This chapter is on the generalized Riemann integral. The Riemann Darboux integral
presended earlier has been obsolete for over 100 years. The integral of this chapter is certainly
not obsolete and is in certain important ways the very best integral currently known. This
integral is called the generalized Riemann integral, also the Henstock Kurzweil integral after
the two people who invented it and sometimes the gauge integral. Other books
which discuss this integral are the books by Bartle [5], Bartle and Sherbert, [6],
Henstock [20], or McLeod [28]. Considerably more is presented in these references. In
what follows, F will be an increasing function, the most important example being
F
(x)
= x.
Definition 12.1.1Let
[a,b]
be a closed and bounded interval. A taggeddivision^{1}of
[a,b]
= I is a set of the form P ≡
{(I,t )}
i i
_{i=1}^{n}where t_{i}∈ I_{i} =
[x ,x]
i−1 i
, anda = x_{i−1}<
⋅⋅⋅
< x_{n} = b. Let the t_{i}be referred to as the tags. A function δ : I →
(0,∞ )
iscalled a gauge function or simply gauge for short.A tagged division, P is called δ fineif
Ii ⊆ [ti − δ(ti),ti + δ(ti)].
A δ fine division, is understood to be tagged. More generally, a collection,
{(Ii,ti)}
_{i=1}^{p}is δfine if the above inclusion holds for each of these intervals and their interiors are disjoint evenif their union is not equal to the whole interval,
[a,b]
.
The following fundamental result is essential.
Proposition 12.1.2Let
[a,b]
be an interval and let δ be a gauge function on
[a,b]
.Then there exists a δ fine tagged division of
[a,b]
.
Proof:Suppose not. Then one of
[ ]
a, a+2b
or
[ ]
a+2b,b
must fail to have a δ fine
tagged division because if they both had such a δ fine division, the union of the
two δ fine divisions would be a δ fine division of
[a,b]
. Denote by I_{1} the interval
which does not have a δ fine division. Then repeat the above argument, dividing I_{1}
into two equal intervals and pick the one, I_{2} which fails to have a δ fine division.
Continue this way to get a nested sequence of closed intervals,
{Ii}
having the
property that each interval in the set fails to have a δ fine division and diam
(Ii)
→ 0.
Therefore,
∩ ∞i=1Ii = {x}
where x ∈
[a,b]
. Now
[x− δ(x),x+ δ (x)]
must contain some I_{k} because the diameters of
these intervals converge to zero. It follows that
{(I,x)}
k
is a δ fine division of I_{k}, contrary to
the construction which required that none of these intervals had a δ fine division. This proves
the proposition.
With this proposition and definition, it is time to define the generalized Riemann integral.
The functions being integrated typically have values in ℝ or ℂ but there is no reason to
restrict to this situation and so in the following definition, X will denote the space in which f
has its values. For example, X could be ℝ^{p} which becomes important in multivariable
calculus. For now, just think ℝ if you like. It will be assumed Cauchy sequences converge and
there is something like the absolute value, called a norm although it is possible to generalize
even further.
Definition 12.1.3Let X be a complete normed vector space. (For example, X = ℝor X = ℂ or X = ℝ^{p}.) Then f :
[a,b]
→ X is generalized Riemann integrable, written asf ∈ R^{∗}
[a,b]
if there exists R ∈ X such that for all ε > 0, there exists a gauge, δ, such that ifP ≡
{(Ii,ti)}
_{i=1}^{n}is δ fine then defining, S
(P,f)
by
∑n
S(P,f) ≡ f (ti)ΔFi,
i=1
where if I_{i} =
[xi−1,xi]
,
ΔFi ≡ F (xi)− F (xi−1)
it follows
|S (P,f )− R| < ε.
Then
∫ ∫
b
I f dF ≡ a fdF ≡ R.
Here
|⋅|
refers to the norm on X for ℝ this is just the absolute value.
Note that if P is δ_{1} fine and δ_{1}≤ δ then it follows P is also δ fine.
How does this relate to the usual Riemann integral discussed above?
To begin with, is there a way to tell whether a given function is in R^{∗}
[a,b]
? The following
Cauchy criterion is useful to make this determination.
Proposition 12.1.4A function f :
[a,b]
→ X is in R^{∗}
[a,b]
if and only if for everyε > 0, there exists a gauge function δ_{ε}such that if P and Q are any two divisions which areδ_{ε}fine, then
|S (P,f)− S (Q, f)| < ε.
Proof:Suppose first that f ∈ R^{∗}
[a,b]
. Then there exists a gauge, δ_{ε}, and an element of X,
R, such that if P is δ_{ε} fine, then
|R − S(P,f)| < ε∕2.
Now let P,Q be two such δ_{ε} fine divisions. Then
Conversely, suppose the condition of the proposition holds. Let ε_{n}→ 0+ as n →∞ and let
δ_{εn} denote the gauge which goes with ε_{n}. Without loss of generality, assume that δ_{εn} is
decreasing in n. Let R_{εn} denote the closure of all the sums, S
(P, f)
where P is δ_{εn} fine. From
the condition, it follows diam
(R εn)
≤ ε_{n} and that these closed sets are nested in the sense
that R_{εn}⊇ R_{εn+1} because δ_{εn} is decreasing in n. Therefore, there exists a unique,
R ∈∩_{n=1}^{∞}R_{εn}. To see this, let r_{n}∈ R_{εn}. Then since the diameters of the R_{εn} are converging
to 0,
{rn}
is a Cauchy sequence which must converge to some R ∈ X. Since R_{εn} is closed, it
follows R ∈ R_{εn} for each n. Letting ε > 0 be given, there exists ε_{n}< ε and for P a δ_{εn} fine
division,
|S(P,f)− R | ≤ εn < ε.
Therefore, R = ∫_{I}f. This proves the proposition.
Are there examples of functions which are in R^{∗}
[a,b]
? Are there examples of functions
which are not? It turns out the second question is harder than the first although it is very
easy to answer this question in the case of the obsolete Riemann integral. The generalized
Riemann integral is a vastly superior integral which can integrate a very impressive collection
of functions. Consider the first question. Recall the definition of the Riemann integral given
above which is listed here for convenience.
Definition 12.1.5A bounded function f defined on
[a,b]
is said to be RiemannStieltjes integrable if there exists a number, I with the property that for every ε > 0, thereexists δ > 0 such that if
P ≡ {x0,x1,⋅⋅⋅,xn}
is any partition having
||P ||
< δ, and z_{i}∈
[xi− 1,xi]
,
| |
|| ∑n ||
||I − f (zi)(F (xi)− F (xi−1))|| < ε.
i=1
The number∫_{a}^{b}f
(x)
dFis defined as I.
First note that if δ > 0 and if every interval in a division has length less than δ then the
division is δ fine. In fact, you could pick the tags as any point in the intervals. Then the
following theorem follows immediately.
Theorem 12.1.6Let f be Riemann Stieltjes integrable according toDefinition12.1.5. Then f is generalized Riemann integrable and the integrals are thesame.
Proof: Just let the gauge functions be constant functions.
In particular, the following important theorem follows from Theorem 9.1.6.
Theorem 12.1.7Let f be continuous on
[a,b]
and let F be any increasingintegrator. Then f ∈ R^{∗}
[a,b]
.
Many functions other than continuous ones are integrable however. In fact, it is
fairly difficult to come up with easy examples because this integral can integrate
almost anything you can imagine, including the function which equals 1 on the
rationals and 0 on the irrationals which is not Riemann integrable. This will be shown
later.
The integral is linear. This will be shown next.
Theorem 12.1.8Suppose α and β are constants and that f and g are in R^{∗}
[a,b]
.Then αf + βg ∈ R^{∗}
[a,b]
and
∫ ∫ ∫
I (αf + βg)dF = α I fdF + β I gdF.
Proof:Let η =
---ε---
|β|+|α|+1
and choose gauges, δ_{g} and δ_{f} such that if P is δ_{g} fine,
|| ∫ ||
||S (P,g) − gdF|| < η
I
and that if P is δ_{f} fine,
|| ∫ ||
||S(P,f)− fdF|| < η.
I
Now let δ = min
(δg,δf)
. Then if P is δ fine the above inequalities both hold. Therefore, from
the definition of S
Since ε > 0 is arbitrary, this shows the number, β∫_{I}gdF + α∫_{I}fdF qualifies in the definition
of the generalized Riemann integral and so αf + βg ∈ R^{∗}
[a,b]
and
∫ ∫ ∫
(αf + βg)dF = β gdF + α fdF.
I I I
The following lemma is also very easy to establish from the definition.
Lemma 12.1.9If f ≥ 0 and f ∈ R^{∗}
[a,b]
, then∫_{I}fdF ≥ 0. Also, if f has complexvalues and is in R^{∗}
[I]
, then bothRef andImf are in R^{∗}
[I]
.
Proof: To show the first part, let ε > 0 be given and let δ be a gauge function such that if
P is δ fine then
|| ∫ ||
||S(f,P)− fdF|| ≤ ε.
I
Since F is increasing, it is clear that S
(f,P )
≥ 0. Therefore,
∫
fdF ≥ S(f,P )− ε ≥ − ε
I
and since ε is arbitrary, it follows ∫_{I}fdF ≥ 0 as claimed.
To verify the second part, note that by Proposition 12.1.4 there exists a gauge, δ such that
if P,Q are δ fine then
|S(f,P)− S (f,Q)| < ε
But
|S(Re f,P )− S (Re f,Q)| = |Re (S(f,P ))− Re (S(f,Q))|
≤ |S (f,P)− S (f,Q )|
and so the conditions of Proposition 12.1.4 are satisfied and you can conclude Ref ∈ R^{∗}
[I]
.
Similar reasoning applies to Imf. This proves the lemma.
|∫ | ∫ ∫
|| fdF || = αfdF = (Re (αf)+ iIm (αf))dF
| I | I I
∫ ∫ ∫
= I Re (αf )dF + i I Im (αf)dF = I Re (αf)dF
∫ ∫
≤ |Re (αf)|dF ≤ |f|dF
I I
Note the assumption that
|f|
∈ R^{∗}
[a,b]
.
The following lemma is also fundamental.
Lemma 12.1.11If f ∈ R^{∗}
[a,b]
and
[c,d]
⊆
[a,b]
, then f ∈ R^{∗}
[c,d]
.
Proof:Let ε > 0 and choose a gauge δ such that if P is a division of
[a,b]
which is δ fine,
then
|S (P,f )− R| < ε∕2.
Now pick a δ fine division of
[c,d]
,
{(Ii,ti)}
_{i=r}^{l} and then let
{(Ii,ti)}
_{i=1}^{r−1} and
{(Ii,ti)}
_{i=l+1}^{n} be fixed δ fine divisions on
[a,c]
and
[d,b]
respectively.
Now let P_{1} and Q_{1} be δ fine divisions of
[c,d]
and let P and Q be the respective δ fine
divisions if
[a,b]
just described which are obtained from P_{1} and Q_{1} by adding in
{(Ii,ti)}
_{i=1}^{r−1} and
{(Ii,ti)}
_{i=l+1}^{n}. Then
ε > |S (Q,f)− S (P,f)| = |S (Q1,f)− S (P1,f)|.
By the above Cauchy criterion, Proposition 12.1.4, f ∈ R^{∗}
[c,d]
as claimed.
Corollary 12.1.12Suppose c ∈
[a,b]
and that f ∈ R^{∗}
[a,b]
. Then f ∈ R^{∗}
[a,c]
andf ∈ R^{∗}
[c,b]
. Furthermore,
∫ ∫ c ∫ b
fdF = fdF + fdF.
I a c
Here∫_{a}^{c}fdF means∫_{}
[a,c]
fdF.
Proof: Let ε > 0. Let δ_{1} be a gauge function on
[a,c]
such that whenever P_{1} is a δ_{1} fine
division of
[a,c]
,
|∫ |
|| cfdF − S(P ,f)||< ε∕3.
| a 1 |
Let δ_{2} be a gauge function on
[c,b]
such that whenever P_{2} is a δ_{2} fine division of
[c,b]
,
||∫ b ||
|| fdF − S(P2,f)||< ε∕3.
| c |
Let δ_{3} be a gauge function on
[a,b]
such that if P is a δ_{3} fine division of
[a,b]
,
||∫ b ||
|| fdF − S (P,f)||< ε∕3.
|a |
Now define a gauge function
{
δ (x) ≡ min (δ1,δ3) on [a,c]
min (δ2,δ3) on [c,b]
Then letting P_{1} be a δ fine division on
[a,c]
and P_{2} be a δ fine division on
[c,b]
, it follows
that P = P_{1}∪ P_{2} is a δ_{3} fine division on
[a,b]
and all the above inequalities hold. Thus
noting that S
(P,f)
= S
(P1,f )
+ S
(P2,f)
,
||∫ ( ∫ c ∫ b ) ||
|| fdF − fdF + fdF ||
| I a c |
||∫ ||
≤ || fdF − (S(P1,f)+ S (P2,f))||
I| (∫ ∫ )|
|| c b ||
+ ||S(P1,f)+ S (P2,f )− a fdF + c fdF ||
||∫ || || ∫ c || || ∫ b ||
≤ || fdF − S(P,f)||+ ||S(P1,f)− fdF ||+ ||S (P2,f)− fdF||
I a | c |
< ε∕3+ ε∕3+ ε∕3 = ε.
Since ε is arbitrary, the conclusion of the corollary follows. This proves the corollary.
The following lemma, sometimes called Henstock’s lemma is of great significance.
Lemma 12.1.13Suppose that f ∈ R^{∗}
[a,b]
and that whenever, Q is a δ fine division ofI,
|| ∫ ||
||S(Q,f) − fdF|| ≤ ε.
I
Then if P =
{(Ii,ti)}
_{i=1}^{n}is any δ fine division, and P^{′} =
{( )}
Iij,tij
_{j=1}^{r}is a subset of P,then
|| r r ∫ ||
||∑ f (t )ΔF − ∑ fdF||≤ ε.
||j=1 ij i j=1 Iij ||