$.

First, we intend to prove that $\displaystyle \sum_{i \mathop \in J} \left({x_i − x_{i - 1}}\right) < \dfrac \epsilon s$.

Let $x^{'}_i$ signify a point in $\left({x_{i - 1} \,.\,.\, x_i}\right) \cap A_s$ whenever $i \in J$.

We have:

Therefore,


 * $\displaystyle \sum_{i \mathop \in J} \left({x_i − x_{i − 1}}\right) < \dfrac \epsilon s$

Next, we intend to prove that $\mu \left({A_s}\right) = 0$.

We have:

We can now define an upper bound for $\mu \left({A_s}\right)$:

Therefore, $\mu \left({A_s}\right) < \dfrac {3 \epsilon} s$.

Since $\epsilon$ is an arbitrary number greater than zero, $\mu \left({A_s}\right) = 0$.

Next, we intend to prove that $\mu \left({A_0}\right) = 0$.

Consider the set sequence $\left({A_{1 / n}}\right)_{n \mathop \in \N_{>0}}$.

This sequence is increasing since $A_{1 / n} \subseteq A_{1 / \left({n + 1}\right)}$ 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 $\left({A_{1 / n}}\right)_{n \mathop \in \N_{>0}}$ 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 $\left({A_{1 / n}}\right)_{n \mathop \in \N_{>0}}$, $A_0$ equals the limit of $\left({A_{1 / n}}\right)_{n \mathop \in \N_{>0}}$.

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 $f$.