Flavors of Integrability

In this post, we explore the various settings in which functions can be integrated. Here, we shorten “Riemann-integrable” by “integrable” for brevity.

Problem 1. Let $f : [a, b] \to \mathbb R$ be a bounded function. Suppose for any $c \in (a, b)$ , $f|_{[a,c]} : [a, c] \to \mathbb R$ is integrable. Prove that $f$ is integrable.

Solution. Since $f$ is bounded, there exists $M > 0$ such that for any $x \in [a, b]$ , $|f(x)| \leq M$ .

Fix $\epsilon > 0$ . Fix $\delta \in (0, b-a)$ to be tuned later. Since $f_{[a,b-\delta]}$ is integrable, for any $k > 0$ there exists a partition $P := \{x_0 = a, x_1,\dots,x_n = b-\delta\}$ such that

$U(f_{[a,b-\delta]}, P) - L(f_{[a,b-\delta]}, P) < k \cdot \epsilon.$

Define the partition $Q := \{x_0, x_1,\dots, x_n, x_{n+1} = b\}$ of $[a, b]$ . Expanding the definition,

$\begin{aligned} &U(f, Q) - L(f, Q)\\ &= U(f, P) - L(f, P) + (M_{n+1}(f, Q) - m_{n+1}(f, Q))\Delta_{n+1} \\ &< k \cdot \epsilon + (M+M) \cdot \delta \\ &< k \cdot \epsilon + 2M \cdot \delta. \end{aligned}$

Hence, set $k := 1/2$ and $\delta := \epsilon/(4M)$ to yield the result.

Henceforth, we will say $f : [a, b] \to \mathbb R$ is integrable on $K \subseteq [a, b]$ to mean that $f|_K : K \to \mathbb R$ is integrable.

Problem 2. Let $f : [a, b] \to \mathbb R$ be bounded. Suppose there exists $c \in (a, b)$ such that $f$ is integrable on $[a, c]$ and $[c, b]$ . Then $f$ is integrable on $[a, b]$ . Furthermore,

$\displaystyle \int_a^b f = \int_a^c f + \int_c^b f.$

Solution. Fix $\epsilon > 0$ . For any $k_1 > 0$ , find a partition $P$ of $[a, c]$ such that

$U(f, P) - L(f, P) < k_1 \cdot \epsilon.$

For any $k_2 > 0$ , find a partition $Q$ of $[c, b]$ such that

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

Then $P \cup Q$ is a partition of $[a, b]$ , and

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

Setting $k_1 = k_2 = 1/2$ yields the desired result.

Finally, we prove the integral identity in two directions. Fix $\epsilon > 0$ . Find a partition $P$ such that

$\displaystyle \int_a^b f < L(f, P) + \epsilon.$

Define the partitions

$\begin{aligned} P_- &:= \{x \in P : x < c\} \cup \{c\} \subseteq [a, c],\\ P_+ &:= \{x \in P : x > c\} \cup \{c\} \subseteq [c, b]. \end{aligned}$

Then

$\displaystyle \int_a^b f < L(f, P) + \epsilon \leq L(f, P_-) + L(f, P_+) + \epsilon \leq \int_a^c f + \int_c^b f+ \epsilon .$

Taking $\epsilon \to 0^+$ yields the inequality

$\displaystyle \int_a^b f \leq \int_a^c f + \int_c^b f.$

For the other direction, fix $\epsilon > 0$ . For $k_1,k_2 > 0$ , find partitions $P_- \subseteq [a, c]$ , $P_+ \subseteq [c, b]$ , such that

$\displaystyle\int_a^c f < L(f, P_-) + k_1 \cdot \epsilon \quad \int_c^b f < L(f, P_+) + k_2 \cdot \epsilon.$

Defining $P := P_- \cup P_+$ ,

$\begin{aligned} \int_a^c f + \int_c^b f & < (L(f, P_-) + L(f, P_+)) + (k_1 + k_2) \cdot \epsilon \\ & = L(f, P) + (k_1 + k_2) \cdot \epsilon \\ &\leq \int_a^b f + (k_1 + k_2) \cdot \epsilon. \end{aligned}$

Setting $k_1 = k_2 = 1/2$ ,

$\displaystyle \int_a^c f + \int_c^b f < \int_a^b f + \epsilon.$

Taking $\epsilon \to 0^+$ ,

$\displaystyle \int_a^c f + \int_c^b f \leq \int_a^b f.$

Combining the results yields the desired result.

Problem 3. Let $f : [a, b] \to \mathbb R$ be continuous except at some $c \in (a, b)$ . Prove that $f$ is integrable.

Solution. For any $u \in (a, c)$ , $f$ is continuous on $[a, u]$ and thus integrable on $[a, u]$ . By Problem 1, $f$ is integrable on $[a, c]$ . Similarly, $f$ is integrable on $[c, b]$ . By Problem 2, $f$ is integrable on $[a, b]$ .

Problem 4. Let $f : [a, b] \to \mathbb R$ be increasing. Prove that $f$ is integrable.

Solution. We prove $[a, b] = [0,1]$ for simplicity. We observe that for any $x \in (a, b)$ , $f(a) < f(x) < f(b)$ so that $f$ is bounded. Fix $\epsilon > 0$ . For any $k > 0$ , use the Archimedean property to find $n \in \mathbb N$ such that $1/n < k \cdot \epsilon$ . Define $P := \{x_0, x_1, \dots, x_n\} \subseteq [0, 1]$ by $x_i = i/n$ . Then

$\begin{aligned} U(f, P) - L(f, P) &= \sum_{i=1}^n (M_i(f, P) - m_i(f, P)) \Delta x_i \\ &= \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \cdot \frac 1n \\ &= \frac 1n \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \\ &= \frac 1n (f(x_n) - f(x_0)) \\ &= \frac {f(1) - f(0)}n \\ & < (f(1) - f(0)) \cdot k \cdot \epsilon. \end{aligned}$

Since $f(1) - f(0) > 0$ , set $k := 1/(f(1) - f(0)) > 0$ to yield the desired result.

—Joel Kindiak, 20 Jan 25, 1941H

KindiakMath

Flavors of Integrability

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Flavors of Integrability

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