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.
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 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 \(m A<\infty\).
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).\(\quad \square\)