Riemann Integrable Functions as Upper Functions

# Riemann Integrable Functions as Upper Functions

We will now classify a bunch of very important upper functions. In the following theorem we will see that the set of Riemann integrable functions are also upper functions.

Theorem 1: Let $f$ be a function defined on the closed and bounded interval $I = [a, b]$. Then if $f$ is bounded and if $f$ is continuous almost everywhere on $I$ then $f$ is an upper function on $I$ and furthermore $\displaystyle{\int_I f(x) \: dx = \int_a^b f(x) \: dx}$. |

**Proof:**For each $n \in \mathbb{N}$ denote the partition $P_n = \{ a_0 = x_0, x_1, ..., x_{2^n} = b \} \in \mathscr{P}[a, b]$ as the partition that subdivides $[a, b]$ into $2^n$ subintervals of equal length, that is, for every $n$, $x_0 = a$ and for every $k \in \{ 1, 2, ..., 2^n \}$:

\begin{align} \quad x_k = a + k \left ( \frac{b - a}{2} \right ) \end{align}

- Then for every $n \in \mathbb{N}$, the next partition $P_{n+1}$ of this form can be obtained by equally subdividing the $2^n$ subintervals created by $P_n$ to obtain $2^{n+1}$ subintervals created by $P_{n+1}$. Now since $f$ is bounded on the interval $[a, b]$ we have that $f$ is also bounded on any subinterval of $[a, b]$. For each fixed $n$ let $k \in \{ 1, 2, ..., 2^n \}$ let:

\begin{align} \quad m_k = \inf \{ f(x) : x \in [x_{k-1}, x_k] \} \end{align}

- Define the step function $f_n(x)$ as follows:

\begin{align} \quad f_n(x) = \left\{\begin{matrix} a & \mathrm{if} \: x = a \\ m_k & \mathrm{if} \: x \in (x_{k-1}, x_k], k \in \{ 1, 2, ..., 2^n\}) \end{matrix}\right. \end{align}

- Then $(f_n(x))_{n=1}^{\infty}$ is a sequence of step functions. Every step function $f_n(x) \leq f(x)$, and the sequence $(f_n(x))_{n=1}^{\infty}$ is clearly an increasing sequence of step functions as illustrated below:

- We need to show that $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ almost everywhere on $I$. Let $x_0$ be any point of continuity of $f$. Since $f$ is continuous at $x_0$ we have that for $\epsilon > 0$ that there exists a $\delta > 0$ such that if $\mid x - x_0 \mid < \delta$ then:

\begin{align} \quad \mid f(x) - f(x_0) \mid < \epsilon \end{align}

- Now choose $N$ sufficiently large such that $\left ( \frac{b - a}{2^N} \right ) < \delta$. Then for $n \geq N$ we see that:

\begin{align} \quad \left ( \frac{b - a}{2^n} \right ) \leq \left ( \frac{b - a}{2^N} \right ) < \delta \end{align}

- So for $n \geq N$ we have that for the partitions $P_n$ that if $x \in (x_{k-1}, x_k]$ for some $k \in \{ 1, 2, ..., 2^n \}$ then $\mid x - x_0 \mid \leq x_k - x_{k-1} < \delta$ and so:

\begin{align} \quad \mid f(x) - f(x_0) \mid < \epsilon \end{align}

- So surely $\mid m_k - f(x_0) \mid \leq \epsilon$. But we defined $f_n(x) = m_k$ for all $x \in (x_{k-1}, x_k]$ and so for all $n \geq N$ we see that $\mid f_n(x_0) - f(x_0) \mid \leq \epsilon$

- Thus $\lim_{n \to \infty} f_n(x_0) = f(x_0)$ at every point of continuity $x_0$ in $I$ of $f$. In other words, the sequence $(f_n(x))_{n=1}^{\infty}$ converges to $f$ at every point of continuity of $f$. But $f$ is continuous almost everywhere which implies that $(f_n(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $I$.

- We now show that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ is finite. Note that for all $n \in \mathbb{N}$ and for $M$ as any upperbound to $f$ on $I$ that:

\begin{align} \quad \int_I f_n(x) \: dx = \sum_{k=1}^{2^n} m_k(x_k - x_{k-1}) \leq \sum_{k=1}^{2^n} M(x_k - x_{k-1}) = M \sum_{k=1}^{2^n} (x_k - x_{k-1}) = M(b - a) \end{align}

- Therefore the increasing sequence $\displaystyle{\left ( \int_I f_n(x) \: dx \right )_{n=1}^{\infty}}$ is bounded above and converges to a finite number.

- Furthermore, if $L(P_n, f, x)$ denotes the lower Riemann-Stieltjes sum of $f$ associated with the partition $P_n$ then:

\begin{align} \quad \int_I f(x) \: dx = \lim_{n \to \infty} \int_I f_n(x) \: dx = \lim_{n \to \infty} \sum_{k=1}^{2^n} m_k(x_k - x_{k-1}) = \lim_{n \to \infty} L(P_n, f, x) \end{align}

- We know that $f$ is Riemann integrable on $[a, b]$ since $f$ is continuous almost everywhere on $I$ and so the set of discontinuities of $f$ on $I$ has measure $0$. So by Riemann's condition $\displaystyle{ \underline{\int_a^b} f(x) \: dx = \int_a^b f(x) \: dx}$. But $\displaystyle{\lim_{n \to \infty} L(P_n, f, x) = \underline{\int_a^b} f(x) \: dx}$ since as $n \to \infty$, $P_n$ gets finer and $\| P_n \| \to 0$. This shows that:

\begin{align} \quad \int_I f(x) \: dx = \int_a^b f(x) \: dx \quad \blacksquare \end{align}