Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

8.7: Integration of Complex and Vector-Valued Functions

( \newcommand{\kernel}{\mathrm{null}\,}\)

I. First we consider functions f:SEn(Cn). For such functions, it is natural (and easy) to define integration "componentwise" as follows.

Definition

A function f:SEn is said to be integrable on AM iff its n (real) components, f1,,fn, are. In this case, we define

Af=Afdm=(Af1,Af2,,Afn)=nk=1¯ekAfk

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+iAfim.

If f:SCn, we use (1), with complex components fk

With this definition, integration of functions f:SEn(Cn) reduces to that of fk:SE1(C), and one easily obtains the same theorems as in §§4-6, as far as they make sense for vectors.

Theorem 8.7.1

A function f:SEn(Cn) is integrable on AM 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 fk are m-measurable; furthermore,

Af+k± and Afk±.

This implies

>Af+k+Afk=A(f+k+fk)=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|=|nk=1¯ekfk|nk=1|¯ek||fk|=nk=1|fk|,

we get

A|f|An1|fk|=n1A|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:SE (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<.

Lemma 8.7.1

If fnf (uniformly) on A (mA<), then

A|fnf|0.

Proof

By assumption,

(ε>0) (k) (n>k)|fnf|<ε on A;

so

(n>k)A|fnf|A(ε)=εmA<.

As ε is arbitrary, the result follows.

Lemma 8.7.2

If

A|f|<(mA<)

and

f=limnfn (uniformly) on AQ (mQ=0)

for some elementary maps fn on A, then all but finitely many fn are elementary and integrable on A, and

limnAfn

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|fnf|<ε and A|fnfk|<ε.

(The latter can be achieved since

limkA|fnfk|=A|fnf|<ε.)

Now, as

|fn||fnf|+|f|,

Problem 7 in §5 yields

(n>k)A|fn|A|fnf|+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)|AfnAfk|=|A(fnfk)|A|fnfk|<ε.

Thus {Afn} is a Cauchy sequence. As E is complete,

limAfn±

exists in E, as asserted.

Finally, suppose gnf (uniformly) on AQ for some other elementary and integrable maps gn. By what was shown above, limAgn exists, and

|limAgnlimAfn|=|limA(gnfn)|limA|gnfn0|=0

by Lemma 1, as gnfn0 (uniformly) on A. Thus

limAgn=limAfn,

and all is proved.

This leads us to the following definition.

Definition

If f:SE is integrable on AM (mA<), we set

Af=Afdm=limnAfn

for any elementary and integrable maps fn such that fnf (uniformly) on AQ,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 AB (AB,BM), we may choose fn=0 on AB in Definition 2. Then

Af=limAfn=limBfn=Bf.

Thus we now define

Af=Bf,

even if mA=, provided f=0 on AB, i.e.,

f=fCB on A

(CB= characteristic function of B), with AB,BM, 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(f0) is σ-finite. Neglecting A(f=0), we may assume that

A=Bn,mBn<, and {Bn}

(if not, replace Bn by nk=1Bk); so BnA.

Lemma 8.7.3

Let ϕ:SE be integrable on A. Let BnA,mBn< and set

fn=ϕCBn,n=1,2,.

Then fnϕ (pointwise) on A, all fn are integrable on A, and

limnAfn

exists in E. Furthermore, this limit does not depend on the choice of {Bn}.

Proof

Fix any xA. As BnA=Bn,

(n0) (n>n0)xBn.

By assumption, fn=ϕ on Bn. Thus

(n>n0)fn(x)=ϕ(x);

so fnϕ (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 ABn(mB<),

Afn

is defined. Since fnϕ (pointwise) and |fn||ϕ| on A, Theorem 5 in §6, with g=|ϕ|, yields

A|fnϕ|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.

Definition

If ϕ:SE is integrable on AM, we set

Aϕ=Aϕdm=limnAfn,

with the fn as in Lemma 3 (even if ϕ has no m-finite support).

Theorem 8.7.2 (linearity)

If f,g:SE are integrable on AM, so is

pf+qg

for any scalars p,q. Moreover,

A(pf+qg)=pAf+qAg.

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,B2A(mB1+mB2<). Let B=B1B2; so

f=g=pf+qg=0 on AB;

additionally,

Af=Bf,Ag=Bg, and A(pf+qg)=B(pf+qg).

Also, mB<; so by Definition 2,

Bf=limBfn and Bg=limBgn

for some elementary and integrable maps

fnf (uniformly) and gng (uniformly) on BQ,mQ=0.

Thus

pfn+qgnpf+qg (uniformly) on BQ.

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

B(pfn+qgn)=pBfn+qBgn.

Hence

A(pf+qg)=B(pf+qg)=limB(pfn+qgn)=lim(pBfn+qBgn)=pBf+qBg=pAf+qAg.

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=limnAfn and Ag=limnAgn,

and so

pAf+qAg=limn(pAfn+qAgn).

As fn,gn have m-finite supports, the first part of the proof yields

pAfn+qAgn=A(pfn+qgn).

Thus as claimed,

pAf+qAg=limA(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?)

Theorem 8.7.3 (additivity)

(i) If f:SE is integrable on each of n disjoint M-sets Ak, it is so on their union

A=nk=1Ak,

and

Af=nk=1Akf.

(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=AkB,k=1,2,,n.

By Definition 2, fix elementary and integrable maps fi (on A) and a set Q (mQ=0) such that fif (uniformly) on BQ (hence also on BkQ), with

Af=Bf=limiBfi and Akf=limiBkfi,k=1,2,,n.

As the fi are elementary and integrable, Theorem 1 in §4 yields

Afi=Bfi=nk=1Bkfi=nk=1Akfi.

Hence

Af=limiBfi=limink=1Bkfi=nk=1(limiAkfi)=nk=1Akf.

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),AkM.

In this case, set gn=fCBn, where Bn=nk=1Ak,n=1,2,. By clause (i), we have

Agn=Bngn=nk=1Akgn=nk=1Akf,

since gn=f on each AkBn.

Also, as is easily seen, |gn||f| on A and gnf (pointwise) on A (proof as in Lemma 3). Thus by Theorem 5 in §6,

A|gnf|0.

As

|AgnAf|=|A(gnf)|A|gnf|,

we obtain

Af=limnAgn,

and the result follows by (3).


8.7: Integration of Complex and Vector-Valued Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.

Support Center

How can we help?