Lebesgue's Criterion Part 1 - Riemann Integrability of a Bnd. Function

# Lebesgue's Criterion Part 1 - Riemann Integrability of a Bounded Function

Recall from the Oscillation and Continuity of a Bounded Function at a Point page that if $f$ is a bounded function on $[a, b]$ then $f$ is continuous at $x \in [a, b]$ if and only if $\omega_f (x) = 0$.

Equivalently, we can say that $f$ is discontinuous at $x \in [a, b]$ if and only if $\omega_f (x) > 0$. We will use this important fact to prove a very important theorem regarding the Riemann integrability of a bounded function $f$ on $[a, b]$ in terms of the set of discontinuities $D$ of $f$ on $[a, b]$ having measure $0$.

The proof is rather lengthy, so we prove one direction here and the second direction on the Lebesgue's Criterion Part 2 - Riemann Integrability of a Bounded Function

Theorem 1 (Lebesgue's Criterion for Riemann Integrability of a Bounded function): Let $f$ be a bounded function on $[a, b]$ and let $D$ be the set of all discontinuities of $f$ on $[a, b]$. Then $f$ is Riemann integrable on $[a, b]$ if and only if $m(D) = 0$. |

**Proof:**$\Rightarrow$ Suppose that $f$ is Riemann integrable on $[a, b]$ and suppose that $m(D) \neq 0$. We will show that a contradiction arises.

- Recall that $f$ is continuous at $x$ if and only if $\omega_f (x) = 0$. So, $f$ is discontinuous at $x$ if and only if $\omega_f (x) > 0$. For each $p \in \mathbb{N}$ let $\displaystyle{D_p = \left \{ x \in [a, b] : \omega_f (x) \geq \frac{1}{p}, p \in \mathbb{N} \right \}}$. Then $D_p$ is the set of all discontinuities of $f$ whose oscillation is greater or equal to $\displaystyle{\frac{1}{p}}$. For each discontinuity $d \in D$ we have that $\omega_f (d) > 0$ and so there will always exist a $p \in \mathbb{N}$ such that $\displaystyle{0 < \frac{1}{n} \leq \omega_f(d)}$. Therefore:

\begin{align} \quad D = \bigcup_{p=1}^{\infty} D_p \end{align}

- Now since $m(D) \neq 0$ we must have that there exists a $p \in \mathbb{N}$ such that $m(D_p) \neq 0$. If not, then each set in $\{ D_p : p \in \mathbb{N} \}$ will has measure $0$ and a countable union of measure $0$ sets is measure $0$ which is a contradiction. Since $m(D_p) \neq 0$, there exists an $\epsilon_0 > 0$ such that for every open covering of intervals $\{ I_k = (a_k, b_k \}_{k \in K}$ (where $K$ is a countable indexing set) of $D_p$ ($D_p \subseteq \bigcup_{k \in K} I_k = \bigcup_{k \in K} (a_k, b_k)$) we have that:

\begin{align} \quad \sum_{k \in K} l(I_k) = \sum_{k \in K} (b_k - a_k) \geq \epsilon_0 \end{align}

- Now let $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$. Since our integrator $\alpha(x) = x$ (an increasing function) we can consider the difference between the upper and lower Riemann sums of $P$:

\begin{align} \quad U(P, f, x) - L(P, f, x) &= \sum_{k=1}^{n} M_k(f) \Delta x_k - \sum_{k=1}^{n} m_k(f) \Delta x_k \\ \quad &= \sum_{k=1}^{n} [M_k(f) - m_k(f)] \Delta x_k \end{align}

- Let $A$ and $B$ be the finite sets defined by:

\begin{align} \quad A = \{ [M_k(f) - m_k(f)] \Delta x_k : k \in \{1, 2, ..., n \} \: \mathrm{and} \: \exists d \in D \: \mathrm{such \: that \:} d \in (x_{k-1}, x_k) \} \end{align}

(5)
\begin{align} \quad \mathrm{and} \quad B = \{ [M_k(f) - m_k(f)] \Delta x_k : k \in \{1, 2, ..., n \} \: \mathrm{and} \forall \: d \in D \: \: d \not \in (x_{k-1}, x_k) \} \end{align}

- In other words, $A$ is the set of terms $[M_k(f) - m_k(f)]\Delta x_k$ in the difference above such that there exists a discontinuity $d \in D$ in the subinterval $[x_{k-1}, x_k]$ and $B$ is the set of terms of the difference above such that there are no discontinuities of $f$ on these subintervals. Let:

\begin{align} \quad S_1 = \sum_{A} [M_k(f) - m_k(f)]\Delta x_k \quad \mathrm{and} \quad S_2 = \sum_{B} [M_k(f) - m_k(f)] \Delta x_k \end{align}

- Then we have that:

\begin{align} \quad U(P, f, x) - L(P, f, x) = S_1 + S_2 \geq S_1 \end{align}

- Note that $\{ (x_{k-1}, x_k) : \exists d \in D \: \mathrm{where \:} d \in (x_{k-1}, x_k) \}$ covers $D_p$ except possibly at a finite number of points (namely if discontinuities occur at $x_1, x_2, ..., x_n$), and so we have that that the sum of the lengths of these intervals is greater or equal to $\epsilon$, i.e., $\sum_{k : \exists d \in D \: s.t. d \in (x_{k-1}, x_k)} \Delta_k \geq \epsilon$. Furthermore since $M_k(f) - m_k(f) \geq \frac{1}{r}$ for each $k$ such that $\exists d \in D$ such that $d \in (x_{k-1}, x_k)$ and so:

\begin{align} \quad U(P, f, x) - L(P, f, x) \geq S_1 \geq \frac{\epsilon}{p} > 0 \end{align}

- So Riemann's condition cannot be satisfied and so $f$ is not Riemann integrable on $[a, b]$.