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# 8.4: Integration of Elementary Functions

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In Chapter 5, integration was treated as antidifferentiation. Now we adopt another, measure-theoretical approach.

Lebesgue's original theory was based on Lebesgue measure (Chapter 7, §8). The more general modern treatment develops the integral for functions $$f : S \rightarrow E$$ in an arbitrary measure space. Henceforth, $$(S, \mathcal{M}, m)$$ is fixed, and the range space $$E$$ is $$E^{1}, E^{*}, C, E^{n},$$ or another complete normed space. Recall that in such a space, $$\sum_{i}\left|a_{i}\right|<\infty$$ implies that $$\sum a_{i}$$ converges and is permutable (Chapter 7, §2).

We start with elementary maps, including simple maps as a special case.

## Definition

Let $$f : S \rightarrow E$$ be elementary on $$A \in \mathcal{M};$$ so $$f=a_{i}$$ on $$A_{i}$$ for some $$\mathcal{M}$$-partition

$A=\bigcup_{i} A_{i} \text { (disjoint).}$

(Note that there may be many such partitions.)

We say that $$f$$ is integrable (with respect to $$m$$), or $$m$$-integrable, on $$A$$ iff

$\sum\left|a_{i}\right| m A_{i}<\infty.$

(The notation "$$|a_{i}| m A_{i}$$" always makes sense by our conventions (2*) in Chapter 4, §4.) If $$m$$ is Lebesgue measure, then we say that $$f$$ is Lebesgue integrable, or L-integrable.

We then define $$\int_{A} f,$$ the $$m$$-integral of $$f$$ on $$A,$$ by

$\int_{A} f=\int_{A} f d m=\sum_{i} a_{i} m A_{i}.$

(The notation "$$dm$$" is used to specify the measure $$m$$.)

The "classical" notation for $$\int_{A} f d m$$ is $$\int_{A} f(x) d m(x)$$.

Note 1. The assumption

$\sum\left|a_{i}\right| m A_{i}<\infty$

implies

$(\forall i) \quad\left|a_{i}\right| m A_{i}<\infty;$

so $$a_{i}=0$$ if $$m A_{i}=\infty,$$ and $$m A_{i}=0$$ if $$|a_{i}|=\infty.$$ Thus by our conventions, all "bad" terms $$a_{i} m A_{i}$$ vanish. Hence the sum in (1) makes sense and is finite.

Note 2. This sum is also independent of the particular choice of $$\{A_{i}\}.$$ For if $$\{B_{k}\}$$ is another $$\mathcal{M}$$-partition of $$A,$$ with $$f=b_{k}$$ on $$B_{k},$$ say, then $$f=a_{i}=b_{k}$$ on $$A_{i} \cap B_{k}$$ whenever $$A_{i} \cap B_{k} \neq \emptyset.$$ Also,

$(\forall i) \quad A_{i}=\bigcup_{k}\left(A_{i} \cap B_{k}\right) \text { (disjoint);}$

so

$(\forall i) \quad a_{i} m A_{i}=\sum_{k} a_{i} m(A_{i} \cap B_{k}),$

and hence (see Theorem 2 of Chapter 7, §2, and Problem 11 there)

$\sum_{i} a_{i} m A_{i}=\sum_{i} \sum_{k} a_{i} m\left(A_{i} \cap B_{k}\right)=\sum_{k} \sum_{i} b_{k} m\left(A_{i} \cap B_{k}\right)=\sum_{k} b_{k} m B_{k}.$

(Explain!)

This makes our definition (1) unambiguous and allows us to choose any $$\mathcal{M}$$-partition $$\{A_{i}\},$$ with $$f$$ constant on each $$A_{i},$$ when forming integrals (1).

## Corollary $$\PageIndex{1}$$

Let $$f : S \rightarrow E$$ be elementary and integrable on $$A \in \mathcal{M}.$$ Then the following statements are true.

(i) $$|f|<\infty$$ a.e. on $$A.$$

(ii) $$f$$ and $$|f|$$ are elementary and integrable on any $$\mathcal{M}$$-set $$B \subseteq A,$$ and

$\left|\int_{B} f\right| \leq \int_{B}|f| \leq \int_{A}|f|.$

(iii) The set $$B=A(f \neq 0)$$ is $$\sigma$$-finite (Definition 4 in Chapter 7, §5), and

$\int_{A} f=\int_{B} f.$

(iv) If $$f=a$$ (constant) on $$A$$,

$\int_{A} f=a \cdot m A.$

(v) $$\int_{A}|f|=0$$ iff $$f=0$$ a.e. on $$A$$.

(vi) If $$m Q=0,$$ then

$\int_{A} f=\int_{A-Q} f$

(so we may neglect sets of measure 0 in integrals).

(vii) For any $$k$$ in the scalar field of $$E, k f$$ is elementary and integrable, and

$\int_{A} k f=k \int_{A} f.$

Note that if $$f$$ is scalar valued, $$k$$ may be a vector. If $$E=E^{*},$$ we assume $$k \in E^{1}.$$

Proof

(i) By Note 1, $$|f|=|a_{i}|=\infty$$ only on those $$A_{i}$$ with $$m A_{i}=0.$$ Let $$Q$$ be the union of all such $$A_{i}.$$ Then $$m Q=0$$ and $$|f|<\infty$$ on $$A-Q,$$ proving (i).

(ii) If $$\{A_{i}\}$$ is an $$\mathcal{M}$$-partition of $$A,\{B \cap A_{i}\}$$ is one for $$B.$$ (Verify!) We have $$f=a_{i}$$ and $$|f|=|a_{i}|$$ on $$B \cap A_{i} \subseteq A_{i}$$.

Also,

$\sum\left|a_{i}\right| m\left(B \cap A_{i}\right) \leq \sum\left|a_{i}\right| m A_{i}<\infty.$

(Why?) Thus $$f$$ and $$|f|$$ are elementary and integrable on $$B,$$ and (ii) easily follows by formula (1).

(iii) By Note 1, $$f=0$$ on $$A_{i}$$ if $$m A_{i}=\infty.$$ Thus $$f \neq 0$$ on $$A_{i}$$ only if $$m A_{i}<\infty$$. Let $$\{A_{i_{k}}\}$$ be the subsequence of those $$A_{i}$$ on which $$f \neq 0;$$ so

$(\forall k) \quad m A_{i_{k}}<\infty.$

Also,

$B=A(f \neq 0)=\bigcup_{k} A_{i_{k}} \in \mathcal{M} \text{ (}\sigma \text {-finite!).}$

By (ii), $$f$$ is elementary and integrable on $$B.$$ Also,

$\int_{B} f=\sum_{k} a_{i_{k}} m A_{i_{k}},$

while

$\int_{A} f=\sum_{i} a_{i} m A_{i}.$

These sums differ only by terms with $$a_{i}=0.$$ Thus (iii) follows.

The proof of (iv)-(vii) is left to the reader.$$\quad \square$$

Note 3. If $$f : S \rightarrow E^{*}$$ is elementary and sign-constant on $$A,$$ we also allow that

$\int_{A} f=\sum_{i} a_{i} m A_{i}=\pm \infty.$

Thus here $$\int_{A} f$$ exists even if $$f$$ is not integrable. Apart from claims of integrability and $$\sigma$$-finiteness, Corollary 1(ii)-(vii) hold for such $$f$$, with the same proofs.

## Example

Let $$m$$ be Lebesgue measure in $$E^{1}.$$ Define $$f=1$$ on $$R$$ (rationals) and $$f=0$$ on $$E^{1}-R ;$$ see Chapter 4, §1, Example (c). Let $$A=[0,1].$$

By Corollary 1 in Chapter 7, §8, $$A \cap R \in \mathcal{M}^{*}$$ and $$m(A \cap R)=0.$$ Also, $$A-R \in \mathcal{M}^{*}$$.

Thus $$\{A \cap R, A-R\}$$ is an $$\mathcal{M}^{*}$$-partition of $$A,$$ with $$f=1$$ on $$A \cap R$$ and $$f=0$$ on $$A-R.$$

Hence $$f$$ is elementary and integrable on $$A,$$ and

$\int_{A} f=1 \cdot m(A \cap R)+0 \cdot m(A-R)=0.$

Thus $$f$$ is L-integrable (even though it is nowhere continuous).

## Theorem $$\PageIndex{1}$$ (additivity)

(i) If $$f : S \rightarrow E$$ is elementary and integrable or elementary and nonnegative on $$A \in \mathcal{M},$$ then

$\int_{A} f=\sum_{k} \int_{B_{k}} f$

for any $$\mathcal{M}$$-partition $$\left\{B_{k}\right\}$$ of $$A$$.

(ii) If $$f$$ is elementary and integrable on each set $$B_{k}$$ of a finite $$\mathcal{M}$$-partition

$A=\bigcup_{k} B_{k},$

it is elementary and integrable on all of $$A,$$ and (2) holds again.

Proof

(i) If $$f$$ is elementary and integrable or elementary and nonnegative on $$A=\bigcup_{k} B_{k},$$ it is surely so on each $$B_{k}$$ by Corollary 2 of §1 and Corollary 1(ii) above.

Thus for each $$k,$$ we can fix an $$\mathcal{M}$$-partition $$B_{k}=\bigcup_{i} A_{k i},$$ with $$f$$ constant $$(f=a_{k i})$$ on $$A_{k i}, i=1,2, \ldots$$. Then

$A=\bigcup_{k} B_{k}=\bigcup_{k} \bigcup_{i} A_{k i}$

is an $$\mathcal{M}$$-partition of $$A$$ into the disjoint sets $$A_{k i} \in \mathcal{M}$$.

Now, by definition,

$\int_{B_{k}} f=\sum_{i} a_{k i} m A_{k i}$

and

$\int_{A} f=\sum_{k, i} a_{k i} m A_{k i}=\sum_{k}\left(\sum_{i} a_{k i} m A_{k i}\right)=\sum_{k} \int_{B_{k}} f$

by rules for double series. This proves formula (2).

(ii) If $$f$$ is elementary and integrable on $$B_{k}(k=1, \ldots, n),$$ then with the same notation, we have

$\sum_{i}\left|a_{k i}\right| m A_{k i}<\infty$

(by integrability); hence

$\sum_{k=1}^{n} \sum_{i}\left|a_{k i}\right| m A_{k i}<\infty.$

This means, however, that $$f$$ is elementary and integrable on $$A,$$ and so clause (ii) follows.$$\quad \square$$

Caution. Clause (ii) fails if the partition $$\{B_{k}\}$$ is infinite.

## Theorem $$\PageIndex{2}$$

(i) If $$f, g : S \rightarrow E^{*}$$ are elementary and nonnegative on $$A,$$ then

$\int_{A}(f+g)=\int_{A} f+\int_{A} g.$

(ii) If $$f, g : S \rightarrow E$$ are elementary and integrable on $$A,$$ so is $$f \pm g,$$ and

$\int_{A}(f \pm g)=\int_{A} f \pm \int_{A} g.$

Proof

Arguing as in the proof of Theorem 1 of §1, we can make $$f$$ and $$g$$ constant on sets of one and the same $$\mathcal{M}$$-partition of $$A,$$ say, $$f=a_{i}$$ and $$g=b_{i}$$ on $$A_{i} \in \mathcal{M};$$ so

$f \pm g=a_{i} \pm b_{i} \text { on } A_{i}, \quad i=1,2, \ldots.$

In case (i), $$f, g \geq 0;$$ so integrability is irrelevant by Note 3, and formula (1) yields

$\int_{A}(f+g)=\sum_{i}\left(a_{i}+b_{i}\right) m A_{i}=\sum_{i} a_{i} m A_{i}+\sum b_{i} m A_{i}=\int_{A} f+\int_{A} g.$

In (ii), we similarly obtain

$\sum_{i}\left|a_{i} \pm b_{i}\right| m A_{i} \leq \sum\left|a_{i}\right| m A_{i}+\sum_{i}\left|b_{i}\right| m A_{i}<\infty.$

(Why?) Thus $$f \pm g$$ is elementary and integrable on $$A.$$ As before, we also get

$\int_{A}(f \pm g)=\int_{A} f \pm \int_{A} g,$

simply by rules for addition of convergent series. (Verify!)$$\quad \square$$

Note 4. As we know, the characteristic function $$C_{B}$$ of a set $$B \subseteq S$$ is defined

$C_{B}(x)=\left\{\begin{array}{ll}{1,} & {x \in B,} \\ {0,} & {x \in S-B.}\end{array}\right.$

If $$g : S \rightarrow E$$ is elementary on $$A,$$ so that

$g=a_{i} \text { on } A_{i}, 1,2, \ldots,$

for some $$\mathcal{M}$$-partition

$A=\bigcup A_{i},$

then

$g=\sum_{i} a_{i} C_{A_{i}} \text { on } A.$

(This sum always exists for disjoint sets $$A_{i}.$$ Why?) We shall often use this notation. If $$m$$ is Lebesgue measure in $$E^{1},$$ the integral

$\int_{A} g=\sum_{i} a_{i} m A_{i}$

has a simple geometric interpretation; see Figure 33. Let $$A=[a, b] \subset E^{1};$$ let $$g$$ be bounded and nonnegative on $$E^{1}.$$ Each product $$a_{i} m A_{i}$$ is the area of a rectangle with base $$A_{i}$$ and altitude $$a_{i}.$$ (We assume the $$A_{i}$$ to be intervals here.) The total area,

$\int_{A} g=\sum_{i} a_{i} m A_{i},$

can be treated as an approximation to the area under some curve $$y=f(x)$$, where $$f$$ is approximated by $$g$$ (Theorem 3 in §1). Integration historically arose from such approximations.

Integration of elementary extended-real functions. Note 3 can be extended to sign-changing functions as follows.

## Definition

If

$f=\sum_{i} a_{i} C_{A_{i}} \quad\left(a_{i} \in E^{*}\right)$

on

$A=\bigcup_{i} A_{i} \quad\left(A_{i} \in \mathcal{M}\right),$

we set

$\int_{A} f=\int_{A} f^{+}-\int_{A} f^{-},$

with

$f^{+}=f \vee 0 \geq 0 \text { and } f^{-}=(-f) \vee 0 \geq 0;$

see §2.

By Theorem 2 in §2, $$f^{+}$$ and $$f^{-}$$ are elementary and nonnegative on $$A;$$ so

$\int_{A} f^{+} \text { and } \int_{A} f^{-}$

are defined by Note 3, and so is

$\int_{A} f=\int_{A} f^{+}-\int_{A} f^{-}$

by our conventions (2*) in Chapter 4, §4.

We shall have use for formula (3), even if

$\int_{A} f^{+}=\int_{A} f^{-}=\infty;$

then we say that $$\int_{A} f$$ is unorthodox and equate it to $$+\infty,$$ by convention; cf. Chapter 4, §4. (Other integrals are called orthodox.) Thus for elementary and (extended) real functions, $$\int_{A} f$$ is always defined. (We further develop this idea in §5.)

Note 5. With $$f$$ as above, we clearly have

$f^{+}=a_{i}^{+} \text { and } f^{-}=a_{i}^{-} \text { on } A_{i},$

where

$a_{i}^{+}=\max \left(a_{i}, 0\right) \text { and } a_{i}^{-}=\max \left(-a_{i}, 0\right).$

Thus

$\int_{A} f^{+}=\sum a_{i}^{+} \cdot m A_{i} \text { and } \int_{A} f^{-}=\sum a_{i}^{-} \cdot m A_{i},$

so that

$\int_{A} f=\int_{A} f^{+}-\int_{A} f^{-}=\sum_{i} a_{i}^{+} \cdot m A_{i}-\sum_{i} a_{i}^{-} \cdot m A_{i}.$

If $$\int_{A} f^{+}<\infty$$ or $$\int_{A} f^{-}<\infty,$$ we can subtract the two series termwise (Problem 14 of Chapter 4, §13) to obtain

$\int_{A} f=\sum_{i}\left(a_{i}^{+}-a_{i}^{-}\right) m A_{i}=\sum_{i} a_{i} m A_{i}$

for $$a_{i}^{+}-a_{i}^{-}=a_{i}.$$ Thus formulas (3) and (4) agree with our previous definitions.