On the fundamental theorem of calculus
Jesper Singh
VT 2015
Examensarbete, 30hp
Magisterexamen i matematik, 240hp
Institutionen f¨or matematik och matematisk statistik
Abstract
The Riemann integral have many flaws, some that becomes visible in the fundamental theorem of calculus. The main point of this essay is to introduce the gauge integral, and prove a much more suitable version of that theorem.
Sammanfattning
Riemannintegralen har m˚anga brister. Vissa utav dessa ser man i integralkalkylens huvudsats.
Huvudm˚alet med denna uppsats ¨ar att introducera gauge integralen och visa en mer l¨amplig version av huvudsatsen.
Contents Abstract
Sammanfattning
1. Introduction 1
2. The Riemann integral 5
2.1. The formal definition and some examples 5
2.2. Properties of the Riemann integral and integrable functions 8
2.3. The fundamental theorem of calculus 16
3. The gauge integral 21
3.1. Gauges and δ-fine partitions 21
3.2. The gauge integral and the relation to the Riemann integral 25 3.3. A little taste of exceptional sets, null sets and null functions 29
3.4. Properties of the gauge integral 32
3.5. The fundamental theorems of calculus for the gauge integral 36
Appendix A. An open letter to authors of calculus books 49
Acknowledgements 55
References 57
1. Introduction
The fundamental theorem of calculus is historically a major mathematical breakthrough, and is absolutely essential for evaluating integrals. In today’s modern society it is simply difficult to imagine a life without it. The history goes way back to sir Isaac Newton long before Riemann made the first sound foundation of the Riemann integral itself. It is argued in [1] that the first published ”proof” of the fundamental theorem of calculus is due to James Gregory. The book titled Geometriae Pars Universalis that was published in 1668 by the Scottish mathematician and astronomer contains a first version of the fundamental theorem of calculus. Later in history it is considered that Newton himself discovered this theorem, even though that version was published at a later date. For further information on the history of the fundamental theorem of calculus we refer to [1].
The main point of this essay is the fundamental theorem of calculus, and in modern notations it is stated as follows.
The fundamental theorem of calculus for the Riemann integral:
Part I Let f, F : [a, b] → R be two functions that satisfies the following a) f is Riemann integrable on [a, b];
b) F is continuous on [a, b], and c) F0(x) = f (x) for all x ∈ [a, b].
Then we have that ˆ b
a
f = F (b) − F (a) .
Part II Let f : [a, b] → R be a continuous function, and define G(x) =
ˆ x a
f .
Then G is differentiable on [a, b], and G0(x) = f (x) for all x ∈ [a, b].
In Theorem 2.11 we give a proof of Part I, and in Corollary 2.1 we have Part II. The Riemann integral have many flaws, some that becomes visible in the above theorem. The main points are that there are too few Riemann integrable functions, and of course it is annoying with all the assumptions. As seen in Example 2.7 there are functions f such that there exists a primitive function F , i.e. F0 = f on [0, 1], but f fails to be Riemann integrable. The French mathematician Arnaud Denjoy was interested in integrating such functions. In 1912, he published a paper [5]
were he defined an integral handling problems like this. Later in 1914, Oskar Perron defined an integral seemingly different to that of Denjoy ([11]). Both these definitions are very technical and difficult to grasp. Surprisingly, these definitions are equivalent. It was not until 1957, that Jaroslav Kurzweil found out an elementary definition of this integral, and later developed by Henstock ([8]).
As we shall see we only need to make some minor changes to the Riemann integral, and we get something that is surprisingly superior to the integrals of Riemann and Lebesgue ([2]). To put things in perspective it can be interesting to know that a function f is Lebesgue integrable if, and only if, f and |f | are integrable in the sense of Denjoy, Perron, Kurzweil, and Henstock ([6]).
There are many names for the things we love. This integral goes under names like for example Denjoy-Perron integral, Henstock–Kurzweil integral, and the generalized Riemann integral. We shall simply call it the gauge integral. One can only speculate why this integral is still in the
1
shadows, and we shall not do that here. For people interested in the history of the Riemann, and Lebesgue integral we refer to [7], and for those interested in the movement in trying to use the gauge integral in freshmen calculus we refer to an open letter attached in Appendix A.
Before we state the fundamental theorem of calculus for the gauge integral let us recall the definition of the Riemann integral, and what changes that needs to be made to get the gauge integral. A partition of the interval [a, b] ⊂ R, is a set P = {x0, x1, ..., xn−1, xn}, xj ∈ I, that satisfies
a = x0< x1< ... < xn−1< xn= b . The norm of a partition P is defined by
kP k = max{x1− x0, x2− x1, ..., xn− xn−1} .
If a point ti ∈ [xi−1, xi] has been chosen from each subinterval for each i = 1, 2, ..., n, then the points ti are called tags of the subinterval [xi−1, xi], and the ordered pairs
P = {([x˙ i−1, xi], ti)}ni=1
of subintervals and corresponding tags is called a tagged partition of [a, b]. Now, for a function f : [a, b] → R and a tagged partition ˙P = {([xi−1, xi], ti)}ni=1, then the Riemann sum of f corresponding to ˙P is defined by
S(f, ˙P ) =
n
X
i=1
f (ti)(xi− xi−1)
Furthermore, we say that the function f is Riemann integrable on [a, b] if ∃ A ∈ R for every ε > 0 there exists a number δε > 0 such that if ˙P = {[xi−1, xi], ti}ni=1 is any tagged partition of [a, b]
with k ˙P k < δε, then we have that
|S(f, ˙P ) − A| < ε . We then say that
ˆ b a
f = A .
We shall for example see that all continuous functions are Riemann integrable (Theorem 2.8), as well as all monotone functions (Theorem 2.9). The transition to gauge integral is surprisingly trivial. We only need to replace the constant δε> 0 above with a function δε: [a, b] → (0, ∞).
Now to the fundamental theorem of calculus for the gauge integral. We now see clearly that the generality of this theorem is far greater than the corresponding one for the Riemann integral.
The fundamental theorem of calculus for the gauge integral:
Part I Let f, F : [a, b] → R be two functions that satisfies the following a) F is continuous on [a, b], and
b) F0(x) = f (x) for all x ∈ [a, b]\E. Here E ⊂ [a, b] is a countable set or the empty set.
Then we have that f is gauge integrable, and ˆ b
a
f = F (b) − F (a) .
2
Part II Let f : [a, b] → R be a function that is gauge integrable, and define G(x) =
ˆ x a
f .
Then G is continuous on [a, b], and G0(x) = f (x) almost everywhere w.r.t. the Lebesgue measure.
Part I in the above theorem is Theorem 3.15, and Part II is Theorem 3.21.
Section 2 is all about the Riemann integral, and this section is based on [3], and section 3 is devoted to the gauge integral and there we have used [2]. We shall follow a pure mathematical path, however the reader with an undergraduate background in mathematical analysis should have no problem to follow the material that is presented. A good prerequisite and information about basic analysis can be found in [10].
3
2. The Riemann integral
The structure of this section about the Riemann integral is as follows. In subsection 2.1, we introduce the Riemann integral, prove that it is a sound definition, and of course give some el- ementary examples. Later in subsection 2.2 we prove all basic properties that is needed to give a proof of the aim of this section, namely the fundamental theorem of calculus. This is done in subsection 2.3.
2.1. The formal definition and some examples. In this subsection we shall introduce the Riemann sum (Definition 2.2), and then the concept of Riemann integral (Definition 2.3). We then prove that the definition of Riemann integral is well-posed (Theorem 2.1). We end this subsection with some examples.
Definition 2.1. If I = [a, b] is a closed bounded interval that belongs to R, then a partition of I is a finite, ordered set P = {x0, x1, ..., xn−1, xn} of points in I such that a = x0 < x1 < ... <
xn−1, xn = b. The points of P is used to divide I = [a, b] into non-overlapping subintervals I1= [x0, x1], I2= [x1, x2], ..., In = [xn−1, xn].
More often than not we will denote the partition P by the notation P = {[xi−1, xi]}ni=1. The norm of P is defined by the number
kP k = max{x1− x0, x2− x1, ..., xn− xn−1}.
Hence the norm is the length of the largest subinterval of I. Many partitions have the same norm, therefore the partition is not a function of the norm. If a point ti ∈ Ii has been chosen from each subinterval Ii = [xi−1, xi], for i = 1, 2, ..., n, then the points ti are called tags of the subinterval Ii. A set of ordered pairs
P = {([x˙ i−1, xi], ti)}ni=1
of subintervals and corresponding tags is called a tagged partition of I. The dot over the P indicates that a tag has been selected from each subinterval. The tags can be chosen in infinitely many ways, for instance we can choose the left endpoints, or the right endpoints, or the midpoints of the subintervals. Note and endpoint of a subinterval can be used as the tag for at most two consecutive subintervals. The norm of a tagged partition is defined in the same matter as for an ordinary partition, so the tags wont depend on the choice of tags.
Definition 2.2. let f : I → R and let ˙P = {([xi−1, xi], ti)}ni=1 is a tagged partition of I, then the sum
S(f, ˙P ) =
n
X
i=1
f (ti)(xi− xi−1), is called the Riemann sum of f corresponding to ˙P .
As we know from calculus the Riemann sum under a positive function f on [a, b] is the sum of the areas of n rectangles whose base are the subintervals Ii= [xi−1, xi] with heights f (ti).
Definition 2.3. A function f : I → R is said to be Riemann-integrable on I if there exists a number A ∈ R such that for every ε > 0 there exists a number δε such that if ˙P = {Ii, ti}ni=1 is any tagged partition of I with k ˙P k < δε, then
|S(f, ˙P ) − A| < ε,
and we write limk ˙P k→0S(f, ˙P ) = A. The set of all Riemann integrable functions on [a, b] will be denoted by R[a, b].
5
Notation 2.1. The interpretation of limk ˙P k→0S(f, ˙P ) = A is that the integral A is ”the limit” of the Riemann sums S(f, ˙P ) when the norm k ˙P k → 0. This is different than the limit of a function since S(f, ˙P ) is not a function of k ˙P k. Many different partitions can share the same norm.
Instead of A we often use the calculus notation A =
ˆ b a
f or
ˆ b a
f (x)dx.
The letter x is just a dummy variable and can be replaced with any other letter as long it does not become ambiguity.
Theorem 2.1. If f ∈ R[a, b], then the value of the integral is uniquely determined.
Proof. Assume that A0 and A00 satisfies Definition 2.3 and let ε > 0 be given. Then there exists δ0ε/2> 0 such that if ˙P1 is any tagged partition with k ˙P1k < δε/20 , we have
|S(f, ˙P1) − A0| < ε/2.
Similarly there exists δ00ε/2> 0 such that if ˙P2 is any tagged partition with k ˙P2k < δε/200 , then
|S(f, ˙P2) − A00| < ε/2.
Put δε = min{δ0ε/2, δε/200 } > 0 and let ˙P be a tagged partition with k ˙P k < δε. Since k ˙P k < δε/20 and k ˙P k < δε/200 , we get
|S(f, ˙P ) − A0| < ε/2 and |S(f, ˙P ) − A00| < ε/2.
Hence by the Triangle Inequality it follows that
|A0− A00| = |L0− S(f, ˙P ) − S(f, ˙P ) − L00|
≤ |L0− S(f, ˙P )| + |S(f, ˙P ) − L00| < ε/2 + ε/2 = ε.
Since ε > 0 is arbitrary, it follows that L0= L00.
If Definition 2.3 is used to show that a function is Riemann integrable we must know (or guess) the value A of the integral, also we must construct a δε that will do for an arbitrary ε > 0. The determination of the value A can be done by calculating Riemann sums and guessing the value A, the construction of a sufficient δε will often give us a hard time. Later on we will prove some theorems that will be very handy when it comes to determine if f ∈ R[a, b]. For now lets have a look at some instructive examples.
Example 2.1. If f : [a, b] → R and f = c, where c is constant, then f ∈ R[a, b] and´b
af = c(b−a).
Let f (c) = c for all x ∈ [a, b]. If ˙P = {([xi−1, xi], ti)}ni=1 is any tagged partition of [a, b], then S(f, ˙P ) =
n
X
i=1
f (ti)(xi− xi−1) =
n
X
i=1
c(xi− xi−1) = c(b − a).
Thus, for any ε > 0 we can choose δεwith ease, lets put δε= 1 so that if k ˙P k < 1, then
|S(f, ˙P ) − A| = |S(f, ˙P ) − c(b − a)| = 0 < ε.
Since ε > 0 is arbitrary, the conclusion is that f ∈ R[a, b] and´b
af = c(b − a).
6
Example 2.2. Let g : [0, 1] → R be defined by g(x) = x for all x ∈ [0, 1]. Then g ∈ R[0, 1] with
´1
0 g = 12. Now the ”trick” of using particular points as tags will be a good guess of the integral value. If Q = {Ii}ni=1is any partition of [0, 1], choose the tag ti∈ Ii= [xi−1, xi] to be the midpoint qi=12(xi−1+ xi) we now have a tagged partition ˙Q = {(Ii, qi)}ni=1. The Riemann sum becomes
S(g, ˙Q) =
n
X
i=1
g(qi)(xi− xi−1=
n
X
i=1
1
2(xi+ xi−1)(xi− xi−1) =
n
X
i=1
1
2(x2i− x2i−1) =1
2(12− 02) =1 2. Now let ˙P = {(Ii, ti)}ni=1be an arbitrary tagged partition of [0, 1] with k ˙P k < δ so that xi−xi−1< δ for i = 1, ..., n. Also let ˙Q have the same partition points, but choose the tag qito be the midpoint of the interval Ii. The tags of ˙P can be anywhere between xi−1and xi. Since ti, qi∈ [xi−1, xi], we have |ti− qi| < δ. Using the Triangle Inequality, we deduce
|S(g, ˙P ) − S(g, ˙Q)| =
n
X
i=1
ti(xi− xi−1) −
n
X
i=1
qi(xi− xi−1)
=
n
X
i=1
(ti− qi)(xi− xi−1)
≤
n
X
i=1
|ti− qi| (xi− xi−1)
< δ
n
X
i=1
(xi− xi−1) = δ(xn− x0) = δ(1 − 0) = δ.
Since S(g, ˙Q) = 12, we infer that if ˙P is any tagged partition with k ˙P k < δ, then
S(g, ˙P ) − 1 2
< δ.
Therefore any δε≤ ε will do. If we choose δε= ε and works backwards we conclude that g ∈ R[0, 1]
and´1 0 g =´1
0 xdx = 12.
Changing a function at a finite number of points does not affect its integrability nor the value of the integral.
Theorem 2.2. Let g : [a, b] → R be integrable on [a, b] and if g(x) = f (x) except for a finite number of points in [a, b], then f is Riemann integrable and ´b
af =´b ag.
Proof. Lets prove the case of one exceptional point, then the extension to a finite number of points can be done by mathematical induction. Assume c ∈ [a, b] and f (x) = g(x) for all x 6= c and let A =´b
a g. For any tagged partition ˙P the Riemann sums S(f, ˙P ) and S(g, ˙P ) are identical with exception of at most two terms, that is the case when c = xi or c = xi−1. Therefore
|S(f, ˙P ) − S(g, ˙P )|
=
n
X
i=1
(f (xi) − g(xi)) (xi− xi−1)
= |[f (c) − g(c)] (xi− xi−1)|
≤ (|f (c)| + |g(c)|) (xi− xi−1) ≤ (|f (c)| + |g(c)|) k ˙P k.
Now, given ε > 0, let δ1> 0 satisfy δ1< ε/2(|f (c)| + |g(c)|), then |S(f, ˙P ) − S(g, ˙P )| < ε/2. Since g ∈ R[a, b] there exists a δ2> 0 such that when k ˙P k < δ2 we have |S(g, ˙P ) − A| < ε/2. Now let δ = min{δ1, δ2}. Then if k ˙P k < δ we obtain
|S(f, ˙P )−A| = |S(f, ˙P )−S(g, ˙P )+S(g, ˙P )−A| ≤ |S(f, ˙P )−S(g, ˙P )|+|S(g, ˙P )−A| < ε/2+ε/2 = ε.
Hence f ∈ R[a, b] with´b
af = A.
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The interpretation of Theorem 2.2 is that a singleton set has no length so the ”area” under the curve is zero.
Example 2.3. Let g : [a, b] → R be defined by g(x) =
( c if x ∈ (a, b) 0 if x ∈ {a, b} .
Since the function f in Example 2.1 is integrable with integral c(b − a), and g = f except at the finite points a or b, then g ∈ R[a, b] with´b
ag = c(b − a) by Theorem 2.2.
2.2. Properties of the Riemann integral and integrable functions. In this subsection we start by proving some basic properties of the Riemann integral, like the linearity in Theorem 2.3, that every Riemann integrable function is bounded on [a, b] (Theorem 2.4), but the most important for the rest of this section on the Riemann integral is that continuous functions defined on a compact interval [a, b] is Riemann integrable (Theorem 2.8).
The obstacle involved when determining the value of the integral and of δεsuggest that it would be very nice and useful to have some general theorem. The next result enables us to perform certain algebraic manipulations of functions that belong to R[a, b].
Theorem 2.3. Suppose f, g ∈ R[a, b] and k ∈ R. Then a) kf ∈ R[a, b] and´b
akf = k´b a f ; b) (f + g) ∈ R[a, b] and´b
a(f + g) =´b af +´b
a g;
c) if f (x) ≤ g(x) for all x ∈ [a, b], then´b a f ≤´b
ag.
Proof. a) For k = 0 there is nothing to prove so we assume k 6= 0. Let ε > 0 be given. If P = {([x˙ i−1, xi], ti)}ni=1 is a tagged partition of [a, b] and since f ∈ R[a, b] there exists a δε > 0 such that
S(f, ˙P ) − ˆ b
a
f
< ε
|k| whenever k ˙P k < δε. Thus, if k ˙P k < δεthen
S(kf, ˙P ) − k ˆ b
a
f
=
kS(f, ˙P ) − k ˆ b
a
f
= |k|
S(f, ˙P ) − ˆ b
a
f
< |k| ε
|k| = ε.
Hence limk ˙P k→0S(kf, ˙P ) = k´b
af , therefore kf ∈ R[a, b] and ´b
akf = k´b
af by Definition 2.3.
b) Let > 0 be given. Since f, g ∈ R[a, b] there exists δε/20 > 0 such that if ˙P1 is any tagged partition with k ˙P1k < δε/20 , then
S(f, ˙P ) − ˆ b
a
f
< ε 2,
and there exists δε/200 > 0 such that if ˙P2 is any tagged partition with k ˙P2k < δ00ε/2, then
S(g, ˙P ) − ˆ b
a
g
< ε 2.
8
Let δε= min{δ0ε/2, δ00ε/2} > 0 and let ˙P be a tagged partition with k ˙P k < δε. Since both k ˙P k < δε/20 and k ˙P k < δε/200 , we get
S(f + g, ˙P ) − ˆ b
a
f + ˆ b
a
g
!
=
S(f, ˙P ) + S(g, ˙P ) − ˆ b
a
f − ˆ b
a
g
≤
S(f, ˙P ) − ˆ b
a
f
+
S(g, ˙P ) − ˆ b
a
g
< ε 2 +ε
2 = ε.
Hence limk ˙P k→0S(f + g, ˙P ) = ´b af +´b
a g, so (f + g) ∈ R[a, b] and ´b
a f + g = ´b a f +´b
ag by Definition 2.3.
c) Let ε > 0 be given. Since f, g ∈ R[a, b] then for any tagged partition ˙P with ˙P < δε we have
S(f, ˙P ) − ˆ b
a
f
< ε
2 and
S(g, ˙P ) − ˆ b
a
g
<ε 2. Removing the absolute values and using the fact that S(f, ˙P ) ≤ S(g, ˙P ) gives us
ˆ b a
f −ε
2 < S(f, ˙P ) ≤ S(g, ˙P ) <
ˆ b a
g +ε 2, adding ε/2 yields
ˆ b a
f < S(f, ˙P ) + ε
2 ≤ S(g, ˙P ) + ε 2 <
ˆ b a
g + ε.
Since´b a f z´b
ag + ε for every ε > 0 the conclusion is that´b a f ≤´b
a g.
Now when some algebraic combinations been established lets look at certain demands on the function f to be required for the Riemann integral to exist. An unbounded function for sure is not Riemann integrable. Let B[a, b] denote the set of all bounded functions on [a, b].
Theorem 2.4. If f ∈ R[a, b], then f ∈ B[a, b].
Proof. The proof is by contradiction. Assume f /∈ B[a, b] and ´b
af = A. Then there exists δ > 0 and ε > 0 such that if ˙P is any tagged partition of [a, b] with k ˙P k < δ, we have |S(f, ˙P ) − A| < ε, which gives
A − ε < S(f, ˙P ) < A + ε, hence
|S(f, ˙P )| < |A + ε| ≤ |A| + ε. (2.1) Now let Q = {[xi−1, xi]}ni=1 be a partition of [a, b] with kQk < δ. Since |f | is unbounded on [a, b], there exists at least one subinterval in Q, say [xk−1, xk], where |f | is not bounded. If |f | where bounded on each [xi−1, xi] ∈ [a, b] by say Mi, then |f | would be bounded on [a, b] by max{M1, ..., Mn}. The strategy now is to tag Q in such way that equation 2.1 is contradicted. To do so tag Q by ti= xi for i 6= k and tk∈ [xk−1, xk] such that
|f (tk)(xk− xk−1)| > |A| + ε +
X
i6=k
f (ti)(xi− xi−1) , this is always possible since f is not bounded, thus
|f (tk)(xk− xk−1)| −
X
i6=k
f (ti)(xi− xi−1)
> |A| + ε.
9
By the inequality form |A + B| ≥ |A| − |B| we find
|S(f, ˙Q)| =
f (tk)(xk− xk−1) +X
i6=k
f (ti)(xi− xi−1)
≥ |f (tk)(xk− xk−1)| −
X
i6=k
f (ti)(xi− xi−1)
> |A| + ε.
Hence we have found a tagged partition of [a, b] which contradicts equation 2.1, therefore an unbounded function cannot be integrable. Since f ∈ R[a, b] the conclusion is that f ∈ B[a, b]. Note that there is no guarantee that a bounded function is Riemann integrable(it may or may not be). Lets have a look at another important steppingstone that will be needed to establish the integrability of several classes of functions such as step, monotone and continuous functions. This result is known as the Cauchy criterion. The major gain by the Cauchy criterion is that it makes it possible to prove that a function is Riemann integrable without the need to know the integrals value. There is a price to be paid thou, we need to consider two Riemann sums instead of just one.
Theorem 2.5 (Cauchy criterion). A function f : [a, b] → R belongs to R[a, b] if and only if for every ε > 0 there exists ηε> 0 such that if ˙P and ˙Q are any tagged partitions of [a, b] with k ˙P k < ηε
and k ˙Qk < ηε, then
|S(f, ˙P ) − S(f, ˙Q)| < ε.
Proof. Let f ∈ R[a, b] with´b
af = A, let ηε= δε/2> 0 be such that if ˙P , ˙Q are tagged partitions of [a, b] such that k ˙P k < ηε and k ˙Qk < ηε, then
|S(f, ˙P ) − A| < ε/2 and |S(f, ˙Q) − A| < ε/2.
Hence
|S(f, ˙P ) − S(f, ˙Q)| = |S(f, ˙P ) − A + A − S(f, ˙Q)|
≤ |S(f, ˙P ) − A| + |A − S(f, ˙P )| < ε/2 + ε/2 = ε.
Conversely for each n ∈ N let δn > 0 be such that if ˙P and ˙Q be tagged partitions of [a, b] with k ˙P k < δn and k ˙Qk < δn, then
|S(f, ˙P ) − S(f, ˙Q)| < 1/n.
We can assume that δn≥ δn+1 for each n ∈ N, if not just substitute δn with δ0n= min{δ1, ..., δn}.
For each n ∈ N, let ˙Pn be a tagged partition of [a, b] with k ˙Pnk < δn. Hence, if m > n then both k ˙Pnk < δn and k ˙Pmk < δn, therefore
|S(f, ˙Pn) − S(f, ˙Pm)| < 1/n if m > n,
consequently {S(f, ˙Pm)}∞m=1 is a Cauchy sequence in R and therefore this sequence converges in R to say A, thus sending m to the limit gives us limm→∞S(f, ˙Pm) = A, and
|S(f, ˙Pn) − L| ≤ 1/n for all n ∈ N.
To be convinced that ´b
a f = A, given ε > 0, let K ∈ N satisfy K > 2/ε. If ˙Q is any tagged partition of [a, b] with k ˙Qk < δK, then
|S(f, ˙Q) − A| = |S(f, ˙Q) − S(f, ˙PK) + S(f, ˙PK) − A|
≤ |S(f, ˙Q) − S(f, ˙PK)| + |S(f, PK) − A| ≤ 1/K + 1/K = 2/K < ε.
Since ε > 0 is arbitrary, f ∈ R[a, b] and´b
a f = A.
10
Theorem 2.5 is a practical tool to determine if either f ∈ R[a, b] or f /∈ R[a, b].
Example 2.4. Let f : [0, 1] → R be defined by f (x) =
( 1 if x ∈ Q ∩ [0, 1]
0 if x ∈ (R\Q) ∩ [0, 1] .
Then f /∈ R[a, b]. Hence we have to show that there exists ε0> 0 such that for any δε0 > 0 there exists partitions ˙P and ˙Q with k ˙P k < ε0and k ˙Qk < ε0 such that
|S(f, ˙P ) − S(f, ˙Q)| ≥ ε0.
In order to do so we choose ε0= 1/2. If ˙P is any partition with tags ti∈ Q ∩ [0, 1] then S(f, ˙P ) = 1, on the other hand if ˙Q is any tagged partition with tags tk∈ (R\Q) ∩ [0, 1] then S(f, ˙P ) = 0. Hence
|S(f, ˙P ) − S(f, ˙Q)| = 1 > 1/2,
since ˙P , ˙Q are tagged partitions with arbitrary small norms, the conclusion is that f /∈ R[a, b]. The definition of the Riemann integral gives us two types of obstacles. The first one is that there is infinitely ways to choose tags ti. Secondly there is infinitely many partitions with norm less than δε. Below we will establish a useful theorem that will ease us from some of the difficulties when proving that a function f is integrable, the theorem is called the squeeze theorem. The basic idea is that a function can be ”squeezed” between two functions known to be integrable. It was this idea that led the French mathematician Gaston Darboux to develop a different approach to integration by introducing upper and lower integrals. For a fully treatment of the Darboux integral we refer to [10].
Theorem 2.6. Let f : [a, b] → R. Then f ∈ R[a, b] if and only if for every ε > 0 there exists functions αε, ωε∈ R[a, b] with
αε(x) ≤ f (x) ≤ ωε(x) for all x ∈ [a, b],
such that ˆ b
a
(ωε− αε) < ε.
Proof. Choose αε= ωε= f for all ε > 0, then´b
a(ωε− αε) = 0 < ε.
Conversely, let ε > 0 be given. Since αε, ωε∈ R[a, b], there exists δε > 0 such that if ˙P is any tagged partition with k ˙P k < δε, then
S(αε, ˙P ) − ˆ b
a
αε
< ε and
S(ωε, ˙P ) − ˆ b
a
ωε
. These inequalities gives
ˆ b a
αε− ε < S(αε, ˙P ) and S(ωε, ˙P ) <
ˆ b a
ωε+ ε.
Since αε≤ f ≤ ωε we have S(αε, ˙P ) ≤ S(ωε, ˙P ), hence ˆ b
a
αε− ε < S(f, ˙P ) <
ˆ b a
ωε+ ε.
If ˙Q is another tagged partition with k ˙Qk < δε, then we also have ˆ b
a
αε− ε < S(f, ˙Q) <
ˆ b a
ωε+ ε,
11
by subtracting these two inequalities and using the assumtion´b
a(ωε− αε) < ε , we conclude that
|S(f, ˙P ) − S(f, ˙Q)| <
ˆ b a
ωε+ ε − ˆ b
a
αε− ε
!
= ˆ b
a
ωε− ˆ b
a
αε+ 2ε = ˆ b
a
(ωε− αε) + 2ε < ε + 2ε = 3ε.
Since ε > 0 is arbitrary f ∈ R[a, b] by Theorem 2.5.
Step functions are important in the nature of Riemann integrability, and will be used to prove that every continuous function over and interval [a, b] is also Riemann integrable over the same interval. First we prove a special case in the following lemma, then we prove the general statement in Theorem 2.7.
Lemma 2.1. If J ⊆ [a, b] having endpoints c < d and if ϕJ(x) = 1 for x ∈ J and ϕJ(x) = 0 for x ∈ [a, b] \ J , then ϕJ∈ R[a, b] and ´b
aϕJ = d − c.
Proof. With c = a and d = b it is just Example 2.1 all over again so assume a < c and d < b. If P = {([x˙ i−1, xi], ti)}ni=1 is any tagged partition of [a, b], then it is clear that
S(ϕJ, ˙P ) =
n
X
i=1
ϕJ(ti)(xi− xi−1)
= ϕJ(t1)(a − c) + ϕJ(t2)(d − c) + ϕJ(t3)(b − d) = ϕJ(t2)(d − c) = d − c.
Hence, for any ε > 0, we can choose δε= 1 so if k ˙P k < δε, then
|S(ϕJ, ˙P ) − (d − c)| = 0 < ε . Since ε > 0 is arbitrary, the conclusion is that ϕJ ∈ R[a, b] and´b
aϕJ = d − c.
Theorem 2.7. If ϕ : [a, b] → R is a step function, then ϕ ∈ R[a, b].
Proof. Step functions like in Lemma 2.1 are called ”elementary step functions”, for further studies see [2]. Any step function ϕ can be written as a linear combination of elementary step functions:
ϕ =
m
X
j=1
kjϕJj,
where Jj has endpoints cj < dj. Lemma 2.1 and Theorem 2.3 implies that ϕ ∈ R[a, b] and that ˆ b
a
ϕ = ˆ b
a
(k1ϕJ1+ ... + kmϕJm) = k1 ˆ b
a
ϕJ1+ ... + km ˆ b
a
ϕJm
= k1(d1− c1) + ... + km(dm− cm) =
m
X
j=1
kj(dj− cj).
Some examples regarding step functions and the squeeze theorem would be appropriate.
Example 2.5. Let g : [0, 1] → R be defined by g(x) =
( 2 if x ∈ [0, 1]
3 if x ∈ (1, 3] .
12
To show that g ∈ R[0, 3] using Definition 2.3 would be half complicated and require some ”tricks”, see [2] Example 7.1.4, page 202. However since
g =
m
X
j=1
kjϕJj =
2
X
j=1
kjϕJj, is a step function and therefore by Theorem 2.7 we have
ˆ 1 0
g =
2
X
j=1
kjϕJj = 2(1 − 0) + 3(3 − 1) = 2 + 6 = 8.
The next example combines step functions with the useful squeeze theorem.
Example 2.6. Let h(x) = x on [0, 1] and let Pn = {0, 1/n, 2/n, ..., (n − 1)/n, n/n = 1}. Define αn: [0, 1] → R by
αn(x) =
( h k−1n = k−1n , if x ∈ [k−1n ,kn), k = 1, 2, ..., n − 1 h n−1n = n−1n , if x ∈ [n−1n , 1], k = n .
Hence αn ≤ h on [0, 1/n), [1/n, 2/n), ..., [(n − 2)/n, (n − 1)/n), [(n − 1)/n, 1]. Similarly we define ωn : [0, 1] → R by
ωn(x) =
( h kn =nk, if x ∈ [k−1n ,kn), k = 1, 2, ..., n − 1 h nn) = 1 = n−1n , if x ∈ [n−1n , 1], k = n .
Hence ωn ≥ h on [0, 1/n), [1/n, 2/n), ..., [(n − 2)/n, (n − 1)/n), [(n − 1)/n, 1]. By Theorem 2.7 we have
ˆ 1 0
αn=
n
X
k=1
(k − 1) n · 1
n = 1 n2
n
X
k=1
k − 1 = 1
n2· (0 + 1 + 2 + ... + n − 1)
= 1
n2 ·n(n − 1)
2 =1
2
1 − 1
n
. Also
ˆ 1 0
ωn=
n
X
k=1
k n· 1
n = 1 n2
n
X
k=1
k = 1
n2 · (1 + 2 + ... + n − 1 + n)
= 1
n2 ·n(n + 1)
2 =1
2
1 + 1
n
. Thus we have αn(x) ≤ h(x) ≤ ωn(x) for x ∈ [0, 1] and
ˆ 1 0
ωn− ˆ 1
0
αn= ˆ 1
0
(ωn− αn) = 1 n.
Therefore for a given ε > 0 we can choose n ∈ N so large that n1 < ε, Theorem 2.6 tells us that h ∈ R[0, 1] and since limn→∞Pn
k=1αn· 1/n = limn→∞Pn
k=1ωn· 1/n = 1/2 we also know that
´1
0 h = 1/2 since´1
0 αn ≤´1 0 h ≤´1
0 ωn.
We are now equipped to prove that all continuous functions over [a, b] belongs to R[a, b], to do so we will use Theorem 2.6 and some properties about continuous functions. Let C[a, b] denote all continuous functions on [a, b].
Theorem 2.8. If f : [a, b] → R and f ∈ C[a, b], then f ∈ R[a, b].
13
Proof. f is continuous on [a, b] and therefore f is uniformly continuous on [a, b]. For a further explanation about uniformly continuity see [10]. Hence, for given ε > 0 there exists a δεsuch that if u, v ∈ [a, b] and |u−v| < δε, then we have |f (u)−f (v)| < ε/(b−a). Let P = {Ii}ni=1be a partition such that kP k < δε. Since f ∈ C[a, b], there exists points u, v ∈ [a, b] such that f (u) ≤ f (x) ≤ f (v) see [10] Theorem 4.5, page 104. Let ui∈ Iibe the point such that f (ui) ≤ f (x) for all x ∈ Ii, and let vi be the point such that f (x) ≤ f (vi) for x ∈ Ii. Define a step function αε: [a, b] → R by
αε(x) =
( f (ui) for x ∈ [xi−1, xi), i = 1, ..., n − 1 f (un) for x ∈ [xn−1, xn] .
Similarly define ωε: [a, b] → R by ωε(x) =
( f (vi) for x ∈ [xi−1, xi), i = 1, ..., n − 1 f (vn) for x ∈ [xn−1, xn] .
This gives us
αε(x) ≤ f (x) ≤ ωε(x) for all x ∈ [a, b].
Hence 0 ≤
ˆ b a
(ωε− αε) =
n
X
i=1
(f (vi) − f (ui)) (xi− xi−1)
<
n
X
i=1
ε b − a
(xi− xi−1) =
ε
b − a
(xn− xo) = ε.
Therefore it follows from Theorem 2.6 that f ∈ R[a, b].
Monotone functions are not always continuous over an interval [a, b], however they are always integrable on [a, b].
Theorem 2.9. If f : [a, b] → R is monotone on [a, b], then f ∈ R[a, b].
Proof. Assume f (xi−1) ≤ f (xi) on [a, b] (the case when f is decreasing is handled similarly).
Partition [a, b] into n subintervals Ik = [xk−1, xk] with equal lenght, thus xk− xk−1= (b − a)/n, k = 1, ..., n. Since f is increasing on Ik, the minimum value is attained at the left endpoint xk−1 and the maximum value is attained at the right endpoint xk. Define αε: [a, b] → R by
αε(x) =
( f (xk−1) for x ∈ [xk−1, xk), k = 1, ..., n − 1 f (xn−1) for x ∈ [xn−1, xn] .
Similarly define ωε: [a, b] → R by ωε(x) =
( f (xk) for x ∈ [xk−1, xk), i = 1, ..., n − 1 f (xn) for x ∈ [xn−1, xn] .
Then we have
αε(x) ≤ f (x) ≤ ωε for all x ∈ [a, b].
Theorem 2.7 gives ˆ b
a
αε=
n
X
k=1
f (xk−1)(xk− xk−1) = b − a
n (f (x0) + f (x1) + ... + f (xn−1)) ,
and ˆ b
a
ωε=
n
X
k=1
f (xk)(xk− xk−1) = b − a
n (f (x1) + f (x2) + ... + f (xn)) .
14
Subtracting and cancellation of terms yields ˆ b
a
(ωε− αε) =
n
X
k=1
(f (xk) − f (xk−1)) (xk− xk−1)
= b − a
n (f (xn) − f (x0)) = b − a
n (f (b) − f (a)) . Hence for a given ε > 0, we choose n ∈ N so large that n > (b − a)(f (b) − f (a))/ε. Consequently
´b
a(ωε− αε) < ε and therefore f ∈ R[a, b] by Theorem 2.6. Lets return to arbitrary functions that belong to R[a, b]. The next result is quite obvious, nevertheless it requires a proof.
Theorem 2.10. Let f : [a, b] → R and let c ∈ (a, b). Then f ∈ R[a, b] if and only if its restriction f1 of f to [a, c] is in R[a, c] and f2 of f to [c, b] is in R[a, b]. In this case
ˆ b a
f = ˆ c
a
f + ˆ b
c
f.
Proof. Suppose that f1 ∈ R[a, c] with ´c
af1 = A1 and f2 ∈ R[c, b] with ´b
c f1 = A2. Then for given ε > 0 there exists δ0> 0 such that if ˙P1 is a tagged partitions of [a, c] with k ˙P1k < δ0, then
|S(f1, ˙P1) − A1| < ε/3. Also there exist δ00> 0 such that if ˙P2is any tagged partition of [c, b] with k ˙P2k < δ00, then |S(f2, ˙P2) − A2| < ε/3. If M is a bound for |f |, define δε= min{δ0, δ00, ε/6M } and let ˙Q be a tagged partition of [a, b] with k ˙Qk < δ. We have to prove that
|S(f, ˙Q) − (A1+ A2)| < ε.
There are two cases to consider
i) If c ∈ ˙Q and we split ˙Q into a partition ˙Q1 of [a, c] and a partition ˙Q2of [c, b]. Since S(f, ˙Q) = S(f, ˙Q1) + S(f, ˙Q2) and since k ˙Q1k < δ0 and k ˙Q2k < δ00, we have
|S(f, ˙Q) − (A1+ A2)| = |S(f, ˙Q1) + S(f, ˙Q2) − (A1+ A2)|
≤ |S(f, ˙Q1) − A1| + |S(f, ˙Q2) − A2| < ε 3 +ε
3 = 2ε 3 < ε.
ii) If c /∈ ˙Q = {(Ik, tk)}mk=1, then there exists k ≤ m such that c ∈ (xk−1, xk). Just add c to ˙Q and let ˙Q1 be the tagged partition of [a, c] defined by
Q˙1= {(I1, t1), ..., (Ik−1, tk−1), ([xk−1, c], c)}.
similarly let ˙Q2be the tagged partition of [c, b] defined by
Q˙2= {([c, xk], c), (Ik+1, tk+1), ..., (Im, tm)}.
Subtracting many terms shows that S(f, ˙Q) − S(f, ˙Q1) − S(f, ˙Q2)
=
m
X
k=1
f (tk)(xk− xk−1) −
k−1
X
k=1
f (tk)(xk− xk−1) −
m
X
k=k−1
f (tk)(xk− xk−1)
= f (tk)(xk− xk−1) − f (c)(c − xk−1) − f (c)(xk− c)
= f (tk)(xk− xk−1) − f (c)(xk− xk−1) = (f (tk) − f (c)) (xk− xk−1).
Also since f (tk) − f (c) ≤ 2M and xk− xk−1< ε/6M it follows that
|S(f, ˙Q) − S(f, ˙Q1) − S(f, ˙Q2)| ≤ 2M (xk− xk−1) < 2M ε 6M =ε
3.
15
Now since k ˙Q1k < δε≤ δ0 and k ˙Q2k < δε≤ δ00, it follows that
|S(f, ˙Q) − (A1+ A2)|
= |S(f, ˙Q) − S(f, ˙Q1) − S(f, ˙Q2+ S(f, ˙Q1) + S(f, ˙Q2− (A1+ A2)|
≤ |S(f, ˙Q) − S(f, ˙Q1) − S(f, ˙Q2| + |S(f, ˙Q1− A1| + |S(f, ˙Q2− A2|
< ε 3 +ε
3 +ε 3 = ε.
Since ε > 0 is arbitrary, we conclude that f ∈ R[a, b] and´b a f =´c
af +´b c f .
Conversely, suppose that f ∈ R[a, b]. For given ε > 0, let ηε> 0, by Theorem 2.5 we have
|S(f, ˙P ) − S(f, ˙Q| < ε,
for any tagged partitions ˙P and ˙Q with norm less than ηε. Let f1 be the restriction of f to [a, c].
Also let ˙P1, ˙Q1be tagged partitions of [a, c] with k ˙P1k, k ˙Q1k < ηε. By adding additional partition points and tags from [c, b], we can extend ˙P1 and ˙Q1 to tagged partitions ˙P and ˙Q of [a, b] that satisfy k ˙P k, k ˙Qk < ηε. If we add the same additional points and tags in [c, b] for both ˙P and ˙Q, then
S(f1, ˙P1) − S(f1, ˙Q1) = S(f, ˙P ) − S(f, ˙Q).
Since k ˙P k, k ˙Qk < ηε, we have
|S(f1, ˙P1) − S(f1, ˙Q1)| < ε.
Therefore the restriction f1 of f to [a, c] belongs to R[a, c] by Theorem 2.5. A repetition of the argument above shows that the restriction f2 of f belongs to R[c, d], only difference is that the partition we are ”borrowing” points from is [a, c], instead of [c, b]. Now since since f ∈ R[a, c] and f ∈ R[c, b] has been established the equality´c
a f +´b c f =´b
af follows from the first part of the
theorem.
2.3. The fundamental theorem of calculus. This is the final part of the Riemann integral.
We shall state and prove the fundamental theorem of calculus. Furthermore, we give one example to show some of its flaws.
Theorem 2.11. Let f, F : [a, b] → R such that a) F ∈ C[a, b];
b) F0(x) = f (x) for all x ∈ [a, b];
c) f ∈ R[a, b].
Then ˆ b
a
f = F (b) − F (a).
Proof. Since f ∈ R[a, b], given ε > 0, there exists δε > 0 such that if ˙P is any tagged partition with k ˙P k < δε, then
S(f, ˙P ) − ˆ b
a
f
< ε.
Since F is continuous on [a, b] and if [xi−1, xi] ∈ ˙P then the mean-value theorem applied to F on [xi−1, xi] implies that there exists ui ∈ (xi−1, xi), such that
F (xi) − F (xi−1) = F0(ui) · (xi− xi−1) for i = 1, ..., n.
16
For more information on the Mean-value theorem of differential caclulus see [10] Theorem 5.6, page 134. Adding, using the fact that F0(ui) = f (ui) and cancellation of terms gives
F (b) − F (a) =
n
X
i=1
(F (xi) − F (xi−1)) =
n
X
i=1
f (ui)(xi− xi−1).
Let ˙Pu= {([xi−1, xi], ui)}ni=1, thus S(f, ˙Pu) =Pn
i=1f (ui)(xi− xi−1), therefore
S(f, ˙Pu) − ˆ b
a
f
=
F (b) − F (a) − ˆ b
a
f
< ε.
Since ε > 0 is arbitrary, we conclude that´b
af = F (b) − F (a).
Even if F0 exist for all x ∈ [a, b] it is not always true that f ∈ R[a, b].
Example 2.7. Let F : [0, 1] → R be defined by F (x) =
( x2cos(1/x2) if x ∈ (0, 1]
0 if x = 0 .
Differentiating yields
F0(x) =
( 2x cos(1/x2) + (2/x) sin(1/x2) for x ∈ (0, 1],
0 for x = 0 .
For x ∈ (0, 1] the chain and product rules applies and for x = 0 we have
F (x) − F (0)
x − 0 − F0(0)
=
x2cos(1/x2) x
=
x cos(1/x2)
≤ x < ε.
Hence for |x − 0| < δ choose ε = δ, this proves that F0(0) = 0. Therefore F is continuous and differentiable at every point x ∈ [0, 1], but since F0 fails to be bounded on [0, 1] it follows that F0 ∈ R[0, 1]. So condition c) cannot be omitted in Theorem 2.11. In section 3 we will see that/ f ∈ G[0, 1] and Theorem 3.15 will even provide us with a antiderivative of f in some sense. We now wish to differentiate an integral with a fluctuating upper limit, the following will be needed.
Definition 2.4. If f ∈ R[a, b], then the function F : [a, b] → R defined by F (x) =
ˆ x a
f for x ∈ [a, b], is called the indefinite integral of f with base point a.
Definition 2.5. A function f : [a, b] → R is said to be a Lipschitz function (or satisfy a Lipschitz condition) on [a, b] if there exists a constant K > 0 such that
|f (x1) − f (x2)| ≤ K|x1− x2| for all x1, x2∈ [a, b].
Every Lipschitz function is uniformly continuous.
Theorem 2.12. If f : [a, b] → R satisfies a Lipschitz condition, then f is uniformly continuous on [a, b].
Proof. Let f be as in Definition 2.5. For given ε > 0, we choose δ = ε/K. Now if |x1− x2| < δ for all x1, x2∈ [a, b], then
|f (x1) − f (x2)| ≤ K|x1− x2| < K ε K = ε.
17
The indefinite integral F of a Riemann integrable function f satisfies a Lipschitz condition and is therefore continuous.
Theorem 2.13. The indefinite integral F in Definition 2.4 is continuous on [a, b]. If |f (x)| ≤ M for all x ∈ [a, b], then |F (x) − F (y)| ≤ M |x − y| for all x, y ∈ [a, b].
Proof. Theorem 2.10 implies that if x, y ∈ [a, b] and y ≤ x, then F (x) =
ˆ x a
f = ˆ y
a
f + ˆ x
y
f = F (y) + ˆ x
y
f, from this we have
F (x) − F (y) = ˆ x
y
f.
By hypothesis if −M ≤ f (x) ≤ M for all x ∈ [a, b], then Theorem 2.3 implies that
−M (x − y) ≤ ˆ x
y
f ≤ M (x − y), since y ≤ x it follows that
|F (x) − F (y)| =
ˆ x y
f
≤ M |x − y|.
Therefore F is uniformly continuous on [a, b] by Theorem 2.12. The derivative of the indefinite integral F exists at every point where f i continuous.
Theorem 2.14. Let f ∈ R[a, b] and let limx→c+f (x) = f (c) for c ∈ [a, b). Then the indefinite integral
F (x) = ˆ x
a
f, has a right hand derivative at c equal to f (c).
Proof. Assume c ∈ [a, b). Since f is continuous at c, given ε > 0 there exists δε> 0 such that if c ≤ x < c + δε, then
f (c) − ε < f (x) < f (c) + ε. (2.2) Let h satisfy 0 < h < δε. Since f ∈ R[a, b] Theorem 2.10 implies that
f ∈ (R[a, c], R[a, c + h], R[c, c + h]) and
F (c + h) − F (c) = ˆ c+h
c
f.
On the interval [c, c + h] inequality 2.2 holds for f , therefore ˆ c+h
c
f (c) − ε <
ˆ c+h c
f (x) <
ˆ c+h c
f (c) + ε.
Hence
(f (c) − ε) · h < F (c + h) − F (c) = ˆ c+h
c
f < (f (c) + ε) · h, divide by h > 0 and subtract f (c) everywhere, we obtain
−ε < F (c + h) − F (c)
h − f (c) < ε,
thus
F (c + h) − F (c)
h − f (c)
< ε,
18
since ε > 0 is arbitrary we conclude that lim
h→0+
F (c + h) − F (c)
h = f (c).
Approaching the limit from the left gives the following result.
Theorem 2.15. Let f ∈ R[a, b] and let limx→c−f (x) = f (c) for c ∈ [a, b). Then the indefinite integral
F (x) = ˆ x
a
f, has a left hand derivative at c equal to f (c).
Proof. Assume c ∈ (a, b]. Since f is continuous at c, given ε > 0 there exists δε> 0 such that if c − δε< x ≤ c, then
f (c) − ε < f (x) < f (c) + ε. (2.3) Let h satisfy 0 < h < δε. Since f ∈ R[a, b] Theorem 2.10 implies that
f ∈ (R[a, c − h], R[a, c], R[c − h, c]) and
F (c) − F (c − h) = ˆ h
c−h
f.
On the interval [c − h, c] inequality 2.3 holds for f , therefore ˆ c
c−h
f (c) − ε <
ˆ c c−h
f (x) <
ˆ c c−h
f (c) + ε.
Hence
(f (c) − ε) · h < F (c) − F (c − h) = ˆ c
c−h
f < (f (c) + ε) · h, divide by h > 0 and subtract f (c) everywhere, we obtain
−ε < F (c) − F (c − h)
h − f (c) < ε,
thus
F (c) − F (c − h)
h − f (c)
< ε, since ε > 0 is arbitrary we conclude that
lim
h→0−
F (c) − F (c − h)
h = f (c).
If f ∈ C[a, b], then we have a familiar result.
Corollary 2.1. If f ∈ C[a, b], then the indefinite integral F (x) = ´x
a f is differentiable on the whole interval [a, b] and F0(x) = f (x) for all x ∈ [a, b].
Proof. Let c ∈ (a, b). Since f is continuous at c we have limx→c+f (x) = limx→c−f (x) = f (c).
Therefore according to Theorems 2.14 and 2.15 above we have F0(c) = lim
h→0+
F (c + h) − F (c)
h = lim
h→0−
F (c) − F (c − h)
h = lim
h→0
F (c + h) − F (c)
h = f (c).
If c = a or c = b we get
F+0(a) = lim
h→0+
F (a + h) − F (a)
h = f (a),
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respectively
F−0(b) lim
h→0−
F (b) − F (b − h)
h = f (b).
The indefinite integral of f need not to be an antiderivative of f , maybe the derivative does not exist at c or F0(c) 6= f (c), however if f ∈ C[a, b], then its indefinite integral F is guaranteed to be an antiderivative of f by Corollary 2.1.
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3. The gauge integral
In this section we shall generalize the Riemann integral. First in subsection 3.1 we introduce some notations, and then in subsection 3.2 we define the gauge integral. There we also prove that every Riemann integrable function is also gauge integrable, and give an example that there is a function that is gauge integrable but not Riemann integrable. To be able to formulate the advanced version of the fundamental theorem of calculus for the gauge integral we need some basic measure theory, this is done in subsection 3.3. Much in the same vein as for the Riemann integral we prove in subsection 3.4 some basic properties. Finally in subsection 3.5 we obtain the aim of this essay ”the fundamental theorem of calculus for the gauge integral”.
3.1. Gauges and δ-fine partitions. In this subsection we shall introduce some necessary nota- tion and background for the gauge integral.
Definition 3.1. The closed neighborhood of x with radius r > 0 is the set B[x, r] = {y ∈ R : |x − y| ≤ r},
which is also called the closed ball with center x and radius r. The open neighborhood of x with radius r > 0 is the set
B(x, r) = {y ∈ R : |x − y| < r}, which is also called the open ball with center x and radius r.
If I = [a, b], and a ≤ b, we define the length of I to be l(I) = b − a.
It is easy to see that l(I) ≥ 0, and that l(I) = 0 if and only if a = b. Similarly, the length of any interval of the forms (a, b), [a, b), (a, b], is also defined to be b − a. The length of the empty set is defined as l(∅) = 0, this lengths coincide with the Lebesgue measure.
We say that an interval I ⊂ R is degenerate if it contains at most one point, and that it is non degenerate if it contains at least two points, in which case it contains infinitely many points. We say that two intervals I1, I2⊂ R are disjoint if I1∩ I2= ∅. We say that two intervals I1, I2⊂ R are non overlapping if I1∩ I2 = ∅ or I1∩ I2 = {p}, where {p} is a singleton set, which is an endpoint of both intervals.
Example 3.1. If J ⊂ R is a bounded open interval and suppose that I1, ..., Imare nonoverlapping closed and bounded intervals contained in J . Then
m
X
i=1
l(Ii) ≤ l(J ).
If I1, ..., Im are degenerate then Pm
i=1l(Ii) = 0 ≤ l(J ). Suppose that the intervals Ii are nondegenerate, so ai< bi. If necessary we relabel such that a1≤ a2≤ ... ≤ am. Since the intervals are nonoverlapping, we have
α < a1< b1≤ a2< b2≤ ... ≤ am< bm< β,
where J = (α, β). It follows that bi− ai ≤ ai+1− ai for i = 1, ..., m − 1 and bm− am< β − am. Adding the inequalities, we get
(b1− a1) + (b2− a2) + ... + (bm− am) =
m
X
i=1
(bi− ai) =
m
X
i=1
l(Ii) < β − a1< β − α = l(J ).
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