8.7: Integration of Complex and Vector-Valued Functions
( \newcommand{\kernel}{\mathrm{null}\,}\)
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.
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.
A function f : S \rightarrow E^{n}\left(C^{n}\right) is integrable on A \in \mathcal{M} iff it iscm-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.
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
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.
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.
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
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).
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?)
(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