# 7.4: Properties of Integrals

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##### 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

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

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