7.3: Integrability Conditions

Proposition $\PageIndex{1}$

If $a<b$ and $f:[a, b] \rightarrow \mathbb{R}$ is monotonic, then $f$ is integrable on $[a, b]$.

Proof

Suppose $f$ is nondecreasing. Given $\epsilon>0,$ let $n \in Z^{+}$ be large enough that

$$\frac{(f(b)-f(a))(b-a)}{n}<\epsilon .$$

For $i=0,1, \ldots, n,$ let

$$x_{i}=a+\frac{(b-a) i}{n}.$$

Let $P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\} .$ Then

$$\begin{aligned} U(f, P)-L(f, P) &=\sum_{i=1}^{n} f\left(x_{i}\right)\left(x_{i}-x_{i-1}\right)-\sum_{i=1}^{n} f\left(x_{i-1}\right)\left(x_{i}-x_{i-1}\right) \\ &=\sum_{i=1}^{n}\left(f\left(x_{i}\right)-f\left(x_{i-1}\right)\right) \frac{b-a}{n} \\ &=\frac{b-a}{n}\left(\left(f\left(x_{1}\right)-f\left(x_{0}\right)\right)+\left(f\left(x_{2}\right)-f\left(x_{1}\right)\right)+\cdots\right.\\ &\left.\quad+\left(f\left(x_{n-1}\right)-f\left(x_{n-2}\right)\right)+\left(f\left(x_{n}\right)-f\left(x_{n-1}\right)\right)\right) \\ &=\frac{b-a}{n}(f(b)-f(a)) \\ &<\epsilon . \end{aligned}$$

Hence $f$ is integrable on $[a, b]$. $\quad$ Q.E.D.

Example $\PageIndex{1}$

Let $\varphi: \mathbb{Q} \cap[0,1] \rightarrow \mathbb{Z}^{+}$ be a one-to-one correspondence. Define $f:[0,1] \rightarrow \mathbb{R}$ by

$$f(x)=\sum_{q \in \underset{q \leq x}{\mathbb{Q} \cap [0,1]}} \frac{1}{2^{\varphi(q)}.}$$

Then $f$ is increasing on $[0,1],$ and hence integrable on $[0,1]$.

Proposition $\PageIndex{2}$

If $a<b$ and $f:[a, b] \rightarrow \mathbb{R}$ is continuous, then $f$ is integrable on $[a, b]$.

Proof

Given $\epsilon>0,$ let

$$\gamma=\frac{\epsilon}{b-a}.$$

Since $f$ is uniformly continuous on $[a, b],$ we may choose $\delta>0$ such that

$$|f(x)-f(y)|<\gamma$$

whenever $|x-y|<\delta .$ Let $P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}$ be a partition with

$$\sup \left\{\left|x_{i}-x_{i-1}\right|: i=1,2, \ldots, n\right\}<\delta .$$

If, for $i=1,2, \dots, n$,

$$m_{i}=\inf \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\}$$

and

$$M_{i}=\sup \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\},$$

then $M_{i}-m_{i}<\gamma .$ Hence

$$\begin{aligned} U(f, P)-L(f, P) &=\sum_{i=1}^{n} M_{i}\left(x_{i}-x_{i-1}\right)-\sum_{i=1}^{n} m_{i}\left(x_{i}-x_{i-1}\right) \\ &=\sum_{i=1}^{n}\left(M_{i}-m_{i}\right)\left(x_{i}-x_{i-1}\right) \\ &<\gamma \sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right) \\ &=\gamma(b-a) \\ &=\epsilon . \end{aligned}$$

Thus $f$ is integrable on $[a, b]$. $\quad$ Q.E.D.

Exercise $\PageIndex{1}$

Suppose $a<b, f:[a, b] \rightarrow \mathbb{R}$ is bounded, and $c \in[a, b] .$ Show that if $f$ is continuous on $[a, b] \backslash\{c\},$ then $f$ is integrable on $[a, b]$.

Exercise $\PageIndex{2}$

Suppose $a<b$ and $f$ is continuous on $[a, b]$ with $f(x) \geq 0$ for all $x \in[a, b] .$ Show that if

$$\int_{a}^{b} f=0,$$

then $f(x)=0$ for all $x \in[a, b]$.

Exercise $\PageIndex{3}$

Suppose $a<b$ and $f$ is continuous on $[a, b] .$ For $i=0,1, \ldots, n$, $n \in \mathbb{Z}^{+},$ let

$$x_{i}=a+\frac{(b-a) i}{n}$$

and, for $i=1,2, \ldots, n,$ let $c_{i} \in\left[x_{i-1}, x_{i}\right] .$ Show that

$$\int_{a}^{b} f=\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{i=1}^{n} f\left(c_{i}\right).$$

In the notation of Exercise $7.3 .3,$ we call the approximation

$$\int_{a}^{b} f \approx \frac{b-a}{n} \sum_{i=1}^{n} f\left(c_{i}\right)$$

a right-hand rule approximation if $c_{i}=x_{i},$ a left-hand rule approximation if $c_{i}=x_{i-1},$ and a midpoint rule approximation if

$$c_{i}=\frac{x_{i-1}+x_{i}}{2}.$$

These are basic ingredients in creating numerical approximations to integrals.