8.7: Integration of Complex and Vector-Valued Functions
( \newcommand{\kernel}{\mathrm{null}\,}\)
I. First we consider functions f:S→En(Cn). For such functions, it is natural (and easy) to define integration "componentwise" as follows.
A function f:S→En is said to be integrable on A∈M iff its n (real) components, f1,…,fn, are. In this case, we define
∫Af=∫Afdm=(∫Af1,∫Af2,…,∫Afn)=n∑k=1¯ek⋅∫Afk
where the ¯ek 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 (f re andf im ) are. Then we also say that ∫Af exists. By (1), we have
∫Af=(∫Afre,∫Afim)=∫Afre+i∫Afim.
If f:S→Cn, we use (1), with complex components fk
With this definition, integration of functions f:S→En(Cn) reduces to that of fk:S→E1(C), and one easily obtains the same theorems as in §§4-6, as far as they make sense for vectors.
A function f:S→En(Cn) is integrable on A∈M iff it iscm-measurable on A and ∫A|f|<∞.
(Alternate definition!)
- Proof
-
Assume the range space is En.
By our definition, if f is integrable on A, then its components fk are. Thus by Theorem 2 and Corollary 1, both in §6, for k=1,2,…,n, the functions f+k and f−k are m-measurable; furthermore,
∫Af+k≠±∞ and ∫Af−k≠±∞.
This implies
∞>∫Af+k+∫Af−k=∫A(f+k+f−k)=∫A|fk|,k=1,2,…,n.
Since |f| is m-measurable by Problem 14 in §3 (|⋅| is a continuous mapping from En to E1), and
|f|=|n∑k=1¯ekfk|≤n∑k=1|¯ek||fk|=n∑k=1|fk|,
we get
∫A|f|≤∫An∑1|fk|=n∑1∫A|fk|<∞.
Conversely, if f satisfies
∫A|f|<∞
then
(∀k)|∫Afk|<∞.
Also, the fk are m-measurable if f is (see Problem 2 in §3). Hence the fk are integrable on A (by Theorem 2 of §6), and so is f.
The proof for Cn is analogous.◻
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→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|<∞ a.e.; thus, dropping a null set, we can make f finite and use the standard metric on E1.
First, we take up the case mA<∞.
If fn→f (uniformly) on A (mA<∞), then
∫A|fn−f|→0.
- Proof
-
By assumption,
(∀ε>0) (∃k) (∀n>k)|fn−f|<ε on A;
so
(∀n>k)∫A|fn−f|≤∫A(ε)=ε⋅mA<∞.
As ε is arbitrary, the result follows.◻
If
∫A|f|<∞(mA<∞)
and
f=limn→∞fn (uniformly) on A−Q (mQ=0)
for some elementary maps fn on A, then all but finitely many fn are elementary and integrable on A, and
limn→∞∫Afn
exists in E; further, the latter limit does not depend on the sequence {fn}.
- Proof
-
By Lemma 1,
(∀ε>0) (∃q) (∀n,k>q)∫A|fn−f|<ε and ∫A|fn−fk|<ε.
(The latter can be achieved since
limk→∞∫A|fn−fk|=∫A|fn−f|<ε.)
Now, as
|fn|≤|fn−f|+|f|,
Problem 7 in §5 yields
(∀n>k)∫A|fn|≤∫A|fn−f|+∫A|f|<ε+∫A|f|<∞.
Thus fn is elementary and integrable for n>k, as claimed. Also, by Theorem 2 and Corollary 1(ii), both in §4,
(∀n,k>q)|∫Afn−∫Afk|=|∫A(fn−fk)|≤∫A|fn−fk|<ε.
Thus {∫Afn} is a Cauchy sequence. As E is complete,
lim∫Afn≠±∞
exists in E, as asserted.
Finally, suppose gn→f (uniformly) on A−Q for some other elementary and integrable maps gn. By what was shown above, lim∫Agn exists, and
|lim∫Agn−lim∫Afn|=|lim∫A(gn−fn)|≤lim∫A|gn−fn−0|=0
by Lemma 1, as gn−fn→0 (uniformly) on A. Thus
lim∫Agn=lim∫Afn,
and all is proved.◻
This leads us to the following definition.
If f:S→E is integrable on A∈M (mA<∞), we set
∫Af=∫Afdm=limn→∞∫Afn
for any elementary and integrable maps fn such that fn→f (uniformly) on A−Q,mQ=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 fn=f(n=1,2,…), we get
∫Af=∫Afn
(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⊇B,B∈M), we may choose fn=0 on A−B in Definition 2. Then
∫Af=lim∫Afn=lim∫Bfn=∫Bf.
Thus we now define
∫Af=∫Bf,
even if mA=∞, provided f=0 on A−B, i.e.,
f=fCB on A
(CB= characteristic function of B), with A⊇B,B∈M, and mB<∞.
If such a B exists, we say that f has m-finite support in A.
*Note 3. By Corollary 1 in §5,
∫A|f|<∞
implies that A(f≠0) is σ-finite. Neglecting A(f=0), we may assume that
A=⋃Bn,mBn<∞, and {Bn}↑
(if not, replace Bn by ∪nk=1Bk); so Bn↗A.
Let ϕ:S→E be integrable on A. Let Bn↗A,mBn<∞ and set
fn=ϕCBn,n=1,2,….
Then fn→ϕ (pointwise) on A, all fn are integrable on A, and
limn→∞∫Afn
exists in E. Furthermore, this limit does not depend on the choice of {Bn}.
- Proof
-
Fix any x∈A. As Bn↗A=∪Bn,
(∃n0) (∀n>n0)x∈Bn.
By assumption, fn=ϕ on Bn. Thus
(∀n>n0)fn(x)=ϕ(x);
so fn→ϕ (pointwise) on A.
Moreover, fn=ϕCBn is m-measurable on A (as ϕ and CBn are); and
|fn|=|ϕ|CBn
implies
∫A|fn|≤∫A|ϕ|<∞.
Thus all fn are integrable on A.
As fn=0 on A−Bn(mB<∞),
∫Afn
is defined. Since fn→ϕ (pointwise) and |fn|≤|ϕ| on A, Theorem 5 in §6, with g=|ϕ|, yields
∫A|fn−ϕ|→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.◻
If ϕ:S→E is integrable on A∈M, we set
∫Aϕ=∫Aϕdm=limn→∞∫Afn,
with the fn as in Lemma 3 (even if ϕ has no m-finite support).
If f,g:S→E are integrable on A∈M, so is
pf+qg
for any scalars p,q. Moreover,
∫A(pf+qg)=p∫Af+q∫Ag.
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 pf+qg is measurable on A (as f and g are), and
|pf+qg|≤|p||f|+|q||g|
yields
∫A|pf+qg|≤|g|∫A|f|+|q|∫A|g|<∞.
Now, as noted above, assume that
f=fCB1 and g=gCB2
for some B1,B2⊆A(mB1+mB2<∞). Let B=B1∪B2; so
f=g=pf+qg=0 on A−B;
additionally,
∫Af=∫Bf,∫Ag=∫Bg, and ∫A(pf+qg)=∫B(pf+qg).
Also, mB<∞; so by Definition 2,
∫Bf=lim∫Bfn and ∫Bg=lim∫Bgn
for some elementary and integrable maps
fn→f (uniformly) and gn→g (uniformly) on B−Q,mQ=0.
Thus
pfn+qgn→pf+qg (uniformly) on B−Q.
But by Theorem 2 and Corollary 1(vii), both of §4 (for elementary and integrable maps),
∫B(pfn+qgn)=p∫Bfn+q∫Bgn.
Hence
∫A(pf+qg)=∫B(pf+qg)=lim∫B(pfn+qgn)=lim(p∫Bfn+q∫Bgn)=p∫Bf+q∫Bg=p∫Af+q∫Ag.
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,… for any functions, and extend the result to any integrable functions.
Using Definition 3, we set
A=∞⋃n=1Bn,{Bn}↑,mBn<∞,
and
fn=fCBn,gn=gCBn,n=1,2,….
Then by definition,
∫Af=limn→∞∫Afn and ∫Ag=limn→∞∫Agn,
and so
p∫Af+q∫Ag=limn→∞(p∫Afn+q∫Agn).
As fn,gn have m-finite supports, the first part of the proof yields
p∫Afn+q∫Agn=∫A(pfn+qgn).
Thus as claimed,
p∫Af+q∫Ag=lim∫A(pfn+qgn)=∫A(pf+qg).◻
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→E is integrable on each of n disjoint M-sets Ak, it is so on their union
A=n⋃k=1Ak,
and
∫Af=n∑k=1∫Akf.
(ii) This holds for countable unions, too, if f is integrable on all of A.
- Proof
-
Let f have m-finite support: f=fCB on A,mB<∞. Then
∫Af=∫Bf and ∫Akf=∫Bkf,
where
Bk=Ak∩B,k=1,2,…,n.
By Definition 2, fix elementary and integrable maps fi (on A) and a set Q (mQ=0) such that fi→f (uniformly) on B−Q (hence also on Bk−Q), with
∫Af=∫Bf=limi→∞∫Bfi and ∫Akf=limi→∞∫Bkfi,k=1,2,…,n.
As the fi are elementary and integrable, Theorem 1 in §4 yields
∫Afi=∫Bfi=n∑k=1∫Bkfi=n∑k=1∫Akfi.
Hence
∫Af=limi→∞∫Bfi=limi→∞n∑k=1∫Bkfi=n∑k=1(limi→∞∫Akfi)=n∑k=1∫Akf.
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=∞⋃k=1Ak (disjoint),Ak∈M.
In this case, set gn=fCBn, where Bn=⋃nk=1Ak,n=1,2,…. By clause (i), we have
∫Agn=∫Bngn=n∑k=1∫Akgn=n∑k=1∫Akf,
since gn=f on each Ak⊆Bn.
Also, as is easily seen, |gn|≤|f| on A and gn→f (pointwise) on A (proof as in Lemma 3). Thus by Theorem 5 in §6,
∫A|gn−f|→0.
As
|∫Agn−∫Af|=|∫A(gn−f)|≤∫A|gn−f|,
we obtain
∫Af=limn→∞∫Agn,
and the result follows by (3).◻