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Show that if f, g ∈ R[a, b] and f and g are real-valued, then max{f, g} ∈R[a, b] and min{f, g} ∈ R[a, b].5. Show thata) if f, g ∈ R[a, b] and f (x) = g(x) almost everywhere on [a, b], thenba f (x) dx = a g(x) dx;b) if f ∈ R[a, b] and f (x) = g(x) almost everywhere on [a, b], then g can failto be Riemann-integrable on [a, b], even if g is defined and bounded on [a, b].b2 P.du Bois-Reymond (1831–1889) – German mathematician.6.2 Linearity, Additivity and Monotonicity of the Integral3496. Integration of vector-valued functions.a) Let r(t) be the radius-vector of a point moving in space, r0 = r(0) the initialposition of the point, and v(t) the velocity vector as a function of time. Show howto recover r(t) from r0 and v(t).b) Does the integration of vector-valued functions reduce to integrating realvalued functions?c) Is the criterion for integrability stated in Proposition 2 valid for vector-valuedfunctions?d) Is Lebesgue’s criterion for integrability valid for vector-valued functions?e) Which concepts and facts from this section extend to functions with complexvalues?6.2 Linearity, Additivity and Monotonicity of the Integral6.2.1 The Integral as a Linear Function on the Space R[a, b]Theorem 1 If f and g are integrable functions on the closed interval [a, b], a linearcombination of them αf + βg is also integrable on [a, b], andabb(αf + βg)(x) dx = αabf (x) dx + βg(x) dx.(6.11)aProof Consider a Riemann sum for the integral on the left-hand side of (6.11), andtransform it as follows:nnn(αf + βg)(ξi )Δxi = αf (ξi )Δxi + βg(ξi )Δxi .i=1i=1(6.12)i=1Since the right-hand side of this last equality tends to the linear combination ofintegrals that makes up the right-hand side of (6.11) if the mesh λ(P ) of the partitiontends to 0, the left-hand side of (6.12) must also have a limit as λ(P ) → 0, and thatlimit must be the same as the limit on the right.

Thus (αf + βg) ∈ R[a, b] andEq. (6.11) holds.If we regard R[a, b] as a vector space over the field of real numbers and the intebgral a f (x) dx as a real-valued function defined on vectors of R[a, b], Theorem 1asserts that the integral is a linear function on the vector space R[a, b].To avoid any possible confusion, functions defined on functions are usuallycalled functionals. Thus we have proved that the integral is a linear functional onthe vector space of integrable functions.3506Integration6.2.2 The Integral as an Additive Function of the Intervalof IntegrationbThe value of the integral a f (x) dx = I (f ; [a, b]) depends on both the integrandand the closed interval over which the integral is taken. For example, if f ∈ R[a, b],βthen, as we know, f |[α,β] ∈ R[α, β] if [α, β] ⊂ [a, b], that is, the integral α f (x) dxis defined, and we can study its dependence on the closed interval [α, β] of integration.Lemma 1 If a < b < c and f ∈ R[a, c], then f |[a,b] ∈ R[a, b], f |[b,c] ∈ R[b, c],and the following equality3 holds; b c cf (x) dx =f (x) dx +f (x) dx.(6.13)aabProof We first note that the integrability of the restrictions of f to the closed intervals [a, b] and [b, c] is guaranteed by Proposition 4 of Sect.

c 6.1.Next, since f ∈ R[a, c], in computing the integral a f (x) dx as the limit ofRiemann sums we may choose any convenient partitions of [a, c]. We shall nowconsider only partitions P of [a, c] that contain the point b. Obviously any suchpartition with distinguished points (P , ξ ) generates partitions (P , ξ ) and (P , ξ )of [a, b] and [b, c] respectively, and P = P ∪ P and ξ = ξ ∪ ξ .But then the following equality holds between the corresponding Riemann sums:σ (f ; P , ξ ) = σ f ; P , ξ + σ f ; P , ξ .Since λ(P ) ≤ λ(P ) and λ(P ) ≤ λ(P ), for λ(P ) sufficiently small, each of theseRiemann sums is close to the corresponding integral in (6.13), which consequentlymust hold.To widen the application of this result slightly, we temporarily revert once againto the definition of the integral.We defined the integral as the limit of Riemann sumsσ (f ; P , ξ ) =nf (ξi )Δxi ,(6.14)i=1corresponding to partitions with distinguished points (P , ξ ) of the closed interval ofintegration [a, b].

A partition P consisted of a finite monotonic sequence of pointsx0 , x1 , . . . , xn , the point x0 being the lower limit of integration a and xn the upperlimit of integration b. This construction was carried out assuming that a < b. If we3 We recall that f |E denotes the restriction of the function f to a set E contained in the domainof definition of f . Formally we should have written the restriction of f to the intervals [a, b] and[b, c], rather than f , on the right-hand side of Eq. (6.13).6.2 Linearity, Additivity and Monotonicity of the Integral351now take two arbitrary points a and b without requiring a < b and, regarding a asthe lower limit of integration and b as the upper, carry out this construction, we shallagain obtain a sum of the form (6.14), in which now Δxi > 0 (i = 1, .

. . , n) if a < band Δxi < 0 (i = 1, . . . , n) if a > b, since Δxi = xi − xi−1 . Thus for a > b the sum(6.14) will differ from the Riemann sum of the corresponding partition of the closedinterval [b, a] (b < a) only in sign.From these considerations we adopt the following convention: if a > b, then b af (x) dx := −f (x) dx.(6.15)abIn this connection, it is also natural to set af (x) dx := 0.(6.16)aAfter these conventions, taking account of Lemma 1, we arrive at the followingimportant property of the integral.Theorem 2 Let a, b, c ∈ R and let f be a function integrable over the largest closedinterval having two of these points as endpoints. Then the restriction of f to each ofthe other closed intervals is also integrable over those intervals and the followingequality holds: b c af (x) dx +f (x) dx +f (x) dx = 0.(6.17)abcProof By the symmetry of Eq.

(6.17) in a, b, and c, we may assume without loss ofgenerality that a = min{a, b, c}.If max{a, b, c} = c and a < b < c, then by Lemma 1 b c cf (x) dx +f (x) dx −f (x) dx = 0,abawhich, when we take account of the convention (6.15) yields (6.17).If max{a, b, c} = b and a < c < b, then by Lemma 1 c b bf (x) dx +f (x) dx −f (x) dx = 0,acawhich, when we take account of (6.15), again yields (6.17).Finally, if two of the points a, b, and c are equal, then (6.17) follows from theconventions (6.15) and (6.16).Definition 1 Suppose that to each ordered pair (α, β) of points α, β ∈ [a, b] a number I (α, β) is assigned so thatI (α, γ ) = I (α, β) + I (β, γ )for any triple of points α, β, γ ∈ [a, b].3526IntegrationThen the function I (α, β) is called an additive (oriented) interval function defined on intervals contained in [a, b].bIf f ∈ R[A, B], and a, b, c ∈ [A, B], then, setting I (a, b) = a f (x) dx, we conclude from (6.17) thatcf (x) dx =abcf (x) dx+af (x) dx,(6.18)bthat is, the integral is an additive interval function on the interval of integration.The orientation of the interval in this case amounts to the fact that we order thepair of endpoints of the interval by indicating which is to be first (the lower limit ofintegration) and which is to be second (the upper limit of integration).6.2.3 Estimation of the Integral, Monotonicity of the Integral,and the Mean-Value Theorema.

A General Estimate of the IntegralWe begin with a general estimate of the integral, which, as will become clear later,holds for integrals of functions that are not necessarily real-valued.Theorem 3 If a ≤ b and f ∈ R[a, b], then |f | ∈ R[a, b] and the following inequality holds: b b≤f(x)dx|f |(x) dx.(6.19)aaIf |f |(x) ≤ C on [a, b] thenb|f |(x) dx ≤ C(b − a).(6.20)aProof For a = b the assertion is trivial, and so we shall assume that a < b.To prove the theorem it now suffices to recall that |f | ∈ R[a, b] (see Proposition 4 of Sect. 6.1), and write the following estimate for the Riemann sumσ (f ; P , ξ ): nnnn f (ξi ) |Δxi | =f (ξi ) Δxi ≤ Cf (ξi )Δxi ≤Δxi = C(b − a).i=1i=1i=1i=1Passing to the limit as λ(P ) → 0, we obtain b bf (x) dx ≤|f |(x) dx ≤ C(b − a).aa6.2 Linearity, Additivity and Monotonicity of the Integral353b.

Monotonicity of the Integral and the First Mean-Value TheoremThe results that follow are specific to integrals of real-valued functions.Theorem 4 If a ≤ b, f1 , f2 ∈ R[a, b], and f1 (x) ≤ f2 (x) at each point x ∈ [a, b],then b bf1 (x) dx ≤f2 (x) dx.(6.21)aaProof For a = b the assertion is trivial. If a < b, it suffices to write the followinginequality for the Riemann sums:nf1 (ξi )Δxi ≤i=1nf2 (ξi )Δxi ,i=1which is valid since Δxi > 0 (i = 1, . . .

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