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5.10: Sufficient Condition of Integrability. Regulated Functions

This page is a draft and is under active development. 

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

In this section, we shall determine a large family of functions that do have antiderivatives. First, we give a general definition, valid for any range space (T,p) (not necessarily E). The domain space remains E1.

Definition 1

A function f:E1(T,p) is said to be regulated on an interval IE1, with endpoints a<b, iff the limits f(p) and f(p+), other than ±, exist at each pI. However, at the endpoints a,b, if in I, we only require f(a+) and f(b) to exist.

Examples

(a) If f is relatively continuous and finite on I, it is regulated.

(b) Every real monotone function is regulated (see Chapter 4, §5, Theorem 1).

(c) If f:E1En(Cn) has bounded variation on I, it is regulated (§7, Theorem 4).

(d) The characteristic function of a set B, denoted CB, is defined by

CB(x)=1 if xB and CB=0 on B.

For any interval JE1,CJ is regulated on E1.

(e) A function f is called a step function on I iff I can be represented as the union, I=kIk, of a sequence of disjoint intervals Ik such that f is constant and ± on each Ik. Note that some Ik may be singletons, {p}.

If the number of the Ik is finite, we call f a simple step function.

When the range space T is E, we can give the following convenient alternative definition. If, say, f=ak± on Ik, then

f=kakCIk on I,

where CIk is as in (d). Note that kakCIk(x) always exists for disjoint Ik. (Why?)

Each simple step function is regulated. (Why?)

Theorem 5.10.1

Let the functions f,g,h be real or complex (or let f,g be vector valued and h scalar valued).

If they are regulated on I, so are f±g,fh, and |f|; so also is f/h if h is bounded away from 0 on I, i.e., (ε>0)|h|ε on I.

Proof

The proof, based on the usual limit properties, is left to the reader.

We shall need two lemmas. One is the famous Heine-Borel lemma.

lemma 5.10.1 (Heine-Borel)

If a closed interval A=[a,b] in E1 (or En) is covered by open sets Gi(iI), i.e.,

AiIGi,

then A can be covered by a finite number of these Gi.

Proof

The proof was sketched in Problem 10 of Chapter 4, §6.

Note 1. This fails for nonclosed intervals A. For example, let

A=(0,1)E1 and Gn=(1n,1).

Then

A=n=1Gn( verify! ), but not Amn=1Gn

for any finite m. (Why?)

The lemma also fails for nonopen sets Gi. For example, cover A by singletons {x},xA. Then none of the {x} can be dropped!

lemma 5.10.2

If a function f:E1T is regulated on I=[a,b], then f can be uniformly approximated by simple step functions on I.

That is, for any ε>0, there is a simple step function g, with ρ(f,g)ε on I; hence

supxIρ(f(x),g(x))ε.

Proof

By assumption, f(p) exists for each p(a,b], and f(p+) exists for p[a,b), all finite.

Thus, given ε>0 and any pI, there is Gp(δ) (δ depending on p) such that ρ(f(x),r)<ε whenever r=f(p) and x(pδ,p), and ρ(f(x),s)<ε whenever s=f(p+) and x(p,p+δ);xI

We choose such a Gp(δ) for every pI. Then the open globes Gp=Gp(δ) cover the closed interval I=[a,b], so by Lemma 1, I is covered by a finite number of such globes, say,

Ink=1Gpk(δk),aGp1,ap1<p2<<pnb.

We now define the step function g on I as follows.

If x=pk, we put

g(x)=f(pk),k=1,2,, n.

If x[a,p1), then

g(x)=f(p1).

If x(p1,p1+δ1), then

g(x)=f(p+1).

More generally, if x is in G¬pk(δk) but in none of the Gpi(δi),i<k, we put

g(x)=f(pk) if x<pk

and

g(x)=f(p+k) if x>pk.

Then by construction, ρ(f,g)<ε on each Gpk, hence on I.

*Note 2. If T is complete, we can say more: f is regulated on I=[a,b] iff f is uniformly approximated by simple step functions on I. (See Problem 2.)

Theorem 5.10.2

If f:E1E is regulated on an interval IE1 and if E is complete, then f exists on I, exact at every continuity point of f in I0.

In particular, all continuous maps f:E1En(Cn) have exact primitives.

Proof

In view of Problem 14 of §5, it suffices to consider closed intervals.

Thus let I=[a,b],a<b, in E1. Suppose first that f is the characteristic function CJ of a subinterval JI with endpoints c and d (acdb), so f=1 on J and f=0 on IJ. We then define F(x)=x on J,F=c on [a,c], and F=d on [d,b] (see Figure 25). Thus F is continuous (why?), and F=f on I{a,b,c,d} (why?). Hence F=f on I; i.e., characteristic functions are integrable.

Screen Shot 2019-06-26 at 2.02.35 PM.png

Then, however, so is any simple step function

f=mk=1akCIk,

by repeated use of Corollary 1 in §5.

Finally, let f be any regulated function on I. Then by Lemma 2, for any εn=1n, there is a simple step function gn such that

supxI|gn(x)f(x)|1n,n=1,2,.

As 1n0, this implies that gnf (uniformly) on I (see Chapter 4, §12, Theorem 1). Also, by what was proved above, the step functions gn have antiderivatives, hence so has f (Theorem 2 in §9); ; i.e., F=f exists on I, as claimed. Moreover, f is exact at continuity points of f in I0 (Problem 10 in §5).

In view of the sufficient condition expressed in Theorem 2, we can now replace the assumption "f exists" in our previous theorems by "f is regulated" (provided E is complete). For example, let us now review Problems 7 and 8 in §5.

Theorem 5.10.3 (weighted law of the mean)

Let f:E1E (E complete) and g:E1E1 be regulated on I=[a,b], with g0 on I. Then the following are true:

(i) There is a finite cE (called the "g-weighted mean of f on I") such that bagf=cbag.

(ii) If f, too, is real and has the Darboux property on I, then c=f(q) for some qI.

Proof

Indeed, as f and g are regulated, so is gf by Theorem 1. Hence by Theorem 2, f and gf exist on I. The rest follows as in Problems 7 and 8 of §5.

Theorem 5.10.4 (second law of the mean)

Suppose f and g are real, f is monotone with f=f on I, and g is regulated on I;I=[a,b]. Then

bafg=f(a)qag+f(b)bqg for some qI.

Proof

To fix ideas, let f; i.e., f0 on I.

The formula f=f means that f is relatively continuous (hence regulated) on I and differentiable on IQ (Q countable). As g is regulated,

xag=G(x)

does exist on I, so G has similar properties, with G(a)=aag=0.

By Theorems 1 and 2, fG=fg exists on I. (Why?) Hence by Corollary 5 in §5, so does Gf, and we have

bafg=bafG=f(x)G(x)|babaGf=f(b)G(b)baGf.

Now G has the Darboux property on I (being relatively continuous), and f0. Also, G and Gf exist on I. Thus by Problems 7 and 8 in §5,

baGf=G(q)baf=G(q)f(x)|ba,qI.

Combining all, we obtain the required result (1) since

fg=f(b)G(b)baGf=f(b)G(b)f(b)G(q)+f(a)G(q)=f(b)bqg+f(a)qag.

We conclude with an application to infinite series. Given f:E1E, we define

af=limx+xaf and af=limxaxf

if these integrals and limits exist.

We say that af and af converge iff they exist and are finite.

Theorem 5.10.5 (integral test of convergence)

If f:E1E1 is nonnegative and nonincreasing on I=[a,+), then

af converges iff n=1f(n) does.

Proof

As f,f is regulated, so f exists on I=[a,+). We fix some natural ka and define

F(x)=xkf for xI.

By Theorem 3(iii) in §5, F on I. Thus by monotonicity,

limx+F(x)=limx+xkf=kf

exists in E; so does kaf. Since

xaf=kaf+xkf,

where kaf is finite by definition, we have

af<+ iff kf<+.

Similarly,

n=1f(n)<+ iff n=kf(n)<+.

Thus we may replace "a" by "k."

Let

In=[n,n+1),n=k,k+1,,

and define two step functions, g and h, constant on each In, by

h=f(n) and g=f(n+1) on In,nk.

Since f, we have gfh on all In, hence on J=[k,+). Therefore,

xkgxkfxkh for xJ.

Also,

mkh=m1n=kn+1nh=m1n=kf(n),

since h=f(n) (constant) on [n,n+1), and so

n+1nh(x)dx=f(n)n+1n1dx=f(n)x|n+1n=f(n)(n+1n)=f(n).

Similarly,

mkg=m1n=kf(n+1)=mn=k+1f(n).

Thus we obtain

mn=k+1f(n)=mkgmkfmkh=m1n=kf(n),

or, letting m,

n=k+1f(n)kfn=kf(n).

Hence kf is finite iff n=1f(n) is, and all is proved.

Examples (continued)

(f) Consider the hyperharmonic series

1np(Problem 2 of Chapter 4, §13).

Let

f(x)=1xp,x1.

If p=1, then f(x)=1/x, so x1f=lnx+ as x+. Hence 1/n diverges.

If p1, then

1f=limx+x1f=limx+x1p1p|x1,

so 1f converges or diverges according as p>1 or p<1, and the same applies to the series 1/np.

(g) Even nonregulated functions may be integrable. Such is Dirichlet's function (Example (c) in Chapter 4, §1). Explain, using the countability of the rationals.


This page titled 5.10: Sufficient Condition of Integrability. Regulated Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform.

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