Technical Integration

In integral calculus, we usually assume that continuous functions $f : [a, b] \to \mathbb R$ can be integrated. Here, we will formally prove this really works, and in doing so, formulate Riemann integration more rigorously.

Lemma 1. Let $f : [a, b] \to \mathbb R$ be continuous and nonnegative (i.e. $f \geq 0$ ). For any partition $P = \{x_0, x_1, ..., x_n\}$ of $[a, b]$ (i.e. $x_0 = a$ , $x_n = b$ , and $P$ is non-decreasing), and for any $x \in [a, b]$ ,

$\displaystyle \sum_{i=1}^n m_i(f, P) \cdot \mathbb{I}_{[x_{i-1}, x_i)} \leq \sum_{i=1}^n f \cdot \mathbb{I}_{[x_{i-1}, x_i)}\leq \sum_{i=1}^n M_i(f, P) \cdot \mathbb{I}_{[x_{i-1}, x_i)},$

where $x_i := i/n$ , $P := \{x_0, x_1, ..., x_n\}$ , and we make the notations

$\displaystyle \begin{aligned} M_i(f, P) := \sup_{x \in [x_{i-1}, x_i] } f(x),\quad m_i (f, P) := \inf_{x \in [x_{i-1}, x_i] } f(x). \end{aligned}$

Recall that for sets $L \subseteq K$ , we define the indicator function $\mathbb{I}_L : \mathbb R \to \{0, 1\}$ by

$\mathbb{I}_L(x) = \begin{cases} 1, & x \in L, \\ 0, & x \notin L. \end{cases}$

Lemma 2. Let $f : [a, b] \to \mathbb R$ be continuous and nonnegative. Then for any $\epsilon > 0$ , there exists a partition $P = \{x_0,x_1,\dots,x_n\}$ such that for any $i = 1,\dots,n$ ,

$M_i (f, P) - m_i(f, P) < \epsilon,$

where we follow the notations in Lemma 1.

Proof. We prove the special case $[a, b] = [0, 1]$ for simplicity and leave the general case as a bookkeeping exercise.

Fix $\epsilon > 0$ . Since $f$ is continuous on $[0, 1]$ , it is uniformly continuous on $[0, 1]$ . Hence, there exists $\delta > 0$ such that for any $u,v \in [0, 1]$ ,

$|u - v| < \delta \quad \Rightarrow \quad |f(u) - f(v)| < k \cdot \epsilon.$

By the Archimedean property, find $n \in \mathbb N$ such that $1/n < \delta$ . Define $P = \{x_0,x_1,\dots,x_n\}$ where $x_i = i/n$ .

Fix $i = 1,\dots,n$ . By the extreme value theorem, find $c_i^{\pm} \in [x_{i-1}, x_i]$ such that $f(c_i^-) = m_i(f, P)$ and $f(c_i^+) = M_i(f, P)$ . Since $|c_i^+ - c_i^-| \leq x_i - x_{i-1} = 1/n < \delta$ ,

$M_i (f, P) - m_i(f, P) = | f(c_i^+) - f(c_i^-) | < k \cdot \epsilon.$

Setting $k = 1$ yields the desired result.

Theorem 1. Let $f : [a, b] \to \mathbb R$ be continuous and nonnegative. Then for any $\epsilon > 0$ , there exists a partition $P = \{x_0, x_1, \dots, x_n\}$ such that

$\displaystyle \sum_{i=1}^{n} (M_i (f, P) - m_i (f, P)) \Delta x_i < \epsilon,$

where $\Delta x_i := x_i - x_{i-1}$ and we follow the notations in Lemma 1. Consequently,

$\displaystyle \sum_{i=0}^n m_{i}(f, P) \Delta x_i \leq \sum_{i=0}^n f(x_i)\Delta x_i\leq \sum_{i=0}^n m_{i}(f,P) \Delta x_i + \epsilon.$

Proof. Fix $\epsilon > 0$ . By Lemma 2, for any $k > 0$ , find a partition $P = \{x_0,x_1,\dots,x_n\}$ such that for any $i = 1,\dots,n$ ,

$M_{i}(f, P) - m_{i}(f, P) < k \cdot \epsilon,$

Summing over $i = 1, \dots, n$ ,

$\displaystyle \begin{aligned} \sum_{i=1}^{n} (M_i (f, P) - m_i (f, P)) \Delta x_i &< \sum_{i=1}^{n} (k \cdot \epsilon) \cdot \Delta x_i \\ &= k \cdot \epsilon \cdot \sum_{i=1}^{n} \Delta x_i \\ &= k \cdot \epsilon \cdot 1 \\ &= k \cdot \epsilon. \end{aligned}$

Setting $k = 1$ yields the desired result.

What has Theorem 1 got to do with integration? Intuitively, $\epsilon \to 0^+$ should induce $n \to \infty$ , so that in the limit, $\displaystyle \sum_{i=0}^n f(x_i)\Delta x_i$ yields the area under the graph of $y = f(x)$ on $[0, 1]$ , yielding the familiar (approximate) definition

$\displaystyle \int_0^1 f(x)\, \mathrm dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x.$

We will now formulate this limiting process precisely using Riemann integration. For the rest of this post, let $f : [a, b] \to \mathbb R$ be bounded. For simplicity, we will occasionally assume $[a, b] = [0, 1]$ to elucidate the underlying analysis.

Definition 1. A partition $P$ of $[a, b]$ is an increasing sequence $P = \{x_0 = a, x_1,\dots, x_n = b\}$ of $n+1$ terms. For any partition $P$ with $n+1$ terms, define

$\displaystyle U(f, P) := \sum_{i = 1}^n M_i(f, P)\Delta x_i,\quad L(f, P) := \sum_{i = 1}^n m_i(f, P)\Delta x_i,$

where $\Delta x_i := x_i - x_{i-1}$ and

$\displaystyle \begin{aligned} M_i(f, P) := \sup_{x \in [x_{i-1}, x_i] } f(x),\quad m_i (f, P) := \inf_{x \in [x_{i-1}, x_i] } f(x). \end{aligned}$

Finally, define the upper and lower integrals by

$\displaystyle \mathcal U_a^b (f) := \inf_P U(f, P),\quad \mathcal L_a^b (f) := \sup_P L(f, P).$

We say that $f$ is Riemann-integrable on $[a, b]$ if $\mathcal U_a^b (f) = \mathcal L_a^b (f)$ . In this case, we write

$\displaystyle \mathcal L_a^b (f) =: \int_a^b f = \int_a^b f(x)\, \mathrm dx,$

and the right-hand side is used when emphasising the underlying variables.

Lemma 3. For partitions $P \subseteq Q$ of $[a, b]$ ,

$U(f, Q) \leq U(f, P),\quad L(f, Q) \geq L(f, P).$

Proof. We will prove the case $Q = P \cup \{c\}$ where $c \notin P$ and leave the general result as an exercise in induction. Find $i = 1,\dots,n$ such that $x_{i-1} < c < x_i$ . Then

$\displaystyle \begin{aligned} &\sup_{x \in [x_{i-1}, c]} f(x) \cdot (c - x_{i-1}) + \sup_{x \in [c, x_i]} f(x) \cdot (x_i - c) \\ &\leq \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (c - x_{i-1}) + \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (x_i - c) \\ &\leq \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (c - x_{i-1} + x_i - c) \\ &\leq \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (x_i - x_{i-1} ). \end{aligned}$

The result $U(f, Q) \leq U(f, P)$ follows by summing on both sides. The result for $L(f, Q) \geq L(f, P)$ follows similarly.

Lemma 4. $\mathcal L_a^b(f) \leq \mathcal U_a^b(f)$ .

Proof. For any partition $P$ ,

$\displaystyle L(f, P) = \sum_{i = 1}^n m_i(f, P) \Delta x_i\leq \sum_{i = 1}^n M_i(f, P)\Delta x_i = U(f, P).$

Now fix partitions $P, Q$ . By Lemma 3,

$L(f, P) \leq L(f, P \cup Q) \leq U(f, P \cup Q) \leq U(f, Q).$

Thus, $U(f, Q)$ is an upper bound for $L(f, P)$ regardless of $P$ . By definition of the supremum,

$\displaystyle \mathcal L_a^b(f) = \sup_P L(f, P) \leq U(f, Q)$

Since $\mathcal L_a^b(f)$ is a lower bound for $L(U, Q)$ regardless of $Q$ , by definition of the infimum,

$\displaystyle \mathcal L_a^b(f) \leq \inf_Q U(f, Q) = \mathcal U_a^b(f),$

as required.

Theorem 2 (Riemann Integrability Criterion). $f$ is Riemann-integrable if and only if for any $\epsilon > 0$ , there exists a partition $P = \{x_0,x_1,\dots,x_n\}$ of $[a, b]$ such that

$\displaystyle U(f, P) - L(f, P) < \epsilon.$

Proof. We will prove in two directions. For $(\Rightarrow)$ , suppose $f$ is Riemann integrable. Then $\mathcal U_a^b (f) = \mathcal L_a^b (f)$ . Expanding the meaning of $\mathcal U_a^b (f)$ , for any $k_1 > 0$ , there exists a partition $Q$ of $[a, b]$ such that

$U(f, Q) < \mathcal L_a^b (f) + k_1 \cdot \epsilon.$

Expanding the definition of $\mathcal L_a^b (f)$ , for any $k_2 > 0$ , there exists a partition $R$ of $[a, b]$ such that

$L(f, R) > \mathcal L_a^b (f) - k_2 \cdot \epsilon.$

Combining the estimates,

$U(f, Q) - L(f, R) < (k_1 + k_2) \cdot \epsilon.$

By Lemma 3, the partition $Q \cup R$ satisfies

$\begin{aligned} U(f, Q \cup R) - L(f, Q \cup R) &\leq U(f, Q) - L(f, R) \\ &< (k_1 + k_2) \cdot \epsilon. \end{aligned}$

Set $k_1 = k_2 = 1/2$ and define the partition $P:= Q \cup R$ .

For $(\Leftarrow)$ . We need to prove that $\mathcal L_a^b(f) = \mathcal U_a^b(f)$ . By Lemma 4, $\mathcal L_a^b(f) \leq \mathcal U_a^b(f)$ unconditionally. To prove $\mathcal U_a^b(f) \leq \mathcal L_a^b(f)$ , the hypothesis yields that, for any $\epsilon > 0$ , there exists a partition $P$ of $[a, b]$ such that

$U(f, P) - L(f, P) < \epsilon.$

Taking the infimum over $P$ on the left-hand side,

$\displaystyle \inf_P (U(f, P) - L(f, P)) \leq 0.$

On the other hand,

$\begin{aligned} \mathcal U_a^b(f) - \mathcal L_a^b(f) &= \inf_P U(f, P) - \sup_P L(f, P) \\ &= \inf_P U(f, P) + \inf_P (- L(f, P)) \\ &= \inf_P (U(f, P) + (- L(f, P))) \\ &= \inf_P (U(f, P) - L(f, P)) \leq 0. \end{aligned}$

Thus, $\mathcal U_a^b(f) \leq \mathcal L_a^b(f)$ . Therefore, $\mathcal L_a^b(f) = \mathcal U_a^b(f)$ , and $f$ is Riemann-integrable, as required.

This tells us, finally, that continuous functions are Riemann-integrable.

Corollary 1. If $f : [a, b] \to \mathbb R$ is continuous, then $f$ is Riemann-integrable.

Proof. Combine the results in Theorem 1 and Theorem 2 to obtain the desired result.

Technically, we have formulated Darboux-integrability that focuses on the “maximal” and “minimal” areas, since Riemann-integrability is more closely connected with general areas. Yet, Darboux integrability basically yields upper and lower estimates on the Riemann integral, and so in the limit, both terms match up and the existence of one implies the other.

—Joel Kindiak, 19 Jan 25, 1554H

KindiakMath

Technical Integration

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Technical Integration

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