7.4: Properties of Integrals

Proposition $\PageIndex{1}$

If $f: D \rightarrow \mathbb{R}$ and $g: D \rightarrow \mathbb{R},$ then

$$\sup \{f(x)+g(x): x \in D\} \leq \sup \{f(x): x \in D\}+\sup \{g(x): x \in D\}$$

and

$$\inf \{f(x)+g(x): x \in D\} \geq \inf \{f(x): x \in D\}+\inf \{g(x): x \in D\}$$

Exercise $\PageIndex{1}$

Prove the previous proposition.

Exercise $\PageIndex{2}$

Find examples for which the inequalities in the previous proposition are strict.

Proposition $\PageIndex{2}$

Suppose $f$ and $g$ are both integrable on $[a, b] .$ Then $f+g$ is integrable on $[a, b]$ and

$$\int_{a}^{b}(f+g)=\int_{a}^{b} f+\int_{a}^{b} g.$$

Proof

Given $\epsilon>0,$ let $P_{1}$ and $P_{2}$ be partitions of $[a, b]$ with

$$U\left(f, P_{1}\right)-L\left(f, P_{1}\right)<\frac{\epsilon}{2}$$

and

$$U\left(g, P_{2}\right)-L\left(g, P_{2}\right)<\frac{\epsilon}{2}.$$

Let $P=P_{1} \cup P_{2} .$ By the previous proposition,

$$U(f+g, P) \leq U(f, P)+U(g, P)$$

and

$$L(f+g, P) \geq L(f, P)+L(g, P).$$

Hence

$$\begin{aligned} U(f+g, P)-L(f+g, P) & \leq(U(f, P)+U(g, P))-(L(f, P)+L(g, P)) \\ &=(U(f, P)-L(f, P))+(U(g, P)-L(g, P)) \\ & \leq\left(U\left(f, P_{1}\right)-L\left(f, P_{1}\right)\right)+\left(U\left(g, P_{2}\right)-L(g, 2 P)\right) \\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon . \end{aligned}$$

Hence $f+g$ is integrable on $[a, b]$.

Moreover,

$$\begin{aligned} \int_{a}^{b}(f+g) & \leq U(f+g, P) \\ & \leq U(f, P)+U(g, P) \\ & \leq\left(\int_{a}^{b} f+\frac{\epsilon}{2}\right)+\left(\int_{a}^{b} g+\frac{\epsilon}{2}\right) \\ &=\int_{a}^{b} f+\int_{a}^{b} g+\epsilon \end{aligned}$$

and

$$\begin{aligned} \int_{a}^{b}(f+g) & \geq L(f+g, P) \\ & \geq L(f, P)+L(g, P) \\ & \geq\left(\int_{a}^{b} f-\frac{\epsilon}{2}\right)+\left(\int_{a}^{b} g-\frac{\epsilon}{2}\right) \\ &=\int_{a}^{b} f+\int_{a}^{b} g-\epsilon . \end{aligned}$$

Since $\epsilon>0$ was arbitrary, it follows that

$$\int_{a}^{b}(f+g)=\int_{a}^{b} f+\int_{a}^{b} g.$$

Q.E.D.

Exercise $\PageIndex{3}$

Suppose $a<b$ and $f:[a, b] \rightarrow \mathbb{R}$ and $g:[a, b] \rightarrow \mathbb{R}$ are both bounded. Show that

$$\overline{\int_{a}^{b}}(f+g) \leq \overline{\int_{a}^{b}} f+\overline{\int_{a}^{b}} g.$$

Find an example for which the inequality is strict.

Exercise $\PageIndex{4}$

Find an example to show that $f+g$ may be integrable on $[a, b]$ even though neither $f$ nor $g$ is integrable on $[a, b]$.

Proposition $\PageIndex{3}$

If $f$ is integrable on $[a, b]$ and $\alpha \in \mathbb{R},$ then $\alpha f$ is integrable on $[a, b]$ and

$$\int_{a}^{b} \alpha f=\alpha \int_{a}^{b} f.$$

Exercise $\PageIndex{5}$

Prove the previous proposition.

Theorem $\PageIndex{4}$

Suppose $a<b, f:[a, b] \rightarrow \mathbb{R}$ is bounded, and $c \in(a, b)$. Then $f$ is integrable on $[a, b]$ if and only if $f$ is integrable on both $[a, c]$ and $[c, b] .$

Proof

Suppose $f$ is integrable on $[a, b] .$ Given $\epsilon>0,$ let $Q$ be a partition of $[a, b]$ such that

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

Let $P=Q \cup\{c\}, P_{1}=P \cap[a, c],$ and $P_{2}=P \cap[c, b] .$ Then

$$\begin{aligned}\left(U\left(f, P_{1}\right)-L\left(f, P_{1}\right)\right)+\left(U\left(f, P_{2}\right)-L\left(f, P_{2}\right)\right) &=\left(U\left(f, P_{1}\right)+U\left(f, P_{2}\right)\right) \\ &-\left(L\left(f, P_{1}\right)+L\left(f, P_{2}\right)\right) \\ &=U(f, P)-L(f, P) \\ & \leq U(f, Q)-L(f, Q) \\ &<\epsilon . \end{aligned}$$

Thus we must have both

$$U\left(f, P_{1}\right)-L\left(f, P_{1}\right)<\epsilon$$

and

$$U\left(f, P_{2}\right)-L\left(f, P_{2}\right)<\epsilon.$$

Hence $f$ is integrable on both $[a, c]$ and $[c, b]$.

Now suppose $f$ is integrable on both $[a, c]$ and $[c, b] .$ Given $\epsilon>0,$ let $P_{1}$ and $P_{2}$ be partitions of $[a, c]$ and $[c, b],$ respectively, such that

$$U\left(f, P_{1}\right)-L\left(f, P_{1}\right)<\frac{\epsilon}{2}$$

and

$$U\left(f, P_{2}\right)-L\left(f, P_{2}\right)<\frac{\epsilon}{2}.$$

Let $P=P_{1} \cup P_{2}$. Then $P$ is a partition of $[a, b]$ and

$$\begin{aligned} U(f, P)-L(f, P) &=\left(U\left(f, P_{1}\right)+U\left(f, P_{2}\right)\right)-\left(L\left(f, P_{1}\right)+L\left(f, P_{2}\right)\right) \\ &=\left(U\left(f, P_{1}\right)-L\left(f, P_{1}\right)\right)+\left(U\left(f, P_{2}\right)-L\left(f, P_{2}\right)\right) \\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2} \\ &=\epsilon . \end{aligned}$$

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

Proposition $\PageIndex{5}$

Suppose $f$ is integrable on $[a, b]$ and $c \in(a, b) .$ Then

$$\int_{a}^{b} f=\int_{a}^{c} f+\int_{c}^{b} f.$$

Proof

If $P$ and $Q$ are partitions of $[a, c]$ and $[c, b],$ respectively, then

$$U(f, P)+U(f, Q)=U(f, P \cup Q) \geq \int_{a}^{b} f.$$

Thus

$$U(f, P) \geq \int_{a}^{b} f-U(f, Q),$$

so

$$\int_{a}^{c} f= \overline{\int_{a}^{c}} f \geq \int_{a}^{b} f-U(f, Q) .$$

Hence

$$U(f, Q) \geq \int_{a}^{b} f-\int_{a}^{c} f,$$

so

$$\int_{c}^{b} f= \overline{\int_{c}^{b}} f \geq \int_{a}^{b} f-\int_{a}^{c} f.$$

Thus

$$\int_{a}^{c} f+\int_{c}^{b} f \geq \int_{a}^{b} f.$$

Similarly, if $P$ and $Q$ are partitions of $[a, c]$ and $[c, b],$ respectively, then

$$L(f, P)+L(f, Q)=L(f, P \cup Q) \leq \int_{a}^{b} f.$$

Thus

$$L(f, P) \leq \int_{a}^{b} f-L(f, Q),$$

so

$$\int_{a}^{c} f= \underline{\int_{a}^{c}} f \leq \int_{a}^{b} f-L(f, Q).$$

Hence

$$L(f, Q) \leq \int_{a}^{b} f-\int_{a}^{c} f,$$

so

$$\int_{c}^{b} f= \underline{\int_{c}^{b}} f \leq \int_{a}^{b} f-\int_{a}^{c} f.$$

Thus

$$\int_{a}^{c} f+\int_{c}^{b} f \leq \int_{a}^{b} f.$$

Hence

$$\int_{a}^{c} f+\int_{c}^{b} f=\int_{a}^{b} f.$$

Q.E.D.

Exercise $\PageIndex{6}$

Suppose $f:[a, b] \rightarrow \mathbb{R}$ is bounded and $B$ is a finite subset of $(a, b) .$ Show that if $f$ is continuous on $[a, b] \backslash B,$ then $f$ is integrable on $[a, b]$.

Proposition $\PageIndex{6}$

If $f$ is integrable on $[a, b]$ with $f(x) \geq 0$ for all $x \in[a, b]$, then

$$\int_{a}^{b} f \geq 0.$$

Proof

The result follows from the fact that $L(f, P) \geq 0$ for any partition $P$ of $[a, b]$. $\quad$ Q.E.D.

Proposition $\PageIndex{7}$

Suppose $f$ and $g$ are both integrable on $[a, b] .$ If, for all $x \in[a, b], f(x) \leq g(x),$ then

$$\int_{a}^{b} f \leq \int_{a}^{b} g.$$

Proof

Since $g(x)-f(x) \geq 0$ for all $x \in[a, b],$ we have, using Propositions 7.4.2, 7.4.3, and 7.4.6,

$$\int_{a}^{b} g-\int_{a}^{b} f=\int_{a}^{b}(g-f) \geq 0.$$

Q.E.D.

Proposition $\PageIndex{8}$

Suppose $f$ is integrable on $[a, b], m \in \mathbb{R}, M \in \mathbb{R},$ and $m \leq f(x) \leq M$ for all $x \in[a, b] .$ Then

$$m(b-a) \leq \int_{a}^{b} f \leq M(b-a).$$

Proof

It follows from the previous proposition that

$$m(b-a)=\int_{a}^{b} m d x \leq \int_{a}^{b} f(x) d x \leq \int_{a}^{b} M d x=M(b-a).$$

Q.E.D.

Exercise $\PageIndex{7}$

Show that

$$1 \leq \int_{-1}^{1} \frac{1}{1+x^{2}} d x \leq 2.$$

Exercise $\PageIndex{8}$

Suppose $f$ is continuous on $[0,1],$ differentiable on $(0,1)$, $f(0)=0,$ and $\left|f^{\prime}(x)\right| \leq 1$ for all $x \in(0,1) .$ Show that

$$-\frac{1}{2} \leq \int_{0}^{1} f \leq \frac{1}{2}.$$

Exercise $\PageIndex{9}$

Suppose $f$ is integrable on $[a, b]$ and define $F:(a, b) \rightarrow \mathbb{R}$ by

$$F(x)=\int_{a}^{x} f.$$

Show that there exists $\alpha \in \mathbb{R}$ such that for any $x, y \in(a, b)$ with $x<y$,

$$|F(y)-F(x)| \leq \alpha(y-x).$$

Proposition $\PageIndex{9}$

Suppose $g$ is integrable on $[a, b], g([a, b]) \subset[c, d],$ and $f:[c, d] \rightarrow \mathbb{R}$ is continuous. If $h=f \circ g,$ then $h$ is integrable on $[a, b]$.

Proof

Let $\epsilon>0$ be given. Let

$$K>\sup \{f(x): x \in[c, d]\}-\inf \{f(x): x \in[c, d]\}$$

and choose $\delta>0$ so that $\delta<\epsilon$ and

$$|f(x)-f(y)|<\frac{\epsilon}{2(b-a)}$$

whenever $|x-y|<\delta .$ Let $P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}$ be a partition of $[a, b]$ such that

$$U(g, P)-L(g, P)<\frac{\delta^{2}}{2 K}.$$

For $i=1,2, \dots, n,$ let

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

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

$$w_{i}=\inf \left\{h(x): x_{i-1} \leq x \leq x_{i}\right\},$$

and

$$W_{i}=\sup \left\{h(x): x_{i-1} \leq x \leq x_{i}\right\}.$$

Finally, let

$$I=\left\{i: i \in \mathbb{Z}, 1 \leq i \leq n, M_{i}-m_{i}<\delta\right\}$$

and

$$J=\left\{i: i \in \mathbb{Z}, 1 \leq i \leq n, M_{i}-m_{i} \geq \delta\right\}.$$

Note that

$$\begin{aligned} \delta \sum_{i \in J}\left(x_{i}-x_{i-1}\right) & \leq \sum_{i \in J}\left(M_{i}-m_{i}\right)\left(x_{i}-x_{i-1}\right) \\ & \leq \sum_{i=1}^{n}\left(M_{i}-m_{i}\right)\left(x_{i}-x_{i-1}\right) \\ &<\frac{\delta^{2}}{2 K}, \end{aligned}$$

from which it follows that

$$\sum_{i \in J}\left(x_{i}-x_{i-1}\right)<\frac{\delta}{2 K}.$$

Then

$$\begin{aligned} U(h, P)-L(h, P) &=\sum_{i \in I}\left(W_{i}-w_{i}\right)\left(x_{i}-x_{i-1}\right)+\sum_{i \in J}\left(W_{i}-w_{i}\right)\left(x_{i}-x_{i-1}\right) \\ &<\frac{\epsilon}{2(b-a)} \sum_{i \in I}\left(x_{i}-x_{i-1}\right)+K \sum_{i \in J}\left(x_{i}-x_{i-1}\right) \\ &<\frac{\epsilon}{2}+\frac{\delta}{2} \\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2} \\ &=\epsilon . \end{aligned}$$

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

Proposition $\PageIndex{10}$

Suppose $f$ and $g$ are both integrable on $[a, b] .$ Then $f g$ is integrable on $[a, b]$.

Proof

Since $f$ and $g$ are both integrable, both $f+g$ and $f-g$ are integrable. Hence, by the previous proposition, both $(f+g)^{2}$ and $(f-g)^{2}$ are integrable. Thus

$$\left.\frac{1}{4}\left((f+g)^{2}-(f-g)^{2}\right)\right)=f g$$

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

Proposition $\PageIndex{11}$

Suppose $f$ is integrable on $[a, b] .$ Then $|f|$ is integrable on $[a, b]$ and

$$\left|\int_{a}^{b} f\right| \leq \int_{a}^{b}|f|.$$

Proof

The integrability of $|f|$ follows from the integrability of $f,$ the continuity of $g(x)=|x|,$ and Proposition $7.4 .9 .$ For the inequality, note that

$$-|f(x)| \leq f(x) \leq|f(x)|$$

for all $x \in[a, b] .$ Hence

$$-\int_{a}^{b}|f| \leq \int_{a}^{b} f \leq \int_{a}^{b}|f|,$$

from which the result follows. $\quad$ Q.E.D.

Exercise $\PageIndex{10}$

Either prove the following statement or show it is false by finding a counterexample: If $f:[0,1] \rightarrow \mathbb{R}$ is bounded and $f^{2}$ is integrable on $[0,1],$ then $f$ is integrable on $[0,1] .$

7.4.1 Extended definitions

Definition

If $f$ integrable on $[a, b],$ then we define

$$\int_{b}^{a} f=-\int_{a}^{b} f.$$

Moreover, if $f$ is a function defined at a point $a \in \mathbb{R},$ we define

$$\int_{a}^{a} f=0.$$

Definition

Suppose $f$ is integrable on a closed interval containing the

points $a, b,$ and $c .$ Show that

$$\int_{a}^{b} f=\int_{a}^{c} f+\int_{c}^{b} f.$$