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We now choose δ = min{δ1 , δ2 }.ω(f ; Δ) < 2(b−a)6.1 Definition of the Integral339nLet P be an arbitrary partition of [a, b] for which λ(P ) < δ. We break the sumi=1 ω(f ; Δi )Δxi corresponding to the partition P into two parts:nω(f ; Δi )Δxi =ω(f ; Δi )Δxi +ω(f ; Δi )Δxi .i=1The sum contains the terms corresponding to intervals Δi of the partition havingno points in common with any of the δ1 -neighborhoods of the points of discontinuε, and soity. For these intervals Δi we have ω(f ; Δi ) < 2(b−a)ω(f ; Δi )Δxi <εεε(b − a) = .Δxi ≤2(b − a)2(b − a)2The sum of the lengths of the remaining intervals of the partition P , as one canεε· k = 2C, and thereforeeasily see, is at most (δ + 2δ1 + δ)k ≤ 4 8C·kω(f ; Δi )Δxi ≤ CΔxi < C ·εε= .2C 2Thus we find that for λ(P ) < δ,nω(f ; Δi )Δxi < ε,i=1that is, the sufficient condition for integrability holds, and so f ∈ R[a, b].Corollary 3 A monotonic function on a closed interval is integrable on that interval.Proof It follows from the monotonicity of f on [a, b] that ω(f ; [a, b]) = |f (b) −εf (a)|.

Suppose ε > 0 is given. We set δ = |f (b)−f(a)| . We assume that f (b) −f (a) = 0, since otherwise f is constant, and there is no doubt as to its integrability.Let P be an arbitrary partition of [a, b] with mesh λ(P ) < δ.Then, taking account of the monotonicity of f , we haveni=1ω(f ; Δi )Δxi < δni=1nf (xi ) − f (xi−1 ) =ω(f ; Δi ) = δi=1 n = δf (xi ) − f (xi−1 ) = δ f (b) − f (a) = ε.i=1Thus f satisfies the sufficient condition for integrability, and therefore f ∈R[a, b].A monotonic function may have a (countably) infinite set of discontinuities on aclosed interval. For example, the function defined by the relations3406!f (x) =1−112n−1for 1 −12n−1≤x <1−12n ,Integrationn ∈ N,for x = 1on [0, 1] is nondecreasing and has a discontinuity at every point of the form 1 −n ∈ N.12n ,Remark We note that, although we are dealing at the moment with real-valued functions on an interval, we have made no use of the assumption that the functions arereal-valued rather than complex-valued or even vector-valued functions of a point ofthe closed interval [a, b], either in the definition of the integral or in the propositionsproved above, except Corollary 3.On the other hand, the concept of upper and lower Riemann sums, to which wenow turn, is specific to real-valued functions.Definition 6 Let f : [a, b] → R be a real-valued function that is defined andbounded on the closed interval [a, b], let P be a partition of [a, b], and let Δi(i = 1, .

. . , n) be the intervals of the partition P . Let mi = infx∈Δi f (x) andMi = supx∈Δi f (x) (i = 1, . . . , n).The sumss(f ; P ) :=nmi Δxii=1andS(f ; P ) :=nMi Δxii=1are called respectively the lower and upper Riemann sums of the function f on theinterval [a, b] corresponding to the partition P of that interval.1 The sums s(f ; P )and S(f ; P ) are also called the lower and upper Darboux sums corresponding tothe partition P of [a, b].If (P , ξ ) is an arbitrary partition with distinguished points on [a, b], then obviouslys(f ; P ) ≤ σ (f ; P , ξ ) ≤ S(f ; P ).(6.7)Lemma 1s(f ; P ) = inf σ (f ; P , ξ ),ξS(f ; P ) = sup σ (f ; P , ξ ).ξ1 Theterm “Riemann sum” here is not quite accurate, since mi and Mi are not always values of thefunction f at some point ξi ∈ Δi .6.1 Definition of the Integral341Proof Let us verify, for example, that the upper Darboux sum corresponding toa partition P of the closed interval [a, b] is the least upper bound of the Riemannsums corresponding to the partitions with distinguished points (P , ξ ), the supremumbeing taken over all sets ξ = (ξ1 , .

. . , ξn ) of distinguished points.In view of (6.7), it suffices to prove that for any ε > 0 there is a set ξ of distinguished points such thatS(f ; P ) < σ (f ; P , ξ ) + ε.(6.8)By definition of the numbers Mi , for each i ∈ {1, . . . , n} there is a point ξ i ∈ Δiεat which Mi < f (ξ i ) + b−a. Let ξ = (ξ 1 , . . .

, ξ n ). Thenni=1Mi Δxi <n i=1nεf (ξ i ) +Δxi =f (ξ i )Δxi + ε,b−ai=1which completes the proof of the second assertion of the lemma. The first assertionis verified similarly.From this lemma and inequality (6.7), taking account of the definition of theRiemann integral, we deduce the following proposition.Proposition 3 A bounded real-valued function f : [a, b] → R is Riemann-integrable on [a, b] if and only if the following limits exist and are equal to each other:I = lim s(f ; P ),λ(P )→0I = lim S(f ; P ).λ(P )→0(6.9)When this happens, the common value I = I = I is the integral bf (x) dx.aProof Indeed, if the limits (6.9) exist and are equal, we conclude by the propertiesof limits and by (6.7) that the Riemann sums have a limit and thatI = lim σ (f ; P , ξ ) = I .λ(P )→0On the other hand, if f ∈ R[a, b], that is, the limitlim σ (f ; P , ξ ) = Iλ(P )→0exists, we conclude from (6.7) and (6.8) that the limit limλ(P )→0 S(f ; P ) = I existsand I = I .Similarly one can verify that limλ(P )→0 s(f ; P ) = I = I .As a corollary of Proposition 3, we obtain the following sharpening of Proposition 2.3426IntegrationProposition 2 A necessary and sufficient condition for a function f : [a, b] → Rdefined on a closed interval [a, b] to be Riemann integrable on [a, b] is the followingrelation:nlimω(f ; Δi )Δxi = 0.(6.10)λ(P )→0i=1Proof Taking account of Proposition 2, we have only to verify that condition (6.10)is necessary for f to be integrable.We remark that ω(f ; Δi ) = Mi − mi , and thereforenω(f ; Δi )Δxi =i=1n(Mi − mi )Δxi = S(f ; P ) − s(f ; P ),i=1and (6.10) now follows from Proposition 3 if f ∈ R[a, b].c.

The Vector Space R[a, b]Many operations can be performed on integrable functions without going outsidethe class of integrable functions R[a, b].Proposition 4 If f, g ∈ R[a, b], thena)b)c)d)e)(f + g) ∈ R[a, b];(αf ) ∈ R[a, b], where α is a numerical coefficient;|f | ∈ R[a, b];f |[c,d] ∈ R[c, d] if [c, d] ⊂ [a, b];(f · g) ∈ R[a, b].We are considering only real-valued functions at the moment, but it is worthwhileto note that properties a), b), c), and d) turn out to be valid for complex-valued andvector-valued functions. For vector-valued functions, in general, the product f · g isnot defined, so that property e) is not considered for them. However, this propertycontinues to hold for complex-valued functions.We now turn to the proof of Proposition 4.Proof a) This assertion is obvious sincennn(f + g)(ξi )Δxi =f (ξi )Δxi +g(ξi )Δxi .i=1i=1i=1b) This assertion is obvious, sincenn(αf )(ξi )Δxi = αf (ξi )Δxi .i=1i=16.1 Definition of the Integral343c) Since ω(|f |; E) ≤ ω(f ; E), we can writenω(|f |; Δi )Δxi ≤i=1nω(f ; Δi )Δxi ,i=1and conclude by Proposition 2 that (f ∈ R[a, b]) ⇒ (|f | ∈ R[a, b]).d) We want to verify that the restriction f |[c,d| to [c, d] of a function f that isintegrable on the closed interval [a, b] is integrable on [c, d] if [c, d] ⊂ [a, b].

Letπ be a partition of [c, d]. By adding points to π , we extend it to a partition P of theclosed interval [a, b] so as to have λ(P ) ≤ λ(π). It is clear that one can always dothis.We can then writeω(f |[c,d] ; Δi )Δxi ≤ω(f ; Δi )Δxi ,πPwhere π is the sum over all the intervals of the partition π and P the sum overall the intervals of P .By construction, as λ(π) → 0 we have λ(P ) → 0 also, and so by Proposition 2we conclude from this inequality that (f ∈ R[a, b]) ⇒ (f ∈ R[c, d]) if [c, d] ⊂[a, b].e) We first verify that if f ∈ R[a, b], then f 2 ∈ R[a, b].If f ∈ R[a, b], then f is bounded on [a, b].

Let |f (x)| ≤ C < ∞ on [a, b]. Then 2 f (x1 ) − f 2 (x2 ) = f (x1 ) + f (x2 ) · f (x1 ) − f (x2 ) ≤ 2C f (x1 ) − f (x2 ),and therefore ω(f 2 ; E) ≤ 2Cω(f ; E) if E ⊂ [a, b]. Hencennω f 2 ; Δi Δxi ≤ 2Cω(f ; Δi )Δxi ,i=1i=1from which we conclude by Proposition 2 that f ∈ R[a, b] ⇒ f 2 ∈ R[a, b] .We now turn to the general case. We write the identity(f · g)(x) =*1)(f + g)2 (x) − (f − g)2 (x) .4From this identity and the result just proved, together with a) and b), which havealready been proved, we conclude that f ∈ R[a, b] ∧ g ∈ R[a, b] ⇒ f · g ∈ R[a, b] .You already know what a vector space is from your study of algebra.

The realvalued functions defined on a set can be added and multiplied by a real number,3446Integrationboth operations being performed pointwise, and the result is another real-valuedfunction on the same set. If functions are regarded as vectors, one can verify thatall the axioms of a vector space over the field of real numbers hold, and the set ofreal-valued functions is a vector space with respect to the operations of pointwiseaddition and multiplication by real numbers.In parts a) and b) of Proposition 4 it was asserted that addition of integrablefunctions and multiplication of an integrable function by a number do not lead outside the class R[a, b] of integrable functions. Thus R[a, b] is itself a vector space –a subspace of the vector space of real-valued functions defined on the closed interval[a, b].d. Lebesgue’s Criterion for Riemann Integrability of a FunctionIn conclusion we present, without proof for the time being, a theorem of Lebesguegiving an intrinsic description of a Riemann-integrable function.To do this, we introduce the following concept, which is useful in its own right.Definition 7 A set E ⊂ R has measure zero or is of measure zero (in the sense ofLebesgue) if for every number ε > 0 there exists a covering of the set E by an atmost countable system {Ik } of intervals, the sum of whose lengths ∞k=1 |Ik | is atmost ε.Since the series ∞k=1 |Ik | converges absolutely, the order of summation of thelengths of the intervals of the covering does not affect the sum (see Proposition 4 ofSect.

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