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# 8.7: Integration of Complex and Vector-Valued Functions


I. First we consider functions $$f : S \rightarrow E^{n}\left(C^{n}\right) .$$ For such functions, it is natural (and easy) to define integration "componentwise" as follows.

## Definition

A function $$f : S \rightarrow E^{n}$$ is said to be integrable on $$A \in \mathcal{M}$$ iff its $$n$$ (real) components, $$f_{1}, \ldots, f_{n},$$ are. In this case, we define

$\int_{A} f=\int_{A} f d m=\left(\int_{A} f_{1}, \int_{A} f_{2}, \ldots, \int_{A} f_{n}\right)=\sum_{k=1}^{n} \overline{e}_{k} \cdot \int_{A} f_{k}$

where the $$\overline{e}_{k}$$ are basic unit vectors (as in Chapter 3, §§1-3, Theorem 2$$)$$.

In particular, a complex function $$f$$ is integrable on $$A$$ iff its real and imaginary parts $$\left(f_{\text { re}} \text{ and} f_{\text { im }}\right)$$ are. Then we also say that $$\int_{A} f$$ exists. By $$(1),$$ we have

$\int_{A} f=\left(\int_{A} f_{\mathrm{re}}, \int_{A} f_{\mathrm{im}}\right)=\int_{A} f_{\mathrm{re}}+i \int_{A} f_{\mathrm{im}}.$

If $$f : S \rightarrow C^{n},$$ we use $$(1),$$ with complex components $$f_{k}$$

With this definition, integration of functions $$f : S \rightarrow E^{n}\left(C^{n}\right)$$ reduces to that of $$f_{k} : S \rightarrow E^{1}(C),$$ and one easily obtains the same theorems as in §§4-6, as far as they make sense for vectors.

## Theorem $$\PageIndex{1}$$

A function $$f : S \rightarrow E^{n}\left(C^{n}\right)$$ is integrable on $$A \in \mathcal{M}$$ iff it isc$$m$$-measurable on $$A$$ and $$\int_{A}|f|<\infty.$$

(Alternate definition!)

Proof

Assume the range space is $$E^{n}$$.

By our definition, if $$f$$ is integrable on $$A,$$ then its components $$f_{k}$$ are. Thus by Theorem 2 and Corollary 1, both in §6, for $$k=1,2, \ldots, n,$$ the functions $$f_{k}^{+}$$ and $$f_{k}^{-}$$ are $$m$$-measurable; furthermore,

$\int_{A} f_{k}^{+} \neq \pm \infty \text { and } \int_{A} f_{k}^{-} \neq \pm \infty.$

This implies

$\infty>\int_{A} f_{k}^{+}+\int_{A} f_{k}^{-}=\int_{A}\left(f_{k}^{+}+f_{k}^{-}\right)=\int_{A}\left|f_{k}\right|, \quad k=1,2, \ldots, n.$

Since $$|f|$$ is $$m$$-measurable by Problem 14 in §3 ($$| \cdot |$$ is a continuous mapping from $$E^{n}$$ to $$E^{1}$$), and

$|f|=\left|\sum_{k=1}^{n} \overline{e}_{k} f_{k}\right| \leq \sum_{k=1}^{n}\left|\overline{e}_{k}\right|\left|f_{k}\right|=\sum_{k=1}^{n}\left|f_{k}\right|,$

we get

$\int_{A}|f| \leq \int_{A} \sum_{1}^{n}\left|f_{k}\right|=\sum_{1}^{n} \int_{A}\left|f_{k}\right|<\infty.$

Conversely, if $$f$$ satisfies

$\int_{A}|f|<\infty$

then

$(\forall k) \quad\left|\int_{A} f_{k}\right|<\infty.$

Also, the $$f_{k}$$ are $$m$$-measurable if $$f$$ is (see Problem 2 in §3). Hence the $$f_{k}$$ are integrable on $$A$$ (by Theorem 2 of §6), and so is $$f.$$

The proof for $$C^{n}$$ is analogous.$$\quad \square$$

Similarly for other theorems (see Problems 1 to 4 below). We have already noted that Theorem 5 of §6 holds for complex and vector-valued functions. So does Theorem 6 in §6. We prove another such proposition (Lemma 1) below.

II. Next we consider the general case, $$f : S \rightarrow E$$ ($$E$$ complete). We now adopt Theorem 1 as a definition. (It agrees with Definition 1 of §4. Verify!) Even if $$E=E^{*},$$ we always assume $$|f|<\infty$$ a.e.; thus, dropping a null set, we can make $$f$$ finite and use the standard metric on $$E^{1}.$$

First, we take up the case $$m A<\infty$$.

## Lemma $$\PageIndex{1}$$

If $$f_{n} \rightarrow f$$ (uniformly) on $$A$$ ($$m A<\infty$$), then

$\int_{A}\left|f_{n}-f\right| \rightarrow 0.$

Proof

By assumption,

$(\forall \varepsilon>0) \text { } (\exists k) \text { } (\forall n>k) \quad\left|f_{n}-f\right|<\varepsilon \text { on } A;$

so

$(\forall n>k) \quad \int_{A}\left|f_{n}-f\right| \leq \int_{A}(\varepsilon)=\varepsilon \cdot m A<\infty.$

As $$\varepsilon$$ is arbitrary, the result follows.$$\quad \square$$

## Lemma $$\PageIndex{2}$$

If

$\int_{A}|f|<\infty \quad(m A<\infty)$

and

$f=\lim _{n \rightarrow \infty} f_{n} \text { (uniformly) on } A-Q \text { }(m Q=0)$

for some elementary maps $$f_{n}$$ on $$A,$$ then all but finitely many $$f_{n}$$ are elementary and integrable on $$A,$$ and

$\lim _{n \rightarrow \infty} \int_{A} f_{n}$

exists in $$E;$$ further, the latter limit does not depend on the sequence $$\left\{f_{n}\right\}$$.

Proof

By Lemma 1,

$(\forall \varepsilon>0) \text { } (\exists q) \text { } (\forall n, k>q) \quad \int_{A}\left|f_{n}-f\right|<\varepsilon \text { and } \int_{A}\left|f_{n}-f_{k}\right|<\varepsilon.$

(The latter can be achieved since

$\lim _{k \rightarrow \infty} \int_{A}\left|f_{n}-f_{k}\right|=\int_{A}\left|f_{n}-f\right|<\varepsilon.)$

Now, as

$\left|f_{n}\right| \leq\left|f_{n}-f\right|+|f|,$

Problem 7 in §5 yields

$(\forall n>k) \quad \int_{A}\left|f_{n}\right| \leq \int_{A}\left|f_{n}-f\right|+\int_{A}|f|<\varepsilon+\int_{A}|f|<\infty.$

Thus $$f_{n}$$ is elementary and integrable for $$n>k,$$ as claimed. Also, by Theorem 2 and Corollary 1(ii), both in §4,

$(\forall n, k>q) \quad\left|\int_{A} f_{n}-\int_{A} f_{k}\right|=\left|\int_{A}\left(f_{n}-f_{k}\right)\right| \leq \int_{A}\left|f_{n}-f_{k}\right|<\varepsilon.$

Thus $$\left\{\int_{A} f_{n}\right\}$$ is a Cauchy sequence. As $$E$$ is complete,

$\lim \int_{A} f_{n} \neq \pm \infty$

exists in $$E,$$ as asserted.

Finally, suppose $$g_{n} \rightarrow f$$ (uniformly) on $$A-Q$$ for some other elementary and integrable maps $$g_{n}.$$ By what was shown above, $$\lim \int_{A} g_{n}$$ exists, and

$\left|\lim \int_{A} g_{n}-\lim \int_{A} f_{n}\right|=\left|\lim \int_{A}\left(g_{n}-f_{n}\right)\right| \leq \lim \int_{A}\left|g_{n}-f_{n}-0\right|=0$

by Lemma 1, as $$g_{n}-f_{n} \rightarrow 0$$ (uniformly) on $$A.$$ Thus

$\lim \int_{A} g_{n}=\lim \int_{A} f_{n},$

and all is proved.$$\quad \square$$

This leads us to the following definition.

## Definition

If $$f : S \rightarrow E$$ is integrable on $$A \in \mathcal{M}$$ $$(m A<\infty),$$ we set

$\int_{A} f=\int_{A} f d m=\lim _{n \rightarrow \infty} \int_{A} f_{n}$

for any elementary and integrable maps $$f_{n}$$ such that $$f_{n} \rightarrow f$$ (uniformly) on $$A-Q, m Q=0$$.

Indeed, such maps exist by Theorem 3 of §1, and Lemma 2 excludes ambiguity.

*Note 1. If $$f$$ itself is elementary and integrable, Definition 2 agrees with that of §4. For, choosing $$f_{n}=f(n=1,2, \ldots),$$ we get

$\int_{A} f=\int_{A} f_{n}$

(the latter as in §4).

*Note 2. We may neglect sets on which $$f=0,$$ along with null sets. For if $$f=0$$ on $$A-B$$ $$(A \supseteq B, B \in \mathcal{M}),$$ we may choose $$f_{n}=0$$ on $$A-B$$ in Definition 2. Then

$\int_{A} f=\lim \int_{A} f_{n}=\lim \int_{B} f_{n}=\int_{B} f.$

Thus we now define

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

even if $$m A=\infty,$$ provided $$f=0$$ on $$A-B,$$ i.e.,

$f=f C_{B} \text { on } A$

$$(C_{B}=$$ characteristic function of $$B),$$ with $$A \supseteq B, B \in \mathcal{M},$$ and $$m B<\infty$$.

If such a $$B$$ exists, we say that $$f$$ has $$m$$-finite support in $$A$$.

*Note 3. By Corollary 1 in §5,

$\int_{A}|f|<\infty$

implies that $$A(f \neq 0)$$ is $$\sigma$$-finite. Neglecting $$A(f=0),$$ we may assume that

$A=\bigcup B_{n}, m B_{n}<\infty, \text { and }\left\{B_{n}\right\} \uparrow$

(if not, replace $$B_{n}$$ by $$\cup_{k=1}^{n} B_{k}$$); so $$B_{n} \nearrow A$$.

## Lemma $$\PageIndex{3}$$

Let $$\phi : S \rightarrow E$$ be integrable on $$A$$. Let $$B_{n} \nearrow A, m B_{n}<\infty$$ and set

$f_{n}=\phi C_{B_{n}}, \quad n=1,2, \ldots.$

Then $$f_{n} \rightarrow \phi$$ (pointwise) on $$A,$$ all $$f_{n}$$ are integrable on $$A,$$ and

$\lim _{n \rightarrow \infty} \int_{A} f_{n}$

exists in $$E.$$ Furthermore, this limit does not depend on the choice of $$\left\{B_{n}\right\}$$.

Proof

Fix any $$x \in A.$$ As $$B_{n} \nearrow A=\cup B_{n}$$,

$\left(\exists n_{0}\right) \text { } \left(\forall n>n_{0}\right) \quad x \in B_{n}.$

By assumption, $$f_{n}=\phi$$ on $$B_{n}.$$ Thus

$\left(\forall n>n_{0}\right) \quad f_{n}(x)=\phi(x);$

so $$f_{n} \rightarrow \phi$$ (pointwise) on $$A$$.

Moreover, $$f_{n}=\phi C_{B_{n}}$$ is $$m$$-measurable on $$A$$ (as $$\phi$$ and $$C_{B_{n}}$$ are); and

$\left|f_{n}\right|=|\phi| C_{B_{n}}$

implies

$\int_{A}\left|f_{n}\right| \leq \int_{A}|\phi|<\infty.$

Thus all $$f_{n}$$ are integrable on $$A$$.

As $$f_{n}=0$$ on $$A-B_{n}(m B<\infty)$$,

$\int_{A} f_{n}$

is defined. Since $$f_{n} \rightarrow \phi$$ (pointwise) and $$\left|f_{n}\right| \leq|\phi|$$ on $$A,$$ Theorem 5 in §6, with $$g=|\phi|,$$ yields

$\int_{A}\left|f_{n}-\phi\right| \rightarrow 0.$

The rest is as in Lemma 2, with our present Theorem 2 below (assuming $$m$$-finite support of $$f$$ and $$g),$$ replacing Theorem 2 of §4. Thus all is proved.$$\quad \square$$

## Definition

If $$\phi : S \rightarrow E$$ is integrable on $$A \in \mathcal{M},$$ we set

$\int_{A} \phi=\int_{A} \phi d m=\lim _{n \rightarrow \infty} \int_{A} f_{n},$

with the $$f_{n}$$ as in Lemma 3 (even if $$\phi$$ has no $$m$$-finite support).

## Theorem $$\PageIndex{2}$$ (linearity)

If $$f, g : S \rightarrow E$$ are integrable on $$A \in \mathcal{M},$$ so is

$p f+q g$

for any scalars $$p, q.$$ Moreover,

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

Furthermore if $$f$$ and $$g$$ are scalar valued, $$p$$ and $$q$$ may be vectors in $$E$$.

Proof

For the moment, $$f, g$$ denotes mappings with $$m$$-finite support in $$A.$$ Integrability is clear since $$p f+q g$$ is measurable on $$A$$ (as $$f$$ and $$g$$ are), and

$|p f+q g| \leq|p||f|+|q||g|$

yields

$\int_{A}|p f+q g| \leq|g| \int_{A}|f|+|q| \int_{A}|g|<\infty.$

Now, as noted above, assume that

$f=f C_{B_{1}} \text { and } g=g C_{B_{2}}$

for some $$B_{1}, B_{2} \subseteq A(m B_{1}+m B_{2}<\infty).$$ Let $$B=B_{1} \cup B_{2};$$ so

$f=g=p f+q g=0 \text { on } A-B;$

additionally,

$\int_{A} f=\int_{B} f, \int_{A} g=\int_{B} g, \text { and } \int_{A}(p f+q g)=\int_{B}(p f+q g).$

Also, $$m B<\infty;$$ so by Definition 2,

$\int_{B} f=\lim \int_{B} f_{n} \text { and } \int_{B} g=\lim \int_{B} g_{n}$

for some elementary and integrable maps

$f_{n} \rightarrow f \text { (uniformly) and } g_{n} \rightarrow g \text { (uniformly) on } B-Q, m Q=0.$

Thus

$p f_{n}+q g_{n} \rightarrow p f+q g \text { (uniformly) on } B-Q.$

But by Theorem 2 and Corollary 1(vii), both of §4 (for elementary and integrable maps),

$\int_{B}\left(p f_{n}+q g_{n}\right)=p \int_{B} f_{n}+q \int_{B} g_{n}.$

Hence

\begin{aligned} \int_{A}(p f+q g)=& \int_{B}(p f+q g)=\lim \int_{B}\left(p f_{n}+q g_{n}\right) \\ &=\lim \left(p \int_{B} f_{n}+q \int_{B} g_{n}\right)=p \int_{B} f+q \int_{B} g=p \int_{A} f+q \int_{A} g. \end{aligned}

This proves the statement of the theorem, provided $$f$$ and $$g$$ have $$m$$-finite support in $$A.$$ For the general case, we now resume the notation $$f, g, \ldots$$ for any functions, and extend the result to any integrable functions.

Using Definition 3, we set

$A=\bigcup_{n=1}^{\infty} B_{n},\left\{B_{n}\right\} \uparrow, m B_{n}<\infty,$

and

$f_{n}=f C_{B_{n}}, g_{n}=g C_{B_{n}}, \quad n=1,2, \ldots.$

Then by definition,

$\int_{A} f=\lim _{n \rightarrow \infty} \int_{A} f_{n} \text { and } \int_{A} g=\lim _{n \rightarrow \infty} \int_{A} g_{n},$

and so

$p \int_{A} f+q \int_{A} g=\lim _{n \rightarrow \infty}\left(p \int_{A} f_{n}+q \int_{A} g_{n}\right).$

As $$f_{n}, g_{n}$$ have $$m$$-finite supports, the first part of the proof yields

$p \int_{A} f_{n}+q \int_{A} g_{n}=\int_{A}\left(p f_{n}+q g_{n}\right).$

Thus as claimed,

$p \int_{A} f+q \int_{A} g=\lim \int_{A}\left(p f_{n}+q g_{n}\right)=\int_{A}(p f+q g). \quad \square$

Similarly, one extends Corollary 1(ii)(iii)(v) of §4 first to maps with $$m$$-finite support, and then to all integrable maps. The other parts of that corollary need no new proof. (Why?)

## Theorem $$\PageIndex{3}$$ (additivity)

(i) If $$f : S \rightarrow E$$ is integrable on each of $$n$$ disjoint $$\mathcal{M}$$-sets $$A_{k},$$ it is so on their union

$A=\bigcup_{k=1}^{n} A_{k},$

and

$\int_{A} f=\sum_{k=1}^{n} \int_{A_{k}} f.$

(ii) This holds for countable unions, too, if $$f$$ is integrable on all of $$A.$$

Proof

Let $$f$$ have $$m$$-finite support: $$f=f C_{B}$$ on $$A, m B<\infty.$$ Then

$\int_{A} f=\int_{B} f \text { and } \int_{A_{k}} f=\int_{B_{k}} f,$

where

$B_{k}=A_{k} \cap B, \quad k=1,2, \ldots, n.$

By Definition 2, fix elementary and integrable maps $$f_{i}$$ (on $$A$$) and a set $$Q$$ $$(m Q=0)$$ such that $$f_{i} \rightarrow f$$ (uniformly) on $$B-Q$$ (hence also on $$B_{k}-Q$$), with

$\int_{A} f=\int_{B} f=\lim _{i \rightarrow \infty} \int_{B} f_{i} \quad \text { and } \int_{A_{k}} f=\lim _{i \rightarrow \infty} \int_{B_{k}} f_{i}, \quad k=1,2, \ldots, n.$

As the $$f_{i}$$ are elementary and integrable, Theorem 1 in §4 yields

$\int_{A} f_{i}=\int_{B} f_{i}=\sum_{k=1}^{n} \int_{B_{k}} f_{i}=\sum_{k=1}^{n} \int_{A_{k}} f_{i}.$

Hence

$\int_{A} f=\lim _{i \rightarrow \infty} \int_{B} f_{i}=\lim _{i \rightarrow \infty} \sum_{k=1}^{n} \int_{B_{k}} f_{i}=\sum_{k=1}^{n}\left(\lim _{i \rightarrow \infty} \int_{A_{k}} f_{i}\right)=\sum_{k=1}^{n} \int_{A_{k}} f.$

Thus clause (i) holds for maps with $$m$$-finite support. For other functions, (i) now follows quite similarly, from Definition 3. (Verify!)

As for (ii), let $$f$$ be integrable on

$A=\bigcup_{k=1}^{\infty} A_{k} \text { (disjoint),} \quad A_{k} \in \mathcal{M}.$

In this case, set $$g_{n}=f C_{B_{n}},$$ where $$B_{n}=\bigcup_{k=1}^{n} A_{k}, n=1,2, \ldots$$. By clause (i), we have

$\int_{A} g_{n}=\int_{B_{n}} g_{n}=\sum_{k=1}^{n} \int_{A_{k}} g_{n}=\sum_{k=1}^{n} \int_{A_{k}} f,$

since $$g_{n}=f$$ on each $$A_{k} \subseteq B_{n}$$.

Also, as is easily seen, $$\left|g_{n}\right| \leq|f|$$ on $$A$$ and $$g_{n} \rightarrow f$$ (pointwise) on $$A$$ (proof as in Lemma 3). Thus by Theorem 5 in §6,

$\int_{A}\left|g_{n}-f\right| \rightarrow 0.$

As

$\left|\int_{A} g_{n}-\int_{A} f\right|=\left|\int_{A}\left(g_{n}-f\right)\right| \leq \int_{A}\left|g_{n}-f\right|,$

we obtain

$\int_{A} f=\lim _{n \rightarrow \infty} \int_{A} g_{n},$

and the result follows by (3).$$\quad \square$$

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