The Classical Theory
Riemann Integration
LetN ∈Z + (throughoutZ + will denote the positive integers) Arectangle in R N is a subset I of R N that can be written as the Cartesian product
1 [ak, bk] of compact intervals [ak, bk], where it is assumed that ak≤bk for each 1 ≤ k ≤ N If I is such a rectangle, its diameter and volume are, respectively, diam(I)≡sup{|y−x|: x, y∈I} v u u t
For the purposes of this exposition, it will be convenient to also take the empty set to be a rectangle with diameter and volume 0.
Given a collection C, 1 I will say that C is non-overlapping if distinct elements ofC have disjoint interiors In that its conclusions seemobvious, the following lemma is surprisingly difficult to prove.
Lemma 1.1.1 IfC is a non-overlapping, finite collection of rectangles each of which is contained in the rectangle J, thenvol (J)≥P
I∈Cvol (I) On the other hand, ifC is any finite collection of rectangles andJ is a rectangle that is covered by C(i.e.,J ⊆S
Proof: Since vol(I∩J)≤vol(I), assume throughout thatJ ⊇S
I∈CI Also, without loss in generality, we will assume that ˚J 6=∅.
The proof is by induction on N Thus, suppose that N = 1 Given a closed interval, use aI and bI to denote its left and right endpoints Choose aJ ≤c00, there is aδ >0 such that
|R(f;C,ξ)−A|< wheneverξ∈Ξ(C) andC is a non-overlapping, finite,exact coverofJ (i.e.,
J =SC) whose mesh size kCk, given bykCk ≡ max diam(I) : I ∈ C , is less than δ Whenf is Riemann integrable onJ, the associated numberA is called theRiemann integral of f on J, and I will use
Any continuous real-valued function f defined on an interval J is Riemann integrable To assess the Riemann integrability of more general bounded functions, it is beneficial to introduce the concept of the Riemann upper sum.
I∈C sup x∈I f(x)vol (I) and theRiemann lower sum
4 1 The Classical Theory for any C and ξ∈Ξ(C) Also, by Cauchy’s convergence criterion, it is clear that a boundedf is Riemann integrable if and only if
L(f;C)≥ lim kCk→0U(f;C), where the limits are taken over non-overlapping, finite, exact covers ofJ My goal now is to show that the preceding can be replaced by the condition 3
C U(f;C) where theC’s run over all non-overlapping, finite, exact covers ofJ.
To this end, partially order the covers C byrefinement That is, say that
C 2 ismore refined thanC 1 and writeC 1 ≤ C 2 , if, for everyI 2 ∈ C 2 , there is an I 1 ∈ C 1 such thatI 2 ⊆I 1 Note that, for every pair C 1 and C 2 , theleast common refinementC 1 ∨C 2 is given byC 1 ∨C 2 ={I 1 ∩I 2 :I 1 ∈ C 1 andI 2 ∈ C 2 }.
Lemma 1.1.5 For any pair of non-overlapping, finite, exact coversC 1 and
C 2 ofJand any bounded functionf :J −→R,L(f;C 1 )≤ U(f;C 2 ) Moreover, if C 1 ≤ C 2 , then L(f;C 1 )≤ L(f;C 2 )andU(f;C 1 )≥ U(f;C 2 ) Finally, if f is bounded, then(1.1.4)holds if and only if for every >0there exists aCsuch that 4
Proof: We will begin by proving the second statement Noting that (1.1.7) L(f;C) =− U(−f;C), one sees that it suffices to check thatU(f;C1)≥ U(f;C2) ifC1≤ C2 But, for eachI1∈ C1, sup x∈I 1 f(x)vol (I1)≥ X
{I 2 ∈C 2 :I 2 ⊆I 1 } sup x∈I 2 f(x)vol (I2), where Lemma 1.1.1 was used to see that vol (I 1 ) = X
In many texts, the condition defined as Riemann integrability is widely accepted, and it aligns with an earlier definition While adopting this definition poses no issues, it is important to note that in the broader context explored in § 1.2, the distinction between these two definitions becomes significant.
4 Here, and elsewhere, sup I f = sup x∈I f (x) and inf I f = inf x∈I f (x). §1.1 Riemann Integration 5
After summing the above overI1∈ C1, one arrives at the required result. Given the preceding, the first assertion is immediate Namely, for any C1 andC2,
Finally, if for each >0 (1.1.6) holds for some C , then, for each >0, infC U(f;C)−sup
L(f;C)≤ U(f;C )− L(f;C )< , and so (1.1.4) holds Conversely, if (1.1.4) holds and >0, chooseC1 andC2 for which sup C L(f;C)≤ L(f;C1) + 2 and inf C U(f;C)≥ U(f;C2)− 2 Then (1.1.6) holds with C =C1∨ C2
Lemma 1.1.5 really depends only on properties of our order relation and not on the properties of vol(I) On the other hand, the next lemma depends on the continuity of volume with respect to the side-lengths of rectangles.
Lemma 1.1.8 Assume that J˚6=∅, and let C be a non-overlapping, finite, exact cover of the rectangle J If f : J −→ Ris a bounded function, then, for each >0, there is aδ >0 such that
U(f;C 0 )≤ U(f;C) + and L(f;C 0 )≥ L(f;C)− whenever C 0 is a non-overlapping, finite, exact cover of J with the property that kC 0 k< δ.
Proof: In view of (1.1.7), we need consider only the Riemann upper sums. Let J = QN
1 [ck, dk] Given a δ > 0, a rectangle I = QN
1≤k≤N, defineI k − (δ) andI k + (δ) to be the rectangles
respectively Then, for any rectangle I 0 ⊆J with diam(I 0 )< δ, eitherI 0 ⊆I for some I∈ Cor, for someI∈ C and 1≤k≤N,I 0 ⊆I k + (δ) orI 0 ⊆I k − (δ). Now letC 0 withkC 0 k< δbe given Then, by an application of Lemma 1.1.1, we can write
I fvol(I∩I 0 ) =U(f;C), where the final step is another application of Lemma 1.1.1 Thus, it remains to estimate
However, by the discussion in the preceding paragraph, for eachI 0 ∈ C 0 , either
I 0 ⊆I for someI∈ C, in which case
I∩I 0 f vol(I∩I 0 ) = 0, or, for someI∈ C and 1≤k≤N,I 0 ⊆I k + (δ) orI 0 ⊆I k − (δ) Thus, if
In this article, we define the notation kfku to represent the uniform norm of a function f, which is the supremum of |f| across its defined set Additionally, Lemma 1.1.1 establishes that for every integer k ranging from 1 to N and for each element I within the set C, certain properties hold true.
, and so we have now proved that, wheneverkC 0 k ≤δ,
As an essentially immediate consequence of Lemma 1.1.8, we have the fol- lowing theorem.
Theorem 1.1.9 Letf : J −→ Rbe a bounded function on the rectangle
C U(f;C), where C runs over non-overlapping, finite, exact covers of J In particular, (1.1.4) is a necessary and sufficient condition that a bounded f on J be Riemann integrable, and so everyf ∈C(J;R)is Riemann integrable.
Proof: First note that, by Lemma 1.1.8, for everyC and >0 there exists a δ >0 for which kC 0 k< δ =⇒ U(f;C 0 )≤inf
The first assertion of L(f;C) is trivially established, demonstrating that (1.1.3) is equivalent to (1.1.4) Consequently, the function f is Riemann integrable if and only if (1.1.4) is satisfied Moreover, if f belongs to C(J;R), for any ε > 0, there exists a δ > 0 such that the maximum difference between the supremum and infimum of f over any collection C is less than ε This implies that the upper and lower sums, U(f;C) and L(f;C), differ by less than the volume of J whenever the norm of C is less than δ.
Exercise 1.1.10 Suppose that f and g are bounded, Riemann integrable functions onJ Show thatf∨g≡max{f, g},f∧g≡min{f, g}, and, for any α, β∈R, αf+βg are all Riemann integrable onJ In addition, check that
Conclude, in particular, that if f and g are Riemann integrable on J and f ≤g, then (R)R
5 Recall that a continuous function on a compact set is uniformly continuous there.
Exercise 1.1.11 Show that if f is a bounded real-valued function on the rectangleJ, thenf is Riemann integrable if and only if, for each >0, there is a δ >0 such that kCk< δ =⇒ X
In fact, show thatf will be Riemann integrable if, for each >0, there exists some C for whichP
Exercise 1.1.12 Show that a boundedf onJ is Riemann integrable if it is continuous on J at all but a finite number of points See Theorem 5.1.4 for more information.
Riemann–Stieltjes Integration
In Section 1.1, I explored classical integration theory based on the standard concept of Euclidean volume In this section, I will broaden the classical theory to encompass more general definitions of volume, specifically focusing on one-dimensional integrals.
LetJ = [a, b] be an interval inRandϕandψa pair of real-valued functions on J Given a non-overlapping, finite, exact coverC of J by closed intervals
I and a choice mapξ∈Ξ(C), define the Riemann sum of ϕoverC with respect toψ relative toξto be
The notation ∆Iψ represents the difference between the values of the function ψ at the right-hand endpoint I + and the left-hand endpoint I − of the interval I When the function ψ(x) equals x for x in the interval J, the Riemann integral R(ϕ|ψ;C,ξ) simplifies to R(ϕ;C,ξ) Therefore, we can state that a function ϕ is Riemann integrable on J with respect to ψ, or ψ-Riemann integrable on J, if there exists a number A such that for every ε > 0, there is a corresponding δ > 0 that satisfies the integration criteria.
|R(ϕ|ψ;C,ξ)−A|< whenever C is a non-overlapping, finite, exact cover ofJ satisfyingkCk< δ. Assuming that ϕ is ψ-Riemann integrable on J, the number A in (1.2.1) is called theRiemann–Stieltjes integral ofϕonJ with respect toψ, and
J ϕ(x)dψ(x) to denote A. §1.2 Riemann–Stieltjes Integration 9
Examples 1.2.2 The following examples may help to explain what is going on here Throughout, J= [a, b] is a compact interval.
(i) Ifϕ∈C(J;R) andψ∈C 1 (J;R) (i.e.,ψis continuously differentiable on
J), then one can use the Mean Value Theorem to check thatϕisψ-Riemann integrable on J and that
(ii) If there exist a=a0 < a1 0, there is aδ >0 for which
U(ϕ|ψ;C)− U(ϕ|ψ;C 0 )≤ U(ϕ|ψ;C)− L(ϕ|ψ;C)< no matter what C 0 is chosen as long as kCk< δ From the above it is clear that infC U(ϕ;C) = lim kCk→0U(ϕ|ψ;C) = lim kCk→0L(ϕ|ψ;C) = sup
L(ϕ;C) and therefore that ϕ is ψ-Riemann integrable on J and (R)R
To extend the previous result, it is important to recognize that if the function ϕ is Riemann integrable on the interval J with respect to both ψ1 and ψ2, then it is also Riemann integrable on J with respect to the difference ψ ≡ ψ2 − ψ1.
(This can be seen directly or as a consequence of Theorem 1.2.3 combined with (iii) in Examples 1.2.2.) In particular, we have the following corollary to Theorem 1.2.6.
Corollary 1.2.7 If ψ = ψ2−ψ1, where ψ1andψ2 are non-decreasing functions onJ, then everyϕ∈C(J;R)is Riemann integrable with respect to ψ and(1.2.5)holds withKψ= ∆Jψ1+ ∆Jψ2. §1.2.2 Functions of Bounded Variation: In this subsection I will carry out a program that will show that, at least amongψ’s that are right-continuous onJ\{J + }and have left limits at each point inJ\{J − }, theψ’s in Corollary 1.2.7 are the only ones with the properties that every ϕ ∈ C(J;R) is ψ- Riemann integrable on J and (1.2.5) holds for someK ψ 0 and for each n ≥ 1, we select a cover {Im,n : m ≥ 1} ⊆ R of Γn that satisfies the condition P∞ m=1V(Im,n) ≤ ˜ à(Γn) + 2 −n It is clear that the collection {Im,n: (m, n) ∈ (Z +)²} forms a countable cover of Γ Consequently, applying Lemma 2.2.1, we conclude that ˜ à(Γ) ≤ X.
As a consequence of Lemma 2.2.3, one has that
The left-hand side is clearly dominated by the right Therefore, it is sufficient to demonstrate that if {I m : m≥1} serves as a cover of Γ using elements of R and >0, then there exists a G(E) with G⊇Γ such that ˜à(G) is less than or equal to the sum of V(I m ) for m from 1 to infinity.
To this end, for each m choose I m 0 ∈ R such that I m ⊆ ˚I m 0 and V(I m 0 ) ≤
V(Im) + 2 −m , and takeG=S∞ m=1˚I m 0 Clearly Γ⊆G∈G(E) and ˜ à(G)≤
A significant implication of (2.2.4) is that for any subset Γ of E, there exists a Gδ(E) set D that includes Γ, satisfying the condition that ˜à(D) equals ˜à(Γ) To establish this, one can select a subset Γ from G n within G(E) such that ˜à(G n) is less than or equal to à(Γ) plus ˜ n 1, and then define D as the infinite intersection of G n Additionally, (2.2.4) proves beneficial in demonstrating the second part of a preliminary additivity result concerning ˜à.
Lemma 2.2.5 If GandG 0 are disjoint open subsets ofE, thenà(G˜ ∪G 0 ) ˜ à(G) + ˜à(G 0 ) Also, if K and K 0 are disjoint compact subsets of E, then ˜ à(K∪K 0 ) = ˜à(K) + ˜à(K 0 ).
Proof: We begin by showing that if {Im : m ≥ 1} is a sequence of non- overlapping elements of R, then
To establish the dominance of the right-hand side, it is sufficient to demonstrate that it is overshadowed by the left-hand side According to Lemmas 2.2.3 and 2.2.2, this holds true for every positive integer n in the set of natural numbers.
42 2 Measures which completes the proof of (2.2.6).
Next, suppose that G and G 0 are disjoint open sets By (5), there exist non-overlapping sequences {Im : m ≥1} and {I m 0 : m ≥ 1} of elements of
R such that G= S∞ m=1I m and G 0 = S∞ m=1I m 0 0 Thus, if I 2m−1 00 = I m and
I 2m 00 =I m 0 form≥1, theI m 00 ’s are non-overlapping elements ofRwhose union isG∪G 0 Hence, by (2.2.6), ˜ à(G∪G 0 ) ∞
To complete the proof, let K andK 0 be given Because they are disjoint and compact, there exist disjoint open sets Gand G 0 such thatK ⊆Gand
= ˜à(H∩G) + ˜à(H∩G 0 )≥à(K) + ˜˜ à(K 0 ), and therefore, by (2.2.4), ˜à(K∪K 0 )≥à(K) + ˜˜ à(K 0 ) Because the opposite inequality always holds, there is nothing more to do
I am now prepared to define the σ-algebra L, although its status as a σ-algebra and the countable additivity of ˜à on it may not be immediately clear Specifically, let L consist of subsets Γ of E that satisfy the condition that for every ε > 0, there exists an open set G containing Γ such that ˜à(G \ Γ) < ε.
Initially, one might assume that every subset Γ is an element of L based on the inequality ˜à(G) ˜ à(G\Γ) + ˜à(Γ) However, the correct general understanding is that ˜à(G) ≤ à(G˜ \Γ) + ˜à(Γ), highlighting the nuance in the definition It is evident that G(E) is a subset of L Additionally, if ˜à(Γ) = 0, then Γ belongs to L, as one can select an open set G that contains Γ, ensuring that ˜à(G\Γ) is less than any ε > 0 Lastly, if Γ is in L, there exists a set D in G δ (E) such that Γ is a subset of D and ˜à(D\Γ) = 0, which can be established by choosing a sequence of sets {G n : n≥1} that contains Γ and satisfies the condition ˜à(G n \Γ) < n⁻¹, ultimately taking D as the limit of these sets.
1 Gn. The next result shows that Lis closed under countable unions.
1 Γ n ∈ L, and, of course (cf Lemma 2.2.3), à(Γ)˜ ≤P∞
Proof: For each n≥1, chooseG(E)3Gn⊇Γn so that ˜à(Gn\Γn)0 be given, and choose an open setG⊇Kso that ˜à(G)≤à(K) +˜ SetH =G\K∈G(E), and choose a non-overlapping sequence{In: n≥1} ⊆R so thatH =S∞