Upper & Lower Riemann Integral

I remember several years ago, in an analysis class that I took, a trivial result that was stated in class was that the upper Riemann integral of a function is always greater than or equal to its lower Riemann integral. When reading Efe A. Ok’s book (Ok, 2007), I stumbled on this exact statement again on page 53, followed by his reminder to prove the statement before continuing. Since I don’t remember proving this in class and Ok doesn’t prove this statement in his book, in this blog post, I try to provide a proof that is as detailed as possible so that there are no doubts as to the validity of the statement.

Definitions to Prove the Result

I devote this entire section to defining terms that will be useful to prove the statement. Some of the symbols I use are from Ok’s book while some definitions are from other sources. This section can be considered a brief review for the expert and, hopefully, a helpful guide for the beginner.

$\textbf{Definition. Supremum \& Infimum}$

A lower bound of a subset $S$ of a partially ordered set $(P, \leq)$ is an element $a \in P$ such that $a \leq x$ for all $x \in S$. A lower bound $a$ of $S$ is called an infimum of $S$ if for all lower bounds $y$ of $S$ in $P$, $y \leq a$ ($a$ is larger than any other lower bound).

An upper bound of a subset $S$ of a partially ordered set $(P, \leq)$ is an element $b \in P$ such that $b \geq x$ for all $x \in S$. An upper bound $b$ of $S$ is called a supremum of $S$ if for all upper bounds $z$ of $S$ in $P$, $z \geq b$ ($b$ is less than any other upper bound).

$\textbf{Definition. Bounded Function}$

A function $f \in \mathbb{R}^{[a, b]}$ (where its domain is on $[a, b]$ and its codomain is on $\mathbb{R}$) is bounded if there exists $B \in \mathbb{R}$ such that $|f(x)| \leq B$ for all $x \in[a, b]$.

$\textbf{Definition. Dissection}$

Let $a$ and $b$ be two real numbers such that $a \leq b$. For any $m \in \mathbb{N}$, dissection of the set $[a, b]$ is denoted by $\left[a_{0}, \ldots, a_{m}\right]$ and is defined by $\left\{\left[a_{0}, a_{1}\right],\left[a_{1}, a_{2}\right], \ldots,\left[a_{m-1}, a_{m}\right]\right\}$ where $a=a_{0}<\cdots<$ $a_{m}=b$. The class of all dissections of $[a, b]$ is defined by $\mathcal{D}[a, b]$.

$\textbf{Definition. Finer Than}$

For any $\boldsymbol{a}:=\left[a_{0}, \ldots, a_{m}\right]$ and $\boldsymbol{b}:=\left[b_{0}, \ldots, b_{k}\right]$ in $\mathcal{D}[a, b]$, we write $\boldsymbol{a} ⋓ \boldsymbol{b}$ for the dissection $\left[c_{0}, \ldots, c_{l}\right] \in \mathcal{D}[a, b]$ where $\left\{c_{0}, \ldots, c_{l}\right\}=\left\{a_{0}, \ldots, a_{m}\right\} \cup\left\{b_{0}, \ldots, b_{k}\right\}$. Moreover, $\boldsymbol{b}$ is finer than $\boldsymbol{a}$ if $\left\{a_{0}, \ldots, a_{m}\right\} \subseteq\left\{b_{0}, \ldots, b_{k}\right\}$.

$\textbf{Definition. \textit{a}-upper Riemann Sum \& \textit{a}-lower Riemann Sum}$

Let $f \in \mathbb{R}^{[a, b]}$ be any bounded function. For any dissection $\boldsymbol{a} \in \mathcal{D}[a, b]$, define $K_{f, \boldsymbol{a}}(i):=\sup \left\{f(t) \mid a_{i-1} \leq t \leq a_{i}\right\}$ and $k_{f, \boldsymbol{a}}(i):=\inf \left\{f(t) \mid a_{i-1} \leq t \leq a_{i}\right\}$ for each $i=1, \ldots, m$. The $a$-upper Riemann sum of $f$ is defined as

$R_{\mathbf{a}}(f):=\sum_{i=1}^{m} K_{f, \mathbf{a}}(i)\left(a_{i}-a_{i-1}\right)$

and the $a$-lower Riemann sum of $f$ is defined as

$r_{\mathbf{a}}(f):=\sum_{i=1}^{m} k_{f, \mathbf{a}}(i)\left(a_{i}-a_{i-1}\right).$

$\textbf{Definition. Upper and Lower Riemann Integrals}$

Let $f \in \mathbb{R}^{[a, b]}$ be any bounded function. $R(f)$ and $r(f)$ are called the upper and lower Riemann integrals of $f$, where they are defined as $R(f):=\inf \left\{R_{\boldsymbol{a}}(f) \mid \boldsymbol{a} \in \mathcal{D}[a, b]\right\}$ and $r(f):=$ $\sup \left\{r_{\boldsymbol{a}}(f) \mid \boldsymbol{a} \in \mathcal{D}[a, b]\right\}$.

A Proof of the Theorem

I now state and prove two lemmas that will be helpful for proving the theorem, and I prove the theorem afterwards. For the following analysis, I only consider functions that are bounded.

$\textbf{Lemma.}$ For any dissection $\boldsymbol{a} \in \mathcal{D}[a, b], R_{\boldsymbol{a}}(f) \geq r_{\boldsymbol{a}}(f)$.

$\textit{Proof.}$ Since for every dissection $\boldsymbol{a} \in \mathcal{D}[a, b], \sup \left\{f(t) \mid a_{i-1} \leq t \leq a_{i}\right\} \geq \inf \left\{f(t) \mid a_{i-1} \leq t \leq a_{i}\right\}$, we have that

$\begin{aligned} \sum_{i=1}^{m} \sup \left\{f(t) \mid a_{i-1} \leq t \leq a_{i}\right\}\left(a_{i}-a_{i-1}\right) & \geq \sum_{i=1}^{m} \inf \left\{f(t) \mid a_{i-1} \leq t \leq a_{i}\right\}\left(a_{i}-a_{i-1}\right) \\ & \Leftrightarrow \\ R_{\boldsymbol{a}}(f) & \geq r_{\boldsymbol{a}}(f). \end{aligned}$

$\textbf{Lemma.}$ If $r(f)>R(f)$, then there exists some $\boldsymbol{a}, \boldsymbol{b} \in \mathcal{D}[a, b]$ such that $r_{\boldsymbol{b}}(f)>R_{\boldsymbol{a}}(f)$.

$\textit{Proof.}$ Suppose the hypothesis is true and suppose by contradiction for all $\boldsymbol{a}, \boldsymbol{b} \in \mathcal{D}[a, b], R_{\boldsymbol{a}}(f) \geq r_{\boldsymbol{b}}(f)$. This implies that either (1) the supremum of $r_b(f)$ is equal to the infimum of $R_a(f)$ or (2) the supremum of $r_{\boldsymbol{b}}(f)$ is less than the infimum of $R_{\boldsymbol{a}}(f)$. In other words, $\sup \left\{r_a(f) \mid \boldsymbol{a} \in \mathcal{D}[a, b]\right\} = r(f) = \inf \left\{R_{\boldsymbol{a}}(f) \mid \boldsymbol{a} \in \mathcal{D}[a, b]\right\} = R(f)$ or $\sup \left\{r_{\boldsymbol{a}}(f) \mid \boldsymbol{a} \in \mathcal{D}[a, b]\right\} = r(f) < \inf \left\{R_{\boldsymbol{a}}(f) \mid \boldsymbol{a} \in \mathcal{D}[a, b]\right\} = R(f)$. In either case, the result contradicts the hypothesis. Hence, the statement is true.

$\textbf{Theorem.}$

$R(f) \geq r(f).$

$\textit{Proof.}$ Suppose by contradiction we had $r(f)>R(f)$. By the lemma above, this implies that there exists some $\boldsymbol{a}, \boldsymbol{b} \in \mathcal{D}[a, b]$ such that $r_{\boldsymbol{b}}(f)>R_{\boldsymbol{a}}(f)$. Then consider the dissection $\boldsymbol{a} ⋓ \boldsymbol{b}$. We know that $\boldsymbol{b} \subseteq \boldsymbol{a} ⋓ \boldsymbol{b} \in \mathcal{D}[a, b]$ and $\boldsymbol{a} \subseteq \boldsymbol{a} ⋓ \boldsymbol{b}$. Hence, since $\boldsymbol{a} ⋓ \boldsymbol{b}$ is finer than $\boldsymbol{a}$ and $\boldsymbol{b}$, we have that $r_{\boldsymbol{a} ⋓ \boldsymbol{b}}(f) \geq r_{\boldsymbol{b}}(f)$ and $R_{\boldsymbol{a} ⋓ \boldsymbol{b}}(f) \leq R_{\boldsymbol{a}}(f)$. This implies that $r_{\boldsymbol{a} ⋓ \boldsymbol{b}}(f) \geq r_{\boldsymbol{b}}(f) > R_{\boldsymbol{a}}(f) \geq R_{\boldsymbol{a} ⋓ \boldsymbol{b}}(f) \Rightarrow r_{\boldsymbol{a} ⋓ \boldsymbol{b}}(f)>R_{\boldsymbol{a} ⋓ \boldsymbol{b}}(f)$. However, this contradicts the fact that $R_{x}(f) \geq r_{x}(f)$ for any dissection $\boldsymbol{x}$, stated in the lemma above.

References

1. Ok, E. A. (2007). Real Analysis with Economic Applications (Vol. 10). Princeton University Press.