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).$\quad \square$