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Integration, Average Behavior: Integrable Functions

Module by: Lawrence Baggett. E-mail the author

Summary: We now wish to extend the definition of the integral to a wider class of functions. This class will consist of those functions that are uniform limits of step functions. The requirement that these limits be uniform is crucial. Pointwise limits of step functions doesn't work, as we will see in the first exercise. The initial step in carrying out this generalization is the following.

We now wish to extend the definition of the integral to a wider class of functions. This class will consist of those functions that are uniform limits of step functions. The requirement that these limits be uniform is crucial. Pointwise limits of step functions doesn't work, as we will see in Exercise 1 below. The initial step in carrying out this generalization is the following.

Theorem 1

Let [a,b][a,b] be a closed bounded interval, and let {hn}{hn} be a sequence of step functions that converges uniformly to a function ff on [a,b].[a,b]. Then the sequence {∫hn}{∫hn} is a convergent sequence of real numbers.

Proof

We will show that {∫hn}{∫hn} is a Cauchy sequence in R.R. Thus, given an ϵ>0,ϵ>0, choose an NN such that for any n≥Nn≥N and any x∈[a,b],x∈[a,b], we have

| f ( x ) - h n ( x ) | < ϵ 2 ( b - a ) . | f ( x ) - h n ( x ) | < ϵ 2 ( b - a ) .
(1)

Then, for any mm and nn both ≥N≥N and any x∈[a,b],x∈[a,b], we have

| h n ( x ) - h m ( x ) | ≤ | h n ( x ) - f ( x ) | + | f ( x ) - h m ( x ) | < ϵ b - a . | h n ( x ) - h m ( x ) | ≤ | h n ( x ) - f ( x ) | + | f ( x ) - h m ( x ) | < ϵ b - a .
(2)

Therefore,

| ∫ h n - ∫ h m | = | ∫ ( h n - h m ) | ≤ ∫ | h n - h m | ≤ ∫ ϵ b - a = ϵ , | ∫ h n - ∫ h m | = | ∫ ( h n - h m ) | ≤ ∫ | h n - h m | ≤ ∫ ϵ b - a = ϵ ,
(3)

as desired.

The preceding theorem provides us with a perfectly good idea of how to define the integral of a function ff that is the uniform limit of a sequence of step functions. However, we first need to establish another kind of consistency result.

Theorem 2

If {hn}{hn} and {kn}{kn} are two sequences of step functions on [a,b],[a,b], each converging uniformly to the same function f,f, then

lim ∫ h n = lim ∫ k n . lim ∫ h n = lim ∫ k n .
(4)

Proof

Given ϵ>0,ϵ>0, choose NN so that if n≥N,n≥N, then |hn(x)-f(x)|<ϵ/(2(b-a))|hn(x)-f(x)|<ϵ/(2(b-a)) for all x∈[a,b],x∈[a,b], and such that |f(x)-kn(x)|<ϵ/(2(b-a))|f(x)-kn(x)|<ϵ/(2(b-a)) for all x∈[a,b].x∈[a,b]. Then, |hn(x)-kn(x)|<ϵ/(b-a)|hn(x)-kn(x)|<ϵ/(b-a) for all x∈[a,b]x∈[a,b] if n≥N.n≥N. So,

| ∫ h n - ∫ k n | ≤ ∫ | h n - k n | ≤ ∫ ϵ b - a = ϵ | ∫ h n - ∫ k n | ≤ ∫ | h n - k n | ≤ ∫ ϵ b - a = ϵ
(5)

if n≥N.n≥N. Taking limits gives

| lim ∫ h n - lim ∫ k n | ≤ ϵ . | lim ∫ h n - lim ∫ k n | ≤ ϵ .
(6)

Since this is true for arbitrary ϵ>0,ϵ>0, it follows that lim∫hn=lim∫kn,lim∫hn=lim∫kn, as desired.

Definition 1:

Let [a,b][a,b] be a closed bounded interval of real numbers. A function f:[a,b]→Rf:[a,b]→R is called integrable on [a,b][a,b] if it is the uniform limit of a sequence {hn}{hn} of step functions.

Let I([a,b])I([a,b]) denote the set of all functions that are integrable on [a,b].[a,b]. If f∈I([a,b]),f∈I([a,b]), define the integral of f,f, denoted ∫f,∫f, by

∫ f = lim ∫ h n , ∫ f = lim ∫ h n ,
(7)

where {hn}{hn} is some (any) sequence of step functions that converges uniformly to ff on [a,b].[a,b].

As in the case of step functions, we use the following notations:

∫ f = ∫ a b f = ∫ a b f ( t ) d t . ∫ f = ∫ a b f = ∫ a b f ( t ) d t .
(8)

REMARK Note that Theorem 2 is crucial in order that this definition be unambiguous. Indeed, we will see below that this critical consistency result is one place where uniform limits of step functions works while pointwise limits do not. See parts (c) and (d) of Exercise 1. Note also that it follows from this definition that ∫aaf=0,∫aaf=0, because ∫aah=0∫aah=0 for any step function. In fact, we will derive almost everything about the integral of a general integrable function from the corresponding results about the integral of a step function. No surprise. This is the essence of mathematical analysis, approximation.

Exercise 1

Define a function ff on the closed interval [0,1][0,1] by f(x)=1f(x)=1 if xx is a rational number and f(x)=0f(x)=0 if xx is an irrational number.

  1. Suppose hh is a step function on [0,1].[0,1]. Prove that there must exist an x∈[0,1]x∈[0,1] such that |f(x)-h(x)|≥1/2.|f(x)-h(x)|≥1/2. HINT: Let (xi-1,xi)(xi-1,xi) be an interval on which hh is a constant c.c. Now use the fact that there are both rationals and irrationals in this interval.
  2. Prove that ff is not the uniform limit of a sequence of step functions. That is, ff is not an integrable function.
  3. Consider the two sequences {hn}{hn} and {kn}{kn} of step functions defined on the interval [0,1][0,1] by hn=χ(0,1/n),hn=χ(0,1/n), and kn=nχ(0,1/n).kn=nχ(0,1/n). Show that both sequences {hn}{hn} and {kn}{kn} converge pointwise to the 0 function on [0,1].[0,1]. HINT: All functions are 0 at x=0.x=0. For x>0,x>0, choose NN so that 1/N<x.1/N<x. Then, for any n≥N,n≥N,hn(x)=kn(x)=0.hn(x)=kn(x)=0.
  4. Let hnhn and knkn be as in part (c). Show that lim∫hn=0,lim∫hn=0, but lim∫kn=1.lim∫kn=1. Conclude that the consistency result in Theorem 2 does not hold for pointwise limits of step functions.

Exercise 2

Define a function ff on the closed interval [0,1][0,1] by f(x)=x.f(x)=x.

  1. For each positive integer n,n, let PnPn be the partition of [0,1][0,1] given by the points {0<1/n<2/n<3/n<...<(n-1)/n<1}.{0<1/n<2/n<3/n<...<(n-1)/n<1}. Define a step function hnhn on [0,1][0,1] by setting hn(x)=i/nhn(x)=i/n if i-1n<x<in,i-1n<x<in, and hn(i/n)=i/nhn(i/n)=i/n for all 0≤i≤n.0≤i≤n. Prove that |f(x)-hn(x)|<1/n|f(x)-hn(x)|<1/n for all x∈[0,1],x∈[0,1], and then conclude that ff is the uniform limit of the hnhn's whence f∈I([0,1]).f∈I([0,1]).
  2. Show that
    ∫hn=∑i=1nin2=n(n+1)2n2.∫hn=∑i=1nin2=n(n+1)2n2.
    (9)
  3. Show that ∫01f(t)dt=1/2.∫01f(t)dt=1/2. The next exercise establishes some additional properties of integrable functions on an interval [a,b].[a,b].

Exercise 3

Let [a,b][a,b] be a closed and bounded interval, and let ff be an element of I([a,b]).I([a,b]).

  1. Show that, for each ϵ>0ϵ>0 there exists a step function hh on [a,b][a,b] such that |f(x)-h(x)|<ϵ|f(x)-h(x)|<ϵ for all x∈[a,b].x∈[a,b].
  2. For each positive integer nn let hnhn be a step function satisfying the conclusion of part (a) for ϵ=1/n.ϵ=1/n. Define kn=hn-1/nkn=hn-1/n and ln=hn+1/n.ln=hn+1/n. Show that knkn and lnln are step functions, that kn(x)<f(x)<ln(x)kn(x)<f(x)<ln(x) for all x∈[a,b],x∈[a,b], and that |ln(x)-kn(x)|=ln(x)-kn(x)=2/n|ln(x)-kn(x)|=ln(x)-kn(x)=2/n for all x.x. Hence, ∫ab(ln-kn)=2n(b-a).∫ab(ln-kn)=2n(b-a).
  3. Conclude from part (b) that, given any ϵ>0,ϵ>0, there exist step functions kk and ll such that k(x)≤f(x)≤l(x)k(x)≤f(x)≤l(x) for which ∫(l(x)-k(x))<ϵ.∫(l(x)-k(x))<ϵ.
  4. Prove that there exists a sequence {jn}{jn} of step functions on [a,b],[a,b], for which jn(x)≤jn+1(x)≤f(x)jn(x)≤jn+1(x)≤f(x) for all x,x, that converges uniformly to f.f. Show also that there exists a sequence {jn'}{jn'} of step functions on [a,b],[a,b], for which jn'(x)≥jn+1'(x)≥f(x)jn'(x)≥jn+1'(x)≥f(x) for all x,x, that converges uniformly to f.f. That is, if f∈I([a,b]),f∈I([a,b]), then ff is the uniform limit of a nondecreasing sequence of step functions and also is the uniform limit of a nonincreasing sequence of step functions. HINT: To construct the jnjn's and jn'jn''s, use the step functions knkn and lnln of part (b), and recall that the maximum and minimum of step functions is again a step function.
  5. Show that if f(x)≥0f(x)≥0 for all x∈[a,b],x∈[a,b], and gg is defined by g(x)=f(x),g(x)=f(x), then g∈I([a,b]).g∈I([a,b]). HINT: Write f=limhnf=limhn where hn(x)≥0hn(x)≥0 for all xx and n.n. Then use part (g) of Exercise 3.28.
  6. (Riemann sums again.) Show that, given an ϵ>0,ϵ>0, there exists a partition PP such that if Q={x0<x1<...<xn}Q={x0<x1<...<xn} is any partition finer than P,P, and {wi}{wi} are any points for which wi∈(xi-1,xi),wi∈(xi-1,xi), then
    |∫abf(t)dt-∑i=1nf(wi)(xi-xi-1)|<ϵ.|∫abf(t)dt-∑i=1nf(wi)(xi-xi-1)|<ϵ.
    (10)
    HINT: Let PP be a partition for which both the step functions kk and ll of part (c) are constant on the open subintervals of P.P. Verify that for any finer partition Q,Q,l(wi)≥f(wi)≥k(wi),l(wi)≥f(wi)≥k(wi), and hence
    ∑il(wi)(xi-xi-1)≥∑if(wi)(xi-xi-1)≥∑ik(wi)(xi-xi-1).∑il(wi)(xi-xi-1)≥∑if(wi)(xi-xi-1)≥∑ik(wi)(xi-xi-1).
    (11)
Definition 2:

A bounded real-valued function ff on a closed bounded interval [a,b][a,b] is called Riemann-integrable if, given any ϵ>0,ϵ>0, there exist step functions kk and l,l, on [a,b][a,b] for which k(x)≤f(x)≤l(x)k(x)≤f(x)≤l(x) for all x,x, such that ∫(l-k)<ϵ.∫(l-k)<ϵ. We denote the set of all functions on [a,b][a,b] that are Riemann-integrable by IR([a,b]).IR([a,b]).

REMARK The notion of Riemann-integrability was introduced by Riemann in the mid nineteenth century and was the first formal definition of integrability. Since then several other definitions have been given for an integral, culminating in the theory of Lebesgue integration. The definition of integrability that we are using in this book is slightly different and less general from that of Riemann, and both of these are very different and less general from the definition given by Lebesgue in the early twentieth century. Part (c) of Exercise 3 above shows that the functions we are calling integrable are necessarily Riemann-integrable. We will see in Exercise 4 that there are Riemann-integrable functions that are not integrable in our sense. In both cases, Riemann's and ours, an integrable function ff must be trapped between two step functions kk and l.l. In our definition, we must have l(x)-k(x)<ϵl(x)-k(x)<ϵ for all x∈[a,b],x∈[a,b], while in Riemann's definition, we only need that ∫l-k<ϵ.∫l-k<ϵ. The distinction is that a small step function must have a small integral, but it isn't necessary for a step function to be (uniformly) small in order for it to have a small integral. It only has to be small on most of the interval [a,b].[a,b].

On the other hand, all the definitions of integrability on [a,b][a,b] include among the integrable functions the continuous ones. And, all the different definitions of integral give the same value to a continuous function. The differences then in these definitions shows up at the point of saying exactly which functions are integrable. Perhaps the most enlightening thing to say in this connection is that it is impossible to make a “good” definition of integrability in such a way that every function is integrable. Subtle points in set theory arise in such attempts, and many fascinating and deep mathematical ideas have come from them. However, we will stick with our definition, since it is simpler than Riemann's and is completely sufficient for our purposes.

Theorem 3

Let [a,b][a,b] be a fixed closed and bounded interval, and let I([a,b])I([a,b]) denote the set of integrable functions on [a,b].[a,b]. Then:

  1. Every element of I([a,b])I([a,b]) is a bounded function. That is, integrable functions are necessarily bounded functions.
  2.  I([a,b])I([a,b]) is a vector space of functions.
  3.  I([a,b])I([a,b]) is closed under multiplication; i.e., if ff and g∈I([a,b]),g∈I([a,b]), then fg∈I([a,b]).fg∈I([a,b]).
  4. Every step function is in I([a,b]).I([a,b]).
  5. If ff is a continuous real-valued function on [a,b],[a,b], then ff is in I([a,b]).I([a,b]). That is, every continuous real-valued function on [a,b][a,b] is integrable on [a,b].[a,b].

Proof

Let f∈I([a,b]),f∈I([a,b]), and write f=limhn,f=limhn, where {hn}{hn} is a sequence of step functions that converges uniformly to f.f. Given the positive number ϵ=1,ϵ=1, choose NN so that |f(x)-hN(x)|<1|f(x)-hN(x)|<1 for all x∈[a,b].x∈[a,b]. Then |f(x)|≤|hN(x)|+1|f(x)|≤|hN(x)|+1 for all x∈[a,b].x∈[a,b]. Because hNhN is a step function, its range is a finite set, so that there exists a number MM for which |hN(x)|≤M|hN(x)|≤M for all x∈[a,b].x∈[a,b]. Hence, |f(x)|≤M+1|f(x)|≤M+1 for all x∈[a,b],x∈[a,b], and this proves part (1).

Next, let ff and gg be integrable, and write f=limhnf=limhn and g=limkn,g=limkn, where {hn}{hn} and {kn}{kn} are sequences of step functions that converge uniformly to ff and gg respectively. If ss and tt are real numbers, then the sequence {shn+tkn}{shn+tkn} converges uniformly to the function sf+tg.sf+tg. See parts (c) and (d) of (Reference). Therefore, sf+tg∈I([a,b]),sf+tg∈I([a,b]), and I([a,b])I([a,b]) is a vector space, proving part (2).

Note that part (3) does not follow immediately from (Reference); the product of uniformly convergent sequences may not be uniformly convergent. To see it for this case, let f=limhnf=limhn and g=limkng=limkn be elements of I([a,b]).I([a,b]). By part (1), both ff and gg are bounded, and we write MfMf and MgMg for numbers that satisfy |f(x)|≤Mf|f(x)|≤Mf and |g(x)|≤Mg|g(x)|≤Mg for all x∈[a,b].x∈[a,b]. Because the sequence {kn}{kn} converges uniformly to g,g, there exists an NN such that if n≥Nn≥N we have |g(x)-kn(x)|<1|g(x)-kn(x)|<1 for all x∈[a,b].x∈[a,b]. This implies that, if n≥N,n≥N, then |kn(x)|≤Mg+1|kn(x)|≤Mg+1 for all x∈[a,b].x∈[a,b].

Now we show that fgfg is the uniform limit of the sequence hnkn.hnkn. For, if n≥N,n≥N, then

| f ( x ) g ( x ) - h n ( x ) k n ( x ) | = | f ( x ) g ( x ) - f ( x ) k n ( x ) + f ( x ) k n ( x ) - h n ( x ) k n ( x ) | ≤ | f ( x ) | | g ( x ) - k n ( x ) | + | k n ( x ) | | f ( x ) - h n ( x ) | ≤ M f | g ( x ) - k n ( x ) | + ( M g + 1 ) | f ( x ) - h n ( x ) | , | f ( x ) g ( x ) - h n ( x ) k n ( x ) | = | f ( x ) g ( x ) - f ( x ) k n ( x ) + f ( x ) k n ( x ) - h n ( x ) k n ( x ) | ≤ | f ( x ) | | g ( x ) - k n ( x ) | + | k n ( x ) | | f ( x ) - h n ( x ) | ≤ M f | g ( x ) - k n ( x ) | + ( M g + 1 ) | f ( x ) - h n ( x ) | ,
(12)

which implies that fg=lim(hnkn).fg=lim(hnkn).

If hh is itself a step function, then it is obviously the uniform limit of the constant sequence {h},{h}, which implies that hh is integrable.

Finally, if ff is continuous on [a,b],[a,b], it follows from (Reference) that ff is the uniform limit of a sequence of step functions, whence f∈I([a,b]).f∈I([a,b]).

Exercise 4

Let ff be the function defined on [0,1][0,1] by f(x)=sin(1/x)f(x)=sin(1/x) if x≠0x≠0 and f(0)=0.f(0)=0.

  1. Show that ff is continuous at every nonzero xx and discontinuous at 0.0. HINT: Observe that, on any interval (0,δ),(0,δ), the function sin(1/x)sin(1/x) attains both the values 1 and -1.-1.
  2. Show that ff is not integrable on [0,1].[0,1]. HINT: Suppose f=limhn.f=limhn. Choose NN so that |f(x)-hN(x)|<1/2|f(x)-hN(x)|<1/2 for all x∈[0,1].x∈[0,1]. Let PP be a partition for which hNhN is constant on its open subintervals, and examine the situation for xx's in the interval (x0,x1).(x0,x1).
  3. Show that ffis Riemann-integrable on [0,1].[0,1]. Conclude that I([a,b])I([a,b]) is a proper subset of IR([a,b]).IR([a,b]).

Exercise 5

  1. Let ff be an integrable function on [a,b].[a,b]. Suppose gg is a function for which g(x)=f(x)g(x)=f(x) for all x∈[a,b]x∈[a,b] except for one point c.c. Prove that gg is integrable and that ∫g=∫f.∫g=∫f. HINT: If f=limhn,f=limhn, define kn(x)=hn(x)kn(x)=hn(x) for all x≠cx≠c and kn(c)=g(c).kn(c)=g(c). Then use (Reference).
  2. Again, let ff be an integrable function on [a,b].[a,b]. Suppose gg is a function for which g(x)=f(x)g(x)=f(x) for all but a finite number of points c1,...,cN∈[a,b].c1,...,cN∈[a,b]. Prove that g∈I([a,b]),g∈I([a,b]), and that ∫g=∫f.∫g=∫f.
  3. Suppose ff is a function on the closed interval [a,b],[a,b], that is uniformly continuous on the open interval (a,b).(a,b). Prove that ff is integrable on [a,b].[a,b]. HINT: Just reproduce the proof to (Reference).

REMARK In view of part (b) of the preceding exercise, we see that whether a function ff is integrable or not is totally independent of the values of the function at a fixed finite set of points. Indeed, the function needn't even be defined at a fixed finite set of points, and still it can be integrable. This observation is helpful in many instances, e.g., in parts (d) and (e) of (Reference).

Theorem 4

The assignment f→∫ff→∫f on I([a,b])I([a,b]) satisfies the following properties.

  1.  (Linearity)  I([a,b])I([a,b]) is a vector space, and ∫(αf+βg)=α∫f+β∫g∫(αf+βg)=α∫f+β∫g for all f,g∈I([a,b])f,g∈I([a,b])and α,β∈R.α,β∈R.
  2.  (Positivity) If f(x)≥0f(x)≥0 for all x∈[a,b],x∈[a,b], then ∫f≥0.∫f≥0.
  3.  (Order-preserving) If f,g∈I([a,b])f,g∈I([a,b]) and f(x)≤g(x)f(x)≤g(x) for all x∈[a,b],x∈[a,b], then ∫f≤∫g.∫f≤∫g.
  4. If f∈I([a,b]),f∈I([a,b]), then so is |f|,|f|, and |∫f|≤∫|f|.|∫f|≤∫|f|.
  5. If ff is the uniform limit of functions fn,fn, each of which is in I([a,b]),I([a,b]), then f∈I([a,b])f∈I([a,b]) and ∫f=lim∫fn.∫f=lim∫fn.
  6. Let {un}{un} be a sequence of functions in I([a,b]).I([a,b]). Suppose that for each nn there is a number mn,mn, for which |un(x)|≤mn|un(x)|≤mn for all x∈[a,b],x∈[a,b], and such that the infinite series ∑mn∑mn converges. Then the infinite series ∑un∑un converges uniformly to an integrable function, and ∫∑un=∑∫un.∫∑un=∑∫un.

Proof

That I([a,b])I([a,b]) is a vector space was proved in part (2) of Theorem 3. Let ff and gg be in I([a,b]),I([a,b]), and write f=limhnf=limhn and g=limkn,g=limkn, where the hnhn's and the knkn's are step functions. Then αf+βg=lim(αhn+βkn),αf+βg=lim(αhn+βkn), so that, by (Reference) and the definition of the integral, we have

∫ ( α f + β g ) = lim ∫ ( α h n + β k n ) = lim ( α ∫ h n + β ∫ k n ) = α lim ∫ h n + β lim ∫ k n = α ∫ f + β ∫ g , ∫ ( α f + β g ) = lim ∫ ( α h n + β k n ) = lim ( α ∫ h n + β ∫ k n ) = α lim ∫ h n + β lim ∫ k n = α ∫ f + β ∫ g ,
(13)

which proves part (1).

Next, if f∈I([a,b])f∈I([a,b]) satisfies f(x)≥0f(x)≥0 for all x∈[a,b],x∈[a,b], let {ln}{ln} be a nonincreasing sequence of step functions that converges uniformly to f.f. See part (d) of Exercise 3. Then ln(x)≥f(x)≥0ln(x)≥f(x)≥0 for all xx and all n.n. So, again by (Reference), we have that

∫ f = lim ∫ l n ≥ 0 . ∫ f = lim ∫ l n ≥ 0 .
(14)

This proves part (2).

Part (3) now follows by combining parts (1) and (2) just as in the proof of (Reference).

To see part (4), let f∈I([a,b])f∈I([a,b]) be given. Write f=limhn.f=limhn. Then |f|=lim|hn|.|f|=lim|hn|. For

| | f ( x ) | - | h n ( x ) | | ≤ | f ( x ) - h n ( x ) | . | | f ( x ) | - | h n ( x ) | | ≤ | f ( x ) - h n ( x ) | .
(15)

Therefore, |f||f| is integrable. Also,

∫ | f | = lim ∫ | h n | ≥ lim | ∫ h n | = | lim ∫ h n | = | ∫ f | . ∫ | f | = lim ∫ | h n | ≥ lim | ∫ h n | = | lim ∫ h n | = | ∫ f | .
(16)

To see part (5), let {fn}{fn} be a sequence of elements of I([a,b]),I([a,b]), and suppose that f=limfn.f=limfn. For each n,n, let hnhn be a step function on [a,b][a,b] such that |fn(x)-hn(x)|<1/n|fn(x)-hn(x)|<1/n for all x∈[a,b].x∈[a,b]. Note also that it follows from parts (3) and (4) that

| ∫ f n - ∫ h n | < b - a n . | ∫ f n - ∫ h n | < b - a n .
(17)

Now {hn}{hn} converges uniformly to f.f. For,

| f ( x ) - h n ( x ) | ≤ | f ( x ) - f n ( x ) | + | f n ( x ) - h n ( x ) | < | f ( x ) - f n ( x ) | + 1 n , | f ( x ) - h n ( x ) | ≤ | f ( x ) - f n ( x ) | + | f n ( x ) - h n ( x ) | < | f ( x ) - f n ( x ) | + 1 n ,
(18)

showing that f=limhn.f=limhn. Therefore, f∈I([a,b]).f∈I([a,b]). Moreover, ∫f=lim∫hn.∫f=lim∫hn. Finally, ∫f=lim∫fn,∫f=lim∫fn, for

| ∫ f - ∫ f n | ≤ | ∫ f - ∫ h n | + | ∫ h n - ∫ f n | ≤ | ∫ f - ∫ h n | + b - a n . | ∫ f - ∫ f n | ≤ | ∫ f - ∫ h n | + | ∫ h n - ∫ f n | ≤ | ∫ f - ∫ h n | + b - a n .
(19)

This completes the proof of part (5).

Part (6) follows directly from part (5) and the Weierstrass M Test ((Reference)). For, part (1) of that theorem implies that the infinite series ∑un∑un converges uniformly, and then ∫∑un=∑∫un∫∑un=∑∫un follows from part (5) of this theorem.

As a final extension of our notion of integral, we define the integral of certain complex-valued functions.

Definition 3:

Let [a,b][a,b] be a fixed bounded and closed interval. A complex-valued function f=u+ivf=u+iv is called integrable if its real and imaginary parts uu and vv are integrable. In this case, we define

∫ a b f = ∫ a b ( u + i v ) = ∫ a b u + i ∫ a b v . ∫ a b f = ∫ a b ( u + i v ) = ∫ a b u + i ∫ a b v .
(20)

Theorem 5

  1. The set of all integrable complex-valued functions on [a,b][a,b] is a vector space over the field of complex numbers, and
    ∫ab(αf+βg)=α∫abf+β∫abg∫ab(αf+βg)=α∫abf+β∫abg
    (21)
    for all integrable complex-valued functions ff and gg and all complex numbers αα and β.β.
  2. If ff is an integrable complex-valued function on [a,b],[a,b], then so is |f|,|f|, and |∫abf|≤∫ab|f|.|∫abf|≤∫ab|f|.

Proof

We leave the verification of part (1) to the exercise that follows.

To see part (2), suppose that ff is integrable, and write f=u+iv.f=u+iv. Then |f|=u2+v2,|f|=u2+v2, so that |f||f| is integrable by Theorem 3 and part (e) of Exercise 3. Now write z=∫abf,z=∫abf, and write zz in polar coordinates as z=reiθ,z=reiθ, where r=|z|=|∫abf|.r=|z|=|∫abf|. (See (Reference).) Define a function gg by g(x)=e-iθf(x)g(x)=e-iθf(x) and notice that |g|=|f|.|g|=|f|. Then ∫abg=e-iθ∫abf=r,∫abg=e-iθ∫abf=r, which is a real number. Writing g=u^+iv^,g=u^+iv^, we then have that r=∫u^+i∫v^,r=∫u^+i∫v^, implying that ∫v^=0.∫v^=0. So,

| ∫ a b f | = r = ∫ a b g = ∫ a b u ^ + i ∫ a b v ^ = ∫ a b u ^ = | ∫ a b u ^ | ≤ ∫ a b | u ^ | ≤ ∫ a b | g | = ∫ a b | f | , | ∫ a b f | = r = ∫ a b g = ∫ a b u ^ + i ∫ a b v ^ = ∫ a b u ^ = | ∫ a b u ^ | ≤ ∫ a b | u ^ | ≤ ∫ a b | g | = ∫ a b | f | ,
(22)

as desired.

Exercise 6

Prove part (1) of the preceding theorem.

HINT: Break α,β,∫f,α,β,∫f, and ∫g∫g into real and imaginary parts.

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